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CBSE Term II, , 2022, , Syllabus, CBSE Term II Class XII, One Paper, No., , Max Marks:, Units, , Marks, , I., , Algebra Cont., , II., , Geometry Cont., , III., , Trigonometry Cont., , IV., , Mensuration Cont., , V., , Statistics, , Probability Cont., , Total, Internal Assessment, Total, , UNIT-I ALGEBRA, ., , Quadratic Equations, , Periods, , Standard form of a quadratic equation ax + bx + c = , a ≠ . Solutions of quadratic equations, only real roots by factorisation, and by using quadratic formula. Relationship between, discriminant and nature of roots. Situational problems based on quadratic equations related to, day to day activities problems on equations reducible to quadratic equations are excluded, , ., , Arithmetic Progressions, Motivation for studying Arithmetic Progression Derivation of the nth term and sum of the first n, terms of AP and their application in solving daily life problems. Applications based on sum to n, terms of an AP. are excluded, , UNIT-II GEOMETRY, ., , Circles, Tangent to a circle at point of contact, . Prove The tangent at any point of a circle is perpendicular to the radius through the, point of contact., . Prove The lengths of tangents drawn from an external point to a circle are equal., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the watermark

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CBSE Term II, , 2022, , ., , Constructions, . Division of a line segment in a given ratio internally ., . Tangents to a circle from a point outside it., , UNIT-III TRIGONOMETRY, ., , Some Applications of Trigonometry, HEIGHTS AND DISTANCES-Angle of elevation, Angle of Depression., Simple problems on heights and distances. Problems should not involve more than two right, triangles. Angles of elevation depression should be only, ,, ,, ., , UNIT-IV MENSURATION, ., , Surface Areas and Volumes, . Surface areas and volumes of combinations of any two of the following: cubes, cuboids,, spheres, hemispheres and right circular cylinders cones., . Problems involving converting one type of metallic solid into another and other mixed, problems. Problems with combination of not more than two different solids be taken ., , UNIT-V STATISTICS, ., , PROBABILITY, , Statistics, Mean, median and mode of grouped data bimodal situation to be avoided . Mean by Direct, Method and Assumed Mean Method only, , Internal Assessment, , Marks, , Periodic Test, , 3, , Multiple Assessments, , 2, , Portfolio, , 2, , Student Enrichment Activities-practical work, , 3, , Total Marks, 10 Marks, for the, Term, , Visit https://telegram.me/booksforcbse for more books., PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the watermark

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CBSE Term II Mathematics X (Standard), , 2, II. Sometimes, quadratic equation involves two unknown constants, and its both roots are given. Then, to find unknowns we put both, roots one-by-one in the quadratic equation and get two linear, equations in two unknowns. On solving these equations, we get, the required values of unknown constants., , So, the quadratic equation has no real roots or, imaginary roots or we can say that roots of, quadratic equation does not exist. This can be, explained using the flow chart., , Solution of a Quadratic Equation, by Factorisation, To find the solution of a quadratic equation by factorisation, method, we use the following steps., Step I Write the given equation in standard form, i.e. ax 2 + bx + c = 0 (if not given in standard form) and, find the value of a , b and c., Step II Find the product of a and c and write it as a sum of its, two factor such that sum is equal to b. i.e. write, ac = p × q and p + q = b where, p and q are factors of ac., Step III Put the value of b obtained from step II in given, equation and write it LHS as product of two linear, factors., Step IV Now, equate each factor equal to zero and get desired, roots of given quadratic equation., , Solution of a Quadratic Equation by, Quadratic Formula, In a quadratic equation ax 2 + bx + c = 0, a ≠ 0, if b 2 − 4 ac ≥ 0 ,, then the roots of the quadratic equation are given by, x=, , −b ± D, − b ± b 2 − 4 ac, or x =, 2a, 2a, , where, D = b 2 − 4 ac is known as discriminant. This result is, known as quadratic formula or Sridharacharya formula., , Relationship between Discriminant and, Nature of Roots, The nature of roots depends upon the value of the discriminant, D, whereas, D can be zero, positive or negative, so three cases, may arise., Case I When D = 0 i.e. b 2 − 4 ac = 0., −b ± 0, b, b, If D = b 2 − 4 ac = 0, then x =, ⇒x= − ,−, 2a, 2a 2a, So, the quadratic equation has two equal real roots or, repeated roots or coincident roots., Case II When D > 0 i.e. b 2 − 4 ac > 0., −b − D, −b + D, and, 2a, 2a, So, the quadratic equation has two distinct real roots., Case III When D < 0 i.e. b 2 − 4 ac < 0., If D = b 2 − 4 ac > 0, then x =, , If D = b 2 − 4 ac < 0, then D can not be evaluated as, , Quadratic equation, ax2+bx+c=0, a ≠ 0, Find discriminant, D, , D=0, ⇒ Roots are real, and equal, , D>0, ⇒ Roots are, real and distinct, , D<0, ⇒ Roots are, imaginary or, real roots, does not exist, , Method to Determine The Value of Unknown, when Nature of Roots is Given, , If nature of roots of a quadratic equation is given and, quadratic equation involves an unknown. Then to find the, value of unknown, first we find the value of discriminant, in terms of unknown. After that use the given condition, i.e. D > 0 or D = 0 or D < 0 and simplify it., , Some Important Points, (i) Three consecutive numbers are x,( x +1 ) and (x + 2),, respectively., (ii) Three consecutive even and odd numbers are, 2 x,(2 x + 2 ), (2 x + 4 ) and (2 x + 1 ), (2 x + 3 ), (2 x + 5 ),, respectively., (iii) Pythagoras theorem,, (Hypotenuse) 2 = (Perpendicular) 2 + (Height) 2, (iv) Area of triangle =, , 1, × Base × Height, 2, , (v) Area of right angled triangle, 1, = × Base × Perpendicular, 2, (vi) Area of rectangle = Length × Breadth, (vii) Perimeter of rectangle = 2 × (Length + Breadth), Distance, (viii) Speed =, Time, (ix) Two-digit number = 10x + y, where x and y are the, digits of ten’s place and unit place, respectively., On reversing the digits, new number = 10y + x, (x) If speed of stream be x km/h and speed of boat in, still water be y km/h. Then speed of boat in, upstream = ( y − x) km/h and speed of boat in, downstream = ( y + x) km/h., , square root of negative value is not defined., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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3, , CBSE Term II Mathematics X (Standard), , Solved Examples, =, , Example 1. Check whether the following equations are, quadratic or not., 3, (i) x + = x 2, x, 1, 2, (iii) x − 2 = 5, x, Sol., , 2, , 1, is a root of the given equation., 3, 1, (iii) On putting x = − in Eq. (i), we get, 2, So, x =, , 2, , (ii) 2 x − 5 x = x − 2 x + 3, (iv) x 2 − 3 x − x + 4 = 0, , 2, , ⎛ 1⎞, ⎛ 1⎞, ⎛ 1⎞, p ⎜ − ⎟ = 3⎜ − ⎟ + 2 ⎜ − ⎟ − 1, ⎝ 2⎠, ⎝ 2⎠, ⎝ 2⎠, 3, = −1−1, 4, 3, 3 − 8 −5, = −2 =, =, ≠0, 4, 4, 4, 1, So, x = − is not a root of the given equation., 2, (iv) On putting x = 2 in Eq. (i), we get, , 3, = x2, x, x2 + 3 = x3, , (i) Given that, x +, ⇒, , x3 − x2 − 3 = 0, , ⇒, , Which is not of the form ax 2 + bx + c, a ≠ 0., Thus, the equation is not a quadratic equation., (ii) Given that, 2 x 2 − 5x = x 2 − 2 x + 3, ⇒, , 2 x 2 − x 2 − 5x + 2 x − 3 = 0, , ⇒, , x 2 − 3x − 3 = 0, , p(2 ) = 3(2 )2 + 2(2 ) − 1, = 12 + 4 − 1 = 15 ≠ 0, So, x = 2 is not a root of the given equation., , Which is of the form ax 2 + bx + c, a ≠ 0., Thus, the equation is a quadratic equation., 1, (iii) Given that, x 2 − 2 = 5, x, ⇒, x 4 − 1 = 5x 2, , Example 3. In each of the following equations, find the, value of unknown constant(s) for which the given, value(s) is (are) solution of the equations., (i) x 2 − k 2 = 0; x = 0. 3, , x 4 − 5x 2 − 1 = 0, , ⇒, , Which is not of the form ax 2 + bx + c, a ≠ 0., , (ii) 3 x 2 + 2 ax − 3 = 0; x =, , Thus, the equation is not a quadratic equation., (iv) Given that, x 2 − 3x − x + 4 = 0, , Sol., , Thus, the equation is not a quadratic equation., , (i) We have, x 2 − k 2 = 0, here k is unknown., , Example 2. Which of the following are the roots of, , ( 0. 3)2 − k 2 = 0, , 3x 2 + 2 x − 1 = 0 ?, , 1, 2, , (iv) x = 2, , (iii) x = −, , k 2 = ( 0. 3)2, , ⇒, , (i) x = − 1, , ⇒, k = ± 0. 3, (ii) We have, 3x 2 + 2 ax − 3 = 0, here a is unknown., 1, is a solution of given equation, so it will, 2, satisfy the given equation., 1, On putting x = − in the given equation, we get, 2, Since, x = −, , Sol. Given equation is of the form p( x ) = 0, where, p( x ) = 3x 2 + 2 x − 1, , −1, 2, , Since, x = 0. 3 is a solution of given equation, so it will, satisfy the given equation,, On putting x = 0. 3 in the given equation, we get, , Which is not of the form ax 2 + bx + c, a ≠ 0., , 1, (ii) x =, 3, , 1+2 −3, =0, 3, , …(i), , (i) On putting x = − 1 in Eq. (i), we get, p( − 1) = 3( − 1)2 + 2( − 1) − 1, = 3−2 −1= 0, So, x = − 1 is a root of the given quadratic equation., 1, (ii) On putting x = in Eq. (i), we get, 3, 2, ⎛ 1⎞, ⎛ 1⎞, ⎛ 1⎞, p ⎜ ⎟ = 3⎜ ⎟ + 2 ⎜ ⎟ − 1, ⎝ 3⎠, ⎝ 3⎠, ⎝ 3⎠, 1 2, = + −1, 3 3, , 2, , ⎛ 1⎞, ⎛ 1⎞, 3⎜ − ⎟ + 2 a ⎜ − ⎟ − 3 = 0, ⎝ 2⎠, ⎝ 2⎠, ⇒, ⇒, ⇒, , 3, −a −3=0, 4, 3, a= −3, 4, 3 − 12, 9, a=, =−, 4, 4, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 4, Example 4. Find the roots of the quadratic equation, 2x 2 +, , 5, x − 2 = 0 by factorisation method., 3, , 5, x −2 = 0, 3, On multiplying by 3 both sides, we get, , Sol. Given equation is 2 x 2 +, , 2, , 6x + ( 9x − 4x ) − 6 = 0, [by splitting the middle term], 6x 2 + 9x − 4x − 6 = 0, , ⇒, ⇒, ⇒, Now,, , x + 5 − ( x − 3) 1, =, ( x − 3)( x + 5), 6, , ⇒, , x+ 5−x+ 3 1, = ⇒ 8 × 6 = ( x − 3) ( x + 5 ), ( x − 3) ( x + 5) 6, , ⇒, , 48 = x 2 + 2 x − 15, , ⇒, , x 2 + 2 x − 63 = 0, , ⇒ x 2 + 9x − 7 x − 63 = 0, , 6x + 5x − 6 = 0, 2, , ⇒, , ⇒, , 3 x ( 2 x + 3 ) − 2 ( 2 x + 3) = 0, (2 x + 3) ( 3x − 2 ) = 0, 2x + 3 = 0, , ⇒ x( x + 9) − 7( x + 9) = 0, ⇒, ( x − 7 ) ( x + 9) = 0, ⇒, x = 7 and x = − 9, ∴Sum of roots = 7 + ( − 9) = − 2, , Example 7. Using the quadratic formula, solve the, , 3, x=−, ⇒, 2, and, 3x − 2 = 0, 2, ⇒, x=, 3, 5, 2, Hence, the roots of the equation 2 x + x − 2 = 0 are, 3, −3, 2, and ., 2, 3, , quadratic equation., x 2 + 2 2 x − 6 = 0., Sol. Given equation is x 2 + 2 2 x − 6 = 0 ., On comparing with ax 2 + bx + c = 0, we get, a = 1, b = 2 2 and c = − 6, By quadratic formula, x =, =, , Example 5. Solve the quadratic equation by, , 3 2 x 2 − ( 6x − x ) − 2 = 0, [by splitting the middle term], 3 2 x 2 − 6x + x − 2 = 0, 3 2 x2 − 3 2 ⋅ 2 x + x − 2 = 0, 3 2 x (x − 2) + 1 (x − 2 ) = 0, (x − 2) (3 2x + 1) = 0, , Now,, , So,, , Sol. Comparing the given quadratic equation, 3x 2 + 4x − 5 = 0, 1, 3 2, , =, , − 2, 6, , 2, , Hence, the roots of the equation 3 2 x − 5x − 2 = 0 are, −, , 2, and 2 ., 6, , Example 6. Find the sum of the roots of the equation,, 1, 1⎤, ⎡ 1, ⎢ x − 3 − x + 5 = 6 ⎥., ⎣, ⎦, ⎡ 1, 1 ⎤ 1, Sol. Given ⎢, −, =, x, −, 3, x, +, 5 ⎥⎦ 6, ⎣, , 2 and − 3 2 are the roots of the given equation., , 3x 2 + 4x − 5 = 0., , 3 2 x+1=0, x=−, , ⇒, , − (2 2 ) ± (2 2 )2 − 4 (1) ( −6), 2 (1 ), , Example 8. Find discriminant of the quadratic equation, , x− 2 =0 ⇒ x= 2, , and, , b 2 − 4ac, 2a, , =, , Sol. Given equation is 3 2 x 2 − 5x − 2 = 0, , ⇒, , −b±, , − 2 2 ± 8 + 24, 2, −2 2 ± 32 − 2 2 ± 4 2, =, =, 2, 2, −2 2 + 4 2 −2 2 − 4 2, =, ,, 2, 2, = 2, − 3 2, , factorisation method., 3 2 x 2 − 5x − 2 = 0, , ⇒, , [by splitting the middle term], , with standard quadratic equation ax 2 + bx + c = 0, we get, a = 3, b = 4 and c = − 5, ∴ Discriminant (D ) = b 2 − 4ac, = ( 4)2 − 4 × ( 3) × ( − 5) = 16 + 60 = 76, , Example 9. Check whether the quadratic equation has, real roots. If real roots exist, find them, 8x 2 + 2x − 3 = 0, Sol. Given equation is 8x 2 + 2 x − 3 = 0., On comparing with ax 2 + bx + c = 0, we get, a = 8, b = 2 and c = − 3, ∴Discriminant, D = b 2 − 4ac, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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5, , CBSE Term II Mathematics X (Standard), On comparing with ax 2 + bx + c = 0, we get, , = (2 ) 2 − 4 ( 8 ) ( − 3 ), = 4 + 96 = 100 > 0, Therefore, the equation 8x 2 + 2 x − 3 = 0 has two distinct real, roots as the discriminant greater than zero., − b ± D − 2 ± 100 − 2 ± 10, Thus roots, x =, =, =, 2a, 16, 16, − 2 + 10 − 2 − 10, =, ,, 16, 16, 8, 12 1, 3, = ,−, = ,−, 16, 16 2, 4, , Example 10. Find the nature of roots of the quadratic, equation 3x 2 − 4 3x + 4 = 0., , If the roots are real, find them. [CBSE 2020 (Standard)], , a = 2 , b = 1 and c = − 1, ∴ Discriminant,, D = b 2 − 4ac = (1 )2 − 4 (2 ) ( − 1 ), = 1 + 8 = 9 > 0 i.e. D > 0, Hence, the equation 2 x 2 + x − 1 = 0 has two distinct, real roots., 9, (iii) Given equation is 2 x 2 − 6x +, = 0., 2, On comparing with ax 2 + bx + c = 0, we get, 9, a = 2 , b = − 6 and c =, 2, ∴ Discriminant, D = b 2 − 4ac, ⎛ 9⎞, = ( − 6 ) 2 − 4 (2 ) ⎜ ⎟, ⎝ 2⎠, , Sol. Given quadratic equation is, 2, , 3 x − 4 3x + 4 = 0, , = 36 − 36 = 0, , Compare with standard quadratic equation, ax 2 + bx + c = 0, we get, a = 3, b = −4 3 and c = 4, , i.e. D = 0, Hence, the equation 2 x 2 − 6x +, , Now, discriminant = b 2 − 4ac, , real roots., , Example 12. The quadratic equation x 2 − 4x + k = 0 has, , = ( − 4 3 )2 − 4 × 3 × 4, , distinct real roots, if k = 4. Why or why not?, , = 48 − 48 = 0, Hence, roots are real and equal., By using Sridharacharya formula,, −b ± D, x=, 2a, − ( −4 3 ) ± 0, =, 2×3, 4 3 2 3, =, =, 2×3, 3, Hence, roots of given quadratic equation are, , Sol. Given quadratic equation is x 2 − 4x + k = 0, Compare with standard equation ax 2 + bx + c = 0, we get, a = 1, b = − 4 and c = k, The condition for distinct real root is b 2 − 4ac > 0, ⇒, , 2 3, 2 3, and, ., 3, 3, , equations have two distinct real roots. Justify your, answer., (i) x 2 − 3 x + 4 = 0, (ii) 2 x 2 + x − 1 = 0, 9, (iii) 2 x 2 − 6 x + = 0, 2, (i) Given equation is x 2 − 3x + 4 = 0., On comparing with ax 2 + bx + c = 0, we get, a = 1, b = − 3 and c = 4, ∴ Discriminant,, D = b 2 − 4ac = ( −3)2 − 4 (1 ) ( 4), = 9 − 16 = − 7 < 0, i.e., D<0, Hence, the equation x 2 − 3x + 4 = 0 has no real root ., (ii) Given equation is 2 x 2 + x − 1 = 0, , ( − 4) 2 − 4 × 1 × k > 0, , ⇒, ⇒, , Example 11. State whether the following quadratic, , Sol., , 9, = 0 has equal and, 2, , 16 − 4k > 0, 16 > 4k, 16, k<, ⇒ k<4, 4, , ⇒, , Example 13. Find the value of k, for which the, , quadratic equation ( k + 4)x 2 + ( k + 1 )x + 1 = 0 has, [CBSE 2020 (Standard)], equal roots., , Sol. Given, quadratic equation is, ( k + 4)x 2 + ( k + 1)x + 1 = 0, Compare with ax 2 + bx + c = 0, we get, a = k + 4, b = k + 1 and c = 1, Condition for equal roots, b 2 − 4ac = 0, ∴, , ( k + 1)2 − 4 × ( k + 4)(1) = 0, , ⇒, , k 2 + 12 + 2 k − 4k − 16 = 0, [Q( a + b )2 = a 2 + b 2 + 2 ab ], k 2 − 2 k − 15 = 0, , ⇒, 2, , ⇒, , k − ( 5 − 3)k − 15 = 0, , ⇒, , k 2 − 5k + 3k − 15 = 0, , ⇒, , k( k − 5) + 3( k − 5) = 0, , [by splitting middle term], , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 6, ⇒, ⇒, , ( k + 3) ( k − 5 ) = 0, k = − 3, 5, , would have been 11 more than five times her actual, age, what is her age now?, , Example 14. The denominator of a fraction is 3 more, than its numerator. The sum of the fraction and its, 29, reciprocal is . Find the fraction., 10, , Sol. Let the actual age of Zeba = x yr, Her age when she was 5 yr younger = ( x – 5) yr, Now, by given condition,, Square of her age = 11 more than five times her actual age, ( x – 5)2 = 5 × actual age + 11, , Sol. Let numerator = x, Then denominator = x + 3, , ⇒, , x, ∴ The fraction is the form of, x+3, , ⇒, , ⇒ x – 14x – x + 14 = 0, , ⇒, , 10 ( x 2 + x 2 + 9 + 6x ) = 29 ( x 2 + 3x ), , x ( x – 14) – 1 ( x – 14) = 0, ( x – 1) ( x – 14) = 0, x = 14, [here, x ≠ 1 because her age is x – 5. So, x – 5 = 1 – 5 = – 4, i.e. age cannot be negative], Hence, required Zeba’s age now is 14 yr., , 20x 2 + 60x + 90 = 29x 2 + 87 x, , ⇒, , 9x 2 + 27 x − 90 = 0, , ⇒, , x 2 + 3x − 10 = 0, , [divide by 9], , ⇒, , x 2 + 5x − 2 x − 10 = 0, , [by splitting middle term], , ⇒, ⇒, ⇒, ⇒, , x ( x + 5) − 2 ( x + 5 ) = 0, ( x + 5) ( x − 2 ) = 0, x + 5 = 0 and x − 2 = 0, x = − 5 and x = 2, , Example 17. A two-digit number is such that the, , Example 15. Find a natural number whose square, diminished by 84 is equal to thrice of 8 more than, the given number., Sol. Let n be a required natural number., Square of a natural number diminished by 84 = n 2 – 84, And thrice of 8 more than the natural number = 3 ( n + 8), Now, by given condition,, n 2 – 84 = 3 ( n + 8), , ⇒, , n 2 – 3n – 108 = 0, , ⇒, ⇒, ⇒, ⇒, , [by splitting the middle term], , ⇒, ⇒, ⇒, , ⇒, , n 2 – 84 = 3n + 24, , x 2 – 15x + 14 = 0, 2, , x 2 + ( x + 3)2 29, =, x ( x + 3), 10, , ⇒, , x 2 + 25 – 10x = 5x + 11, , ⇒, , According to the question,, x, x + 3 29, +, =, x+3, x, 10, , ⇒, , ( x – 5)2 = 5x + 11, , n 2 – 12 n + 9n – 108 = 0, [by splitting the middle term], n ( n – 12 ) + 9 ( n – 12 ) = 0, ( n – 12 ) ( n + 9) = 0, n = 12, [Q n ≠ – 9 because n is a natural number], , Hence, the required natural number is 12., , Example 16. If Zeba were younger by 5 yr than what, she really is, then the square of her age (in years), , product of its digit is 35. When 18 is added to the, number the digits interchange their places. Find, the number., Sol. Let the ten’s digit number be x., According to the question,, Product of the digits = 35, i.e. Ten’s digits × Unit digit = 35, 35, ⇒ Units digit =, x, 35, ∴ Two digit number = 10x +, x, Also it is given that if 18 is added to the number, the digits, gets interchange., 35, 35, 10x +, + 18 = 10 ×, +x, ∴, x, x, ⇒, , 10x 2 + 35 + 18x 350 + x 2, =, x, x, , ⇒, , 9x 2 + 18x − 315 = 0, , ⇒, , x 2 + 2 x − 35 = 0, , ⇒, , [divide by 9], , 2, , x + 7 x − 5x − 35 = 0, , ⇒, x (x + 7) − 5 (x + 7) = 0, ⇒, ( x − 5) ( x + 7 ) = 0, ⇒, x = 5, − 7, But a digit can never be negative., So, x = − 7 is rejected., ∴ The required number is, 10 × 5 + 5 = 50 + 5, = 55, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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7, , CBSE Term II Mathematics X (Standard), , Chapter, Practice, PART 1, Objective Questions, ●, , Multiple Choice Questions, , 8. A quadratic equation with integral coefficient has, integral roots., , 1. Which of the following is a quadratic equation?, [NCERT Exemplar], (a) x 2 + 2 x + 1 = ( 4 − x )2 + 3, 2⎞, ⎛, (b) − 2 x 2 = ( 5 − x ) ⎜ 2 x − ⎟, ⎝, 5⎠, 3, 2, (c) ( k + 1 ) x + x = 7, where k = − 1, 2, (d) x 3 − x 2 = ( x − 1)3, , (c) ( 2 x +, , 2, , 2, , (a) True, (c) Can’t determined, , 3 ) = 3x – 5x, , [NCERT Exemplar], (b) 2 x – x 2 = x 2 + 5, 2, , 2, , 4, , (d) ( x + 2 x ) = x + 3 + 4x, , 2, , 3. If a number x is added to twice its square, then the, resultant is 21. Then the quadratic representation of, this statement is, 2, , 2, , (a) 2x − x + 21 = 0, , (b) 2x + x − 21 = 0, , (c) 2x 2 − x − 20 = 0, , (d) None of these, , (a) x − 4x + 5 = 0, (c) 2 x 2 − 7 x + 6 = 0, , 2, , (b) x + 3x − 12 = 0, (d) 3x 2 − 6x − 2 = 0, , 1, 5, 5. If is a root of the equation x 2 + kx − = 0, then, 2, 4, [NCERT Exemplar], the value of k is, (a) 2, , (b) −2, , (c), , 1, 4, , (d), , 1, 2, , 6. Which of the following equation has root as 3?, (a) x 2 – 5x + 6 = 0, 3, (c) 2 x 2 –, x+1=0, 2, , (b) – x 2 + 3x – 3 = 0, 2, , (d) 3x – 3x + 3 = 0, , 7. 0.2 is a root of the equation x 2 − 0. 4 = 0?, [NCERT Exemplar], (a) True, (c) Can’t determined, , (b) False, (d) None of these, , [NCERT Exemplar], (b) False, (d) None of these, , 10. The roots of the quadratic equation, x 2 − 8 x − 20 = 0 are, (a) 5, − 4, , (b) − 4, 5, , (c) 10, − 2, , (d) − 10, 2, , 11. Which constant must be added and subtracted to, 3, solve the quadratic equation 9x 2 + x − 2 = 0., 4, 1, (a), 8, , 1, (b), 64, , [NCERT Exemplar], 9, (d), 64, , 1, (c), 4, , 12. Solve 12 x 2 + 5x − 3 = 0., , 4. Which of the following equations has 2 as a root?, 2, , (b) False, (d) None of these, , 9. If b = 0, c < 0, then the roots of x 2 + bx + c = 0 are, numerically equal and opposite in sign., , 2. Which of the following is not a quadratic equation?, (a) 2 ( x – 1 )2 = 4x 2 – 2 x + 1, , (a) True, (c) Can’t determined, , 1 4, (a) ,, 3 3, , 1 3, (b) ,, 2 4, , (c) −, , 1 3, ,, 3 4, , 13. The discriminant of the quadratic, equation x 2 − 4x + 1 = 0 is, (a) 2 3, , (b) 4, , (c) 12, , 1, 3, (d) , −, 3, 4, [CBSE 2013], (d) 16, , 14. If the discriminant of the equation, 6x 2 − bx + 2 = 0 is 1, then the value of b is, [CBSE 2012], (a) 7, (b) − 7, (c) Both (a) and (b), (d) None of these, , 15. Value(s) of k for which the quadratic equation, 2 x 2 − kx + k = 0 has equal roots is/are, [NCERT], (a) 0, , (b) 4, , (c) 8, , (d) 0, 8, , 2, , 16. The quadratic equation 2 x − 5x + 1 = 0 has, (a) two distinct real roots, (c) no real roots, , (b) two equal real roots, (d) more than 2 real roots, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 8, 17. If the discriminant of the equation, kx 2 − 3 2 x + 4 2 = 0 is 14, then the value of k is, (a) 2, , (b), , 1, 3 2, , 1, (c), 2, , (d), , 1, 4 2, , 18. Which of the following equations has two distinct, [NCERT Exemplar], real roots?, 9, =0, 4, 2, (c) x + 3x + 2 2 = 0, (a) 2 x 2 − 3 2 x +, , (b) x 2 + x − 5 = 0, (d) 5x 2 − 3x + 1 = 0, , 19. Which of the following equations has no real roots?, (a) x 2 − 4x + 3 2 = 0, , (b) x 2 + 4x − 3 2 = 0, , (c) x 2 − 4x − 3 2 = 0, , (d) 3x 2 + 4 3x + 4 = 0, , 20. ( x 2 + 1 )2 − x 2 = 0 has, (a) four real roots, (c) no real roots, , [NCERT Exemplar], (b) two real roots, (d) one real root, , (ii) Which of the following quadratic equation, describe the speed of Raj’s car?, (a) x 2 − 5x − 500 = 0, 2, , (c) x + 5x − 500 = 0, , (b) x 2 + 4x − 400 = 0, (d) x 2 − 4x + 400 = 0, , (iii) What is the speed of Raj’s car?, (a) 20 km/h, (c) 25 km/h, , (b) 15 km/h, (d) 10 km/h, , (iv) How much time took Ajay to travel 400 km?, (a) 20 h, , (b) 40 h, , (c) 25 h, , (d) 16 h, , (v) How much time took Raj to travel 400 km?, (a) 15 h, , (b) 20 h, , (c) 18 h, , (d) 22 h, , 25. The speed of a motor boat is 20 km/h. For covering, the distance of 15 km the boat took 1 h more for, [CBSE Question Bank], upstream than downstream., , 21. The sum of the squares of three consecutive, integers is 110, then the smallest positive integer is, (a) 6, , (b) 5, , (c) 7, , [NCERT Exemplar], (d) 4, , 22. A line segment AB is 8 cm in length. AB is, produced to P such that BP 2 = AB ⋅ AP. Then, the, [NCERT Exemplar], length of BP is, (a) 5( 5 + 1), (c) 4( 5 + 1), , ●, , Upstream (b), , (b) 5 + 1, (d) 3 + 1, , 23. One year ago, a man was 8 times as old as his son., Now, his age is equal to the square of his son’s age., Present age of man is, (a) 49 yr, (c) 59 yr, , Downstream (a), , (b) 37 yr, (d) 39 yr, , Direction of boat, Direction of stream, , (i) Let speed of the stream be x km/h, then speed of, the motorboat in upstream will be, (a) 20 km/h, (c) (20 − x ) km/h, , Case Based MCQs, 24. Raj and Ajay are very close friends. Both the, families decide to go to Ranikhet by their own cars., Raj’s car travels at a speed of x km/h while Ajay’s, car travels 5 km/h faster than Raj’s car. Raj took, 4 h more than Ajay to complete the journey of, [CBSE Question Bank], 400 km., , Direction of boat, Direction of stream, , (b) (20 + x ) km/h, (d) 2 km/h, , (ii) What is the relation between speed, distance and, time?, Distance, Time, Speed, (b) Distance =, Time, (c) Time = Speed × Distance, (d) Speed = Distance × Time, (a) Speed =, , (iii) Which is the correct quadratic equation for the, speed of the current ?, (a) x 2 + 30x − 200 = 0, , (b) x 2 + 20x − 400 = 0, , (c) x 2 + 30x − 400 = 0, , (d) x 2 − 20x − 400 = 0, , (iv) What is the speed of current ?, (i) What will be the distance covered by Ajay’s car in, two hours?, (a) 2 ( x + 5) km, (c) 2 ( x + 10) km, , (b) ( x – 5) km, (d) (2 x + 5) km, , (a) 20 km/h, (c) 15 km/h, , (b) 10 km/h, (d) 25 km/h, , (v) How much time boat took in downstream?, (a) 90 min, (c) 30 min, , (b) 15 min, (d) 45 min, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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9, , CBSE Term II Mathematics X (Standard), , 26. By quadratic formula, the roots of the quadratic, equation ax 2 + bx + c = 0, a ≠ 0 are given by, −b ± b 2 − 4ac, , −b ± D, 2a, 2a, where, D = b 2 − 4ac is called discriminant., x=, , or x =, , (i) The roots of the quadratic equation, 8 x 2 − 22 x − 21 = 0 are, 7, 3, ,−, 2, 4, 7, 3, (c) , −, 2, 4, (a) −, , 7 3, (b) ,, 2 4, 7 3, (d) − ,, 2 4, 2, , (ii) The discriminant of x + x + 7 = 0 is, (a) 27, (c) 27, , (b) − 27, (d) −27, , (iii) Roots of 4 x 2 − 2 x = 3 are, (a) Real and distinct (b) Real and equal, (c) Imaginary, (d) More than two real roots, , (iv) The value of k for which, 4 x 2 + kx + 9 = 0 has real and equal roots is, (a) 12, (c) Both (a) and (b), , (b) − 12, (d) None of these, , (v) The least positive value of k for which, x 2 + kx + 16 = 0 has real roots, is, (a) 18, (c) 2, , 27. Seven years ago, Varun’s age was five times the, square of Swati’s age. Three years hence, Swati’s, age will be two-fifth of Varun’s age., (i) If seven years ago, Swati’s age be, x yr, then Varun’s age is, 2, , (c) ( 5x + 7 ) yr, , (b) 5x 2 yr, (d) ( 5x 2 − 7 ) yr, , (ii) After three years, Swati’s age is, (a) ( x + 3) yr, (c) ( x + 7 ) yr, , (b) ( x − 3) yr, (d) ( x + 10) yr, , (iii) The quadratic equation related to the given, problem is, (a) 2 x 2 − x − 6 = 0, , (b) 5x 2 − x + 6 = 0, , 2, , 2, , (c) 3x − 2 x + 5 = 0 (d) 7 x − 3x + 1 = 0, , (iv) Present age of Varun’s is, (a) 27 yr, (c) 30 yr, , (b) 20 yr, (d) 37 yr, , (v) If Swati’s present age 10 yr, then present age of, Varun’s is, (a) 40 yr, (c) 45 yr, , ●, , Short Answer Type Questions, 1. Check whether the following are quadratic, equations or not., (i) ( x − 1 ) ( x + 2 ) = ( x − 3 ) ( x + 1 ), (ii) ( x + 2 ) 2 = 4 ( x + 3 ), , 2. If x =, , 1, 3, , is root of the equation, , Px 2 + ( 3 − 2 ) x − 1 = 0, then find the value of, P 2 + 1., , (b) 47 yr, (d) 52 yr, , [NCERT Exemplar], , 3. In each of the following equations, determine the, value of k for which the given value is a solution of, the equation., (i) kx 2 + 2 x − 3 = 0 , x = 2, (ii) x 2 + 2 ax − k = 0 , x = − a, , 4. Find the value of k in the following equations, (i) x 2 − 2 kx − 6 = 0, when x = 3, (ii) x 2 − kx −, , (b) 4, (d) 8, , (a) ( 5x − 7 )2 yr, , PART 2, Subjective Questions, , 1, 5, = 0, when x =, 2, 4, , −1, 1, , x = are the solutions, 2, 3, of the given equation 6x 2 − x − 2 = 0, or not., , 5. Determine whether x =, , 6. Solve the quadratic equation by factorisation, method., 4 3x 2 + 5x − 2 3 = 0, 16, 15, 7. Solve for x : − 1 =, ; x ≠ 0 , − 1., x, x +1, 8. Find the roots of the equation, ax 2 + a = a 2 x + x., , [CBSE 2012], , 9. Solve for x , 6x + 7 − (2 x − 7 ) = 0, , [CBSE 2016], , 10. Find the numerical difference of the, roots of equation x 2 − 7 x − 18 = 0., , [CBSE 2015], , 11. Using the quadratic formula, solve the quadratic, equation., 3x 2 + 11 x + 6 3 = 0, 12. If the discriminant of the equation 5x 2 − sx + 4 = 0, is 1, then find the value of s., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 10, 13. Show that ( x 2 + 1 )2 − x 2 = 0 has no real roots., , 27. The sum of the reciprocals of Anjali’s age 3 yr ago, 1, and 5 yr from now is . Find the present age of, 3, Anjali., 28. ‘A two-digit number is such that the product of the, digits is 12. When 36 is added to the number the, digits interchange their places. Find the two-digit, number., 29. ‘‘John and Janvi together have 45 marbles. Both of, them lost 5 marbles each and the product of the, number of marbles they now have, is 124. Find out, how many marbles they had to start with?’’, 30. The hypotenuse of right angled triangle is 6 m more, than twice the shortest side. If the third side is 2 m, less than the hypotenuse, then find all sides of the, [CBSE 2020 (Standard)], triangle., 31. At present Asha’s age (in years) is 2 more than the, square of her daughter Nisha’s age. When Nisha, grows to her mother’s present age. Asha’s age, would be one year less than 10 times the present, age of Nisha. Find the present ages of both Asha, [NCERT Exemplar], and Nisha., 32. The speed of a boat in still water is 15 km/h. It can, go 30 km upstream and return downstream to the, original point in 4 h and 30 min. Find the speed of, stream., 7, 33. Two water taps together can fill a tank in 1 h., 8, The tap with longer diameter takes 2 h less than, the tap with smaller one to fill the tank separately., Find the time in which each tap can fill the tank, [CBSE 2019], separately., , [NCERT Exemplar], , 14. Find the value of k for which the quadratic equation, 2 x 2 − kx + k = 0 has equal roots. [NCERT Exemplar], 15. Find the values of k for which the equation, 9x 2 + 3kx + 4 = 0 has real roots., 16. If the equation (1 + m 2 )x 2 + (2 mc )x + ( c 2 − a 2 ) = 0, has equal roots, then prove that c 2 = a 2 (1 + m 2 )., 17. The sum of two numbers is 11 and the sum of their, 11, reciprocals is . Find the numbers., [CBSE 2013], 28, 18. In a cricket match. Harbhajan took three wickets, less than twice the number of wickets taken by, Zaheer. The product of the numbers of wickets, taken by these two is 20. Represent the above, situation in the form of a quadratic equation., [CBSE 2015], ●, , Long Answer Type Questions, 19. If x = 2 and x = 3 are roots of the equation, 3x 2 − 2 ax + 2 b = 0, then find the values of a and b., 20. Find the nature of roots of the following quadratic, equations. If the real roots exist, then also find the, roots., (i) 4 x 2 + 12 x + 9 = 0, , (ii) 3 x 2 + 5 x − 7 = 0, , 21. Find the value of k for which the given equation has, equal roots., ( k − 12 ) x 2 + 2( k − 12 ) x + 2 = 0, 22. If x = − 2 is a root of the equation 3x 2 + 7 x + p = 0., Find the values of k, so that the roots of the equation, x 2 + k ( 4x + k − 1 ) + p = 0 are equal., [CBSE 2015], 23. Find two consecutive odd natural numbers, sum of, whose squares is 130., [CBSE 2013], 24. A piece of cloth costs ` 200. If the piece was 5 m, longer and each metre of cloth costs ` 2 less, the, cost of the piece would have remained unchanged., How long is the piece and what is the original rate, [CBSE 2015], per metre?, 25. The difference of two numbers is 4. If the, 4, difference of their reciprocals is , the find the, 21, [CBSE 2008], two numbers., 26. The perimeter of a right angled triangle is 70 units, and its hypotenuse is 29 units we would like to find, the length of the other sides., , ●, , Case Based Questions, 34. In the centre of a rectangular lawn of dimensions, 50 m × 40 m, a rectangular pond has to be, constructed, so that the area of the grass, surrounding the pond would be 1184 m 2, , (i) If the distance between pond and lawn is x m. Find, the length and breadth of rectangular pond., (ii) Find the quadratic equation related to the given, problem., (iii) Find the length and breadth of the pond., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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11, , CBSE Term II Mathematics X (Standard), , SOLUTIONS, Objective Questions, 1. (d) (a) Given that,, x 2 + 2 x + 1 = ( 4 − x )2 + 3, x 2 + 2 x + 1 = 16 + x 2 − 8x + 3, , ⇒, , ⇒, 10x − 18 = 0, which is not of the form ax 2 + bx + c = 0, a ≠ 0., Thus, the equation is not a quadratic equation., 2⎞, ⎛, (b) Given that, − 2 x 2 = ( 5 − x ) ⎜ 2 x − ⎟, ⎝, 5⎠, − 2 x 2 = 10x − 2 x 2 − 2 +, , ⇒, , 2x, 5, , ⇒ 50x + 2 x − 10 = 0, ⇒, 52 x − 10 = 0, which is also not a quadratic equation., 3, (c) Given that, x 2 ( k + 1) + x = 7, 2, Given,, k = −1, 3, 2, ⇒, x ( − 1 + 1) + x = 7, 2, ⇒, 3x − 14 = 0, which is also not a quadratic equation., (d) Given that, x 3 − x 2 = ( x − 1 )3, x 3 − x 2 = x 3 − 3 x 2 (1 ) + 3x ( 1 ) 2 − ( 1 ) 3, , ⇒, , [Q ( a − b )3 = a 3 − b 3 + 3ab 2 − 3a 2b], x 3 − x 2 = x 3 − 3x 2 + 3x − 1, , ⇒, ⇒, , 2, , − x + 3 x 2 − 3 x + 1 = 0 ⇒ 2 x 2 − 3x + 1 = 0, , which represents a quadratic equation because it has, the quadratic form ax 2 + bx + c = 0, a ≠ 0., 2. (d) (a) Given that, 2 ( x – 1)2 = 4x 2 – 2 x + 1, ⇒, , 2 ( x 2 + 1 – 2 x ) = 4x 2 – 2 x + 1, , ⇒, , 2 x 2 + 2 – 4x = 4 x 2 – 2 x + 1, , ⇒, , 2 x2 + 2 x – 1 = 0, , which represents a quadratic equation because it has, the quadratic form ax 2 + bx + c = 0, a ≠ 0., (b) Given that, 2 x – x 2 = x 2 + 5, ⇒, , 2 x2 – 2 x + 5 = 0, , which also represents a quadratic equation because it, has the quadratic form ax 2 + bx + c = 0, a ≠ 0., (c) Given that, ( 2 ⋅ x +, ⇒, ⇒, , 3 ) 2 = 3 x 2 – 5x, , 2 ⋅ x 2 + 3 + 2 6 ⋅ x = 3 x 2 – 5x, x2 – ( 5 + 2 6 ) x – 3 = 0, , which also represents a quadratic equation because it, has the quadratic form ax 2 + bx + c = 0, a ≠ 0., (d) Given that, ( x 2 + 2 x )2 = x 4 + 3 + 4x 2, ⇒, , 4x 3 – 3 = 0, , ⇒, , x 4 + 4x 2 + 4x 3 = x 4 + 3 + 4x 2, , which is not of the form ax 2 + bx + c = 0, a ≠ 0., Thus, the equation is not quadratic., This is a cubic equation., 3. (b) Let the number be x., Then according to the given condition,, 2 x 2 + x = 21, 2, , 2 x + x − 21 = 0, , ⇒, , 4. (c) (a) Substituting x = 2 in x 2 − 4x + 5 , we get, (2 )2 − 4 (2 ) + 5 = 4 − 8 + 5 = 1 ≠ 0., So, x = 2 is not a root of x 2 − 4x + 5 = 0., (b) Substituting x = 2 in x 2 + 3x − 12, we get, (2 )2 + 3(2 ) − 12, = 4 + 6 − 12 = −2 ≠ 0, So, x = 2 is not a root of x 2 + 3x − 12 = 0., (c) Substituting x = 2 in 2 x 2 − 7 x + 6, we get, 2 (2 )2 − 7 (2 ) + 6 = 2 ( 4) − 14 + 6, = 8 − 14 + 6 = 14 − 14 = 0, So, x = 2 is root of the equation 2 x 2 − 7 x + 6 = 0., (d) Substituting x = 2 in 3x 2 − 6x − 2 , we get, 3(2 )2 − 6(2 ) − 2 = 12 − 12 − 2 = −2 ≠ 0, So, x = 2 is not a root of 3x 2 − 6x − 2 = 0., 1, 5, 5. (a) Since, is a root of the quadratic equation x 2 + kx − = 0., 2, 4, 2, , ⎛ 1⎞ 5, ⎛ 1⎞, Then, ⎜ ⎟ + k ⎜ ⎟ − = 0, ⎝ 2⎠ 4, ⎝ 2⎠, ⇒, ⇒, , 1 k 5, + − =0, 4 2 4, 1 + 2k − 5, =0, 4, , ⇒, 2k − 4 = 0, ⇒, 2k = 4, ⇒, k =2, 6. (a) (a) Given that, x 2 – 5x + 6 = 0, Put x = 3, we get, ( 3)2 − 5( 3) + 6 = 9 − 15 + 6 = 0, Hence, x = 3 is a root of the equation., (b) – x 2 + 3x – 3 = 0, Put x = 3, we get, −( 3)2 + 3( 3) − 3 = −9 + 9 − 3 = −3 ≠ 0, Hence, x = 3 is not a root of the equation., 3, (c) 2 x 2 –, x+1=0, 2, Put x = 3, we get, 3( 3), 2 ( 3) 2 −, +1, 2 ( 3), , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 12, 9, 9, +1=, +1≠ 0, 2, 2, Hence, x = 3 is not a root of the equation., (d) 3x 2 – 3x + 3 = 0, , ⎛, ⎜y +, ⎝, , ⇒9 2 −, , 1, must be added and subtracted to solve the given, 64, equation., 12. (d) Given quadratic equation is 12 x 2 + 5x − 3 = 0., On comparing the given equation with, ax 2 + bx + c = 0, we get a = 12 , b = 5 and c = −3, Thus,, , Put x = 3, we get, 3( 3)2 − 3( 3) + 3 = 27 − 9 + 3 = 21 ≠ 0., Hence, x = 3 is not a root of the equation., 7. (b) False, since 0.2 does not satisfy the, equation i.e. ( 0.2 )2 − 0. 4 = 0. 04 − 0. 4 ≠ 0., 8. (b) False, consider the quadratic equation 2 x 2 + x − 6 = 0, with integral coefficient. The roots of the given quadratic, 3, equation are −2 and which are not integrals., 2, 9. (a) Given that, b = 0 and c < 0 and quadratic equation, x 2 + bx + c = 0, , ...(i), , Put b = 0 in Eq. (i), we get, x2 + 0 + c = 0, ⎡here, c > 0⎤, ⎢∴ − c > 0 ⎥, ⎣, ⎦, , x2 = – c, , ⇒, ∴, , x=±, , −c, , 2, , So, the roots of x + bx + c = 0 are numerically equal and, opposite in sign., 10. (c) Given, quadratic equation is x 2 − 8x − 20 = 0, which is, already in its standard form., On comparing it with ax 2 + bx + c = 0, we get, a = 1, b = −8 and c = −20, Here,, ac = 1 × ( −20) = −20., Here, ac has −ve sign,, So, let p = −10 and q = 2 as p × q = −20 and p + q = −8, ∴, x 2 − 8x − 20 = 0, ⇒, , x 2 − 10x + 2 x − 20 = 0, , ⇒ x ( x − 10) + 2( x − 10) = 0, ⇒, ( x − 10)( x + 2 ) = 0, Now, put x − 10 = 0 or x + 2 = 0, ⇒, x = 10 or x = −2, Thus, 10 and −2 are the required roots of a given quadratic, equation., 3, 11. (b) Given equation is 9x 2 + x − 2 = 0., 4, 1, 2, ( 3 x ) + ( 3x ) − 2 = 0, 4, On putting 3x = y, we have, 1, y2 + y − 2 = 0, 4, 2, , 2, , 1, ⎛ 1⎞, ⎛ 1⎞, y2 + y + ⎜ ⎟ − ⎜ ⎟ − 2 = 0, ⎝ 8⎠, ⎝ 8⎠, 4, ⎛, ⎜y +, ⎝, , 2, , 1⎞, 1, + 2, ⎟ =, 8⎠, 64, 2, , 2, , 1⎞, 1 + 64⋅ 2, ⎟ =, 8⎠, 64, , 2, , 2, , On substituting the values of a = 12 , b = 5 and c = −3 in, quadratic formula,, x=, , − b ± b 2 − 4ac, , we get, 2a, , −5 ± ( 5)2 − 4 × 12 × ( −3), 2 × 12, −5 ± 25 + 144, ⇒, x=, 24, − 5 ± 169, =, 24, − 5 ± 13, =, 24, −5 + 13 8 1, Now,, [taking +ve sign], x=, =, =, 24, 24 3, −5 − 13, 18, 3, Or, [taking −ve sign], x=, = − =−, 24, 24, 4, 1, 3, Hence, the roots of the given equation are and − ., 3, 4, 13. (c) Given quadratic equation is x 2 − 4x + 1 = 0., x=, , On comparing with ax 2 + bx + c = 0, we get, a = 1, b = −4 and c = 1, Now, discriminant (D ) = b 2 − 4ac, = ( − 4) 2 − 4 × 1 × 1, = 16 − 4 = 12, 14. (c) Given, 6x 2 − bx + 2 = 0, On comparing with Ax 2 + Bx + C = 0, we get, A = 6, B = −b and C = 2, We know that,, Discriminant, D = B 2 − 4AC, ⇒, , 1 = ( − b )2 − 4 × 6 × 2, , ⇒, , 1 = b 2 − 48, , [given, D = 1], , ⇒, b 2 = 49, [taking square root on both sides], ⇒, b=±7, Hence, the required value of b is −7 or 7., 15. (d) Given equation is 2 x 2 − kx + k = 0, On comparing with ax 2 + bx + c = 0, we get, a = 2, b = − k and c = k, For equal roots, the discriminant must be zero., i.e., D = b 2 − 4ac = 0, ⇒, , ( − k ) 2 − 4 (2 ) k = 0, , [Q ( a + b ) = a + b + 2 ab ], , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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13, , CBSE Term II Mathematics X (Standard), k 2 − 8k = 0, , ⇒, , ⇒, k ( k − 8) = 0, ∴, k = 0, 8, Hence, the required values of k are 0 and 8., 16. (c) Given equation is 2 x 2 − 5x + 1 = 0., On comparing with ax 2 + bx + c = 0, we get, , (d) Given equation is, 5x 2 − 3x + 1 = 0, On comparing with ax 2 + bx + c = 0, we get, a = 5, b = − 3, c = 1, Now, D = b 2 − 4ac = ( −3)2 − 4( 5)(1) = 9 − 20 <0, Hence, roots of the equation are not real., 19. (a) (a) The given equation is x 2 − 4x + 3 2 = 0., On comparing with ax 2 + bx + c = 0, we get, , a = 2 , b = − 5 and c = 1, , a = 1, b = − 4 and c = 3 2, , ∴ Discriminant,, D = b 2 − 4ac, , The discriminant of x 2 − 4x + 3 2 = 0 is, D = b 2 − 4ac, , = ( − 5 )2 − 4 × (2 ) × (1) = 5 − 8 = − 3 < 0, Since, discriminant is negative, therefore, quadratic equation 2 x 2 − 5x + 1 = 0 has no real roots i.e., imaginary roots., 17. (d) Given quadratic equation is, kx 2 − 3 2 x + 4 2 = 0, On comparing with, ax 2 + bx + c = 0, we get, a = k , b = − 3 2 and c = 4 2, ⇒, , 14 = ( − 3 2 )2 − 4 × k × 4 2, , ⇒, , 14 = 18 − 16 2 k, 16 2 k = 4, , ⇒, , k=, , = 16 − 12 2 = 16 − 12 × (1 . 41), ⇒, , 1, 4 2, , 18. (b), (a) Given equation is 2 x 2 − 3 2 x + 9 / 4 = 0, we get, On comparing with ax 2 + bx + c = 0, a = 2, b = −3 2 and c = 9/4, Now, D = b 2 − 4ac, = ( −3 2 )2 − 4(2 )( 9 / 4), , (b) The given equation is x 2 + 4x − 3 2 = 0, On comparing the equation with ax 2 + bx + c = 0, we get, a = 1, b = 4 and c = −3 2, Then, D = b 2 − 4ac = ( 4)2 − 4(1)( −3 2 ), Hence, the equation has real roots., (c) Given equation is x 2 − 4x − 3 2 = 0, On comparing the equation with ax 2 + bx + c = 0,, we get, a = 1, b = − 4 and c = – 3 2, Then, D = b 2 − 4ac = ( −4)2 − 4(1)( −3 2 ), = 16 + 12 2 > 0, Hence, the equation has real roots., (d) Given equation is 3x 2 + 4 3x + 4 = 0., On comparing the equation with ax 2 + bx + c = 0,we get, a = 3, b = 4 3 and c = 4, Then, D = b 2 − 4ac = ( 4 3 )2 − 4( 3)( 4), = 48 − 48 = 0, Thus, the equation has real roots., Hence, x 2 − 4x + 3 2 = 0 has no real roots., , = 18 − 18 = 0, Thus, the equation has real and equal roots., (b) The given equation is x 2 + x − 5 = 0, On comparing with ax 2 + bx + c = 0, we get, a = 1, b = 1 and c = − 5, The discriminant of x 2 + x − 5 = 0 is, 2, , 2, , D = b − 4ac = (1) − 4 (1) ( −5), ⇒, So,, , = 16 − 16. 92 = − 0. 92, b 2 − 4ac < 0, , = 16 + 12 2 > 0, , Q Discriminant, D = b 2 − 4ac, , ⇒, , = ( − 4 ) 2 − 4 ( 1) ( 3 2 ), , = 1 + 20 = 21, b 2 − 4ac > 0, x 2 + x − 5 = 0 has two distinct real roots., , (c) Given equation is x 2 + 3x + 2 2 = 0, On comparing with ax 2 + bx + c = 0, we get, a = 1, b = 3 and c = 2 2, Now, D = b 2 − 4ac = ( 3)2 − 4(1)(2 2 ), = 9−8 2 < 0, ∴ Roots of the equation are not real., , 20. (c) Given equation is ( x 2 + 1 )2 − x 2 = 0, ⇒, , x4 + 1 + 2 x2 − x2 = 0, [Q( a + b )2 = a 2 + b 2 + 2 ab], , ⇒, , 4, , 2, , x + x +1=0, x2 = y, , Let, 2 2, , 2, , ∴, , (x ) + x + 1 = 0, , ⇒, , y2 + y + 1 = 0, , On comparing with ay 2 + by + c = 0, we get, a = 1, b = 1 and c = 1, Discriminant, D = b 2 − 4ac, = (1 ) 2 − 4 ( 1 ) ( 1 ), =1− 4= −3, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 14, Since,, ∴, , which is the required quadratic equation., Now, x 2 − 7 x − x + 7 = 0, , D<0, y + y + 1 = 0 i.e. x 4 + x 2 + 1 = 0, 2, , or ( x 2 + 1)2 − x 2 = 0 has no real roots., 21. (b) Let the smallest integer be x. Then, three consecutive, integers are x, x + 1, x + 2., From the question,, x 2 + ( x + 1)2 + ( x + 2 )2 = 110, ⇒, , x 2 + x 2 + 1 + 2 x + x 2 + 4 + 4x = 110, , ⇒, , 3x 2 + 6x − 105 = 0, , ∴, , −6±, , x=, , 62 − 4 ⋅ 3 ⋅ ( − 105), 2×3, , [using formula], , −6±, , 36 + 1260, 6, − 6 ± 1296 − 6 ± 36, =, =, 6, 6, − 6 + 36 − 6 − 36 30 − 42, ,, = ,, = 5, − 7, =, 6, 6, 6, 6, =, , When x = 5,, x+1 = 5+1 = 6, x+2 = 5+2 =7, When x = −7, x + 1 = −7 + 1 = −6, x + 2 = −7 + 2 = −5, ∴Three consecutive integers are 5, 6, 7 or − 7, − 6, − 5., Hence, smallest positive integer is 5., , x cm, , 8 cm, A, , B, , P, , Now, BP 2 = AB ⋅ AP ⇒ x 2 = 8 ⋅ ( 8 + x ), ⇒, , Time taken by Ajay and Raj to complete the 400 km, journey, 400, 400, and t 2 =, t1 =, x+5, x, According to the question,, t 2 = t1 + 4, 400, 400, =, +4, ∴, x, x+5, 100, 100, ⇒, =, + 1 (divide by 4), x, x+5, ⇒, ⇒, , 22. (c) Let BP = x cm, Then, AP = AB + BP = ( 8 + x ) cm, , ⇒, , 100( x + 5) = 100x + x( x + 5), 100x + 500 = 100x + x 2 + 5x, x 2 + 5x − 500 = 0, , (iii) (a) Consider the quadratic equation x 2 + 5x − 500 = 0, On comparing with ax 2 + bx + c = 0, we get, , 2, , a = 1, b = 5 and c = − 500, , x − 8x − 64 = 0, 2, , ∴, , − ( − 8) ± ( − 8) − 4 ⋅ 1 ⋅ ( − 64), x=, 2, , or, , x=, , 8±, , [by factorisation], , ⇒ x( x − 7 ) − 1( x − 7 ) = 0, ⇒, ( x − 7 )( x − 1) = 0, ⇒ x − 7 = 0 or x − 1 = 0, ⇒, x = 7 or x = 1, But x = 1 is not possible because if x = 1, then present age of, the son and father are same., So, x = 7., Hence, present age of his son = 7 yr, and present age of man = 8 × 7 − 7 = 49 yr., 24. (i) (a) Given, Raj’s car travel at a speed of x km/h. Then, Ajay’s car travels a distance in one hour is ( x + 5) km., Therefore, Ajay’s car travels a distance in two hours is, 2( x + 5) km., Distance, (ii) (c)Q Time =, Speed, , 64 × 5, 2, , 8± 8 5, =4± 4 5, 2, But the length of BP is positive., =, , So, x = ( 4 + 4 5 ) cm = 4( 5 + 1) cm, 23. (a) Let present age of his son = x yr, One year ago, his son’s age = ( x − 1) yr, One year ago, man’s age = 8( x − 1) yr, = ( 8x − 8) yr, Present age of man = ( 8x − 8 + 1) yr, = ( 8x − 7 ) yr, According to the question,, 8x − 7 = x 2 ⇒ x 2 − 8x + 7 = 0, , Q, , x=, =, , − b ± b 2 − 4ac, 2a, − 5 ± ( 5)2 − 4 × (1)( − 500), 2 ×1, , − 5 ± 25 + 2000 − 5 ± 2025, =, 2, 2, −5 ± 45 −50 40, =, =, ,, = − 25, 20, 2, 2, 2, Since, speed cannot be negative, so we consider only,, x = 20., Hence, speed of Raj’s car is 20 km/h., (iv) (d) To travel 400 km, time taken by Ajay, 400, 400, 400, t1 =, =, =, = 16 h, ( x + 5) 20 + 5 25, =, , (v) (b) To travel 400 km, time taken by Raj,, 400 400, =, = 20 h, t2 =, x, 20, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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15, , CBSE Term II Mathematics X (Standard), , 25., , (i) (c) Since, the speed of stream be x km/h and speed of, motorboat is 20 km/h. Therefore, the speed of, motorboat in upstream will be (20 − x ) km/h., (ii) (a) The relation between speed, distance and time is, Distance, Speed =, Time, Distance, (iii) (c)Q Time =, Speed, , Now,, or, , On comparing with ax 2 + bx + c, we get, a = 1, b = 1 and c = 7, Now, discriminant (D ) = b 2 − 4ac = 12 − 4 × 1 × 7, = 1 − 28 = − 27, (iii) (a) Given equation is 4x 2 − 2 x − 3 = 0, On comparing with ax 2 + bx + c = 0, we get, a = 4, b = − 2 and c = − 3, ∴ Discriminant (D ) = b 2 − 4ac, = ( −2 )2 − 4 × 4 × ( −3) = 4 + 48 = 52 > 0, So, 4x 2 − 2 x = 3 has two distinct real roots., (iv) (c) Given equation is 4x 2 + kx + 9 = 0., On comparing with ax 2 + bx + c = 0, we get, , 15 (20 + x − 20 + x ) = (20 + x )(20 − x ), 15(2 x ) = 400 − x 2, , Now,, , 2, , x + 30x − 400 = 0, , x 2 + ( 40 − 10)x − 400 = 0, , ⇒, , x 2 + 40x − 10x − 400 = 0, , Since, roots of given equation are real and equal., ∴, D=0, ⇒ k 2 − 144 = 0 ⇒ k 2 = 144, ⇒, k = ± 12, , ⇒, x( x + 40) − 10 ( x + 40) = 0, ⇒, ( x − 10) ( x + 40) = 0, ⇒, x = 10, − 40, Since, speed cannot be negative, so we consider only, positive value., ∴, x = 10, Hence, speed of current is 10 km/h., (v) (c) The time taken by motorboat in downstream, 15, t1 =, 20 + x, 15, 15 1, =, =, = h = 30 min, 20 + 10 30 2, 26., , (i) (c) Given quadratic equation is, 8x 2 − 22 x − 21 = 0, On comparing the given equation with, ax 2 + bx + c = 0, we get, a = 8, b = − 22 and c = − 21, By quadratic formula,, x=, , 484 + 672 22 ± 1156, =, 16, 16, 22 ± 34 11 ± 17, x=, =, 16, 8, =, , ⇒, , − ( − 22 ) ± ( − 22 )2 − 4 × 8 × ( − 21), 2×8, 22 ±, , a = 4, b = k and c = 9, D = b 2 − 4ac, = k 2 − 4 × 4 × 9 = k 2 − 144, , (iv) (b) Consider quadratic equation,, x 2 + 30x − 400 = 0, ⇒, , [taking − ve sign], , 7, 3, and − ., 4, 2, (ii) (b) Given quadratic equation is x 2 + x + 7 = 0., , According to the question,, t 2 = 1 + t1, 15, 15, =1+, ∴, 20 − x, 20 + x, 15, 15, ⇒, −, =1, 20 − x 20 + x, , ⇒, , [taking + ve sign], , Hence, the roots of the given equation are, , Here, distance = 15 km/h, Speed of motorboat in downstream = (20 + x ) km/h, and speed of motorboat in upstream = (20 − x ) km/h, Time taken by motorboat in downstream and upstream, 15, 15, are t1 =, h and t 2 =, h., 20 + x, 20 − x, , ⇒, ⇒, , 11 + 17 28 7, =, =, 8, 8 2, 11 − 17, 6, 3, x=, =− =−, 8, 8, 4, x=, , (v) (d) Given equation is x 2 + kx + 16 = 0, On comparing with ax 2 + bx + c = 0, we get, a = 1, b = k and c = 16, Now, D = b 2 − 4ac, = k 2 − 4 × 1 × 16 = k 2 − 64, Since, roots of given equation are real., ∴, D≥0, ⇒ k 2 − 64 ≥ 0, ⇒, , 27., , k 2 ≥ 64, , ⇒, k ≥ 8 and k ≤ −8, Hence, positive least value of k is 8., (i) (b) Seven years ago,, Swati’s age = x yr, Varun’s age = 5x 2 yr, (ii) (d) Swati’s present age = ( x + 7 ) yr, and Varun’s present age = ( 5x 2 + 7 ) yr, After three years, we have, Swati’s age = ( x + 7 + 3) = ( x + 10) yr, Varun’s age = ( 5x 2 + 7 + 3) = ( 5x 2 + 10)yr, (iii) (a) According to the question,, 2, x + 10 = ( 5x 2 + 10), 5, ⇒, 2 x2 − x − 6 = 0, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 16, 2 x2 − x − 6 = 0, , (iv) (a) Now,, ⇒, , x=−, , ⇒, , ⇒, 3, ,2, 2, , 3., , ⇒, ⇒, , 4k + 4 − 3 = 0, 4k + 1 = 0, −1, ⇒, k=, 4, (ii) We have, x 2 + 2 ax − k = 0, here k is unknown., Since, x = − a is a solution of given equation, so it will, satisfy the given equation., On putting x = − a in the given equation, we get, ( − a )2 + 2 a ( − a ) − k = 0, , = 5 ( 3) 2 + 7, = 45 + 7 = 52 yr, Subjective Questions, (i) Given, ( x − 1)( x + 2 ) = ( x − 3)( x + 1), LHS = ( x − 1) ( x + 2 ) = x 2 + 2 x − x − 2, , ...(i), , = x2 + x − 2, RHS = ( x − 3) ( x + 1) = x 2 + x − 3x − 3, 4., , On substituting these values in Eq. (i), we get, x2 + x − 2 = x2 − 2 x − 3, x2 − x2 + x + 2 x − 2 + 3 = 0, , ⇒, 3x + 1 = 0, It is not of the form ax 2 + bx + c = 0, a ≠ 0., , ...(ii), , Hence, given equation represents a quadratic equation., 2. Given equation is, Px 2 + ( 3 − 2 ) x − 1 = 0, 1, is a root of the equation., and x =, 3, 2, , ⇒, ⇒, , P, +, 3, , 3− 2 − 3, =0, 3, P, 2, −, =0, 3, 3, , ⇒, , − a2 − k = 0, k = − a2, , (i) Given quadratic equation is, …(i), x 2 − 2 kx − 6 = 0, Since, x = 3 is one of the root of the given quadratic, equation. Then, it satisfies the given equation., So, put x = 3 in Eq. (i), we get, ⇒, ⇒, , [Q( a + b )2 = a 2 + b 2 + 2 ab ], x 2 + 4x − 4x + 4 − 12 = 0, x 2 − 8 = 0 or x 2 + 0x − 8 = 0, , 1, ⎛ 1 ⎞, −1= 0, P⎜ ⎟ + ( 3 − 2), ⎝ 3⎠, 3, , a2 − 2a2 − k = 0, , ( 3)2 − 2 k( 3) − 6 = 0, , It is of the form ax 2 + bx + c = 0, a ≠ 0., , ∴, , ⇒, ⇒, , = x2 − 2 x − 3, , As a = 0 and it is an equation of degree 1., Hence, the given equation does not represent a, quadratic equation., (ii) Given, ( x + 2 )2 = 4( x + 3), x 2 + 4 + 4x = 4x + 12, , = 6 + 1=7, (i) We have,, kx 2 + 2 x − 3 = 0, here k is unknown., Since, x = 2 is a solution of given equation, so it will, satisfy the given equation., On putting x = 2 in the given equation, we get, k(2 )2 + 2(2 ) − 3 = 0, , = ( 5 × 4 + 7 ) yr, = (20 + 7 ) yr, = 27 yr, (v) (d) Here, Swati’s present age = 10 yr, ⇒, x + 7 = 10, ⇒, x=3, So, Varun’s present age = ( 5x 2 + 7 ) yr, , ⇒, ⇒, , P 2 + 1 = ( 6 )2 + 1, , ∴, , ∴, x =2, [Q age can’t negative], ⇒ Present age of Varun’s = ( 5x 2 + 7 ) yr, , ⇒, , P=, , 2 x − 4x + 3x − 6 = 0, , ⇒2x (x − 2) + 3 (x − 2) = 0, ⇒, (2 x + 3) ( x − 2 ) = 0, , 1., , 2, ×3, 3, P= 2 × 3= 6, , ⇒, , 2, , 9 − 6k − 6 = 0, 6k = 3, 1, ⇒, k=, 2, (ii) Given quadratic equation is, 5, x 2 − kx − = 0, 4, 1, Put x = , we get, 2, 2, , ⎛ 1⎞ 5, ⎛ 1⎞, ⎜ ⎟ − k⎜ ⎟ − = 0, ⎝ 2⎠ 4, ⎝ 2⎠, 1 k 5, − − =0, ⇒, 4 2 4, 1 − 2k − 5, ⇒, =0, 4, ⇒, 2k = − 4, ⇒, k = −2, 5. Given equation is in the form p( x ) = 0, where, p( x ) = 6x 2 − x − 2, , ...(i), , −1, On putting x =, in Eq. (i), we get, 2, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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17, , CBSE Term II Mathematics X (Standard), 2, , ⇒, , On squaring both sides, we get, 6 x + 7 = (2 x − 7 ) 2, , 1 1, 6 1, − −2 = − − 2, 9 3, 9 3, 6 − 3 − 18 −15, =, =, ≠0, 9, 9, =6×, , ⇒, ⇒, , 6x + 7 = 4x 2 + 49 − 28x, 4x 2 − 34x + 42 = 0, 2 x 2 − 17 x + 21 = 0, , ⇒, , 2 x 2 + 14x + 3x + 21 = 0, , 2 x ( x + 7 ) + 3( x + 7 ) = 0, (2 x + 3) ( x + 7 ) = 0, 3, ⇒, x = − ,−7, 2, 10. Given equation is x 2 − 7 x − 18 = 0, , 1, is not a solution of the given equation., 3, 6. Given, 4 3x 2 + 5x − 2 3 = 0., So, x =, , ⇒ x 2 − 9x + 2 x − 18 = 0, , On comparing with standard form of quadratic equation, i.e., ax 2 + bx + c = 0, we get, a = 4 3, b = 5 and c = −2 3, Here, ac = 4 3 × ( −2 3 ) = −24, Then, factors of ac are 8 and −3., , ⇒ x ( x − 9) + 2 ( x − 9 ) = 0, ⇒, ( x + 2 ) ( x − 9) = 0, ⇒, x = − 2, 9, So, the roots of given equation are − 2 and 9., ∴ Required numerical difference of the roots, = 9 − ( − 2 ) = 11, 11. The given equation is 3x 2 + 11x + 6 3 = 0., On comparing with ax 2 + bx + c = 0, we get, a = 3, b = 11 and c = 6 3, , ⇒ 4x( 3x + 2 ) − 3 ( 3x + 2 ) = 0, ⇒, , ( 4x − 3 ) ( 3 x + 2 ) = 0, , ⇒, , 4x − 3 = 0, , and, , 3, 4, −2, x=, 3, , Hence, roots of equation 4 3x 2 + 5x − 2 3 = 0 are, , ⇒, , ⇒, , 16, 15, −1 =, x, x+1, 16 15, 16( x + 1) − 15x, −, =1 ⇒, =1, x x+1, x ( x + 1), 16x + 16 − 15x = x 2 + x, , ⇒, , x 2 = 16, , ⇒, , x2 = ± 4, , Hence, the roots are 4 and −4., , x=, =, , 3, 4, , −2, ., 3, , 7. Given,, ⇒, , x=, , x=, , or, , and, , On substituting the values of a, b and c in the quadratic, formula,, , 3x + 2 = 0, , ⇒, , [divide by 2], , ⇒, ⇒, , ⎛ 1⎞, p⎜ ⎟ ≠ 0, ⎝ 3⎠, , 4 3x 2 + 8x − 3x − 2 3 = 0, , 6x + 7 = 2 x − 7, , ⇒, , ⎛ 1⎞, ⎛ 1⎞, ⎛ 1⎞, p⎜ ⎟ = 6⎜ ⎟ − ⎜ ⎟ − 2, ⎝ 3⎠, ⎝ 3⎠, ⎝ 3⎠, , ⇒, , ax − a x − x + a = 0, , ax ( x − a ) − 1 ( x − a ) = 0, ( ax − 1) ( x − a ) = 0, 1, x= ,a, ⇒, a, 9. Given that, 6x + 7 − (2 x − 7 ) = 0, , 2, , 4 3x 2 + ( 8 − 3)x − 2 3 = 0, , 2, , ⇒, ⇒, , −1, is a solution of the given equation., 2, 1, Now, on putting x = in Eq. (i), we get, 3, , ∴, , 2, , ⇒, , So, x =, , ⇒, , ax 2 + a = a 2x + x, , 8. Given that,, , ⎛ −1⎞, ⎛ −1⎞, ⎛ −1⎞, p ⎜ ⎟ = 6⎜ ⎟ − ⎜ ⎟ − 2, ⎝2⎠, ⎝2⎠, ⎝2⎠, 6 1, 6 + 2 − 8 8−8, = + −2=, =, 4 2, 4, 4, ⎛ −1⎞, p⎜ ⎟ = 0, ⎝2⎠, , ⇒, and, , − b ± b 2 − 4ac, 2a, − 11 ± (11)2 − 4( 3 )( 6 3 ), 2( 3 ), − 11 ± 121 − 72, , 2 3, −11 ± 49, =, 2 3, −11 ± 7, =, 2 3, −11 + 7, −4, −2, =, =, x=, 2 3, 2 3, 3, −11 − 7 −18 −9, x=, =, =, 2 3, 2 3, 3, , [taking +ve sign], [taking −ve sign], , −2, −9, −2 3, and, (or, and −3 3) are the required, 3, 3, 3, solutions of the given equation., Hence,, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 18, 12. Given equation is 5x 2 − sx + 4 = 0, , 16. Given equation is, (1 + m 2 )x 2 + (2 mc )x + ( c 2 − a 2 ) = 0, , 2, , On comparing with ax + bx + c = 0, we get, , On comparing with Ax 2 + Bx + C = 0, we get, , a = 5, b = − s and c = 4, ∴ Discriminant (D ) = b 2 − 4ac, , A = (1 + m 2 ), B = 2 mc and C = ( c 2 − a 2 ), , = ( − s )2 − 4 × 5 × 4, , Since, the given equation has equal roots., ∴ Discriminant, D = 0 ⇒ B 2 − 4AC = 0, , 2, , = s − 80, Given,, ⇒, , ⇒, , s 2 = 81, , ⇒, ⇒, 13. Given that,, , s=± 9, , ⇒, ⇒, , ( x 2 + 1) 2 = x 2 ⇒ x, 2, , x +1=±x, x2 m x + 1 = 0, , a = 1, b = m 1 and c = 1, D = b 2 − 4ac, = ( m 1) 2 − 4 × 1 × 1, , =1 − 4= − 3 < 0, ∴ It has no real roots., 14. Given equation is 2 x 2 − kx + k = 0, , ⇒, ⇒, , − c 2 + a 2 + m 2a 2 = 0, , ⇒, , − c 2 + a 2(1 + m 2 ) = 0, , a = 2, b = − k and c = k, D = b 2 − 4ac, , Since, the given equation has equal roots., ∴, D=0, ⇒, k 2 − 8k = 0, ⇒, k ( k − 8) = 0, ⇒, k = 0, 8, 15. Given quadratic equation is, 9x 2 + 3kx + 4 = 0, On comparing with ax 2 + bx + c = 0, we get, a = 9, b = 3k and c = 4, Now, D = b 2 − 4ac = ( 3k )2 − 4( 9) ( 4), = 9k − 144, Since, roots of given equation are real., , ⇒, , k 2 − ( 4) 2 ≥ 0, , ⇒, , ( k − 4) ( k + 4) ≥ 0, , ⇒, , k ≤ − 4 or k ≥ 4, , Hence proved., , x(11 − x ) = 28, x 2 − 11x + 28 = 0, x 2 − (7 + 4)x + 28 = 0, , 2 x 2 − 3x − 20 = 0, 2, , 19. Given, 3x − 2 ax + 2 b = 0, , ∴ D ≥ 0 ⇒ 9k 2 − 144 ≥ 0, k 2 − 16 ≥ 0, , c 2 = a 2(1 + m 2 ), , ⇒, , 2, , ∴, , ⇒, , ⇒, x 2 − 7 x − 4x + 28 = 0, ⇒, x( x − 7 ) − 4( x − 7 ) = 0, ⇒, ( x − 7 )( x − 4) = 0, ⇒, x = 4 or x = 7, When x = 4, then 11 − x = 11 − 4 = 7, When x = 7, then 11 − x = 11 − 7 = 4, Hence, the numbers are 4 and 7., 18. Let the number of wickets taken by Zaheer in a cricket, match are x, then number of wickets taken by Harbhajan, = 2x − 3, According to the question,, x (2 x − 3) = 20, ⇒, 2 x 2 − 3x = 20, , = k 2 − 8k, , 9( k 2 − 16) ≥ 0, , − c 2 = − a 2(1 + m 2 ), , ⇒, , = ( − k )2 − 4 × 2 × k, , ⇒, , ⇒, , ⇒, ⇒, , On comparing with ax 2 + bx + c = 0, we get, ∴, , 4m c − 4 ( c 2 − a 2 + m 2c 2 − m 2a 2 ) = 0, , 17. Let one number be x., Then, another number = (11 − x ), [Q sum of two numbers = 11, given], According to the question,, 1, 1, 11, +, =, x (11 − x ) 28, 11 − x + x 11, ⇒, =, x(11 − x ) 28, , On comparing with ax 2 + bx + c, we get, ∴, , 2 2, , m 2c 2 − ( c 2 − a 2 + m 2c 2 − m 2a 2 ) = 0 [dividing by 4], m 2c 2 − c 2 + a 2 − m 2c 2 + m 2a 2 = 0, , ⇒, , ( x 2 + 1) 2 − x 2 = 0, ⇒, , (2 mc )2 − 4(1 + m 2 ) ( c 2 − a 2 ) = 0, , ⇒, , D =1, s 2 − 80 = 1, , [Q9 ≠ 0], [Q a 2 − b 2 = ( a − b )( a + b )], , ...(i), , Here, a and b are unknown constants. Since, x = 2 and x = 3, are the solutions of given equation, so it will satisfy the given, equation., On putting x = 2 and x = 3 one-by-one,, in Eq. (i), we get, 3(2 )2 − 2 a × (2 ) + 2 b = 0, ⇒, ⇒, , 3 × 4 − 4a + 2 b = 0, 12 − 4a + 2 b = 0, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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19, , CBSE Term II Mathematics X (Standard), , ⇒, ⇒, and, , − 2 (2 a − b − 6 ) = 0, 2a − b = 6, 3( 3)2 − 2 a × 3 + 2 b = 0, , 21. Given quadratic equation is, On comparing with ax 2 + bx + c = 0 , we get, , ⇒, 27 − 6a + 2 b = 0, … (iii), ⇒, 6a − 2 b = 27, On multiplying Eq. (ii) by 2 and then subtract it from, Eq. (iii), we get, 6a − 2 b − 4a + 2 b = 27 − 12, 15, ⇒, 2 a = 15 ⇒ a =, 2, 15, in Eq. (ii), we get, On substituting a =, 2, 15, −b=6, 2×, 2, ⇒, 15 − b = 6, ⇒, b = 15 − 6 = 9, ⇒, b=9, Hence, the required values of a and b are 15/2 and 9,, respectively., 20., , (i) Given quadratic equation is, 4x 2 + 12 x + 9 = 0, On comparing with ax 2 + bx + c = 0, we get, Now,, , ( k − 12 ) x 2 + 2( k − 12 ) x + 2 = 0, , [Q − 2 ≠ 0] ...(ii), , a = 4, b = 12 and c = 9, D = b 2 − 4ac = (12 )2 − 4( 4) ( 9), , = 144 − 144 = 0, Since, D = 0, so given quadratic equation has two equal, and real roots which are given by, − b ± D −12 ± 0, =, x=, 2( 4), 2a, −12 + 0, −12 − 0, or x =, ⇒, x=, 8, 8, 3, 3, x = − or x = −, ⇒, 2, 2, −3, 3, and − ., Hence, the roots are, 2, 2, (ii) Given quadratic equation is, 3x 2 + 5x − 7 = 0, On comparing with ax 2 + bx + c = 0, we get, , a = k − 12 , b = 2( k − 12 ) and c = 2, Now, D = b 2 − 4ac, = [2( k − 12 )]2 − 4( k − 12 ) (2 ), = 4( k − 12 )2 − 8( k − 12 ), = ( k − 12 ) [ 4( k − 12 ) − 8], = ( k − 12 ) ( 4k − 48 − 8), = ( k − 12 ) ( 4k − 56), Since, roots of given equation are equal., ∴, D=0, ⇒, , ( k − 12 ) ( 4k − 56) = 0, , ⇒, , k − 12 = 0 or 4k − 56 = 0, 56, k = 12 or k =, 4, , ⇒, ⇒, , Since, x = − 2 is a root of the given equation, so it will satisfy, the given equation., On putting x = − 2 in the given equation, we get, 3 ( − 2 )2 + 7 ( − 2 ) + p = 0, ⇒, 12 − 14 + p = 0, ⇒, −2 + p = 0, ⇒, p =2, On putting p = 2 in x 2 + k ( 4x + k − 1) + p = 0, we get, x 2 + k ( 4x + k − 1 ) + 2 = 0, ⇒, , a = 1, b = 4k and c = k 2 − k + 2, ∴, , − b ± D −5 ± 109, =, 2( 3), 2a, −5 + 109, 6, , ⇒, , x=, , or, , −5 − 109, x=, 6, , Hence, the roots are, , D = b 2 − 4ac, = ( 4k ) 2 − 4 × 1 × ( k 2 − k + 2 ), = 16k 2 − 4k 2 + 4k − 8, , Now, D = b 2 − 4ac = ( 5)2 − 4( 3) ( −7 ) = 25 + 84 = 109, , x=, , x 2 + 4kx + ( k 2 − k + 2 ) = 0, , On comparing with ax 2 + bx + c = 0, we get, , a = 3, b = 5 and c = − 7, Since, D > 0, so given quadratic equation has two, distinct real roots which are given by, , k = 12 or k = 14, , But k = 12 does not satisfy the given equation because if, k = 12, then coefficients of x 2 and x become zero., Hence, required value of k is 14., 22. Given equation is 3x 2 + 7 x + p = 0, , = 12 k 2 + 4k − 8, Since, roots are equal., ∴, D=0, ⇒, 12 k 2 + 4k − 8 = 0, 3k + k − 2 = 0, , ⇒, [taking +ve sign], [taking −ve sign], , −5 + 109, −5 − 109, and, ., 6, 6, , ⇒, , [divide by 4], , 2, , 2, , 3k + 3k − 2 k − 2 = 0, , ⇒ 3k ( k + 1) − 2 ( k + 1 ) = 0, ⇒, ( 3 k − 2 ) ( k + 1) = 0, 2, k = ,−1, ⇒, 3, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 20, 23. Let two consecutive odd natural numbers are x and x + 2., Then according to the given condition,, x 2 + ( x + 2 )2 = 130, ⇒, , x 2 + x 2 + 4x + 4 = 130, , ⇒, , 2 x 2 + 4x − 126 = 0, x 2 + 2 x − 63 = 0, , ⇒, ⇒, , C, , [divide by 2], , 2, , x + 9x − 7 x − 63 = 0, , ⇒, x ( x + 9 ) − 7 ( x + 9) = 0, ⇒, ( x − 7 ) ( x + 9) = 0, ⇒, x = 7, − 9, Since, natural number cannot be negative., So, we neglect x = − 9., Thus, x = 7 and x + 2 = 7 + 2 = 9, Hence, two consecutive odd numbers are 7 and 9., 24. Let the length of piece be x m., 200, Then, rate = `, per m, x, Now, new length = ( x + 5) m, Since, the cost remains same., 200, per m, ∴ New rate = `, x+5, According to the given condition,, 200, 200, =, −2, x+5, x, 200, ⎛ 100 − x ⎞, =2⎜, ⇒, ⎟, ⎝ x ⎠, x+5, ⇒, ⇒, , 100x = ( x + 5) (100 − x ), 100x = 100x − x 2 + 500 − 5x, x 2 + 5x − 500 = 0, , ⇒, ⇒, , 2, , x + 25x − 20x − 500 = 0, , ⇒, x ( x + 25) − 20 ( x + 25) = 0, ⇒, ( x − 20) ( x + 25) = 0, ⇒, x = 20, − 25, Since, length of piece cannot be negative, so neglect x = − 25., Thus, x = 20, 200 200, Now, rate =, =, = ` 10, x, 20, Hence, length of piece is ` 20 m and rate per metre is ` 10., 25. Let first number be x., Then, second number = x + 4, [Q difference of two numbers = 4], According to the question,, 1, 1, 4, −, =, x x + 4 21, ( x + 4) − x, 4, 4, 4, =, ⇒, =, ⇒ 2, x ( x + 4), 21, x + 4x 21, ⇒, ⇒, ⇒, ⇒, , When x = − 7, then second number = − 7 + 4 = − 3, When x = 3, then second number = 3 + 4 = 7, Hence, two numbers are − 7 , − 3 or 3, 7., 26. Let one side = x., , x 2 + 4x = 21 ⇒ x 2 + 4x − 21 = 0, x 2 + (7 − 3)x − 21 = 0 ⇒ x 2 + 7 x − 3x − 21 = 0, x ( x + 7 ) − 3 ( x + 7 ) = 0 ⇒ ( x − 3) ( x + 7 ) = 0, x = − 7, 3, , x, , 29, , A, , B, , Now, perimeter of a triangle,, 70 = x + 29 + AB, ⇒, AB = 70 − 29 − x = 41 − x., In right ΔABC, use Pythagoras theorem,, BC 2 = AC 2 + AB 2, ⇒, , (29)2 = x 2 + ( 41 − x )2, 841 = x 2 + 1681 + x 2 − 82 x, , ⇒, , 2 x 2 − 82 x + 840 = 0, , ⇒, , x 2 − 41x + 420 = 0, , ⇒, ⇒, , [divide by 2], , 2, , x − 21x − 20x + 420 = 0, , ⇒, x ( x − 21) − 20 ( x − 21) = 0, ⇒, ( x − 20) ( x − 21) = 0, ⇒, x = 20, 21, Hence, length of other sides of a ΔABC are 20 units, 21 units., 27. Let present age of Anjali be x yr., ∴ Anjali’s age 3 yr ago = ( x − 3) yr, and Anjali’s age 5 yr from now = ( x + 5) yr, According to the question,, 1, 1, 1, +, =, x−3 x+ 5 3, x+ 5+ x−3 1, =, ⇒, ( x − 3)( x + 5) 3, 2x + 2, 1, ⇒, =, x 2 − 3x + 5x − 15 3, ⇒, , 3(2 x + 2 ) = x 2 + 2 x − 15, , ⇒, , 6x + 6 = x 2 + 2 x − 15, , ⇒ x 2 + 2 x − 15 − 6x − 6 = 0, ⇒, , x 2 − 4x − 21 = 0,, , which is the required quadratic equation., Now, by factorisation method, we get, x 2 − 7 x + 3x − 21 = 0, ⇒, x( x − 7 ) + 3( x − 7 ) = 0, ⇒, ( x − 7 )( x + 3) = 0, ⇒, x − 7 = 0 or x + 3 = 0, ⇒, x = 7 or x = − 3, But x = − 3 is not possible because age cannot be negative., ∴ x =7, Hence, Anjali’s present age is 7 yr., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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21, , CBSE Term II Mathematics X (Standard), , 28. Let the ten’s digit of the number be x., According to the question,, Product of the digits = 12, i.e. Ten’s digit × Unit’s digit = 12, 12, [Q ten’s digit = x], ⇒, Unit’s digit =, x, 12, Two-digit number = 10x +, ∴, x, Also, it is given that if 36 is added to the number, the digits, get interchange., 12, 12, 10x +, + 36 = 10 ×, +x, ∴, x, x, 10x 2 + 12 + 36x = 120 + x 2, , ⇒, , 9x 2 − 108 + 36x = 0, x 2 + 4x − 12 = 0, , ⇒, ⇒, , 30. Let length of the shortest side = x m., Then, hypotenuse = (2 x + 6) m and, third side = (2 x + 6 − 2 ) m = (2 x + 4) m, By Pythagoras theorem,, ( 2 x + 6 ) 2 = x 2 + ( 2 x + 4) 2, [Q (Hypotenuse) 2 = (Perpendicular) 2 + (Base) 2], ⇒ 4x 2 + 24x + 36 = x 2 + 4x 2 + 16x + 16, [Q( a + b )2 = a 2 + 2 ab + b 2], ⇒ x 2 + 4x 2 + 16x + 16 − 4x 2 − 24x − 36 = 0, By quadratic formula,, x=, , [divide both sides by 9], , which is the required quadratic equation., By factorisation method, we get, x 2 + 6x − 2 x − 12 = 0, ⇒, x( x + 6) − 2( x + 6) = 0, ⇒, ( x + 6)( x − 2 ) = 0, ⇒, x + 6 = 0 or x − 2 = 0, ⇒, x = − 6 or x = 2, But a digit can never be negative., , ⇒, , So, x = 2., , ⇒, , ⇒, ⇒, , Hence, the required two-digit number, 12, = 10 × 2 +, = 20 + 6 = 26, 2, 29. Given, John and Janvi together have 45 marbles., Let John has x marbles., Then, number of marbles Janvi has = 45 − x, Q Both of them lost 5 marbles each., ∴ The number of marbles John has = x − 5, and the number of marbles Janvi has = 45 − x − 5 = 40 − x, Now, product of the number of marbles = 124, ∴, ( x − 5) ( 40 − x ) = 124, ⇒, 40x − x 2 − 200 + 5x = 124, ⇒, , − x 2 + 45x − 200 − 124 = 0, , ⇒, , − x 2 + 45x − 324 = 0, , ⇒, , x 2 − 45x + 324 = 0, , [multiplying by ( −1)], , which is the required quadratic equation., Now, by factorisation method, we get, x 2 − 36x − 9x + 324 = 0, ⇒, x( x − 36) − 9( x − 36) = 0, ⇒, ( x − 36)( x − 9) = 0, ⇒, x − 36 = 0 or x − 9 = 0, ⇒, x = 36 or x = 9, when John has 36 marbles, then, Janvi has = 45 − 36 = 9 marbles., when John has 9 marbles, then, Janvi has = 45 − 9 = 36 marbles., , x 2 − 8x − 20 = 0, , ⇒, , − ( −8) ± ( − 8)2 − 4 × 1 × ( −20), 2 ×1, , ⎡, − b ± b 2 − 4ac, ; here a = 1, b = −8 and c = −20], ⎢Q x =, 2a, ⎢⎣, 8 ± 64 + 80, 8 ± 144, ⇒ x=, x=, 2, 2, 8 ± 12, x=, 2, 8 + 12, 8 − 12, or x =, x=, 2, 2, −4, 20, or x =, x=, 2, 2, , ⇒, x = 10 or x = − 2, But length of side cannot be negative., ∴, x = 10, Hence, shortest side is 10 m, hypotenuse is 2 × 10 + 6 = 26 m, and third side = 2 × 10 + 4 = 24 m., 31. Let Nisha’s present age be x yr., Then, Asha’s present age = x 2 + 2, , [by given condition], , Now, when Nisha grows to her mother’s present age., Then, Asha’s age will be [( x 2 + 2 ) – x ] yr., Again by given condition,, Age of Asha = One year less than 10 times the present age of, Nisha, ( x 2 + 2 ) + {( x 2 + 2 ) – x} = 10x – 1, ⇒, 2 x 2 – x + 4 = 10x – 1, ⇒, 2 x 2 – 11x + 5 = 0, 2, ⇒, 2 x − 10x – x + 5 = 0, ⇒, 2 x ( x – 5) – 1 ( x – 5) = 0, ⇒, ( x – 5 ) (2 x – 1 ) = 0, ∴, x=5, 1, 1, [here, x = cannot be possible, because at x = ,, 2, 2, 1, Asha’s age is 2 yr which is not possible], 4, Hence, required age of Nisha = 5 yr, and required age of Asha = x 2 + 2, = ( 5)2 + 2 = 25 + 2 = 27 yr, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 22, 32. Let speed of the stream = x km/h, Given, speed of boat in still water = 15 km/h, ∴ Speed of boat upstream = (15 − x ) km/h, and speed of boat downstream = (15 + x ) km/h, According to the question,, 30, 30, 1, +, =4, 15 − x 15 + x, 2, ⎡, distance, ⎢Q time = speed and distance = 30 km, ⎣, 30⎞, 1 ⎤, ⎛, and also, 4h 30 min = ⎜ 4 + ⎟ h = 4 h, ⎝, 60⎠, 2 ⎥⎦, 30(15 + x ) + 30(15 − x ) 9, ⇒, =, (15 − x )(15 + x ), 2, ⇒, , 450 + 30x + 450 − 30x 9, =, 2, (15)2 − x 2, , When, x = 5, Time taken by smaller tap = 5 h, Time taken by larger tap = x − 2 = 5 − 2 = 3 h, 3, When, x =, 4, 3, Time taken by smaller tap = h, 4, Time taken by larger tap = x − 2, 3, −5, = − 2 = , which is not solution., 4, 4, Hence, time taken by smaller tap = 5 h and time taken by, larger tap = 3 h., 34. (i) Given that a rectangular pond has to be constructed in the, centre of a rectangular lawn of dimensions 50 m × 40 m ., x, , 2, , 2, , x, , [Q ( A − B )( A + B ) = A − B ], 900, 9, =, 225 − x 2 2, , ⇒, , 900 × 2, = 225 − x 2 ⇒ 200 = 225 − x 2 ⇒ x 2 = 25, 9, [taking square root on both sides], ⇒, x=±5, But speed cannot be negative., ∴ x=5, Hence, speed of stream is 5 km/h., 33. Let the time taken by smaller tap to fill tank completely = x h, 1, So, volume of tank filled by smaller tap in 1 h =, x, 1, Volume of tank filled by larger tap in 1 h =, x −2, 7 15, h, Now, time taken by both taps to fill = 1 =, 8 8, 15, 1 15 15, h= ×, Tank filled by smaller tap in, =, 8, x, 8 8x, 15, 1, 15, 15, Tank filled by larger tap in, h=, ×, =, x −2, 8 8( x − 2 ), 8, , x, x, , ⇒, , Therefore,, ⇒, , 15, 15, 15 ⎡ 1, 1 ⎤, +, =1 ⇒, +, =1, 8x 8( x − 2 ), 8 ⎢⎣ x x − 2 ⎥⎦, 2( x − 1) 8, =, ⇒ 15( x − 1) = 4( x 2 − 2 x ), x 2 − 2 x 15, 2, , ⇒, , 15x − 15 = 4x − 8x, 23x = 4x 2 + 15, , ⇒, ⇒, , 2, , 4x − 23x + 15 = 0, , By using quadratic formula, − ( −23) ± ( −23)2 − 4 ⋅ 4 ⋅ 15, 23 ± 529 − 240, x=, ⇒ x=, 2⋅4, 8, 23 ± 17, 23 ± 289, x=, ⇒ x=, ⇒, 8, 8, 23 + 17 40, Taking positive sign, x =, =, =5, 8, 8, 23 − 17 6 3, Taking negative sign, x =, = =, 8, 8 4, , 40 m, , 50 m, , Now, length of rectangular lawn, ( l1 ) = 50 m, and breadth of rectangular lawn, ( b1 ) = 40 m, ∴ Length of rectangular pond, ( l2 ) = 50 – ( x + x ) = 50 – 2 x, and breadth of rectangular pond, ( b 2 ) = 40 – ( x + x ) = 40 – 2 x, (ii) Given, area of the grass surrounding the pond = 1184 m2, ∴ Area of rectangular lawn – Area of rectangular pond, = Area of grass surrounding the pond, l1 × b1 – l2 × b 2 = 1184, [Q area of rectangle = length × breadth], ⇒, 50 × 40 − ( 50 − 2 x ) ( 40 − 2 x ) = 1184, ⇒ 2000 − (2000 − 80x − 100x + 4x 2 ) = 1184, 80x + 100x – 4x 2 = 1184, , ⇒, , 4x 2 − 180x + 1184 = 0, , ⇒, , x 2 – 45x + 296 = 0, , ⇒, , [divide by 4], , 2, , (iii) Now, x − 4.5x + 296 = 0 ⇒ x 2– 37 x – 8x + 296 = 0, [by splitting the middle term], ⇒ x ( x – 37 ) – 8 ( x – 37 ) = 0, ⇒, ( x – 37 ) ( x – 8) = 0, ∴, x=8, [at x = 37, length and breadth of pond are –24 and –34,, respectively but length and breadth cannot, be negative. So, x = 37 cannot be possible], ∴ Length of pond = 50 – 2 x = 50 – 2 ( 8), = 50 – 16 = 34 m, and breadth of pond = 40 – 2 x, = 40 – 2( 8) = 40 – 16 = 24 m, Hence, required length and breadth of pond are 34 m, and 24 m, respectively., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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Chapter Test, Multiple Choice Questions, , (ii) 15 y 2 − 41y − 14 = 0, , 1. Is − 8 a solution of the equation, 3 x 2 + 8 x + 2 = 0?, , [NCERT Exemplar], , (a) Yes, (b) No, (c) Cannot be determined, (d) None of the above, , x − 14 x + 24 = 0, , [CBSE 2013], , (a) 2, 12, , (b) 3, 8, , (c) 8, 3, , (d) None of these, 2, , 3. The roots of a equation 2x + 5 2x + 5 = 0 are, , (c), , −5 2 ± 10, 5, , −5 2 ± 10, (b), 4, (d) None of these, , 4. The quadratic equation 7 y 2 − 4 y + 5 = 0 has, (a) Real and distinct, (b) Real and equal, (c) Imaginary, (d) More than 2 real roots, , (c) − 21,, , 7, 2, 7, 3, , 1, =0, 21, 1, ,, 21, 1, (d) ,, 21, (b), , 1, 21, , 1, 21, 3, 21, , (iv) 6 x 2 − 31x + 40 = 0, 5 8, ,−, 2 3, 2 5, (d) ,, 3 8, , 5 8, ,, 2 3, 5 3, (c) − ,, 2 8, , (b) −, , (a), , (v) 3 x 2 + 2 5x − 5 = 0, (a) 2 5 , − 5, , (b) 5 , −, , 5, 3, , (c) − 5 , 3 5, , (d) − 5 ,, , 5, 3, , Short Answer Type Questions, , 5. If a number is added to twice its square, then, the resultant is 21. The quadratic representation, of this situation is, [CBSE 2014, 15], (a) 2 x2 + x − 21 = 0, (b) 2 x2 + x + 21 = 0, (c) 2 x2 − x + 21 = 0, (d) 2 x2 − x − 21 = 0, , 7. Find the roots of the equation x 2 + 182 = 27 x, 8. Find the roots of the quadratic equation, a2b 2x 2 + b 2x − a2x − 1 = 0 [CBSE 2012, 11], , 9. If the roots of the equation x 2 + 2cx + ab = 0 are, real and unequal, then prove that the equation, x 2 − 2(a + b) x + a2 + b 2 + 2c2 = 0 has no real roots., , Case Study MCQs, , Long Answer Type Questions, , 6. Sohan is preparing for UPSC exam. For this, he, has to practice the chapter of quadratic, equations. So, he started with factorisation, method., Let two roots of ax 2 + bx + c be p and q., ∴, , 3, ,, 5, 2, (d) ,, 5, (b), , (a) 21, 3, , 2, , 5, , 2, 3, 7, 5, , (iii) 21x 2 − 2x +, , 2. Solve the quadratic equation, , −5 2 ±, (a), 2, , 3, (a) ,, 7, 2, (c) ,, 3, , Now, factorize each of the following quadratic, equations and find the roots., (i) x 2 − 10x + 21 = 0, (b) 21, 1, , (c) 3, 7, , (d) 3, 9, , 2x 2 + px − 15 = 0 and the quadratic equation, p(x 2 + x) + k = 0 has equal roots, then find the, value of k., [CBSE 2016], , 11. A rectangular park is to be designed whose, , ac = p × q and p + q = b, , (a) 9, 3, , 10. If x = − 5 is a root of the quadratic equation, , breadth is 3 m less than its length. Its area is to, be 4 sq m more than the area of a park that has, already been made in the shape of an isosceles, triangle with its base as the breadth of the, rectangular park and of altitude 12 m. Find its, length and breadth of the rectangular park., , Answers, 1. (b), 7. 13 ,14, , 2. (a), , 3. (b), 8. x = +, , 4. (c) 5. (a), 1, 1, ,−, b2, a2, , 6. (i) (c) (ii) (d) (iii) (b) (iv) (a) (v) (d), 10. 7/4, , For Detailed Solutions, Scan the code, , 11. Length = 7 m and Breadth = 4 m, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 24, , CHAPTER 02, , Arithmetic, Progressions, In this Chapter..., !, , Arithmetic Progression, , !, , nth Term of an AP, , !, , Sum of n-Term of an AP, , !, , Arithmetic Mean, , Sequence Some numbers arranged in definite order,, according to a definite rule are said to form a sequence., Progression Sequences which follow a definite pattern are, called progressions., , Arithmetic Progression, An Arithmetic Progression (AP) is a list of numbers in which, each term is obtained by adding a fixed number to the, preceding term except the first term., This fixed number is called the common difference ( d ) of the, AP. It can be positive, negative or zero., In other words, a list of numbers a 1 , a 2 , a 3 ,. . . , a n is called, an arithmetic progression (AP), if there exists a constant, number d (called common difference) such that, a 2 − a1 = d, a3 − a2 = d, a4 − a3 = d, M, a n − a n − 1 = d and so on., Each of the number in this list is called a term., , In general, a , a + d , a + 2 d , a + 3 d , … represent an arithmetic, progression, where a is the first term and d is the common, difference. This is called general form of an AP., If number of terms in an AP is finite, then it is called a finite, AP, otherwise it is called an infinite AP and such AP ’s do not, have a last term., Method to Check an AP, When a List of Numbers is Given, , Sometimes, a list of numbers or sequence is given and we, have to check that this sequence is an AP or not. For this, we, find the differences of consecutive terms. If these differences, are same, then given list of numbers or sequence is an AP,, otherwise not., Method to Write an AP When First Term and, Common Difference are Given, , To write an AP, the minimum information required to know, the first term a and the common difference d of the, arithmetic progression. Then, we put the values of a and d in, a , a + d , a + 2 d , a + 3 d , … to get the required AP., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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25, , CBSE Term II Mathematics X (Standard), , nth Term of an AP, , Sum of First n-Terms of an AP, , If the first term of an AP is ‘a’ and its common difference is, ‘d’, then its nth term is given by the formula, , If first term of an AP is ‘a’ and its common difference is ‘d’,, then the sum of its first n terms S n , is given by the formula, , a n = a + ( n − 1 )d, The nth term of an AP is also called its general term., In an AP, nth term is known as last term of an AP and it is, denoted by l, which is given by the formula, l = a + ( n − 1 )d, , Let ‘a’ be the first term, ‘d’ be the common difference and ‘l’, be the last term of an AP, then nth term from the end can be, found by the formula, nth term from the end = l − ( n − 1 )d, Selection of Terms in an AP, Terms, , n, [2 a + ( n − 1 ) d ], 2, , Sn =, , n, [a + a n ], 2, , where, a n = nth term of an AP., , nth Term from the End of an AP, , Number of terms, , or, , Sn =, , (i) If l is the last term of an AP having n terms, then sum of, all the terms is given by this formula, n, Sn = [a + l ], 2, (ii) If S n and S n −1 are the sums of first n and ( n −1 ) terms of, an AP respectively, then its nth term a n is given by, a n = Sn − Sn − 1, , Common difference, , 3, , a − d, a , a + d, , 4, , a − 3 d, a − d, a + d, a + 3 d, , 2d, , 5, , a − 2 d, a − d, a , a + d, a + 2 d, , d, , d, , Arithmetic Mean, If a , b and c are in AP, then b is known as arithmetic mean of, a+c, ., a and c, i.e. b =, 2, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 26, , Solved Examples, Example 1. Examine that the sequence 13, 10, 7, 4,... is, an AP., , 3, 6, 9, 12, ..., 111?, , Sol. Given, AP is 13, 10, 7, 4, ……, Here, a1 = 13, a 2 = 10, a 3 = 7, a 4 = 4, ………, Here, we have a 2 − a1 = 10 − 13 = −3,, a 3 − a 2 = 7 − 10 = −3,, a 4 − a 3 = 4 − 7 = −3 and so on., Since, difference of any two consecutive terms is same., So, the given sequence is an AP., , Example 2. Find the common difference of the following, AP’s ., (i) 3 , − 2 , − 7 , − 12 ,. . ., 1, 1, 1 1, (iii) 5 , 9 ,13 ,17 ……, 2, 2, 2 2, , Example 5. How many terms are there in the sequence, Sol. Given, sequence is 3, 6, 9, 12, ..., 111., Here, 6 − 3 = 9 − 6 = 12 − 9 ... = 3, So, it is an AP with first term, a = 3 and common difference,, d = 3. Let there be n terms in the given sequence., Then,, , nth term = 111, , ⇒, , a + ( n − 1)d = 111, , ⇒, , 3 + ( n − 1) × 3 = 111, , [Q a n = ( a + ( n − 1)d)], , ⇒, , (ii) 11, 11, 11, 11, ..., , Sol. (i) Given, AP is 3, − 2 , − 7 , − 12 ,..., Here, a1 = 3, a 2 = −2 , a 3 = −7, a 4 = −12 and so on., ∴ Common difference ( d) = a 2 − a1 = −2 − 3 = −5, (ii) Given, AP is 11, 11, 11, 11, ..., Here, a1 = 11, a 2 = 11, a 3 = 11, a 4 = 11 and so on., ∴Common difference ( d) = a 2 − a1 = 11 − 11 = 0, 1 1, 1, 1, (iii) Given, AP is 5 , 9 ,13 ,17 ,……, 2 2, 2, 2, 1, 1, 1, 1, Here, a1 = 5 , a 2 = 9 , a 3 = 13 , a 4 = 17 and so on., 2, 2, 2, 2, 1, 1, ∴ Common difference (d) = a 2 − a1 = 9 − 5, 2, 2, 19 11 8, =, −, = =4, 2 2 2, , Example 3. Write an AP having 4 as the first term and, − 3 as the common difference., Sol. Given, first term ( a ) = 4 and common difference ( d) = − 3, On putting the values of a and d in general form, a , a + d, a + 2 d, a + 3d,... , we get, 4, 4 − 3, 4 + 2( −3), 4 + 3( −3), ..., 4, 1, 4 – 6, 4 – 9, … or 4, 1, − 2 , − 5,..., Which is the required AP., , Example 4. Find the 20th term of the sequence, 7, 3, −1 , −5 . . ., Sol. Given, sequence is 7 , 3, − 1, − 5,... ., Here, 3 − 7 = − 4, −1 − 3 = − 4, −5 + 1 = − 4 and so on., So, given sequence is an AP, in which a = 7 and d = − 4., Since, nth term, a n = a + ( n − 1)d, On putting n = 20, we get, [Q a = 7 , d = −4], a 20 = a + (20 − 1)d = 7 + 19 ( −4), = 7 − 19 × 4 = 7 − 76 = − 69, Hence, 20th term of given sequence is − 69., , 3(1 + n − 1) = 111, 111, ⇒, n=, ⇒ n = 37, 3, Hence, the given sequence contains 37 terms., , Example 6. Which term of the AP: 21 , 18 , 15,. . . is −81 ?, Sol. Given, AP is 21, 18, 15,... ., Here, a = 21 and d = 18 − 21 = −3, Let nth term of given AP be − 81, Then,, a n = −81, [Q a n = a + ( n − 1)d ], ⇒, a + ( n − 1)d = −81, On putting the values of a and d, we get, 21 + ( n − 1)( −3) = −81 ⇒ 21 − 3n + 3 = −81, ⇒, 24 − 3n = −81 ⇒ −3n = −81 − 24 = −105, − 105, n=, = 35, ⇒, −3, Hence, 35th term of given AP is − 81., , Example 7. How many numbers of two digits are, divisible by 7?, Sol. Two-digits numbers are 10, 11, 12 , 13, 14, 15,... , 97 , 98, 99, in which only 14, 21,28,..., 98 are divisible by 7., Here, 21 − 14 = 28 − 21... = 7., So, this list of numbers forms an AP, whose first term, ( a ) = 14, common difference ( d) = 7., Let there are n terms in the above sequence, then a n = 98, [Q a n = a + ( n − 1)d], ⇒, a + ( n − 1)d = 98, ⇒, 14 + ( n − 1)7 = 98 ⇒ 14 + 7 n − 7 = 98, 91, ⇒, 7 n = 91 ⇒ n =, = 13, 7, Hence, 13 numbers of two digits are divisible by 7., , Example 8. Determine the 10 th term from the end of, the AP : 4, 9, 14, ..., 254., Sol. Given, AP is 4, 9, 14,..., 254., Here,, l = last term = 254, d = common difference = 9 − 4 = 5, ∴ 10th term from the end = l − (10 − 1)d = l − 9d, = 254 − 9 × 5 = 254 − 45 = 209, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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27, , CBSE Term II Mathematics X (Standard), , Alternate Method, On reversing the given AP, new AP is 254, ..., 14, 9, 4., Here, first term ( a ) = 254 and, common difference ( d) = 4 − 9 = −5, Now, 10th term of new AP = a10, = 254 + (10 − 1)( −5), = 254 − 9 × 5 = 209, Hence, 10th term from the end of given AP is 209., , Example 9. Determine the general term of an AP, whose 7th term is − 1 and 16 th term is 17., Sol. Let a be the first term and d be the common difference of, the AP, whose 7th term is −1 and 16th term is 17., Since, a 7 = − 1 and a16 = 17, ...(i), ∴ We have, a + (7 − 1)d = − 1 ⇒ a + 6d = − 1, and, ...(ii), a + (16 − 1)d = 17 ⇒ a + 15d = 17, [Q a n = a + ( n − 1)d], On subtracting Eq. (i) from Eq. (ii), we get, a + 15d − a − 6d = 17 + 1, ⇒, 9d = 18 ⇒ d = 2, On substituting d = 2 in Eq. (i), we get, a + 6 × 2 = −1, ⇒, a + 12 = − 1, ⇒, a = − 13, Hence, general term,, a n = a + ( n − 1 )d, = − 13 + ( n − 1)2 [Q a = − 13 and d = 2], = − 13 + 2 n − 2 = 2 n − 15, , Example 10. Find four numbers in AP whose sum is 20, and the sum of whose squares is 120., Sol. Let the numbers be a − 3d, a − d, a + d and a + 3d., Then, according to the given condition, we have, ( a − 3d) + ( a − d) + ( a + d) + ( a + 3d) = 20, and ( a − 3d)2 + ( a − d)2 + ( a + d)2 + ( a + 3d )2 = 120, , preceding year. If it grew by 1m in the first year,, then in how many years will it have ceased, [CBSE 2015], growing?, Sol. Given that, tree grow 5 cm or 0.05 m less than preceding, year., ∴The following sequence can be formed., 1, (1 − 0. 05), (1 − 2 × 0. 05), ... ,0, i.e., 1, 0.95, 0.90, ... ,0 which is an AP., Here,, a = 1, d = 0. 95 − 1 = − 0. 05 and l = 0, Let, l = a n = a + ( n − 1)d, Then,, 0 = 1 + ( n − 1) ( − 0. 05), , …(i), …(ii), , 2, , ⇒, 4a + 20d = 120, ⇒, a 2 + 5d2 = 30, ⇒, 25 + 5d2 = 30 [Q a = 5], 2, 2, ⇒, 5d = 5 ⇒ d = 1 ⇒ d = ±1, If d = 1, then the numbers are 2, 4, 6, 8 and if d = −1, then, the numbers are 8, 6, 4, 2., Hence, the numbers are 2, 4, 6, 8 or 8, 6, 4, 2., , Example 11. A sum of ` 2000 is invested at 7% simple, interest per year. Calculate the interest at the end, of each year. Do these interest form an AP? If so,, then find the interest at the end of 20th year, making use of this fact., , ( n − 1) ( 0. 05) = 1, , 1, 0. 05, 1, n − 1 = × 100, ⇒, 5, ⇒, n − 1 = 20, ⇒, n = 21, Hence, in 21 yr, tree will have ceased growing., ⇒, , 4a = 20 ⇒ a = 5, From Eq. (ii), we get, a 2 + 9d2 − 6da + a 2 + d2 − 2 ad + a 2 + d2 + 2 ad, + a 2 + 9d2 + 6ad = 120, , Sol. Given, initial money P = ` 2000, , Example 12. Each year, a tree grow 5 cm less than the, , ⇒, , From Eq. (i), we get, , 2, , Rate of interest, R = 7% per year; Time, T = 1, 2, 3, 4,..., We know that, simple interest is given by the following formula, PRT, SI =, 100, 2000 × 7 × 1, ∴SI at the end of 1st year =, = ` 140, 100, 2000 × 7 × 2, SI at the end of 2nd year =, = ` 280, 100, 2000 × 7 × 3, SI at the end of 3rd year =, = ` 420, 100, Thus, required list of numbers is 140, 280, 420, ... ., Here, 280 − 140 = 420 − 280 K = 140, So, above list of numbers forms an AP, whose first term, ( a ) = 140 and common difference ( d) = 140., Now, SI at the end of 20th year will be equal to 20th term of, the above AP., Q a 20 = a + (20 − 1) d = 140 + 19 × 140 = 140 + 2660 = 2800, Hence, the interest at the end of 20th year will be ` 2800., , n −1=, , Example 13. The eighth term of an AP is half its second, term and the eleventh term exceeds one-third of its, fourth term by 1. Find the 15th term., [NCERT Exemplar], Sol. Let a and d be the first term and last term of an AP. Then,, 1, 1, a 8 = a 2 and a11 = a 4 + 1, 2, 3, 1, ⇒ a + ( 8 − 1) d = [ a + (2 − 1) d], 2, 1, and a + (11 − 1) d = [( a + ( 4 − 1) d) + 1], 3, 1, a + 7 d = ( a + d), ⇒, 2, 1, and, a + 10d = [( a + 3d) + 1], 3, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 28, ⇒ 2 a + 14d − a − d = 0, and, 3a + 30d = a + 3d + 1, ⇒, a + 13d = 0, and 2 a + 27 d − 1 = 0, On solving Eqs. (i) and (ii), we get, a = − 13, d = 1, ∴, a15 = a + (15 − 1) (1), = − 13 + 14 = 1, , ⇒, …(i), …(ii), , Example 14. The fourth term of an AP is 11. The sum, of the fifth and seventh terms of the AP is 24. Find, its common difference., [CBSE 2015], Sol. Let a be the first term and d be the common difference., Then,, a 4 = 11 ⇒ a + ( 4 − 1) d = 11, …(i), ⇒, a + 3d = 11, Also, given a 5 + a 7 = 24, ⇒, a + ( 5 − 1) d + a + (7 − 1) d = 24, ⇒, a + 4d + a + 6d = 24, ⇒, 2 a + 10d = 24, [divide by 2] …(ii), ⇒, a + 5d = 12, On subtracting Eq. (i) from Eq. (ii), we get, 1, 2d = 1 ⇒ d =, 2, 1, Hence, common difference is ., 2, , Example 15. Find the sum of the first 22 terms of the, AP : 8, 3, − 2, ..., Sol. Given, AP is 8, 3, − 2, ..., Here, first term, ( a ) = 8, Common difference, ( d) = 3 − 8 = − 5 and n = 22, n, Q Sum of first n terms, (Sn ) = [2 a + ( n − 1) d], 2, 22, ∴ Sum of first 22 terms, (S22 ) =, [2 × 8 + (22 − 1) × ( − 5)], 2, = 11 [16 + 21 × ( − 5)], = 11 [16 − 105], = 11 ( − 89) = − 979, Hence, sum of first 22 terms of an AP is − 979., , Example 16. Find the sum of first 24 terms of an AP,, whose nth term is given by a n = 3 + 2 n., Sol. Given, nth term of an AP, a n = 3 + 2 n, Clearly, sum of first 24 terms, (S24 ), 24, =, ( a + a 24 ) = 12( 5 + 51), 2, [Q a1 = 3 + 2 = 5 and a 24 = 3 + 2 × 24 = 3 + 48 = 51], = 12 × 56 = 672, , Example 17. If the sum of first 10 terms of an AP is 140, and the sum of first 16 terms is 320, then find the, sum of first m terms., Sol. Let the first term of this AP be a and common difference be d., Given, sum of first 10 terms, (S10 ) = 140, , 10, 140, [2 a + (10 − 1)d] = 140 ⇒ 2 a + 9d =, 2, 5, n, ⎡, ⎤, Q S = [2 a + ( n − 1)d], ⎢⎣ n 2, ⎥⎦, , ...(i), ⇒, 2 a + 9d = 28, Also, given sum of first 16 terms, (S16 ) = 320, 16, ⇒, [2 a + (16 − 1)d] = 320, 2, 320, ...(ii), 2 a + 15d =, ⇒ 2 a + 15d = 40, ⇒, 8, On subtracting Eq. (i) from Eq. (ii), we get, 6d = 12 ⇒ d = 2, On putting d = 2 in Eq. (i), we get, 2 a + 9(2 ) = 28 ⇒ 2 a = 28 − 18, 10, ⇒, a=, =5, 2, Thus, a = 5 and d = 2., m, Hence, sum of first m terms, (Sm ) = [2 a + ( m − 1)d], 2, m, = [2( 5) + ( m − 1)2 ] = m [ 5 + ( m − 1)], 2, = m ( 5 + m − 1 ) = m ( m + 4 ) = m 2 + 4m, , Example 18. Find the sum of all three-digit natural, [CBSE 2009], numbers, which are multiples of 11., Sol. All three-digit natural numbers, multiples of 11 are 110, 121,, 132, …, 990., Here, common difference, 121 − 110 = 132 − 121 = ... = 11., So, it is an AP with first term, a = 110, common difference,, d = 11 and last term, l = 990., Let, l = a n = a + ( n − 1)d, ∴, 990 = 110 + ( n − 1) × 11, ⇒, 990 = 110 + 11n − 11, ⇒, 11n = 891 ⇒ n = 81, n, Q, Sn = [ a + l ], 2, 81, S81 = [110 + 990], ∴, 2, 81, =, × 1100 = 81 × 550 = 44550, 2, , Example 19. If S n , the sum of first n terms of an AP is, given by S n = 3n 2 − 4n, find the nth term.[CBSE 2019], Sol. Given,, , S n = 3n 2 − 4n, , …(i), , On replacing n by ( n − 1) in Eq. (i), we get, Sn −1 = 3( n − 1)2 − 4( n − 1), nth term of the AP a n = Sn − S n −1, ∴, a n = ( 3n 2 − 4n ) − [ 3( n − 1)2 − 4( n − 1)], ⇒, , a n = 3[ n 2 − ( n − 1)2 ] − 4[ n − ( n − 1)], , ⇒, , a n = 3[ n 2 − n 2 + 2 n − 1] − 4[ n − n + 1], , ⇒, a n = 3 (2 n − 1) − 4, ⇒, a n = 6n − 3 − 4 ⇒ a n = 6 n − 7, Thus, the nth term of the AP = 6n − 7., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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29, , CBSE Term II Mathematics X (Standard), , Chapter, Practice, PART 1, Objective Questions, ●, , Multiple Choice Questions, 1. Which of the following form of an AP?, (a) − 1, − 1 , − 1 , − 1 , ..., (c) 1 , 1 , 2 , 2 , 3, 3, K, , [NCERT Exemplar], (b) 0, 2 , 0, 2, …, 1 1 1, (d) , , , K, 2 3 4, , 2. Which of the following is not an AP?, [CBSE 2020 (Standard)], (a) −1.2 , 0.8, 2.8, ..., (b) 3, 3 + 2 , 3 + 2 2 , 3 + 3 2 , …, 4 7 9 12, (c) , , , , ...., 3 3 3 3, −1 −2 −3, (d), ,, , , ...., 5 5 5, , 9, (b), 14, 19, (d), 14, , 4. The common difference of an AP, whose nth term is, a n = ( 3n + 7 ), is, (b) 7, (d) 6, , 5. The value of x for which 2 x ,( x + 10) and ( 3x + 2 ) are, the three consecutive terms of an AP, is, (a) 6, (c) 18, , [CBSE 2020 (Standard)], (b) − 6, (d) −18, , 6. The value of p for which (2 p + 1 ), 10 and ( 5p + 5) are, three consecutive terms of an AP is, (a) − 1, (c) 1, , [NCERT Exemplar], (b) − 2 , 4, − 8, 16, (d) − 2 , − 4, − 8, − 16, , 8. Let a be a sequence defined by a1 = 1, a 2 = 1 and, a, a n = a n − 1 + a n − 2 for all n > 2, then the value of 4 is, a3, 2, 3, , (b), , 5, 4, , (c), , 4, 5, , (d), , 3, 2, , 9. If an AP have 8 as the first term and −5 as the, common difference and its first three terms are, 8, A , B, then ( A + B ) is equal to, , [CBSE 2020 (Standard)], , (a) 3, (c) 10, , (a) − 2 , 0, 2 , 4, (c) − 2 , − 4, − 6, − 8, , (a), , 5, 3. If − , a, 2 are consecutive terms in an Arithmetic, 7, Progression, then the value of ‘a’ is, 9, (a), 7, 19, (c), 7, , 7. The first four terms of an AP whose first term is − 2, and the common difference is −2, are, , (b) − 2, (d) 2, , (a) 0, (c) 1, , (b) −1, (d) 2, , 10. In an AP, if d = −4, n = 7 and a n = 4, then a is equal to, (a) 6, (c) 20, , (b) 7, (d) 28, , 11. The 11th term of an AP − 5,, , −5, 5, , 0, , . . ., 2, 2, [NCERT Exemplar], , (a) − 20, (c) −30, , (b) 20, (d) 30, , 12. The 21st term of an AP whose first two terms are, [NCERT Exemplar], − 3 and 4, is, (a) 17, (c) 143, , (b) 137, (d) − 143, , 13. If the 2nd term of an AP is 13 and 5th term is 25,, what is its 7th term?, (a) 30, (c) 37, , (b) 33, (d) 38, , 14. Which term of an AP : 21, 42, 63, 84, ... is 210?, [NCERT Exemplar], (a) 9th, (c) 11th, , (b) 10th, (d) 12th, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 30, 15. If the common difference of an AP is 5, then what, is a18 − a13 ?, (a) 5, (c) 25, , (b) 20, (d) 30, , 16. What is the common difference of an AP in which, [NCERT Exemplar], a18 − a14 = 32 ?, (b) − 8, (d) 4, , (a) 8, (c) − 4, , 17. Two APs have the same common difference., The first term of one of these is − 1 and that of, the other is − 8. The difference between their 4th, [NCERT Exemplar], terms is, (a) − 1, (c) 7, , (b) − 8, (d) − 9, , (b) 11, (d) 0, , 19. The 4th term from the end of an AP, − 11 , − 8 , − 5, . . . , 49 is, (a) 37, (c) 43, , (b) 40, (d) 58, , 20. Which term of the AP 5, 15, 25, ... will be 130 more, than its 31st term?, (a) 42, (c) 46, , (b) 44, (d) 48, , 21. The number of terms of an AP 5, 9, 13, ..., 185 is, (a) 31, (c) 41, , [NCERT Exemplar, CBSE 2020 (Standard)], (b) 51, (d) 46, , 22. The sum of AP, sequence −37 , − 33, − 29, . . . . . . . . ., upto 12 term is, (b) −180, (d) −170, , (a) 180, (c) 170, , 23. The sum of first 16 terms of the AP 10, 6, 2, ... is, [NCERT Exemplar], (a) − 320, (c) − 352, , (b) 320, (d) − 400, , (b) 5, (d) 15, , 25. In an AP, if a = 1 , a n = 20 and S n = 399, then n is, equal to, (a) 19, (c) 38, , (b) 21, (d) 42, , 26. The sum of first five multiples of 3 is, (a) 45, , (b) 55, , (c) 65, , 27. In a flower bed, there are 43 rose plants in the first, row, 41 in the second, 39 in the third and so on., (i) If there are 11 rose plants in the last row, then, number of rose required are, (a) 16, (c) 17, , (b) 15, (d) 10, , (ii) Difference of rose plants in 7th row and 13th row is, (a) 11, (c) 13, , (b) 12, (d) 14, , (iii) If there are x rose plants in 15 rose, then x is, equal to, (b) 12, (d) 15, , (iv) The rose plants in 6th row is, (a) 35, (c) 33, , (b) 37, (d) 31, , (v) The total number of rose plants in 5th and, 8th row is, (a) 64, (c) 46, , (b) 54, (d) 45, , 28. The sum of the first five terms of an AP and the, sum of the first seven terms of the same AP is 167., The sum of the first ten terms of this AP is 235., (i) Let first term and common difference of an AP be, a and d, respectively. Then pair of linear equations, for given problem is, (a) 13a + 31d = 167, 2 a + 9d = 47, (b) 13a + 31d = 169, 2 a + 9d = 45, (c) 12 a + 31d = 167 , 2 a + 9d = 47, (d) 12 a + 31d = 169, 2 a + 9d = 45, , (ii) Common difference of given AP is, (a) 5, (c) 9, , (b) 7, (d) 11, , (iii) First term of given AP is, (a) 3, (c) 2, , (b) 4, (d) 1, , (iv) Fourth term of the AP is, , 24. If the first term of an AP is − 5 and the common, difference is 2, then the sum of the first 6 terms is, (a) 0, (c) 6, , Case Based MCQs, , (a) 10, (c) 13, , 18. If 7 times the 7th term of an AP is equal to 11 times, its 11th term, then its 18th term will be, (a) 7, (c) 18, , ●, , [NCERT Exemplar], (d) 75, , (a) 15, (c) 17, , (b) 16, (d) 18, , (v) Sum of first twenty terms is, (a) 970, (c) 950, , (b) 990, (d) 980, , 29. India is competitive manufacturing location due to, the low cost of manpower and strong technical and, engineering capabilities contributing to higher, quality production runs. The production of TV sets, in a factory increases uniformly by a fixed number, every year. It produced 16000 sets in 6th year and, 22600 in 9th year., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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31, , CBSE Term II Mathematics X (Standard), , (ii) What is the minimum number of days he needs to, practice till his goal is achieved?, (a) 10, (c) 11, , (b) 12, (d) 9, , (iii) Which of the following term is not in the AP of the, above given situation?, (a) 41, (c) 37, , (b) 30, (d) 39, , (iv) If nth term of an AP is given by a n = 2 n + 3, then, common difference of an AP is, , Based on the above information, answer the, following questions:, (i) Find the production during first year., (a) 4000 sets, (c) 6000 sets, , (b) 5000 sets, (d) 7000 sets, , (ii) Find the production during 8th year., (a) 48000 sets, (c) 43000 sets, , (b) 20400 sets, (d) None of these, , (iii) Find the production during first 3 years., (a) 20000 sets, (c) 31000 sets, , (b) 25000 sets, (d) 21600 sets, , (a) 2, (c) 5, , (b) 3, (d) 1, , (v) The value of x, for which 2 x, x + 10 , 3 x + 2 are three, consecutive terms of an AP, is, (a) 6, (c) 18, , (b) − 6, (d) − 18, , 31. Your elder brother wants to buy a car and plans to, take loan from a bank for his car. He repays his total, loan of ` 118000 by paying every month starting, with the first installment of ` 1000. If he increases, the installment by ` 100 every month, answer the, following :, , (iv) In which year, the production is ` 29200., (a) 11, (c) 10, , (b) 12, (d) 8, , (v) Find the difference of the production during, 7th year and 4th year., (a) 5500, (c) 5400, , (b) 6700, (d) 6600, , 30. Your friend Veer wants to participate in a 200 m, race. He can currently run that distance in 51, seconds and with each day of practice it takes him, 2 seconds less. He wants to do in 31 seconds., , [CBSE Question Bank], , (i) The amount paid by him in 30th installment is, (a) 3900, (c) 3700, , (b) 3500, (d) 3600, , (ii) The amount paid by him in the 30 installments is, (a) 37000, (c) 75300, , (b) 73500, (d) 75000, , (iii) What amount does he still have to pay after 30th, installment?, (a) 45500, (c) 44500, , (b) 49000, (d) 54000, , (iv) If total installments are 40, then amount paid in, the last installment?, [CBSE Question Bank], , (i) Which of the following terms are in AP for the, given situation?, (a) 51, 53, 55…., (c) − 51, − 53, − 55…., , (b) 51, 49, 47…., (d) 51, 55, 59…, , (a) 4900, (c) 5900, , (b) 3900, (d) 9400, , (v) The ratio of the 1st installment to the last, installment is, (a) 1 : 49, (c) 10 : 39, , (b) 10 : 49, (d) 39 : 10, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 32, , PART 2, Subjective Questions, ●, , Short Answer Type Questions, 1. Justify whether it is true to say that, 5, −3, , − 2 , , . . . forms an AP as a 2 − a1 = a 3 − a 2 ., −1 ,, 2, 2, [NCERT Exemplar], , 2. Find the values of a, b and c if it is given that the, numbers a, 7, b, 23, c are in AP. [CBSE 2020 (Standard)], 3. The angles of a triangle are in AP. The greatest angle, is twice the least. Find all the angles of the triangle., [NCERT Exemplar], , 4. The taxi fare after each km, when the fare is ` 15 for, the first kilometre and ` 8 for each additional, kilometre, does not form an AP as the total fare, (in `) after each kilometre is 15, 8, 8, 8, … . Is the, statement true? Give reasons., 5. Determine k, so that k 2 + 4k + 8, 2 k 2 + 3k + 6 and, 3k 2 + 4k + 4 are three consecutive terms of an AP., [NCERT Exemplar], , 6. Show that ( a − b )2 ,( a 2 + b 2 ) and ( a + b )2 are in AP., [CBSE 2020 (Standard)], , 7. Find the 11th term from the last term (towards the, first term) of the AP 12, 8, 4, .., − 84., [CBSE 2020 (Standard)], , 8. For the AP −3, − 7 , − 11 ,K can we find directly, a 30 − a 20 without actually finding a 30 and a 20 ?, Give reason for your answer., [NCERT Exemplar], 9. Is 0 a term of the AP 31, 28, 25,…? Justify your, answer., 10. If four numbers are in AP such that their sum is 50, and the greatest number is 4 times the least, then, find the numbers., 11. Find the 20th term of the AP whose 7th term is 24, less than the 11th term, first term being 12., [NCERT Exemplar], , 12. If the 9th term of an AP is zero, then prove that its, 29th term is twice its 19th term., 13. The 16th term of an AP is 1 more than twice its 8th, term. If the 12th term of an AP is 47, then find its, nth term., 14. Find the 19th term of the following sequence., ⎧ n 2 , where n is even, tn = ⎨ 2, ⎩n − 1, where n is odd, [CBSE 2015], , 15. Split 207 into three parts such that these are in AP, and the product of the two smaller parts is 4623., [NCERT Exemplar], , 16. Find the 12th term from the end of the AP, [NCERT Exemplar], − 2 , − 4, − 6, . . . , − 100., 17. How many numbers lie between 10 and 300, which, divided by 4 leave a remainder 3?, 18. If m times the mth term of an AP is equal to n, times its nth term, show that the ( m + n )th term of, the AP is zero., 19. Find the sum of first 20 terms of the following AP, sequence 1, 4, 7, 10, ……, 20. Which term of the AP : 120, 116, 112, … is first, [CBSE 2012], negative term?, 21. How many terms of AP 18, 16, 14, ... should be taken,, [CBSE 2013], so that their sum is zero?, 22. Find the sum of first 8 multiples of 3. [CBSE 2018], 23. Subha Rao started work in 1995 at an annual salary, of ` 5000 and received an increment of ` 200 each, year. In which year did his income reach ` 7000?, [NCERT Exemplar], , 24. Ramkali saves ` 5 in the first week of a year and, then increased her weekly savings by ` 1.75. If in, the nth week, her weekly saving becomes ` 20.75., [NCERT Exemplar], Find n., 1 + 3 + 5 + Kupto n terms, 25. If, = 9, then find the, 2 + 5 + 8 + . . . upto 8 terms, value of n., 26. In an AP, if S n = 3n 2 + 5n and a k = 164, then find, [NCERT Exemplar], the value of k., 27. Find the sum ( −5) + ( −8 ) + ( −11 ) + K + ( −230)., [CBSE 2020 (Standard)], , 28. Sum of the first n terms of an AP is 5n 2 − 3n. Find, [CBSE 2010], the AP and also find its 16th term., 29. The sum of the first n terms of an AP whose first, term is 8 and the common difference is 20, is equal, to the sum of first 2n terms of another AP whose, first term is −30 and the common difference is 8., [NCERT Exemplar], Find the value of n., 30. Find the sum of 10 terms of an AP., 2 , 8 , 18 , 32 , …,, , 31. Find the sum of all multiples of 7 lying between, 500 and 900., 32. Find the sum of all the two digit numbers which leave, [CBSE 2019], the remainder 2 when divided by 5., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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33, , CBSE Term II Mathematics X (Standard), , ⎛, (ii) ⎜ 4 −, ⎝, , 33. If S n denotes the sum of first n terms of an AP, then, [NCERT Exemplar], prove that S12 = 3 (S 8 − S 4 )., 34. Find the sum of last ten terms of the AP 8, 10,, 12,..., 126., [NCERT Exemplar], 35. Find the sum of the first 100 natural numbers., , 39. If the sum of first 14 terms of an Arithmetic, Progression is 1050 and its fourth term is 40, find, [CBSE 2020 (Standard)], its 20th term., , Long Answer Type Questions, 40. If the nth terms of the two AP’s 9, 7, 5, ... and 24,, 21, 18, ... are the same, then find the value of n., Also, that term., [NCERT Exemplar], 41. The 26th, 11th and the last terms of an AP are, 0, 3, 1, and − , respectively. Find the common difference, 5, [NCERT Exemplar], and the number of terms., 42. The 4th term of an AP is zero. Prove that the 25th, term of the AP is three times its 11th term., [CBSE 2016], , 43. If the mth term of an AP is, , 44. In an AP given that the first term ( a) = 54, the, common difference ( d) = − 3 and the n th term, ( a n ) = 0, find n and the sum of first nterms (S n ) of, [CBSE 2020 (Standard)], the AP., 45. Solve 1 + 4 + 7 + 10 + . . . + x = 287., [CBSE 2020 (Standard)], , 46. Solve the equation:, 1 + 5 + 9 + 13 + K + x = 1326, [CBSE 2020 (Standard)], , 47. Find the sum, (i) 1 + ( − 2 ) + ( − 5 ) + ( − 8 ) + . . . + ( − 236 ), , 3⎞, ⎟ + . . . upto n terms., n⎠, , 48. Find the sum of the two middle most terms of an, 4, 2, 1, AP − , − 1 , − ,. . . , 4 ., 3, 3, 3, 49. Find the sum of first 17 terms of an AP whose 4th, and 9th terms are − 15 and − 30, respectively., 50. The sum of first n terms of three AP’s are S1 , S 2, and S 3 . The first term of each AP is unity and their, common differences are 1, 2 and 3, respectively., Prove that, [CBSE 2016], S 1 + S 3 = 2S 2 ., 51. If the sum of first four terms of an AP is 40 and that, of first 14 terms is 280. Find the sum of its first n, terms., [CBSE 2019], 52. The ratio of the 11th term to the 18th term of an AP is, 2 : 3. Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first five terms to, [NCERT Exemplar], the sum of the first 21 terms., 53. The sum of four consecutive numbers in AP is, 32 and the ratio of the product of the first and, last terms to the product of two middle terms is, [CBSE 2020 (Standard)], 7 : 15. Find the numbers., 54. Show that the sum of an AP whose first term is a,, the second term b and the last term c, is equal to, ( a + c ) ( b + c − 2 a), ., [NCERT Exemplar], 2 ( b − a), 1, 2, 55. How many terms of the AP 20, 19 , 18 , ... must, 3, 3, be taken, so that their sum is 300?, , 36. Find the sum of first seven numbers which are, multiples of 2 as well as of 9., [NCERT Exemplar], 37. For an AP, it is given that the first term ( a) = 5,, common difference ( d) = 3, and the nth term, ( a n ) = 50. Find n and sum of first n terms (S n ) of, the AP., [CBSE 2020 (Standard)], 38. Find the sum of first 16 terms of an Arithmetic, Progression whose 4th and 9th terms are −15 and, − 30, respectively., [CBSE 2020 (Standard)], , 1, 1, and nth term is ,, n, m, then show that its mnth term is 1., , 2⎞ ⎛, ⎟ + ⎜4 −, n⎠ ⎝, , [NCERT Exemplar], , [CBSE 2020 (Standard)], , ●, , 1⎞ ⎛, ⎟ + ⎜4 −, n⎠ ⎝, , ●, , Case Base Questions, 56. Kanika was given her pocket money on Jan 1st,, 2008. She puts ` 1 on day 1, ` 2 on day 2, ` 3 on day, 3 and continued doing so till the end of the month., From this money into her piggy bank, she also, spent ` 204 of her pocket money and found that at, the end of the month she still had ` 100 with her., [NCERT Exemplar], , (i) How much Kanika take till the end of the month, from pocket money?, (ii) How much was pocket money for the month?, (iii) What is the amount saved by Kanika, till, January 13th, 2008?, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 34, , SOLUTIONS, Objective Questions, 1. (a) (a) Here, t1 = − 1, t 2 = − 1, t 3 = − 1 and t 4 = − 1, Now, t 2 − t1 = − 1 + 1 = 0, t3 − t2 = − 1 + 1 = 0, t4 − t3 = − 1 + 1 = 0, Clearly, the difference of successive terms is same,, therefore given list of numbers forms an AP., (b) Here, t1 = 0, t 2 = 2 , t 3 = 0 and t 4 = 2, Now, t 2 − t1 = 2 − 0 = 2, t3 − t2 = 0 − 2 = − 2, t4 − t3 = 2 − 0 = 2, Clearly, the difference of successive terms is not same,, therefore given list of numbers does not form an AP., (c) Here, t1 = 1, t 2 = 1, t 3 = 2 and t 4 = 2, Now, t 2 − t1 = 1 − 1 = 0, t3 − t2 = 2 − 1 = 1, t4 − t2 = 2 − 2 = 0, Clearly, the difference of successive terms is not same,, therefore given list of numbers does not form an AP., 1 1 1, (d) , , , …, 2 3 4, 1, 1, 1, Here, t1 = , t 2 = and t 3 =, 2, 3, 4, 1 1 2−3, 1, Now,, t 2 − t1 = − =, =−, 3 2, 6, 6, 1 1 3−4, 1, =−, t3 − t2 = − =, 4 3, 12, 12, Clearly, the difference of successive terms is not same,, therefore given list of numbers does not form an AP., 2. (c) The condition for given series is not AP is the common, difference of two consecutive terms is not constant., (a) We have, − 1.2 , 0.8, 2.8, ..., Here, a1 = − 1.2, a 2 = 0.8, a 3 = 2.8, Now, a 2 − a1 = 0.8 − ( −1.2 ) = 2.0, and a 3 − a 2 = 2.8 − 0.8 = 2, Thus, given series is an AP., (b) We have, 3, 3 + 2 , 3 + 2 2 , 3 + 3 2 , …, Here, a1 = 3, a 2 = 3 + 2 , a 3 = 3 + 2 2, Now, a 2 − a1 = 3 + 2 − 3 = 2, and a 3 − a 2 = 3 + 2 2 − ( 3 + 2 ) = 2, Thus, given series is an AP., 4 7 9 12, (c) We have,, , , , ...., 3 3 3 3, 4, 7, 9, Here, a1 = , a 2 = , a 3 =, 3, 3, 3, 7 4 3, Now, a 2 − a1 = − = = 1, 3 3 3, 9 7 2, a3 − a2 = − =, 3 3 3, Thus, given series is not an AP, as common difference is, not constant., , −1 −2 −3, ,, , , ...., 5 5 5, 1, 2, −3, Here, a1 = − , a 2 = − , a 3 =, 5, 5, 5, 2 ⎛ −1⎞, 1, Now, a 2 − a1 = − − ⎜ ⎟ = −, 5 ⎝ 5⎠, 5, 1, −3 ⎛ 2 ⎞ −3 2, a3 − a2 =, − ⎜− ⎟ =, + =−, 5, 5 ⎝ 5⎠, 5, 5, Thus, given series is an AP., Hence, in the given options, option (c) is not an AP., 5, 3. (b) Given, − , a , 2 are consecutive terms in AP., 7, ⎛ 5⎞, [Q In AP, a 2 − a1 = a 3 − a 2], ∴, a − ⎜− ⎟ = 2 − a, ⎝ 7⎠, 5, 9, 9, 2 a = 2 − ⇒2 a = ⇒ a =, ⇒, 7, 7, 14, 4. (a) Given, nth term of an AP is, a n = ( 3n + 7 ), ∴ The common difference of an AP = a n − a n − 1, = ( 3n + 7 ) − [ 3( n − 1) + 7 ], = 3n + 7 − ( 3n + 4) = 7 − 4 = 3, a+c, [Q b − a = c − b], 5. (a) If a , b , c are in AP, then b =, 2, Given, 2 x , ( x + 10) and ( 3x + 2 ) are in AP., 2 x + ( 3x + 2 ), ∴, x + 10 =, 2, 5x + 2, ⇒, x + 10 =, ⇒ 2 x + 20 = 5x + 2, 2, ⇒, 5x − 2 x = 20 − 2 ⇒ 3x = 18, 18, x=, =6, ⇒, 3, 6. (d) Let a1 = 2 p + 1, a 2 = 10 and a 3 = 5p + 5., (d) We have,, , Given that three consecutive terms are in AP., ∴, a 2 − a1 = a 3 − a 2, ⇒, 10 − (2 p + 1) = 5p + 5 − 10, ⇒, 10 + 10 = 5p + 5 + 2 p + 1, ⇒, 20 = 7 p + 6, ⇒, 7 p = 20 − 6, ⇒, 7 p = 14, 14, p=, =2, ⇒, 7, 7. (c) Let the first four terms of an AP are, a , a + d, a + 2 d and a + 3d., Given, that first term, a = − 2 and common difference,, d = − 2, then we have an AP as follows, − 2 , − 2 − 2 , − 2 + 2 ( −2 ), − 2 + 3( −2 ) i.e. − 2 , − 4 , − 6, − 8, 8. (d) We have, a1 = 1, a 2 = 1 and a n = a n −1 + a n − 2 for all n > 2, On putting n = 3 and 4, we get, a 3 = a 2 + a1 = 1 + 1 = 2, a4 = a3 + a2 = 2 + 1 = 3, a4 3, Now,, =, a3 2, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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35, , CBSE Term II Mathematics X (Standard), , 9. (c) Given, first term ( a ) = 8, common difference ( d) = − 5, On putting the values of a and d in general form,, a , a + d, a + 2 d, a + 3d, ..., we get, 8, 8 − 5, 8 + 2( −5), 8 + 3( −5), ... or 8, 3, −2 , − 7 , ..., On comparing with given terms 8, A ,B , ... , we get, A = 3, B = − 2, ∴, A + B = 3 + ( −2 ) = 3 − 2 = 1, 10. (d) In an AP,, an = a + ( n − 1) d, ⇒, ⇒, ∴, , 4 = a + ( 7 − 1 ) ( − 4) [by given condition], 4 = a + 6 ( − 4) ⇒ 4 + 24 = a, a = 28, 5, 5, 11. (b) Given AP, − 5, − , 0 , , ……, 2, 2, −5, 5, Here, a = − 5, d =, + 5=, 2, 2, [Q a n = a + ( n − 1 ) d ], ∴, a11 = a + (11 − 1 ) d, 5, = − 5 + (10 ) × = − 5 + 25 = 20, 2, 12. (b) Given, first two terms of an AP are a = − 3 and a + d = 4., ⇒, −3 + d = 4, Common difference, d = 7, ∴, a 21 = a + (21 − 1 ) d[Q a n = a + ( n − 1 ) d ], = − 3 + (20 ) 7, = − 3 + 140 = 137, 13. (b) Given, a 2 = 13 and a 5 = 25, ⇒, a + (2 − 1 ) d = 13, [Q a n = a + ( n − 1 ) d ], and, a + ( 5 − 1 ) d = 25, …(i), ⇒, a + d = 13, and, …(ii), a + 4 d = 25, On subtracting Eq. (i) from Eq. (ii), we get, 3 d = 25 − 13 = 12 ⇒ d = 4, From Eq. (i), a = 13 − 4 = 9, ∴, a 7 = a + ( 7 − 1 ) d = 9 + 6 × 4 = 33, 14. (b) Let nth term of the given AP be 210., Here, first term, a = 21, and common difference,, d = 42 − 21 = 21 and a n = 210, Q, a n = a + (n − 1) d, ⇒, 210 = 21 + ( n − 1 ) 21, ⇒, 210 = 21 + 21 n − 21, ⇒, 210 = 21 n ⇒ n = 10, Hence, the 10th term of an AP is 210., 15. (c) Given, the common difference of AP i.e. d = 5, Now, a18 − a13 = a + (18 − 1 ) d − [ a + (13 − 1 ) d], [Q a n = a + ( n − 1 ) d ], = a + 17 × 5 − a − 12 × 5 = 85 − 60 = 25, 16. (a) Given,, a18 − a14 = 32, ⇒ a + (18 − 1 ) d − [ a + (14 − 1 ) d] = 32 [Q a n = a + ( n − 1 ) d ], ⇒, a + 17 d − a − 13 d = 32, ⇒, 4 d = 32, ∴, d=8, Which is the required common difference of an AP., , 17. (c) Let the common difference of two APs are d1 and d2,, respectively., ...(i), By condition,, d1 = d2 = d, Let the first term of first AP ( a1 ) = − 1, and the first term of second AP ( a 2 ) = − 8, We know that, the nth term of an AP, Tn = a + ( n − 1 ) d, ∴ 4th term of first AP, T4 = a1 + ( 4 − 1) d = − 1 + 3 d, and 4th term of second AP, T4ʹ = a 2 + ( 4 − 1 ) d = − 8 + 3 d, Now, the difference between their 4th terms is, |T4 − T4ʹ| = ( − 1 + 3 d ) − ( − 8 + 3 d ), = − 1 + 3 d + 8 − 3d = 7, Hence, the required difference is 7., 18. (d) According to the question,, 7 a 7 = 11 a11, ⇒, 7 [ a + ( 7 − 1 ) d ] = 11 [ a + (11 − 1 ) d ], [Q a n = a + ( n − 1 ) d], ⇒, 7 ( a + 6 d) = 11 ( a + 10 d), ⇒, 7 a + 42 d = 11a + 110 d, ⇒, 4a + 68 d = 0, ⇒, 4 ( a + 17 d ) = 0, …(i), ⇒, a + 17 d = 0, ∴ 18th term of an AP, a18 = a + (18 − 1 ) d, [from Eq. (i)], = a + 17 d = 0, 19. (b) We know that, the n th term of an AP from the end is, …(i), a n = l − ( n − 1) d, Here, l = Last term and l = 49, [given], Common difference, d = − 8 − ( − 11 ), = − 8 + 11 = 3, From Eq. (i), a 4 = 49 − ( 4 − 1 ) 3 = 49 − 9 = 40, 20. (b) We have, a = 5 and d = 10, ∴, a 31 = a + 30d = 5 + 30 × 10 = 305, Let nth term of the given AP be 130 more than its 31st term., Then,, a n = 130 + a 31, ⇒, a + ( n − 1)d = 130 + 305, ⇒, 5 + 10( n − 1) = 435, ⇒, 10( n − 1) = 430, ⇒, n − 1 = 43, ⇒, n = 44, Hence, 44th term of the given AP is 130 more than its, 31st term., 21. (d) Given, AP sequence is 5, 9, 13, ..., 185., Here, first term a = 5, Common difference, d = 9 − 5 = 4, and last term, l = 185, Q, l = a + ( n − 1) d, ∴, 185 = 5 + ( n − 1)4, ⇒, 180 = ( n − 1)4, 180, ( n − 1) =, ⇒, 4, ⇒, ( n − 1) = 45, ⇒, n = 45 + 1 ⇒ n = 46, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 36, 22. (b) Given sequence is, −37 , − 33, − 29, …… upto 12 terms, Here a = −37 , d = −33 − ( −37 ) = 4, n, Q Sn = [2 a + ( n − 1)d], 2, 12, ∴ S12 = [2 × ( −37 ) + (12 − 1)4], 2, = 6[ −74 + 44], = 6 × ( −30) = −180, 23. (a) Given, AP is 10, 6, 2 , ..., Here, first term a = 10, common difference, d = − 4, 16, [2 a + (16 − 1 ) d ], S16 =, ∴, 2, n, ⎡, ⎤, Q Sn = {2 a + ( n − 1) d}, 2, ⎣⎢, ⎦⎥, , 24. (a) Given,, ∴, , = 8 [2 × 10 + 15 ( − 4)], = 8 (20 − 60 ) = 8 ( − 40 ) = − 320, a = − 5 and d = 2, 6, S 6 = [ 2 a + ( 6 − 1) d ], 2, n, ⎡, ⎤, Q Sn = {2 a + ( n − 1) d}, 2, ⎦⎥, ⎣⎢, , = 3 [ 2 ( − 5) + 5 (2 )], = 3 ( − 10 + 10 ) = 0, n, 25. (c)Q, S n = [2 a + ( n − 1 ) d ], 2, n, 399 = [2 × 1 + ( n − 1 ) d ], 2, …(i), 798 = 2 n + n ( n − 1 ) d, and, a n = 20, ⇒ a + ( n − 1 ) d = 20, [Q a n = a + ( n − 1 ) d ], …(ii), ⇒ 1 + ( n − 1 ) d = 20 ⇒ ( n − 1) d = 19, Using Eq. (ii) in Eq. (i), we get, 798 = 2 n + 19 n, ⇒, 798 = 21 n, 798, n=, = 38, ∴, 21, 26. (a) The first five multiples of 3 are 3, 6, 9, 12 and 15., Here, first term, a = 3, common difference, d = 6 − 3 = 3 and, number of terms, n = 5, 5, S 5 = [2 a + ( 5 − 1 ) d ], ∴, 2, n, ⎡, ⎤, Q Sn = {2 a + ( n − 1) d}, 2, ⎣⎢, ⎦⎥, 5, [ 2 × 3 + 4 × 3], 2, 5, = ( 6 + 12 ) = 5 × 9 = 45, 2, =, , 27. (i) (c) Number of rose plants in 1st, 2nd and 3rd row ...... are, 43, 41, 39, ......, So, it forms an AP with first term,, a = 43 and common difference,, d = 41 − 43 = − 2, , Let n be the number of rows required., ∴, a n = 11, ⇒, a + ( n − 1) d = 11, ⇒ 43 + ( n − 1) ( − 2 ) = 11, ⇒, − 2 ( n − 1) = − 32, ⇒, n − 1 = 16 ⇒ n = 17, (ii) (b) Number of rose plants in 7th row = a 7, = a + 6d = 43 + 6 ( − 2 ) = 43 − 12 = 31, Number of rose plants in 13th row = a 13, = a + 12 d = 43 + 12 ( − 2 ) = 43 − 24 = 19, ∴ Required difference = 31 − 19 = 12, (iii) (d) Here, n = 15, ∴, a 15 = a + 14d = 43 + 14 ( − 2 ) = 43 − 28 = 15, (iv) (c) Number of rose plants in 6th row, = a 6 = a + 5d, = 43 + 5 ( − 2 ), = 43 − 10 = 33, (v) (a) Number of rose plants in 5th row, = a 5 = a + 4d, = 43 + 4 ( − 2 ), = 43 − 8 = 35, Number of rose plants in 8th row, = a 8 = a + 7d, = 43 + 7 ( − 2 ), = 43 − 14 = 29, ∴ Required sum = 35 + 29 = 64, 28. (i) (c) Let the number of terms of AP be n., Q Sum of first n terms of an AP,, n, ...(i), S n = [2 a + ( n − 1) d ], 2, ∴ Sum of first five terms of an AP,, 5, [from Eq.(i)], S 5 = [ 2 a + ( 5 − 1) d ], 2, 5, = (2 a + 4 d ) = 5 ( a + 2 d ), 2, ...(ii), ⇒, S 5 = 5a + 10 d, and sum of first seven terms of an AP,, 7, S 7 = [2 a + ( 7 − 1 ) d ], 2, 7, = [2 a + 6 d ] = 7 ( a + 3 d ), 2, ...(iii), ⇒, S 7 = 7 a + 21 d, Now, by given condition,, S 5 + S 7 = 167, ⇒, 5a + 10 d + 7 a + 21 d = 167, ...(iv), ⇒, 12 a + 31 d = 167, Given that, sum of first ten terms of this AP is 235., ∴, S10 = 235, 10, [2 a + (10 − 1 ) d ] = 235, ⇒, 2, ⇒, 5 (2 a + 9 d ) = 235, …(v), ⇒, 2 a + 9 d = 47, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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37, , CBSE Term II Mathematics X (Standard), , (ii) (a) On multiplying Eq. (v) by 6 and then subtracting it, into Eq. (iv), we get, 12 a + 54 d = 282, 12 a + 31 d = 167, −, −, −, 23 d = 115, ⇒, d=5, (iii) (d) Now, put the value of d in Eq. (v), we get, 2 a + 9 ( 5) = 47 ⇒ 2 a + 45 = 47, ⇒, 2 a = 47 − 45 = 2 ⇒ a = 1, (iv) (b) a 4 = a + 3d, = 1 + 3 ( 5) = 1 + 15 = 16, (v) (a) Sum of first twenty terms of this AP,, 20, [2 a + (20 − 1 ) d ], S20 =, 2, = 10 [2 × ( 1 ) + 19 × ( 5)] = 10 (2 + 95), = 10 × 97 = 970, Hence, the required sum of its first twenty terms is 970., 29. (i) (b) Let the production of TV sets in first year be ‘a’ units., Then, production in the next consecutive years are, a + d, a + 2 d, …., Thus, we get the sequence, a , a + d, a + 2 d, …, This is an AP sequence, whose first term = a, and common difference = d., Given, T6 = 16000 and T9 = 22600, ∴, a + ( 6 − 1)d = 16000, and, [Q Tn = a + ( n − 1) d], a + ( 9 − 1)d = 22600, …(i), ⇒, a + 5d = 16000, and, …(ii), a + 8 d = 22600, On subtracting Eq. (i) from Eq. (ii), we get, 3d = 22600 − 16000, ⇒, 3d = 6600, ⇒, d = 2200, Put d = 2200 in Eq. (i), we get, a + 5 × 2200 = 16000, ⇒, a = 16000 − 11000 = 5000, Hence, the production during first year is 5000 sets., (ii) (b) The production during 8th year is, T8 = a + ( 8 − 1) d, = 5000 + 7 × 2200, = 5000 + 15400 = 20400, Hence, production during 8th year is 20400 sets., (iii) (d) The production during first 3 years,, 3, S 3 = [ 2 a + ( 3 − 1) d ], 2, 3, = [2 × 5000 + 2 × 2200], 2, = 3[ 5000 + 2200], = 3 × 7200 = 21600, (iv) (b) Let in nth year, the production is 29200, Q, Tn = a + ( n − 1) d, ∴, 29200 = 5000 + ( n − 1)2200, ⇒, ( n − 1)2200 = 24200, , 24200, 2200, ⇒, n − 1 = 11 ⇒ n = 12, Hence, production is ` 29200 in 12th year., (v) (d) The difference of the production during 7th year and, 4th year = T7 − T4, = a + (7 − 1 ) d − [ a + ( 4 − 1 ) d ], = 6d − 3d = 3d = 3 × 2200 = 6600, 30. (i) (b) In first day, Veer takes 51 seconds to complete the, 200 m race. But in each day he takes 2 seconds lesser, than the previous days., Thus, AP series will formed, 51, 49, 47, …, (ii) (c) Since, Veer wants to achieve the race in 31 seconds., Let Veer takes n days to achieve the target., ∴, Tn = a + ( n − 1)d, Here, a = 51, d = 49 − 51 = − 2, ∴, 31 = 51 + ( n − 1)( −2 ), ⇒, ( n − 1)2 = 20 ⇒ ( n − 1) = 10 ⇒ n = 11, Hence, he needs minimum 11 days to achieve the goal., (iii) (b) In an AP series, we get the series of odd terms., Hence, term 30 is not an AP., (iv) (a) Given, a n = 2 n + 3, ∴Common difference = a n + 1 − a n, = 2 ( n + 1 ) + 3 − ( 2 n + 3), = 2n + 2 + 3 − 2n − 3 = 2, (v) (a) Given, terms 2 x , x + 10, 3x + 2 are in AP., 2 x + ( 3x + 2 ), ∴, x + 10 =, 2, ⇒, 2 x + 20 = 5x + 2 ⇒ 3x = 18 ⇒ x = 6, 31. (i) (a) Since, he pays first installment of ` 1000 and next, consecutive months he pay the installment are, 1100, 1200, 1300, … ., Thus, we get the AP sequence,, 1000, 1100, 1200, …, Here, a = 1000, d = 1100 − 1000 = 100, Now, T30 = a + ( 30 − 1) d, = 1000 + 29 × 100 = 1000 + 2900 = 3900, Hence, the amount paid by him in 30th installment is, ` 3900., 30, (ii) (b) Now, S30 = [2 a + ( 30 − 1) d], 2, = 15(2 × 1000 + 29 × 100), = 15(2000 + 2900), = 15 × 4900 = ` 73500, (iii) (c) After 30th installment, he still have to pay, = 118000 − 73500 = 44500, (iv) (a) The amount in last 40th installment is, T40 = a + ( 40 − 1) d, = 1000 + 39 × 100, = 1000 + 3900 = ` 4900, (v) (b) The ratio of 1st installment to the last installment, 1000, 10, is, i.e. ., 4900, 49, ⇒, , ( n − 1) =, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 38, Subjective Questions, 5, −3, 1. Here, a1 = − 1, a 2 =, , a 3 = − 2 and a 4 =, 2, 2, 1, −3, Now,, a 2 − a1 =, +1=−, 2, 2, 3, 1, a3 − a2 = − 2 + = −, 2, 2, 5, 9, a4 − a3 = + 2 =, 2, 2, Clearly, the difference of successive terms is not same,, although, a 2 − a1 = a 3 − a 2 but a 3 − a 2 ≠ a 4 − a 3., therefore it, does not form an AP., 2. Given, sequence a, 7, b, 23, c is an AP., Since, a, 7, b is in AP., a+b, x + z⎤, [Q If x , y , z in AP, then, y =, 7=, ∴, 2 ⎦⎥, 2, ...(i), ⇒, a + b = 14, Since, 7, b, 23 is in AP., 7 + 23, 30, ...(ii), ∴, ⇒b =, ⇒ b = 15, b=, 2, 2, Since, b, 23, c is in AP., b+c, 23 =, ∴, 2, [from Eq. (ii), b = 15], ⇒, 23 × 2 = 15 + c, ⇒, c = 46 − 15, ⇒, c = 31, Put b = 15 in Eq. (i), we get, a + 15 = 14, ⇒, a = 14 − 15, ⇒, a = −1, Hence, values of a, b and c are respectively −1, 15 and 31., 3. Let the angles are ( a − d)° , a ° , ( a + d)°., Then, we get ( a − d) + a + ( a + d) = 180, and, a + d = 2( a − d), ⇒, 3a = 180° ⇒ a = 60°, and, 60° + d = 2 ( 60° − d), ⇒, 60° + d = 120° − 2 d, ⇒, 3d = 60°, ⇒, d = 20°, ∴ The angles of an AP are, a − d = 60° − 20° = 40°, a = 60°, and, a + d = 60° + 20° = 80°, Hence, angles of an AP are 40°, 60°, 80°., 4. No, because the total fare (in `) after each kilometre is, 15, (15 + 8), (15 + 2 × 8), (15 + 3 × 8),… or 15, 23, 31, 39,…, Let, t1 = 15, t 2 = 23, t 3 = 31 and t 4 = 39, Now,, t 2 − t1 = 23 − 15 = 8, t 3 − t 2 = 31 − 23 = 8, t 4 − t 3 = 39 − 31 = 8, Since, all the successive terms of the given list have same, difference i.e. common difference = 8, Hence, the total fare after each killometre form an AP., , 5. Since, k 2 + 4k + 8, 2 k 2 + 3k + 6 and 3k 2 + 4k + 4 are, consecutive terms of an AP., ∴ 2 k 2 + 3k + 6 − ( k 2 + 4 k + 8 ) = 3 k 2 + 4 k + 4, − (2 k 2 + 3k + 6 ) = Common difference, 2, ⇒ 2 k + 3k + 6 − k 2 − 4k − 8 = 3k 2 + 4k + 4 − 2 k 2 − 3k − 6, ⇒, k2 − k − 2 = k2 + k − 2, ⇒, − k = k ⇒ −2 k = 0 ⇒ k = 0, 6. Let a1 = ( a − b )2, a 2 = a 2 + b 2 and a 3 = ( a + b )2., Now,, , a 2 − a1 = a 2 + b 2 − ( a − b )2, = a 2 + b 2 − ( a 2 + b 2 − 2 ab ) = 2ab, , …(i), , a 3 − a 2 = ( a + b )2 − ( a 2 + b 2 ), …(ii), = a 2 + b 2 + 2 ab − ( a 2 + b 2 ) = 2ab, From Eqs. (i) and (ii), we get, a 2 − a1 = a 3 − a 2, Hence, given terms are in AP., 7. Given, sequence of an AP is 12, 8, 4, ..., − 84., Here, first term is a = 12, Common difference is d = 8 − 12 = − 4 and last term, l = − 84, The nth term from the last term of an AP is l − ( n − 1)d., ∴The 11th term from the last term of an AP, = l − (11 − 1)d, = − 84 − (10) × ( − 4), = − 84 + 40 = − 44, 8. Q nth term of an AP, a n = a + ( n − 1 ) d, ∴, a 30 = a + ( 30 − 1 ) d = a + 29 d, ...(i), and, a 20 = a + (20 − 1 ) d = a + 19 d, Now,, a 30 − a 20 = ( a + 29 d ) − ( a + 19 d ) = 10 d, and from given AP common difference,, d = − 7 − ( − 3) = − 7 + 3 = − 4, [from Eq. (i)], ∴, a 30 − a 20 = 10 ( − 4) = − 40, 9. Let 0 be the nth term of given AP. i.e. a n = 0., and, , Given that, first term a = 31,, Common difference, d = 28 − 31 = − 3, The n th term of an AP, is a n = a + ( n − 1 ) d, ⇒, 0 = 31 + ( n − 1 ) ( −3), 31, ⇒, 3 ( n − 1 ) = 31 ⇒ n − 1 =, 3, 31, 34, 1, n=, + 1=, = 11, ∴, 3, 3, 3, Since, n should be positive integer. So, 0 is not a term of the, given AP., 10. Let four numbers in AP are a , a + d, a + 2 d, a + 3d. Then,, ( a ) + ( a + d) + ( a + 2 d) + ( a + 3d) = 50 ⇒ 4a + 6d = 50, …(i), ⇒, 2 a + 3d = 25, and, ( a + 3d ) = 4 ( a ), …(ii), ⇒, a=d, On solving Eq. (i) and Eq. (ii), we get, a=d=5, ∴ The four numbers in AP are, a = 5, a + d = 5 + 5 = 10, a + 2 d = 5 + 10 = 15, a + 3d = 5 + 15 = 20, Hence, four numbers in AP are 5, 10, 15 and 20., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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39, , CBSE Term II Mathematics X (Standard), , 11. Let the first term, common difference and number of terms, of an AP are a , d and n, respectively., Given that, first term ( a ) = 12., Now by condition,, 7th term (T7 ) = 11th term (T11 ) − 24, [Q nth term of an AP, Tn = a + (n − 1) d], ⇒, a + (7 − 1) d = a + (11 − 1) d − 24, ⇒, a + 6 d = a + 10 d − 24, ⇒, 24 = 4 d ⇒ d = 6, ∴ 20th term of AP, T20 = a + (20 − 1) d = 12 + 19 × 6 = 126, Hence, the required 20th term of an AP is 126., 12. Let the first term, common difference and number of terms, of an AP are a , d and n, respectively., Given that, 9th term of an AP, T9 = 0, [Q nth term of an AP, Tn = a + ( n − 1) d], ... (i), ⇒ a + ( 9 − 1) d = 0 ⇒ a + 8 d = 0 ⇒ a = − 8 d, Now, its 19th term, T19 = a + (19 − 1) d, [from Eq. (i)], = − 8 d + 18 d, ... (ii), ⇒, T19 = 10 d, and its 29th term, T29 = a + (29 − 1) d, [from Eq. (i)], = − 8 d + 28 d, = 20d = 2 × (10 d), [from Eq. (ii)], ⇒, T29 = 2 × T19, Hence, its 29th term is twice its 19th term. Hence proved., 13. Let first term and common difference of an AP are a and d., According to the given condition,, a12 = 47, ...(i), ⇒, a + 11d = 47, and, [by given condition], a16 = 1 + 2 a 8, ⇒ [ a + (16 − 1) d] = 1 + 2 [ a + ( 8 − 1) d], ...(ii), ⇒, a − d = −1, On solving Eqs. (i) and (ii), we get, d = 4 and a = 3, ∴, a n = 3 + ( n − 1) 4 = 4 n − 1, ⎧ n 2 , where n is even, 14. We have, t n = ⎨ 2, ⎩n − 1, where n is odd, For 19th term, i.e. for n = 19 which is odd, we take, t n = n 2 − 1 = (19)2 − 1 = 360, 15. Let the three parts of the number 207 are ( a − d) , a and, ( a + d), which are in AP., Now, by given condition,, Sum of these parts = 207, ⇒, a − d + a + a + d = 207, ⇒, 3a = 207, a = 69, Given that, product of the two smaller parts = 4623, ⇒, a ( a − d) = 4623, ⇒, 69 ⋅ ( 69 − d) = 4623 ⇒ 69 − d = 67, ⇒, d = 69 − 67 = 2, So,, first part = a − d = 69 − 2 = 67,, second part = a = 69, and, third part = a + d = 69 + 2 = 71,, Hence, required three parts are 67, 69, 71., , 16. Given AP, −2 , − 4, − 6,... , − 100, Here, first term ( a ) = − 2, common difference, ( d ) = − 4 − ( − 2 ) = − 2 and the last term ( l ) = − 100., We know that, the nth term a n of an AP from the end is, a n = l − ( n − 1) d, where l is the last term and d is the, common difference., ∴ 12th term from the end,, a12 = − 100 − (12 − 1) ( − 2 ), = − 100 + (11) (2 ) = − 100 + 22 = − 78., Hence, the 12th term from the end is – 78., 17. Here, the first number is 11, which divided by 4 leave, remainder 3 between 10 and 300. Last term before 300 is, 299, which divided by 4 leave remainder 3., ∴ Required AP is 11, 15, 19, 23,... , 299, Here, first term ( a ) = 11, common difference d = 15 − 11 = 4, [last term], Q, n th term, a n = a + ( n − 1 ) d = l, ⇒, 299 = 11 + ( n − 1 ) 4, ⇒, 299 − 11 = ( n − 1 ) 4, ⇒, 4 ( n − 1 ) = 288 ⇒ ( n − 1 ) = 72, ∴, n = 73, 18. Let first term of an AP is a and common difference is d., The n th term of an AP is, a n = a + ( n − 1 )d, According to the given condition,, m × am = n × an, ∴ m × [ a + ( m − 1 ) d ] = n × [ a + ( n − 1 )d ], ⇒, a ( m − n ) = [( n 2 − m 2 ) + ( − n + m )]d, ⇒, ⇒, , a( m − n ) = [( n − m ) ( n + m ) + ( m − n )]d, ...(i), a = [ − ( n + m ) + 1] d, [divide both sides by m − n ], Now, ( m + n )th term of an AP is, a m + n = a + ( m + n − 1 )d, = [ − ( n + m ) + 1 ] d + ( m + n − 1) d, Hence proved., =0, , 19. Given, AP sequence is 1, 4, 7, 10, .... whose first term is a =1, and common difference, d = 4 − 1 = 3., Q Sum of n terms of an AP is, n, S n = [2 a + ( n − 1) d ], 2, 20, [put, d = 3], S20 = [2 × 1 + (20 − 1) × 3], ∴, 2, = 10 [2 + 19 × 3] = 10 [2 + 57 ] = 590, Hence, sum of the first 20 terms of an AP is 590., 20. Given sequence in AP is 120, 116, 112, ..., Here, a = 120, d = 116 − 120 = − 4, The nth term of an AP is, a n = a + ( n − 1) d, a n = 120 + ( n − 1) ( − 4), For, first negative term, a n < 0, ∴, 120 + ( n − 1) ( − 4) < 0, ⇒, 4 ( n − 1) > 120, ⇒, ( n − 1) > 30, ⇒, n > 31, ∴ The first negative term is 32., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 40, 21. Here, a = 18 and d = −2, Let n terms are taken, so that their sum is zero., Then, we have, Sn = 0, n, ⇒, [2 a + ( n − 1)d] = 0, 2, ⇒, 2 a + ( n − 1 )d = 0, ⇒ 2 × 18 + ( n − 1)( −2 ) = 0, ⇒, n − 1 = 18 ⇒ n = 19, 22. First 8 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24., ∴The sum of first 8 multiples of 3, 8, n, = [ a + l ] = [ 3 + 24] = 4 × 27 = 108, 2, 2, 23. The annual salary received by Subha Rao in the years, 1995, 1996, 1997 etc., is ` 5000, ` 5200, ` 5400, …, ` 7000, Hence, the list of numbers 5000, 5200, 5400, …, 7000 forms, an AP, Q, a 2 − a1 = a 3 − a 2 = 200, Let nth term of an AP, a n = 7000, ⇒, 7000 = a + ( n − 1)d [Q a n = a + ( n − 1)d], ⇒, 7000 = 5000 + ( n − 1)(200), ⇒, 200( n − 1) = 7000 − 5000 = 2000, 2000, n −1 =, = 10, ⇒, 200, ⇒, n = 10 + 1 = 11, Thus, 11th year of his service or in 2005 Subha Rao received, an annual salary ` 7000., 24. Ramkali’ savings in the subsequent weeks are respectively, ` 5, ` 5 + ` 1.75, ` 5 + 2 × ` 1.75, ` 5 + 3 × 1.75 …, In nth week her saving will be ` 5 + (n − 1) × ` 1.75, [given], ⇒, 5 + ( n − 1) × 1.75 = 20.75, ⇒, ( n − 1) × 1.75 = 20.75 −5 = 15.75, 15.75, ⇒, n −1 =, =9, 1.75, ⇒, , n = 9 + 1 = 10, 1 + 3 + 5 + K upto n terms, 25. Given,, =9, 2 + 5 + 8 + ... upto 8 term, n, [2(1) + ( n − 1)2 ], 2, ⇒, =9, 8, [ 2 ( 2 ) + ( 8 − 1) 3], 2, n (2 n ), =9, ⇒, 8(25), ⇒ n 2 = 9 × 100 ⇒ n 2 = 900 ⇒ n = 30, 26. Q n th term of an AP,, a n = Sn − Sn −1, = 3n 2 + 5n − 3 ( n − 1 ) 2 − 5 ( n − 1 ), [Q Sn = 3n 2 + 5n (given)], 2, , or, ⇒, ∴, , 2, , = 3n + 5n − 3n − 3 + 6 n − 5n + 5, …(i), a n = 6n + 2, [Q a k = 164 (given)], a k = 6 k + 2 = 164, 6k = 164 − 2 = 162, k = 27, , 27. Given series is ( −5) + ( −8) + ( −11) + .... + ( −230), Here, first term, a = − 5 and common difference,, d = − 8 − ( − 5) = − 8 + 5 = − 3, Q, a n = a + ( n − 1)d, ∴, ( −230) = − 5 + ( n − 1)( −3), ⇒, ( n − 1)( −3) = − 230 + 5, −225, ⇒, ( n − 1) =, −3, ⇒, n − 1 = 75, ⇒, n = 75 + 1 = 76, Q The sum of n th term is, n, Sn = [ a + l ], 2, 76, Sn = [ −5 + ( −230)], ∴, 2, = 38 [ −235] = − 8930, 28., S n = 5n 2 − 3n, Now,, a n = Sn − Sn −1, = 5n 2 − 3n − [ 5( n − 1)2 − 3( n − 1)], = 5n 2 − 3n − [ 5( n 2 + 1 − 2 n ) − 3n + 3], ⇒, a n = 10n − 8, Clearly, a16 = 10 × 16 − 8 = 160 − 8 = 152, Now, for finding AP, put n = 1, 2, 3, 4 …… in Eq. (i)., So, from Eq. (i), we have, a1 = 2 , a 2 = 12 , a 3 = 22, The AP is 2, 12, 22, ...... ., 29. Given, a1 = 8, d1 = 20, a 2 = −30, d2 = 8, , …(i), , Sn = S2n, 2n, n, [2 × 8 + ( n − 1) × 20] =, [2 × ( −30) × 30 + (2 n − 1) × 8], 2, 2, ⇒, [16 + ( n − 1)20] = 2 [ −60 + (2 n − 1)8], ⇒, 16 + 20n − 20 = −120 + 32 n − 16, ⇒, 12 n = 132 ⇒ n = 11, 30. Here,, a1 = 2 , a 2 = 8 = 2 2 , a 3 = 3 2, ∴, ∴, , a = 2 , d = a 2 − a1 = 2 2 − 2 = 2, 10, S10 = [2 × 2 + (10 − 1)( 2 )], 2, = 5[2 2 + 9 2 ] = 55 2, , 31. The multiples of 7 lying between 500 and 900 are 504, 511,, 518, ..., 896., Clearly, it forms an AP., Here, a = 504 and d = 511 − 504 = 7, Let there are n terms, i.e. a n = 896, ⇒, a + ( n − 1) d = 896, ⇒, 504 + ( n − 1)7 = 896, ⇒, ( n − 1)7 = 392, ⇒, n − 1 = 56, ⇒, n = 57, n, 57, Now,, S57 = ( a + l ) =, ( 504 + 896), 2, 2, 57, × 1400 = 39900, =, 2, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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41, , CBSE Term II Mathematics X (Standard), , 32. The sequence of two digit number which divided by 5 and, leave the remainder 2 is, 12, 17, 22, ..., 97 which is an AP, Here, a = 12 , d = 17 − 12 = 5 and l = 97, ∴, l = a + ( n − 1) d, ∴, 97 = 12 + ( n − 1)5, ⇒, 85 = ( n − 1) 5, ⇒, ( n − 1) = 17, ⇒, n = 17 + 1 = 18, ∴ Required sum of two digit number which divided by, n, 5 and leave the remainder 2 is ( a + l ), 2, 18, =, (12 + 97 ) = 9 × 109 = 981, 2, 33. Q Sum of n terms of an AP,, n, …(i), S n = [2 a + ( n − 1) d ], 2, 8, ∴, S 8 = [ 2 a + ( 8 − 1) d ], 2, = 4 (2 a + 7 d ) = 8a + 28d, 4, and, S 4 = [ 2 a + ( 4 − 1) d ], 2, = 2 (2 a + 3 d ) = 4 a + 6 d, Now,, …(ii), S8 − S4 = 8a + 28d − 4a − 6d = 4a + 22 d, 12, and, [2 a + (12 − 1 ) d ] = 6 (2 a + 11 d ), S12 =, 2, = 3 ( 4a + 22 d) = 3 (S8 − S4 ) [from Eq. (ii)], Hence proved., ∴, S12 = 3(S8 − S4 ), 34. For finding, the sum of last ten terms, we write the given AP, in reverse order., i.e. 126, 124, 122 , ... , 12 , 10, 8, Here, first term ( a ) = 126,, common difference,, ( d) = 124 − 126 = − 2, 10, n, ⎡, ⎤, S10 =, [2 a + (10 − 1 ) d] Q Sn = [2 a + ( n − 1) d], ∴, ⎢⎣, ⎥⎦, 2, 2, = 5 {2 (126) + 9 ( −2 )}, = 5 (252 − 18), = 5 × 234, = 1170, 35. Let the sequence of 100 natural numbers be 1, 2, 3, ...., 100, Here, a = 1, d = 2 − 1 = 3 − 2 = 1, Thus, natural number sequence is an AP., Now, sum of first 100 natural number is, n, 100, ⎡, ⎤, S100 =, [2 × 1 + (100 − 1)1] Sn = [2 a + ( n − 1)d], 2, 2, ⎣⎢, ⎦⎥, = 50 [2 + 99], = 50 × 101 = 5050, 36. For finding, the sum of first seven numbers which are multiples, of 2 as well as of 9. Take LCM of 2 and 9 which is 18., So, the series becomes 18, 36, 54,..., Here, first term ( a ) = 18,, , common difference ( d) = 36 − 18 = 18, n, S 7 = [2 a + ( n − 1 ) d ], ∴, 2, 7, = [2 (18) + (7 − 1) 18], 2, 7, = [ 36 + 6 × 18], 2, = 7 (18 + 3 × 18), = 7 (18 + 54), = 7 × 72 = 504, 37. Given, first term of an AP, a = 5, Common difference, d = 3, nth term of an AP, a n = 50, ∴, a + ( n − 1)d = 50, ⇒, 5 + ( n − 1)3 = 50, ⇒, ( n − 1)3 = 50 − 5, 45, n −1=, ⇒, 3, ⇒, n − 1 = 15, ⇒, n = 15 + 1 = 16, ∴ The sum of nth term of an AP is, n, Sn = [2 a + ( n − 1)d], 2, 16, S16 = [2 × 5 + (16 − 1) × 3 ], ∴, 2, = 8 [10 + 15 × 3], = 8 [10 + 45], = 8 × 55 = 440, 38. Let a and d be the first term and common difference of an, AP. Then,, a 4 = − 15 and a 9 = − 30, ⇒, a + ( 4 − 1) d = − 15, and, a + ( 9 − 1)d = − 30, …(i), ⇒, a + 3d = − 15, and, …(ii), a + 8d = −30, On subtracting Eq. (i) from Eq. (ii), we get, 8d − 3d = − 30 − ( −15), ⇒, 5d = − 30 + 15, 15, d=−, ⇒, 5, ⇒, d = −3, Put d = − 3 in Eq. (i), we get, a + 3( −3) = − 15, ⇒, a = − 15 + 9, ⇒, a = −6, ∴The sum of first 16 terms of an AP is, n, S n = [ 2 a + ( n − 1) d ], 2, 16, [2 ( −6) + 15 ( −3)], S16 =, ⇒, 2, = 8 [−12 − 45], = 8 × ( −57 ) = − 456, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 42, 39. Let a and d be the first term and common difference of an, AP. Then,, [given], S14 = 1050 and T4 = 40, 14, [2 a + (14 − 1) d] = 1050 and a + ( 4 − 1) d = 40, ⇒, 2, ⇒, 7 [2 a + 13d] = 1050 and a + 3d = 40, …(i), ⇒, 2 a + 13d = 150, and, …(ii), a + 3d = 40, Multiply Eq. (ii) by 2 and subtract Eq. (ii) from Eq. (i),, 13d − 6d = 150 − 80, ⇒, 7 d = 70, ⇒, d = 10, Put d = 10 in Eq. (i), we get, 2 a + 13 × 10 = 150, ⇒, 2 a = 150 − 130, ⇒, 2 a = 20, ⇒, a = 10, ∴The 20th term of an AP is, a 20 = a + (20 − 1) d, = 10 + 19 × 10, = 10 + 190 = 200, 40. Let the first term, common difference and number of terms, of the AP 9, 7, 5, ... are a1 , d1 and n1, respectively., i.e. first term ( a1 ) = 9 and common difference ( d1 ), = 7 − 9 = − 2., ∴ Its nth term, Tʹ n = a1 + ( n − 1) d1, ⇒, Tʹn = 9 + ( n − 1) ( − 2 ), ⇒, Tʹn = 9 − 2 n + 2, ...(i), ⇒, Tʹn = 11 − 2 n, Let the first term, common difference and the number of, terms of the AP 24, 21, 18, ... are a 2 , d2 and n 2, respectively., i.e. first term, ( a 2 ) = 24 and common difference, ( d2 ) = 21 − 24 = − 3., ∴ Its nth term, Tnʹʹ = a 2 + ( n − 1) d2, ⇒, , Tnʹʹ = 24 + ( n − 1) ( − 3), , ⇒, , Tnʹʹ = 24 − 3n + 3, , ...(ii), ⇒, Tnʹʹ = 27 − 3n, Now, by given condition,, nth terms of the both APs are same,, i.e., Tʹn = Tʹʹ, n, [from Eqs. (i) and (ii)], 11 − 2 n = 27 − 3n, ⇒, n = 16, ∴ nth term of first AP,, Tʹ n = 11 − 2 n = 11 − 2 (16), = 11 − 32 = − 21, and nth term of second AP,, Tʹʹ n = 27 − 3n, = 27 − 3 (16) = 27 − 48 = − 21, Hence, the value of n is 16 and that term i.e. nth term is –21., 41. Let the first term, common difference and number of terms, of an AP are a, d and n, respectively., We know that, if last term of an AP is known, then, ...(i), l = a + ( n − 1) d, , and nth term of an AP is, Tn = a + ( n − 1) d, Given that, 26th term of an AP = 0, ⇒, T26 = a + (26 − 1) d = 0, ⇒, a + 25 d = 0, 11th term of an AP = 3, ⇒, T11 = a + (11 − 1) d = 3, ⇒, a + 10 d = 3, and last term of an AP = − 1 / 5, ⇒, l = a + ( n − 1) d, ⇒, − 1 / 5 = a + ( n − 1) d, Now, subtracting Eq. (iv) from Eq. (iii),, a + 25 d = 0, a + 10 d = 3, −, , −, , ...(ii), [from Eq. (i)], ...(iii), [from Eq. (ii)], ... (iv), [from Eq. (i)], ...(v), , −, , 15 d = − 3, 1, d=−, ⇒, 5, Put the value of d in Eq. (iii), we get, ⎛ 1⎞, a + 25 ⎜ − ⎟ = 0, ⎝ 5⎠, ⇒, a −5=0, ⇒, a=5, Now, put the value of a , d in Eq. (v), we get, − 1 / 5 = 5 + ( n − 1) ( − 1 / 5), ⇒, − 1 = 25 − ( n − 1), ⇒, − 1 = 25 − n + 1, ⇒, n = 25 + 2 = 27, Hence, the common difference and number of terms are, − 1 / 5 and 27, respectively., 42. Let a and d be the first term and common difference of the, given AP, respectively., Given, a4 = 0, ⇒, a + 3d = 0, …(i), ⇒, a = − 3d, Now,, a 25 = a + 24d, [from Eq. (i)], = − 3d + 24d, …(ii), ⇒, a 25 = 21d, Also,, a11 = a + 10d, [From Eq. (i)], = − 3d + 10d, ⇒, a11 = 7 d, …(iii), ⇒, 3a11 = 21d, From Eqs. (ii) and (iii), we get, a 25 = 3a11, , Hence proved., , 43. Let a and d be the first term and common difference of an, AP., 1, Then, Tm =, n, 1, …(i), ⇒, a + ( m − 1 )d =, n, 1, and, Tn =, m, 1, …(ii), a + ( n − 1 )d =, ⇒, m, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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43, , CBSE Term II Mathematics X (Standard), , Subtracting Eq. (ii) from Eq. (i), we get, 1 1, [( m − 1) − ( n − 1)]d = −, n m, m −n, ⇒, ( m − n )d =, mn, 1, d=, ⇒, mn, 1, in Eq (i), we get, Put d =, mn, 1, 1, a + ( m − 1), =, mn n, 1, 1, 1, a= −, ( m − 1) =, ⇒, n mn, mn, ∴, a mn = a + ( mn − 1)d, 1, 1, =, + ( mn − 1) ×, mn, mn, 1, mn, =, =1, [1 + mn − 1] =, mn, mn, 44. Given first term of an AP, a = 54, Common difference, d = − 3, and nth term of an AP,, an = 0, ⇒, a + ( n − 1 )d = 0, ⇒, 54 + ( n − 1)( −3) = 0, ⇒, 54 − 3n + 3 = 0, ⇒, 3n = 57, 57, n=, ⇒, 3, ⇒, n = 19, Now, sum of first 19 terms of given AP is, n, Sn = [2 a + ( n − 1)d], 2, 19, S19 = [2 × 54 + (19 − 1)( −3)], ∴, 2, 19, = [108 − 54], 2, = 513, 45. Given equation is 1 + 4 + 7 + 10 + .... + x = 287, Consider series, 1 + 4 + 7 + 10 + .... + x, Here, a1 = 1, a 2 = 4, a 3 = 7, Now, a 2 − a1 = 4 − 1 = 3, a3 − a2 = 7 − 4 = 3, It implies that common difference is constant say 3. So, it is, an AP series, whose first term is a = 1 and common, difference d = 3., Here, last term of an AP is l = x, Q, l = a + ( n − 1) d, ∴, x = 1 + ( n − 1) × 3, ⇒, x = 1 + 3n − 3, ⇒, x = 3n − 2, x+2, ⇒, n=, 3, , Now, sum of AP series is, n, S n = [2 a + ( n − 1 ) d ], 2, 1 ⎛ x + 2⎞ ⎡, ⎛x+2 ⎞ ⎤, − 1⎟ 3⎥, = ⎜, ⎟ 2 ×1 + ⎜, ⎝ 3, ⎠ ⎦, 2 ⎝ 3 ⎠ ⎢⎣, x+2, =, [2 + x − 1 ], 6, x+2, ⇒, 287 =, × ( x + 1), 6, 2, ⇒, 1722 = x + 3x + 2, ⇒, , x 2 + 3x − 1720 = 0, , ⇒, , x 2 + ( 43 − 40)x − 1720 = 0, , ⇒, , [splitting middle term], , 2, , x + 43x − 40x − 1720 = 0, , ⇒, x ( x + 43) − 40( x + 43) = 0, ⇒, ( x − 40) ( x + 43) = 0, ⇒, x = 40, − 43, But x = − 43 is not possible, because it is an increasing AP., Hence, required value of x is 40., 46. Given, equation is 1 + 5 + 9 + 13 + K + x = 1326, Here, first term is a1 = 1, last term is a n = l = x, Difference of two consecutive terms,, 5 − 1 = 4 and 9 − 5 = 4, which is same., Thus, given series is an AP., Then, nth term of given AP is, a n = a + ( n − 1 )d, ∴, x = 1 + ( n − 1 )4, ⇒, ( n − 1) 4 = x − 1, x −1, ⇒, n −1=, 4, x −1, n=, +1, ⇒, 4, x+3, ⇒, n=, 4, Now, sum of given AP is, n, S n = [2 a + ( n − 1 ) d ], 2, x+ 3⎡, x+3, ⎤, 1326 =, 2 ×1+, ×4, ∴, ⎥⎦, 4 × 2 ⎢⎣, 4, ⇒, ⇒, ⇒, , 1326 × 8 = ( x + 3) [2 + x + 3], 10608 = ( x + 3) ( x + 5), x 2 + 8x + 15 − 10608 = 0, x 2 + 8x − 10593 = 0, , ⇒, 2, , ⇒, , x + (107 − 99)x − 10593 = 0, , ⇒, , [by splitting middle term], x 2 + 107 x − 99x − 10593 = 0, , ⇒, ⇒, ⇒, , x( x + 107 ) − 99( x + 107 ) = 0, ( x − 99) ( x + 107 ) = 0, x = 99, − 107, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 44, 47. (i) Here, first term ( a ) = 1, and common difference, ( d) = ( − 2 ) − 1 = − 3, Q Sum of n terms of an AP,, n, S n = [2 a + ( n − 1) d ], 2, n, Sn = [2 × 1 + ( n − 1) × ( − 3)], ⇒, 2, n, ⇒, S n = (2 − 3 n + 3), 2, n, ...(i), ⇒, S n = ( 5 − 3n ), 2, We know that, if the last term ( l) of an AP is known, then, l = a + ( n − 1) d, ⇒, − 236 = 1 + ( n − 1) ( − 3) [Q l = − 236, given ], ⇒, − 237 = − ( n − 1) × 3, ⇒, n − 1 = 79, ⇒, n = 80, Now, put the value of n in Eq. (i), we get, 80, Sn =, [ 5 − 3 × 80], 2, = 40 ( 5 − 240), = 40 × ( − 235), = − 9400, Hence, the required sum is − 9400., Alternate Method, Given, a = 1, d = − 3 and l = − 236, ∴ Sum of n terms of an AP,, n, Sn = [ a + l], 2, 80, =, [1 + ( − 236)], [Q n = 80], 2, = 40 × ( − 235), = − 9400, 1, (ii) Here, first term, a = 4 −, n, Common difference,, 2⎞ ⎛, 1 ⎞ − 2 1 −1, ⎛, d = ⎜4 − ⎟ − ⎜4 − ⎟ =, + =, ⎝, n⎠ ⎝, n⎠, n, n, n, Q Sum of n terms of an AP,, n, S n = [2 a + ( n − 1 ) d ], 2, n ⎡ ⎛, 1⎞, ⎛ −1⎞ ⎤, ⇒, Sn = ⎢2 ⎜ 4 − ⎟ + ( n − 1 ) ⎜ ⎟ ⎥, ⎝ n ⎠⎦, n⎠, 2 ⎣ ⎝, n ⎧, 2, 1⎫, ⎨8 − − 1 + ⎬, 2 ⎩, n, n⎭, n ⎛, 1⎞, = ⎜7 − ⎟, 2 ⎝, n⎠, n ⎛ 7 n − 1⎞, = ×⎜, ⎟, 2 ⎝ n ⎠, 7n − 1, =, 2, =, , 4, 2, 1, 48. Given AP sequence is − , − 1, − , .... , 4, 3, 3, 3, 4, Here, first term ( a ) = − ,, 3, 4 1, common difference ( d ) = − 1 + =, 3 3, 1 13, and the last term ( l ) = 4 =, 3 3, Q nth term of an AP,, l = an = a + (n − 1) d, 13, 4, 1, = − + (n − 1), ⇒, 3, 3, 3, ⇒, 13 = − 4 + ( n − 1 ), ⇒, n − 1 = 17, [even], ⇒, n = 18, ⎛n, ⎞, ⎛ n⎞, So, the two middle most terms are ⎜ ⎟ th and ⎜ + 1⎟ th., ⎝2, ⎠, ⎝2⎠, ⎛ 18, ⎞, ⎛ 18⎞, i.e. ⎜ ⎟ th and ⎜ + 1⎟ th terms, ⎝2, ⎠, ⎝2⎠, i.e. 9th and 10th terms., 4, ⎛ 1 ⎞ −4 + 8 4, =, + 8⎜ ⎟ =, ⎝ 3⎠, 3, 3, 3, −4, ⎛ 1 ⎞ −4 + 9 5, =, = a + 9d =, + 9⎜ ⎟ =, ⎝ 3⎠, 3, 3, 3, , ∴, , a9 = a + 8d = −, , and, , a10, , So, sum of the two middle most terms, = a 9 + a10, 4 5 9, = + = =3, 3 3 3, 49. Let the first term, common difference and the number of, terms in an AP are a, d and n, respectively., We know that, the n th term of an AP,, ... (i), Tn = a + ( n − 1) d, ∴ 4th term of an AP,, [given], T4 = a + ( 4 − 1 ) d = − 15, ⇒, a + 3 d = − 15, and 9th term of an AP,, T9 = a + ( 9 − 1 ) d = − 30, ⇒, a + 8 d = − 30, Now, subtract Eq. (ii) from Eq. (iii), we get, a + 8 d = − 30, a + 3d = − 15, − −, , ...(ii), [given], ...(iii), , +, , 5 d = − 15, ⇒, d=−3, Put the value of d in Eq. (ii), we get, a + 3 ( − 3) = − 15, ⇒, a − 9 = − 15, ⇒, a = − 15 + 9, ⇒, a=−6, Q Sum of first n terms of an AP,, n, S n = [2 a + ( n − 1) d ], 2, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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45, , CBSE Term II Mathematics X (Standard), , ∴ Sum of first 17 terms of an AP,, 17, S17 = [2 × ( − 6) + (17 − 1) ( − 3)], 2, 17, =, [ − 12 + (16) ( − 3)], 2, 17, =, ( − 12 − 48), 2, 17, =, × ( − 60 ), 2, = 17 × ( − 30 ), = − 510, Hence, the required sum of first 17 terms of an AP is − 510., 50. Given first term of each sum of an AP is 1 and common ratio, of each sum are 1, 2 and 3, respectively., n, ∴, S 1 = [2(1) + ( n − 1)1], 2, n, = [2 + n − 1 ], 2, n, = ( n + 1), 2, n, S 2 = [2(1) + ( n − 1)2 ], ∴, 2, n, = [2 + 2 n − 2 ], 2, , and, , = n2, n, S 3 = [ 2 ( 1 ) + ( n − 1) 3], 2, n, = [2 + 3 n − 3], 2, n, = ( 3n − 1), 2, LHS = S1 + S 3, n, n, = ( n + 1) + ( 3n − 1 ), 2, 2, n, = [ n + 1 + 3n − 1], 2, n, = × 4n, 2, = 2 n 2 = 2S 2, , Hence proved., , 51. Given, S4 = 40 and S14 = 280, Let a be the first term and d be the common difference of, 4, [2 a + ( 4 − 1)d] = 40, 2, n, ⎡, ⎤, Q S = {2 a + ( n − 1) d}, ⎢⎣ n 2, ⎥⎦, , given AP. Then, S4 = 40 ⇒, , ⇒, and, , 2[2 a + 3d] = 40 ⇒ 2 a + 3d = 20, S14 = 280, 14, ⇒, [2 a + (14 − 1) d] = 280, 2, ⇒, 2 a + 13d = 40, On subtracting Eq. (i) from Eq. (ii), we get, 10d = 20 ⇒ d = 2, , …(i), , …(ii), , On substituting d = 2 in Eq. (i), we get, ⇒, 2 a + 3 × 2 = 20, ⇒, 2 a = 14, a =7, n, n, Now, Sn = [2 a + ( n − 1)d] = [2(7 ) + ( n − 1)2 ], 2, 2, n, = [14 + 2 n − 2 ] = n[ 6 + n ] = 6n + n 2, 2, Hence, the sum of first n terms is n 2 + 6n ., 52. Let a and d be the first term and common difference of an AP., Given that,, a11 : a18 = 2 : 3, ⇒, , a + 10 d 2, =, a + 17 d 3, , ⇒, 3a + 30 d = 2 a + 34 d, …(i), ⇒, a = 4d, Now,, a 5 = a + 4 d = 4 d + 4 d = 8 d [from Eq. (i)], and, a 21 = a + 20 d = 4 d + 20 d = 24 d [from Eq. (i)], ∴, a 5 : a 21 = 8d : 24 d = 1 : 3, Now, sum of the first five terms,, 5, S 5 = [ 2 a + ( 5 − 1) d ], 2, 5, [from Eq. (i)], = [2 ( 4 d ) + 4 d ], 2, 5, 5, = ( 8 d + 4 d ) = × 12 d = 30 d, 2, 2, and sum of the first 21 terms,, 21, [2 a + (21 − 1 ) d ], S21 =, 2, 21, [from Eq. (i)], =, [2 ( 4 d ) + 20 d ], 2, 21, =, (28 d ) = 294 d, 2, So, ratio of the sum of the first five terms to the sum of the, first 21 terms, S 5 : S21 = 30 d : 294 d = 5 : 49, 53. Let the four consecutive number of an AP be, a , a + d, a + 2 d and a + 3d., Since, sum of four consecutive number in AP is 32., ∴ a + a + d + a + 2 d + a + 3d = 32, ⇒, 4a + 6d = 32, [divide by 2], ⇒, 2 a + 3d = 16, 16 − 3d, …(i), a=, ⇒, 2, According to the question,, Product of first and last terms 7, =, Product of two middle terms 15, a ( a + 3d ), 7, =, ( a + d) ( a + 2 d) 15, ⇒, ⇒, , 15a ( a + 3d) = 7 ( a + d) ( a + 2 d), 15a 2 + 45ad = 7 a 2 + 21ad + 14d2, , ⇒, , 8a 2 + 24ad − 14d2 = 0, , ⇒, , 4a 2 + 12 ad − 7 d2 = 0, , [divide by 2], , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 46, ⇒, ⇒, , 4a 2 + (14 − 2 )ad − 7 d2 = 0, [by splitting middle term], 4a 2 + 14ad − 2 ad − 7 d2 = 0, , ⇒, ⇒, , 2 a(2 a + 7 d) − d(2 a + 7 d) = 0, (2 a + 7 d ) ( 2 a − d ) = 0, 7d, d, …(ii), and a =, ⇒, a=−, 2, 2, 7d, in Eq. (i), we get, Put a = −, 2, 7 d 16 − 3d, −, =, 2, 2, ⇒, −7 d = 16 − 3d, ⇒, 4d = − 16 ⇒ d = − 4, d, Now, put a = in Eq. (i), we get, 2, d 16 − 3d, =, 2, 2, ⇒, d = 16 − 3d, ⇒, 4d = 16 ⇒ d = 4, 7d, For a = −, and d = − 4, then, 2, 7 × ( − 4), a=−, = 14, 2, d, 4, For a = and d = 4, then a = = 2, 2, 2, Therefore, the four consecutive numbers in an AP are, 14, [14 + ( −4)] , [14 + (2 × − 4)] , [14 + ( 3 × − 4)], or 2 , (2 + 4), (2 + 2 × 4), (2 + 3 × 4)., Hence, the numbers are 14, 10, 6, 2 or 2, 6, 10, 14., 54. Given that, the AP is a , b , c., Here, first term = a, common difference = b − a, and last term, l = a n = c, Q, an = l = a + (n − 1) d, ⇒, c = a + ( n − 1) ( b − a ), c−a, ⇒, (n − 1) =, b−a, c−a, ⇒, n=, +1, b−a, c − a + b − a c + b − 2a, …(i), n=, =, ⇒, b−a, b−a, ∴ Sum of an AP,, n, [2 a + ( n − 1) d ], 2, ⎤, ⎫, ⎧b + c − 2a, (b + c − 2a) ⎡, =, − 1 ⎬ ( b − a )⎥, 2a + ⎨, ⎢, 2 (b − a) ⎣, ⎭, ⎩ b−a, ⎦, , Sn =, , ⎡, ⎤, c−a, ⎢2 a + b − a ⋅ ( b − a )⎥, ⎣, ⎦, ( b + c − 2a), =, (2 a + c − a ), 2 (b − a), , =, , (b + c − 2a), 2 (b − a), , =, , (b + c − 2a), ⋅ ( a + c), 2( b − a ), , Hence proved., , 1, 2, 55. Given AP is 20, 19 , 18 ,... ., 3, 3, 1, 58, 58 − 60, −2, Here, a = 20 and d = 19 − 20 =, − 20 =, =, 3, 3, 3, 3, Let n terms of given AP be required to get sum 300., n, We know that, Sn = [2 a + ( n − 1)d], 2, n ⎡, ⎛ −2 ⎞ ⎤, ⇒, 300 = ⎢2(20) + ( n − 1)⎜ ⎟ ⎥, ⎝ 3 ⎠⎦, 2 ⎣, , ⇒, , [Qa = 20 and d = − 2 / 3], 2, 2⎤, ⎡, 600 = n 40 − n +, ⎢⎣, 3, 3 ⎥⎦, 1, 600 = [120n − 2 n 2 + 2 n ], 3, 600 × 3 = 122 n − 2 n 2, , ⇒, , 1800 + 2 n 2 − 122 n = 0, , ⇒, , 2 [ n 2 − 61n + 900 ] = 0, , ⇒, ⇒, , n 2 − 61n + 900 = 0, , ⇒, ⇒, , [divide by 2], , 2, , n − 36n − 25n + 900 = 0, , ⇒, n ( n − 36) − 25( n − 36) = 0, ⇒, ( n − 36)( n − 25) = 0 ⇒ n = 36 or 25, Since, a is positive and d is negative, so both values of n, are possible., Hence, sum of 25 terms of given AP, = Sum of 36 terms of given AP = 300., 56. (i) Now, she takes ` 1 on day 1, ` 2 on day 2, ` 3 on day 3, and so on till the end of the month, from this money., i.e. 1 + 2 + 3 + 4 + ... + 31., which form an AP in which terms are 31 and first term, ( a ) = 1, common difference ( d) = 2 − 1 = 1, ∴ Sum of first 31 terms = S31, Sum of n terms,, n, S n = [2 a + ( n − 1) d ], 2, 31, ∴, S31 =, [2 × 1 + ( 31 − 1) × 1], 2, 31, 31 × 32, =, (2 + 30) =, = 31 × 16 = 496, 2, 2, So, Kanika takes ` 496 till the end of the month from this, money., (ii) Let her pocket money be ` x., Now, she spent ` 204 of her pocket money and found that, at the end of the month she still has ` 100 with her., Now, according to the condition,, ( x − 496) − 204 = 100, ⇒, x − 700 = 100, ∴, x = ` 800, Hence, ` 800 was her pocket money for the month., (iii) Here, a = 1, d = 1, n = 13, Now, a n = a + ( n − 1) d, ⇒ a 13 = 1 + 12(1) = 1 + 12 = 13, So, Kanika, saved ` 13 till January 13th, 2008., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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Chapter Test, Multiple Choice Questions, , (iv) Total amount paid in 13th and 17th, installment is, , 1. The list of numbers − 10, − 6, − 2, 2, ... is, [NCERT Exemplar], (a) an AP with d = − 16, (c) an AP with d = − 4, , (b) an AP with d = 4, (d) not an AP, , 2. In an AP, if a = 3.5, d = 0 and n = 101, then an will be, (a) 0, (c) 103.5, , (b) 3.5, (d) 104.5, , 3. Is an sequence defined by an = 2n2 + 1 forms an, AP?, , a = 1 and 20th term = 58 is, (b) 580, (d) 560, , 5. The 10th term of an AP is 52 and 16th term is 82,, then 32nd term of the AP is, , [NCERT Exemplar], , (b) 159, (d) 156, , Short Answer Type Questions, , The difference between their 10th terms is the, same as the difference between their 21st, terms, which is the same as the difference, between any two corresponding, [NCERT, terms?, Exemplar], Why?, , 8. Determine the AP whose fifth term is 19 and, the difference of the eighth term from the, thirteenth term is 20., whose middle most term is 30., , 6. Kartik starts repaying a loan as first installment, of, ` 100. He increases the installment by ` 5 every, month., (i) AP formed from the given situation is, , Long Answer Type Questions, , 10. An AP consists of 37 terms. The sum of the, three middle most terms is 225 and the sum, of the last three terms is 429. Find the AP., , 11. If sum of first 6 terms of an AP is 36 and that, of the first 16 terms is 256, then find the sum, of first 10 terms., , (b) 100, 105, 110, ......, (d) 110, 115, 120, ......, , (ii) The amount Kartik will pay in 30th, installment is, (a) ` 265, (c) ` 255, , (b) ` 360, (d) ` 416, , 9. Find the sum of all the 11 terms of an AP, , Case Based MCQs, , (a) 105, 110, 115, ......, (c) 95, 100, 105, ......, , (a) ` 314, (c) ` 412, , The first term of one AP is 2 and that of the, other is 7., , 4. The sum of first 20 terms of an AP in which, , (a) 152, (c) 162, , (v) If he increases the installment by ` 6 every, month, then the amount he will pay in, 53th installment is, , 7. Two AP’s have the same common difference., , (a) Yes, (b) Not, (c) Cannot be determined, (d) None of the above, , (a) 590, (c) 570, , (a) ` 380, (b) ` 300, (c) ` 360, (d) ` 340, , 12. Which term of the AP : 121, 117, 113, ... is its, second negative term?, , (b) ` 235, (d) ` 245, , 13. The sum of the third and the seventh terms of, an AP is 6 and their product is 8. Find the sum, of first sixteen terms of the AP., , (iii) If Kartik pays ` 795, then it is, (a) 140th installment, (b) 150th installment, (c) 160th installment, (d) 170th installment, , 14. Solve the equation −4 + (−1) + 2 + ..... + x = 437., , Answers, 1. (b), , 2. (b), , 3. (b), , 7. (37), , 8. 3, 7, 11, 15, , 11. 100 12. 33rd term, , 4. (c), , 5. (a), , 6. (i) (b) (ii) (d) (iii) (a) (iv) (d) (v) (c), , 9. 330 10. 3, 7, 11, 15, 13. 20 or 76, , 14. x =, , For Detailed Solutions, Scan the code, , 50, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 48, , CHAPTER 03, , Circles, In this Chapter..., !, , Circle, , !, , Tangent of a Circle, , !, , Theorem Related to Tangent of a Circle, , A circle is a collection of all points in a plane which are at a, constant distance i.e. radius from a fixed point i.e. centre., In the given figure, O is the centre of circle and OA is the radius of, the circle. Also, AB is the diameter of the circle., , Semi-circle, A diameter of a circle divides it into two equal parts or, in two equal arcs. Each of these two arcs is called a, semi-circle., , Circumference, B, , O, , A, , Two or more circles having the same centre are called concentric, circles., , Some Important Terms Related to Circle, Chord, A line segment joining any two points on the circumference of the, circle is called a chord of the circle. If this chord passes through the, centre, then this chord (or diamter) is the longest chord of the circle., , The length of the complete circle is called the, circumference of the circle., , Arc, A continuous piece of a circle is called an arc. In, adjoining figure, P and Q are two points on a circle, which divide it into two parts, called the arcs. The, larger part is called the major arc QRP and the smaller, part is called the minor arc PMQ., R, Major arc, , O, , P, A, , B, , Minor arc, , Q, , M, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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49, , CBSE Term II Mathematics X (Standard), , Segment, The region between a chord and either of its arcs is called a, segment of the circular region or simply a segment of the, circle. The segment formed by minor arc along with chord, is, called minor segment and the segment formed by major arc, is, called the major segment., , Tangent to a Circle, A line which touches the circle at a point, is called tangent to, a circle., In the figure, O is the centre of circle, AB is a tangent line, and P is a point of contact., There is only one tangent at a point of the circle., A circle can have maximum two parallel tangents which, can be drawn to the opposite sides of the centre., ●, , ●, , Major, segment, B, , A, , O, , Minor segment, , A, , Sector, The region between an arc and the two radii, joining the ends, of the arc to the centre, is called a sector., , O, Minor, sector, , Length of a Tangent, , B, , O, , The sector formed by minor arc, is called minor sector and, the sector formed by major arc, is called major sector., , Important Results Related to Circle, (i) The perpendicular drawn from the centre of a circle to, a chord bisects it and vice-versa., (ii) Equal chords of a circle are equidistant from the centre., (iii) The angle subtended by an arc (or corresponding, chord) at the centre of the circle is twice the angle, subtended by the same arc at any point on the, remaining part of the circle., C, , A, B, Segment of the tangent between, the given point B and point of contact A, , Tangent line, , In the above figure, AB is called the length of tangent., ∴ Length of tangent to the circle from an exterior point,, AB =, (Distance of exterior point from centre) 2 − (Radius) 2, , Number of Tangent from a Point on a Circle, (i) If point P lies outside the circle, then two tangents can, be drawn to the circle, i.e. PT1 and PT2 ., , θ, , a ng, T1 (T, , O, 2θ, A, , B (Tangent line), , The length of the segment of the tangent, between the given, point (on the tangent) and the point of contact, is called the, length of tangent from the given point., , Major sector, , A, , P, (Point of contact), , B, , (iv) Equal chords of a circle subtend equal angles at, the centre., (v) The angle in a semi-circle is a right angle., (vi) Angles in the same segment of a circle are equal., (vii) The sum of any pair of opposite angles of a cyclic, quadrilateral is 180°., (viii) If two circles intersect at two points, then the line, through the centres is the perpendicular bisector of the, common chord., , ent), , P, T2 (T, ang, , ent), , (ii) If point P lies on the circle, then there is one and only, one tangent to a circle passing through point P., , P Tangent, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 50, (iii) If a point P lies inside the circle, then there is no, tangent to a circle passing through a point lying, inside the circle., , Important Results Related to Tangent to a Circle, (i) If two circles touch internally or externally, then point of, contact lies on the straight line through the two centres., , No Tangent, , O, , P, , Theorems Related to Tangent of Circle, Theorem 1 The tangent at any point of a circle is, perpendicular to the radius through the point of contact., O, , A, , B, , P, , Oʹ, , P, , O, , P, , Oʹ, , (ii) A pair of tangents drawn at two points of a circle are either, parallel or they intersect each other at a point outside the, circle., (iii) If two tangents drawn to a circle are parallel to each other,, then the line segment joining their point of contact is a, diameter of the circle., (iv) If two tangents are drawn to a circle from an external point,, then, (a) They subtend equal angles at the centre,, i.e. ∠POA = ∠POB., A, , Here, O is centre of circle and AB is tangent of circle at P, and it is point of contact and OP is radius., ∴, OP ⊥ AB., Theorem 2 A perpendicular drawn from the end point of, radius is tangent to the circle. If OP ⊥ AB, then AB is, tangent to circle., , P, , O, B, , (b) They are equally inclined to the segment joining, the centre to that point,, i.e. ∠APQ = ∠BPQ., , O, , A, , P, , B, , Theorem 3 The lengths of two tangents drawn from an, external point to a circle are equal., A, O, , (v) The opposite sides of a quadrilateral circumscribing a circle, subtend supplementary angles at the centre of the circle., P, , D, , C, , B, , O, , Here, P is exterior point and PA and PB are tangents., ∴, , PA = PB, , A, , B, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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51, , CBSE Term II Mathematics X (Standard), , Solved Examples, Example 1. Prove that a tangent to a circle is, perpendicular to the radius through the point of, [CBSE 2020 (Standard)], contact., Sol. Given A circle with centre O and a tangent AB at a point, P on the circle., To prove OP ⊥ AB, Construction Take any point Q, other than P on the tangent, AB and join OQ., , From Eqs. (i) and (ii), we get, TP = TQ, Hence, T is the mid-point of the line segment PQ., , Example 3. In figure, PQ is tangent to the circle with, centre O, at the point B. If ∠AOB = 100°, then find, [CBSE 2020 (Standard)], ∠ABP., O, A, , 10, , 0°, , O, , Q, B, , R, Q, , P, , A, , Proof Here, Q is a point on the tangent AB, other than the, point of contact P. So, Q lies outside the circle (if Q lies, inside the circle, then AB becomes a secant and not a, tangent to the circle)., Let OQ intersects the circle at R ., Then,, [radii of the circle], OP = OR, Now,, OQ = OR + RQ, ⇒, OQ > OR, [Q OP = OR ], ⇒, OQ > OP or OP < OQ, Thus, OP is shorter than any other segment joining O to any, point of AB. Also, we know that the shortest distance, between a point and a line is perpendicular distance from, the point to the line., So, OP is perpendicular to AB., i.e., Hence proved., OP ⊥ AB, , Example 2. In given figure, two circles touch each, other at the point C. Prove that the common, tangent to the circles at C, bisects the common, tangent at P and Q., P, , A, , T, C, , Q, B, , Sol. We know that, tangents drawn from an external point are of, equal length. Therefore, according to the given figure,, TP = TC …(i) [Q point T is external], and, TQ = TC …(ii) [Q point T is external], P, , T, , P, , B, , Sol. Given, ∠AOB = 100° ,, A, 100°, , O, , B, Q, , P, , In ΔOAB,, ⇒, , [radii of the circle], OA = OB, ∠OBA = ∠OAB, [angles opposite to equal sides are equal] …(i), , In ΔOAB,, ∠AOB + ∠OAB + ∠OBA = 180°, [by angle sum property of triangle], ⇒, 100° + ∠OBA + ∠OBA = 180° [from Eq. (i)], ⇒, 2 ∠OBA = 180° − 100°, 80°, ∠OBA =, ⇒, 2, …(ii), ⇒, ∠OBA = 40°, We know that, radius of circle is perpendicular to the, tangent., ∴, ∠OBP = 90° ⇒ ∠OBA + ∠ABP = 90°, ⇒, 40° + ∠ABP = 90°, ⇒, ∠ABP = 90° − 40° ⇒ ∠ABP = 50°, , Example 4. In below figure, PA is a tangent from, an external point P to a circle with centre O., If ∠POB = 115°, find ∠APO. [CBSE 2020 (Standard)], A, , Q, , O, , P, , A, , B, C, , 115°, B, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 52, Sol. Given, ∠POB = 115°, , ⇒, A, , AP 2 = OA 2 − OQ2, , [Q OP = OQ = radii of a circle] …(ii), Now, in right angled ΔOQA,, OA 2 = OQ2 + AQ2, , P, O, 115°, , ⇒, B, , Since, AB is a straight line., ∴, ∠POB + ∠AOP = 180°, ⇒, 115° + ∠AOP = 180°, ⇒, ∠AOP = 180° − 115°, ⇒, ∠AOP = 65°, We know that, radius line is perpendicular to the tangent., ∴ ∠PAO = 90°, In ΔAOP,, ∠PAO + ∠AOP + ∠APO = 180°, [Q sum of all angles of a triangle is 180°], ∴, 90° + 65° + ∠APO = 180°, ⇒, ∠APO = 180° − ( 65° + 90° ), = 180° − 155° = 25°, , AQ2 = OA 2 − OQ2, , ...(iii), , From Eqs. (ii) and (iii), we get, AP 2 = AQ2, ⇒, , AP = AQ, , Hence proved., , Example 6. Prove that the angle between the two, tangents drawn from an external point to a circle is, supplementary to the angle subtended by the line, segment joining the points of contact at the centre., Sol. Let PQ and PR be two tangents drawn from an external point, P to a circle with centre O., Q, , P, , O, , Example 5. Prove that the length of tangents drawn, from an external point to a circle are equal., [CBSE 2020 (Standard)], Sol. Let AP and AQ are two tangents drawn from a point A to a, circle with centre O., , R, , To prove ∠QOR = 180° − ∠QPR, or ∠QOR + ∠QPR = 180°, Proof In Δ OQP and ΔORP,, , P, , PQ = PR, , O, , A, , Q, , To prove AP = AQ, Construction, , Join OP, OQ and OA., , Proof We know that, a tangent at any point of a circle is, perpendicular to the radius through the point of contact., Here, AP is a tangent and OP is the radius of the circle, through P., ∴, OP ⊥ AP, Similarly,, OQ ⊥ AQ, ...(i), ⇒, ∠OPA = ∠OQA = 90°, First Method, In ΔOPA and ΔOQA, we have, [radii of a circle], OP = OQ, [from Eq. (i)], ∠OPA = ∠OQA = 90°, [common sides], OA = OA, So,, ΔOPA ≅ ΔOQA [by RHS congruence rule], [by CPCT], ∴, AP = AQ, Second Method, In right angled ΔOPA,, OA 2 = OP 2 + AP 2, [by Pythagoras theorem], ⇒, AP 2 = OA 2 − OP 2, , ∴, Then,, and, ⇒, and, , [Q tangents drawn from an, external point are equal in length], [radii of circle], OQ = OR, [common sides], OP = OP, [by SSS congruence rule], ΔOQP ≅ ΔORP, [by CPCT], ∠ QPO = ∠ RPO, [by CPCT], ∠ POQ = ∠ POR, ∠ QPR = 2 ∠ OPQ⎫, …(i), ⎬, ∠ QOR = 2 ∠ POQ ⎭, , Now, in right angled ΔOQP,, ∠ QPO + ∠ QOP = 90°, ⇒, ∠ QOP = 90° − ∠ QPO, ⇒, 2 ∠QOP = 180° − 2 ∠QPO, [multiplying both sides by 2], [from Eq. (i)], ⇒, ∠ QOR = 180° − ∠ QPR, Hence proved., ⇒, ∠ QOR + ∠ QPR = 180°, , Example 7. In figure, find the perimeter of ΔABC,, if AP = 12 cm., A, , B, P, , D, , C, Q, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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53, , CBSE Term II Mathematics X (Standard), , ∴, ⇒, ⇒, Now,, , AC = AR + RC, 11 = 4 + RC, RC = 11 − 4 = 7 cm, BC = BQ + QC, = 3 + 7 = 10 cm, Hence, length of BC is 10 cm., , Sol. Given, AP = 12 cm, ⇒, , AQ = AP = 12 cm, A, , B, , D, , C, Q, , P, , Example 9. In the given figure, from an external point, P, two tangents PQ and PR are drawn to a circle of, radius 4 cm with centre O. If ∠QPR = 90°, then, find the length of PQ., , Example 8. In below figure, ΔABC is circumscribing a, circle, the length of BC is …… cm., [CBSE 2020 (Standard)], 4c, m, , A, , 11, cm, , R, , 3c, m, , P, , B, , C, , Q, , Sol. We know that, the tangents drawn from an external point to, a circle are equal. Therefore,, , Q, 4 cm, 45°, O, , P, 90°, , R, , ∠QPR, 2, 90°, =, = 45°, 2, Also, radius of circle OQ is perpendicular to the tangent line, QP., Now, in right angled ΔOQP,, OQ, tan 45° =, QP, 4, 1=, ⇒, QP, , ∴, , ∠OPQ =, , ⇒, QP = 4 cm, Hence, length of PQ is 4 cm., , Example 10. In the given figure, if tangents PA and PB, , A, , R, , 80º, , B 3 cm Q 7 cm C, , and, , R, , Sol. We know that, if pair of tangents are drawn from an external, point P, then line joining from centre O to the point P,, bisects the angle P., , m, 7c, , 3c, m, , P, , O, , P, , from an external point P to a circle with centre O,, are inclined to each other at an angle of 80°, then, [CBSE 2020 (Standard)], find ∠AOB., , m, 4c, , 4c, m, , A, , Q, 4 cm, , [Q tangents drawn from an external point, are equal in lengths], Also,, BD = BP, [Q B is an external point] …(i), and, CD = CQ, [Q C is an external point] …(ii), Now,, AP = AB + BP, [from Eq. (i)] …(iii), ⇒, 12 = AB + BD, and, AQ = AC + CQ, [from Eq. (i)] …(iv), ⇒, 12 = AC + CD, Perimeter of ΔABC, = AB + BC + AC, = AB + BD + DC + AC, = 12 + 12, [from Eqs. (iii) and (iv)], = 24 cm, Hence, perimeter of a ΔABC is 24 cm., , P, , O, , BP = BQ ,, AP = AR, , [point B is an external], [point A is an external], , CQ = CR, , [point C is an external], , B, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 54, Sol. Given, ∠APB = 80°., , Example 12. In given figure, two tangents TP and TQ, , We know that, line drawn from centre of a circle to the, tangent is perpendicular., Since, OA⊥PA and OB⊥PB., Then, ∠OAP = ∠OBP = 90°, …(i), , are drawn to a circle with centre O from an external, point T. Prove that ∠PTQ = 2 ∠OPQ., [CBSE 2020 (Standard)], P, , A, 80º, , T, , P, , O, , O, Q, , B, , Sol. Given, TP and TQ are two tangents of a circle with centre O, and points P and Q are point of contact., To prove ∠ PTQ = 2 ∠OPQ, , Since, OAPB is a quadrilateral., By using angle sum property of a quadrilateral,, ∠AOB + ∠OBP + ∠APB + ∠OAP = 360°, ⇒, ∠AOB + 90° + 80° + 90° = 360°, ⇒, ∠AOB = 360° − 260°, ⇒, ∠AOB = 100°, , Example 11. In given figure, PA and PB are tangents to, the circle with centre O, such that ∠APB = 50°,, then the measure of ∠OAB is ......, , Proof Let ∠PTQ = θ, As we know that, the length of tangents drawn from an, external point to a circle are equal., So, ΔTPQ is an isosceles triangle., Therefore, according to the given figure,, 1, θ, ∠TPQ = ∠TQP = (180° − θ ) = 90°−, 2, 2, As we know that, the tangents at any point of a circle is, perpendicular to the radius through the point of contact., P, , A, T, , 50°, , P, , O, Q, , B, , ∴, ∠OPT = 90°, Now, ∠OPQ = ∠OPT − ∠TPQ, θ ⎞ θ ∠PTQ, ⎛, = 90° − ⎜ 90° − ⎟ = =, ⎝, 2⎠ 2, 2, , Sol. Given, ∠APB = 50°, A, , P, , !, , O, , 50°, , O, B, , Since, P is an external point of a circle., Therefore,, PA = PB, [Q tangents drawn from an external to a circle are equal], ⇒, ∠PBA = ∠PAB, [Q angles opposite to equal sides are equal] ...(i), In ΔAPB,, ∠APB + ∠PBA + ∠PAB = 180°, [Q sum of all angles of a triangle is 180°], [from Eq. (i)], ∴, 50° + 2 ∠PAB = 180°, ⇒, 2 ∠PAB = 130°, ...(ii), ⇒, ∠PAB = 65°, Also, radius OA is perpendicular to the tangent of a circle., Therefore,, ∠OAP = 90°, ⇒, ∠OAB + ∠PAB = 90°, [from Eq. (ii)], ⇒, ∠OAB + 65° = 90°, ⇒, ∠OAB = 90° − 65° = 25°, , ⇒, , ∠PTQ = 2 ∠OPQ, , Hence proved., , Example 13. In figure, a quadrilateral ABCD is drawn, to circumscribe a circle. Prove that, AB + CD = BC + AD, [CBSE 2020 (Standard)], A, , B, , D, C, , Sol. Given A quadrilateral ABCD is circumscribing a circle., To prove AB + CD = AD + BC, Proof Let P , Q, R and S be the point of contact., A, , P, , S, , B, Q, , D, R, , C, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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55, , CBSE Term II Mathematics X (Standard), , We know that, the length of tangents drawn from an external, point to a circle are equal., ∴, AP = AS, [Q both are tangents to a circle from point A] …(i), Similarly, BP = BQ,, …(ii), …(iii), CR = CQ, and, …(iv), DR = DS, Adding Eqs. (i), (ii), (iii) and (iv), we get, ( AP + BP ) + ( CR + DR ) = ( AS + BQ) + ( CQ + DS), ⇒, AB + CD = ( AS + DS) + (BQ + CQ), ⇒, AB + CD = AD + BC Hence proved., , Example 14. Prove that the tangents at the extremities, of any chord of a circle make equal angles with the, [CBSE 2020 (Standard)], chord., Sol. Let AB be a chord of a circle having centre O. Let AP and, BP be the tangents at A and B, which intersect at point P., To prove ∠PAC = ∠PBC, Construction Join points C and P., Proof We know that, tangents drawn from an external point, are equal., , Now, ∠OPQ = 90° − 60° = 30°, [Q ∠OPT = 90°, as radius line OP is perpendicular, to the tangent], ⇒, ∠OQP = 30°, [angles opposite to equal sides are equal], In ΔOPQ, using angle sum property of a triangle,, ∠POQ + ∠OPQ + ∠OQP = 180°, ⇒, ∠POQ + 30° + 30° = 180°, ⇒, ∠POQ = 180° − 60° = 120°, ⇒, ∠PQʹ Q = 60°, [angle subtended by an arc at centre is twice the angle, subtended at remaining part of circle], ⇒ ∠PRQ = 180°−∠PQʹ Q = 120°, [Q opposite angles are supplementary in a cyclic, quadrilateral PQʹ QR ], , Example 16. In given figure, AB is a chord of circle, with centre O, AOC is diameter and AT is tangent, at A. Prove that ∠BAT = ∠ACB., C, , A, , O, O, , C, , B, , P, , A, B, , ∴ In ΔPCA and ΔPCB,, [QP is an external point of a circle], PA = PB,, ∠APC = ∠BPC, [Q AP and BP are equally inclined to OP], and, [common sides], PC = PC, [by SAS similarity rule], ∴, ΔPAC ~ ΔPBC, [by CPCT], ⇒, ∠PAC = ∠PBC, , Example 15. In given figure, PQ is a chord of a circle, and PT is tangent at P such that ∠QPT = 60°, then, the measure of ∠PRQ is ....... [CBSE 2020 (Standard)], Q, , O, 60°, , R, , P, T, , Sol. Take a point Qʹ on circle and join PQʹ and QQʹ., Qʹ, Q, O, , º, 60 R, A, , P, , T, , Sol. Given AB is a chord of a circle, AOC is a diameter of the, circle having centre O and line AT is tangent at A., To prove ∠BAT = ∠ACB, C, B, , O, , A, , T, , Proof We know that, diameter of a circle subtends 90° to the, semi-circle., ∴, ∠ABC = 90°, Let ∠ACB = θ,, then ∠CAB = 180° − ( 90° + θ), [by using angle sum property of triangle], ...(i), ⇒, ∠CAB = 90° − θ, We know that, radius of a circle is perpendicular to the, tangent., ∴, ∠OAT = 90°, ⇒, ∠OAB + ∠BAT = 90°, [Q ∠OAT = ∠CAT], ⇒, ∠CAB + ∠BAT = 90°, ⇒, 90° − θ + ∠BAT = 90°, ⇒, ∠BAT = θ, Hence proved., ⇒, ∠BAT = ∠ACB, , T, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 56, , Chapter, Practice, PART 1, Objective Questions, , A, O, 115°, , ●, , Multiple Choice Questions, , B, , 1. If radii of two concentric circles are 4 cm and 5 cm,, then length of each chord of one circle, which is, [NCERT Exemplar], tangent to the other circle, is, (a) 3 cm, , (b) 6 cm, , (c) 9 cm, , (d) 1 cm, , 2. The length of tangent from an external point P on a, circle with centre O is always less than OP., , 3. The length of the tangents to the circle from a point, at any distance of 5 cm from centre of the circle of, radius 3 cm is, (a) 2 cm, (c) 8 cm, , (b) 4 cm, (d) None of these, , 4. The length of the tangent drawn from a point 8 cm, away from the centre of circle of radius 6 cm is, (b) 2 7 cm, , (c) 10 cm, , (d) 5 cm, , 5. PQ is a tangent to a circle with centre O at the, point P. If ΔOPQ is an isosceles triangle, then, ∠OPQ is equal to, (a) 30°, , (a) 25°, , (b) 20°, , (b) 45°, , (c) 60°, , (d) 90°, , 6. In figure, if O is the centre of a circle, PQ is a chord, and the tangent PR at P makes an angle of 50° with, [NCERT Exemplar], PQ, then ∠POQ is equal to, P, , R, , 50°, , (c) 30°, , (d) 65°, , 8. In figure, AT is a tangent to the circle with centre O, such that OT = 4 cm and ∠ OTA = 30° . Then, AT is, [NCERT Exemplar], equal to, , [NCERT Exemplar], (b) False, (d) None of these, , (a) True, (c) Can’t determined, , (a) 7 cm, , P, , O, 4 cm, , 30°, , T, , A, , (a) 4 cm, , (b) 2 cm, , (c) 2 3 cm, , (d) 4 3 cm, , 9. PQ is a tangent drawn from a point P to a circle, with centre O and QOR is a diameter of the circle, such that ∠POR = 135°, then ∠OPQ is, (a) 60°, , (b) 45°, , (c) 30°, , (d) 90°, , 10. A tangent PQ at a point P of a circle of radius 6 cm, meets a line through the centre O at a point Q, so, that OQ = 14 cm, then length of PQ is, (a) 4 10 cm, (c) 5 10 cm, , (b) 6 10 cm, (d) 7 10 cm, , 11. In figure, if PA and PB are tangents to the circle, with centre O such that ∠ APB = 50°, then ∠OAB is, equal to, [NCERT Exemplar], A, , O, P, , Q, , (a) 100°, , (b) 80°, , (c) 90°, , 50°, , O, , (d) 75°, , 7. In the given figure, PA is a tangent from an external, point P to a circle with centre O. If ∠POB = 115°,, then ∠APO is, , B, , (a) 25°, (c) 40°, , (b) 30°, (d) 50°, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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57, , CBSE Term II Mathematics X (Standard), , 12. If angle between two tangents drawn from a point P, to a circle of radius a and centre O is 90°, then, OP = a 2., [NCERT Exemplar], (a) True, (c) Can’t say, , (b) False, (d) Partially true or false, , (b) 41 cm, (d) 40 cm, , (a) 41 cm, (c) 40 cm, , 18. PA is a tangent to the circle with centre O., If BC = 3 cm, AC = 4 cm and ΔACB ~ΔPAO, then, [CBSE 2013], OA is equal to, B, , 13. In the given figure, find the value of x°., A, , C, O, , x°, , 30°, , B, , O, , P, , A, , (a) 130°, , (b) 75°, , (c) 120°, , (d) 60°, , 14. From the given figure, find the value of x ° + y°., P, , (b) 5 cm, 5, (d) cm, 2, , (a) 2.7 cm, (c) 5 cm, , 19. In the given figure, if ∠ACB = 50°, then ∠ATO is, A, , y°, , x°, , Q, , O, , R, C, , (a) 270°, (c) 90°, , O, , T, , (b) 180°, (d) None of these, , 15. At one end A of a diameter AB of a circle of radius, 5 cm, tangent XAY is drawn to the circle. The, length of the chord CD parallel to XY and at a, [NCERT Exemplar], distance 8 cm from A is, (a) 4 cm, , (b) 5 cm, , (c) 6 cm, , (d) 8 cm, , 16. In figure, AB is a chord of the circle and AOC is its, diameter such that ∠ ACB = 50°. If AT is the tangent, to the circle at the point A, then ∠BAT is equal to, , B, , (a) 30°, (b) 50°, (c) 40°, (d) Can’t be determined, , 20. In the adjoining figure, PQ is a chord of a circle, with centre O and PT is a tangent at P such that, ∠QPT = 60°, then ∠PRQ =, Q, , [NCERT Exemplar], C, , O, R, 60°, , B, , O, , P, , A, , (a) 120°, A, , (a) 45°, , T, , (b) 60°, , (c) 50°, , (d) 55°, , 17. In the adjoining figure, AD = 8 cm, AC = 6 cm and, TB is the tangent at B to the circle with centre O., [CBSE 2013], If BT is 4 cm, then OT =, , (b) 160°, , T, , (c) 130°, , (d) 150°, , 21. In figure, if PQR is the tangent to a circle at Q,, whose centre is O, AB is a chord parallel to PR and, ∠BQR = 70°, then ∠AQB is equal to, [NCERT Exemplar], B, , D, , A, , A, , O, C, , O, , D, , 70°, P, , B 4 cm, , T, , (a) 20°, , Q, , (b) 40°, , R, , (c) 35°, , (d) 45°, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 58, 22. From an external point P, tangents PA and PB are, drawn to a circle with centre O. If CD is the, tangent to the circle at a point E and PA = 14 cm,, then perimeter of ΔPCD is, (a) 14 cm, , (b) 21 cm, , (c) 28 cm, , 28. In figure, if ∠AOB = 125°, then ∠COD is equal to, [NCERT Exemplar], A, B, , (d) 35 cm, 125°, O, , 23. Tangents AP and AQ are drawn to circle with, centre O from an external point A, then ∠PAQ is, equal to, (a) 2∠OPQ, , (b), , ∠OPQ, 2, , (c), , ∠OPQ, 3, , (d), , ∠OPQ, 4, , 24. In the given figure, two tangents AB andAC are, drawn to a circle with centre O such that, ∠BAC = 120°, then OA is equal to, , (b) 45°, (d) 55°, , (a) 62.5°, (c) 35°, ●, , Case Based MCQs, 29. A playground is in the shape of a triangle with right, angle at B , AB = 3 m and BC = 4 m. A pit was dig, inside it such that it touches the walls AC , BC and, AB at P , Q and R, respectively such that AP = x m., , O, , C, 120°, , B, , C, , D, , A, , (a) 2AB, , (b) 3AB, , (c) 4AB, , (d) 5AB, , A, , 25. In the given figure, O is the centre of a circle, BOA, is its diameter and the tangent at the point P meets, BA extended at T. If ∠PBO = 30°, then ∠PTA =, [CBSE 2016], P, 30°, O, , B, , (a) 40°, , (b) 50°, , (c) 30°, , (d) 20°, , 26. In adjoining figure, PQ and PR are tangents to the, circle with centre O and S is a point on the circle, such that ∠SQL = 50° and ∠SRM = 60°. Then,, [NCERT Exemplar], ∠QSR, L, , P, , O, Q, , C, , Based on the above information, answer the, following questions., (i) The value of AR =, (a) 2x m, (c) x m, , (b) x / 2 m, (d) 3x m, , (ii) The value of BQ =, 50°, , S, , Q, , P, , O, 60°, , R, , (b) 50°, , (c) 60°, , (a) 2x m, , (b) ( 3 − x ) m, , (c) (2 − x ) m, , (d) 4x m, , (iii) The value of CQ =, , M, , (a) 40°, , R, B, , T, , A, , r, , (d) 70°, , 27. In given figure, AB is diameter of a circle with, centre O and AT is tangent. If ∠AOQ = 58 ° , then, [CBSE 2015], ∠ATQ =, B, , (a) ( 4 + x ) m, , (b) ( 5 − x ) m, , (c) (1 + x ) m, , (d) Both (b) and (c), , (iv) Which of the following is correct?, (a) Quadrilateral AROP is a square, (b) Quadrilateral BROQ is a square, (c) Quadrilateral CQOP is a square, (d) None of the above, , (v) Radius of the pit is, (a) 1 m, (c) 4 m, , O, 58°, A, , (a) 52°, , (b) 58°, , Q, T, , (c) 61°, , (d) 62°, , (b) 3 m, (d) 5 m, , 30. A student draws two circles that touch each other, externally at point K with centres A and B and radii, 6 cm and 4 cm, respectively as shown in the figure., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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59, , CBSE Term II Mathematics X (Standard), , T, K, 4 cm B, , X, , P, , 6 cm A, , 3c, m, Y, , Q, , 5. If a number of circles touch a given line segment, PQ at a point A, then their centres lie on the, perpendicular bisector of PQ. Why or why not?, , 8, , [NCERT Exemplar], , cm, S, , Based on the above information, answer the following, questions., (i) The value of PA =, (a) 10 cm, (c) 13 cm, , (b) 5 cm, (d) Can’t be determined, , (ii) The value of BQ =, (a) 4 cm, (c) 6 cm, , (b) 5 cm, (d) 18 cm, , (iii) The value of PK =, (a) 13 cm, (c) 16 cm, , (b) 15 cm, (d) 18 cm, , (iv) The value of QY =, (a) 2 cm, (c) 1 cm, , (b) 5 cm, (d) 3 cm, , (v) If two circles touch externally, then the number of, common tangents can be drawn is, (a) 1, (c) 3, , (b) 2, (d) None of these, , PART 2, Subjective Questions, ●, , 6. Out of the two concentric circles, the radius of, the outer circle is 5 cm and the chord AC of, length 8 cm is a tangent to the inner circle. Find, the radius of the inner circle., 7. If a chord AB subtends an angle of 60° at the, centre of a circle, then find the angle between the, [NCERT Exemplar], tangents at A and B., 8. From an external point P, two tangents, PA and, PB are drawn to a circle with centre O. At one, point E on the circle tangent is drawn, which, intersects PA and PB at C and D, respectively., If PA = 10 cm, find the perimeter of the trianlge, PCD., 9. Prove that the centre of a circle touching two, intersecting lines lies on the angle bisector of the, [NCERT Exemplar], lines., 10. If from an external point B of a circle with centre, O, two tangents BC and BD are drawn, such that, ∠ DBC = 120°, prove that BC + BD = BO i.e., BO = 2 BC., 11. In figure, AB and CD are common tangents to, two circles of equal radii. Prove that AB = CD., [NCERT Exemplar], B, , A, , Short Answer Type Questions, 1. If PQ is a tangent to a circle with centre O and radius, 6 cm such that ∠PQO = 60°, then find the length of a, tangent PQ and a line OQ., 2. The tangent to the circumcircle of an isosceles ΔABC, at A, in which AB = AC, is parallel to BC., [NCERT Exemplar], , 12. In figure, common tangents AB and CD to two, circles intersect at E. Prove that AB = CD., [NCERT Exemplar], A, , 3. If AB is a chord of a circle with centre O, AOC is a, diameter and AT is the tangent at A as shown in, figure. Prove that ∠ BAT = ∠ ACB. [NCERT Exemplar], C, , A, , D, , E, , C, , O, , D, , C, , B, , 13. If PA and PB are two tangents drawn from a point, P to a circle with centre O touching it at A and B,, prove that OP is perpendicular bisector of AB., , B, , T, , 4. Prove that a diameter AB of a circle bisects all, those chords, which are parallel to the tangent at, [NCERT Exemplar], the point A., , [CBSE 2008], , 14. Tangents AP and AQ are drawn to circle with, centre O from an external point A. Prove that, [CBSE 2013, 12,11, 09], ∠PAQ = 2 ∠OPQ., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 60, 15. In the given figure, ∠ADC = 90°, BC = 38 cm,, CD = 28 cm and BP = 25 cm, then find the radius of, [CBSE 2011], the circle., B, , Q, C, O, , R, , 21. If a hexagon ABCDEF circumscribe a circle,, prove that, AB + CD + EF = BC + DE + FA [NCERT Exemplar], 22. In the given figure, AD is a diameter of a circle with, centre O and AB is a tangent at A. C is a point on the, circle such that DC produced intersects the, tangent at B and ∠ABD = 50°. Find ∠COA., [CBSE 2015], D, , P, A, , D, S, , C, , O, , 16. ΔABC is a right angled triangle with ∠B = 90°,, BC = 3 cm and AB = 4 cm. A circle with centre O, and radius r cm has been inscribed in ΔABC. Find, the radius of the incircle., 17. The radii of two concentric circles are 13 cm and, 8 cm. AB is a diameter of the bigger circle. BD is a, tangent to the smaller circle touching it at D., Find the length of AD., [CBSE 2010], 18. A circle is inscribed in a ΔABC having sides, AB = 8 cm, BC = 10 cm and CA = 12 cm, as shown in, [CBSE 2012], figure. Find AD, BE and CF., , 50°, , B, , A, , 23. Tangents PQ and PR are drawn to a circle such that, ∠RPQ = 30°. A chord RS is drawn parallel to the, tangent PQ. Find ∠RQS., [CBSE 2015], 24. PA and PB are the tangents to a circle, which, circumscribes an equilateral ΔABQ. If ∠PAB = 60°,, as shown in the figure, prove that QP bisects AB at, [CBSE 2015], right angle., A, , A, , 60°, Q, , P, , M, , D, , F, , B, C, , ●, , B, , E, , Long Answer Type Questions, 19. Let s denotes the semi-perimeter of a ΔABC, in, which BC = a, CA = b and AB = c. If a circle touches, the sides BC , CA , AB at D , E, F , respectively. Prove, [NCERT Exemplar], that BD = s − b., 20. AC and AD are tangents at C and D, respectively., If ∠BCD = 44° , then find∠CAD , ∠ADC , ∠CBD, and ∠ACD., C, , 25. Two circles with centres O and Oʹ of radii 3 cm and, 4 cm, respectively intersect at two points P and Q,, such that OP and O ʹ P are tangents to the two, circles. Find the length of the common chord PQ., [NCERT Exemplar], , 26. If an isosceles ΔABC in which AB = AC = 6 cm, is, inscribed in a circle of radius 9 cm, find the area of, the triangle., 27. In a figure, the common tangents AB and CD of, two circles with centres O and Oʹ intersect at E., Prove that the points O , E and O ʹ are collinear., [NCERT Exemplar], , A, , A, , 44°, , D, , O, D, , B, , O, , Oʹ, , E, C, , B, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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61, , CBSE Term II Mathematics X (Standard), , 28. In figure, O is the centre of a circle of radius 5 cm,, T is a point such that OT = 13 and OT intersects the, circle at E, if AB is the tangent to the circle at E,, [NCERT Exemplar], find the length of AB., , (ii) A circle of radius 3 cm is inscribed in a right, angled ΔBAC such that BD = 9 cm and DC = 3 cm., Find the length of AB., A, , P, A, , F, O, , T, , E, , O, , R, , E, , B, Q, ●, , (a) 6 cm, (c) 15 cm, , Case Based Questions, 29. Dheeraj loves geometry. So, he was curious to know, more about the concepts of circles. His grand father, is a mathematicians. So, he reached to his grand, father to learn something interesting about tangents, and circles. His grand father gave him knowledge, on circles and tangents and ask him to solve the, following questions., , C, , D, , B, , (b) 12 cm, (d) 10 cm, , (iii) In the given figure, what is the length of CD?, A, , 4 cm, , 2 cm, , B, , P, 10 cm, , 5 cm, S, , Q, R, , C, , D, , (a) 11 cm, (c) 7 cm, , (b) 9 cm, (d) 13 cm, , (iv) If PA and PB are two tangents to a circle with, centre O from an external point P such that, ∠OPB = 50 °, then find ∠BPA, (a) 60°, (c) 120°, , (i) In the given figure, AP , AQ and BC are tangents to, the circle such that AB = 7 cm, BC = 4 cm and, AC = 9 cm. Find AP, , (b) 50°, (d) 100°, , (v) In the given figure, P is an external point from,, which tangents are drawn to two externally, touching circles. If PA = 11 cm, then find PC., P, , A, , B, P, , (a) 12 cm, , (b) 15 cm, , D, , C, , A, Q, , (c) 13 cm, , B, , (d) 10 cm, , (a) 3.5 cm, (c) 11 cm, , C, , (b) 4 cm, (d) Can’t be determined, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 62, , SOLUTIONS, Objective Questions, 1. (b) Let O be the centre of two concentric circles C1 and C 2,, whose radii are r1 = 4 cm and r2 = 5 cm . Now, we draw a, chord AC of circle C 2, which touches the circle C1 at B., Also, join OB, which is perpendicular to AC., [Q tangent at any point of circle is perpendicular to radius, through the point of contact], , 4. (b) Since, tangent to a circle is perpendicular to the radius, through the point of contact., ∴, ∠OTP = 90°, T, 6 cm, P 8 cm, , C2, , O, , C1, O, , A, , In ΔOTP, we have, OP 2 = OT 2 + PT 2, , C, , B, , Now, in right angled ΔOBC, by using Pythagoras theorem,, OC 2 = BC 2 + BO 2, [Q(hypotenuse) 2 = (base) 2 + (perpendicular) 2], ⇒, , 5 2 = BC 2 + 4 2, , ⇒, , BC 2 = 25 − 16 = 9, , T, , PT is a tangent drawn from external point P. Join OT., Q, OT ⊥ PT, So, ΔOPT is a right angled triangle formed., In right angled triangle, hypotenuse is always greater than, any of the two sides of the triangle., ∴, OP > PT, or, PT < OP, 3. (b) Given, OB = 5 cm and radius OA = 3 cm, By Pythagoras theorem, in right angled ΔOAB,, O, 3, A, , OB 2 = OA 2 + AB 2, AB 2 = 25 − 9, AB 2 = 16, AB = 4 cm, , PT 2 = 64 − 36 = 28, , ⇒, , PT = 28 = 2 7 cm, , 5. (b) Since, PQ is a tangent to a circle from a point P and, centre of circle is O., ∴ ΔOPQ is an isosceles triangle., , O, , ∠OQP = 90°, OP = QP, ∠POQ = ∠OPQ, , P, , AB 2 = ( 5)2 − ( 3)2, , ⇒, , P, , O, , B, , ( 8)2 = ( 6)2 + PT 2, , Q, , ⇒, BC = 3 cm, ∴ Length of chord AC = 2 BC, = 2 × 3 = 6 cm, 2. (a), , 5, , ⇒, , [Q OB = 5, AB = 3], , ∴, In ΔOPQ,, ∠POQ + ∠OQP + ∠OPQ = 180°, ⇒, 2 ∠OPQ = 180° − 90° [Q ∠POQ = ∠OPQ], ⇒, 2 ∠OPQ = 90° ⇒ ∠OPQ = 45°, 6. (a) Given, ∠ QPR = 50°, We know that, the tangent at any point of a circle is, perpendicular to the radius through the point of contact., ∴, ∠OPR = 90°, [from figure], ⇒, ∠OPQ + ∠QPR = 90°, [Q ∠QPR = 50°], ⇒, ∠OPQ = 90° − 50° = 40°, Now,, OP = OQ = Radius of circle, ∴, ∠OQP = ∠ OPQ = 40°, [since, angles opposite to equal sides are equal], In Δ OPQ, ∠ O + ∠ P + ∠ Q = 180°, [since, sum of all angles of a triangle = 180°], ⇒, ∠ O = 180° − ( 40° + 40° ) [Q ∠P = 40° = ∠Q], = 180° − 80° = 100°, 7. (a) Here, ∠OAP = 90° [Q tangent at any point of a circle is, perpendicular to the radius], Now, ∠AOP + ∠BOP = 180°, ⇒, ∠AOP + 115° = 180°, ⇒, ∠AOP = (180° − 115° ) = 65°, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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63, , CBSE Term II Mathematics X (Standard), , And also, ∠OAP + ∠AOP + ∠APO = 180°, [angle sum property of triangle], ⇒ 90°+ 65°+∠APO = 180°, ⇒, 155° + ∠APO = 180°, ⇒ ∠APO = 180°−155° = 25°, 8. (c) Join OA., , O, 4 cm, , 30°, , T, , A, , We know that, the tangent at any point of a circle is, perpendicular to the radius through the point of contact., ∴, ∠OAT = 90°, AT, In Δ OAT,, cos 30° =, OT, 3 AT, ⇒, =, 2, 4, ⇒, AT = 2 3 cm, 9. (b) Given, PQ is a tangent from point P, centre O and QOR, as diameter., R, , PQ2 = 160, , ⇒, , PQ = 16 × 10 = 4 10 cm, , ⇒, , 11. (a) Given, PA and PB are tangent lines., ∴, PA = PB, [since, the length of tangents drawn from an, external point to a circle is equal], [say], ⇒, ∠PBA = ∠ PAB = θ, In ΔPAB, ∠ P + ∠ A + ∠ B = 180°, [since, sum of all angles of a triangle = 180°], ⇒, 50° + θ + θ = 180°, ⇒, 2θ = 180° − 50° = 130°, ⇒, θ = 65°, Also,, OA ⊥ PA, [since, tangent at any point of a circle is perpendicular, to the radius through the point of contact], ∴, ∠ PAO = 90°, ⇒, ∠ PAB + ∠ BAO = 90°, ⇒, 65° + ∠ BAO = 90°, ⇒, ∠ BAO = 90° − 65° = 25°, 12. (a) From point P, two tangents are drawn., Given,, OT = a, Also, line OP bisects the ∠RPT., ∴, ∠ TPO = ∠ RPO = 45°, Also,, OT ⊥ PT, T, , 135°, , O, , a, 45°, , In ΔPQO,, , P, , P, , Q, , ∠ROP = ∠OPQ + 90°, [Q exterior angle of a triangle is equal to the sum, of opposite angles], ∠OPQ = 135° − 90°, = 45°, , 10. (a) Here, OP = 6 cm and OQ = 14 cm, , R, , In right angled Δ OTP,, , 6 cm, , ⇒, , 13. (c) Given, ∠OBA = 30°, 14, , P, , cm, Q, , We know that, tangent at any point of a circle is, perpendicular to the radius through the point of contact., So, OP ⊥ PQ, Now, in right angled ΔOPQ,, OQ2 = OP 2 + PQ2, ⇒, , (14)2 = ( 6)2 + PQ2, , ⇒, , PQ2 = 196 − 36, , OT, OP, 1, a, =, 2 OP, OP = a 2, , sin 45° =, , ⇒, O, , O, , 90°, , [by Pythagoras theorem], , In ΔABO,, x ° = ∠ABO + 90°, [Q external angle = sum of opposite internal angles], x° = 30° + 90° = 120°, 14. (a) In ΔPOQ,, x ° = ∠PQO + 90°, [Q external angle = sum of opposite internal angles], = (180° − y ° ) + 90°, = 270° − y °, x ° + y ° = 270°, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 64, 15. (d) First, draw a circle of radius 5 cm having centre O., A tangent XY is drawn at point A., C, B, , 5 cm, , X, , 5, , cm O 3, , E, cm, , 8, , A, , D, , cm, , Y, , A chord CD is drawn, which is parallel to XY and at a, distance of 8 cm from A., Also,, AE = 8 cm. Join OC, Now, in right angled ΔOEC,, OC 2 = OE 2 + EC 2, ⇒, , EC 2 = OC 2 − OE 2, [by Pythagoras theorem], = 52 − 32, , [Q OC = radius = 5 cm, OE = AE − AO = 8 − 5 = 3 cm], = 25 − 9 = 16, ⇒, EC = 4 cm, Hence, length of chord CD = 2 CE, = 2 × 4 = 8 cm, [since, perpendicular from centre to the, chord bisects the chord], 16. (c) In figure, AOC is a diameter of the circle. We know that,, diameter subtends an angle 90° at the circle., So,, ∠ ABC = 90°, In ΔACB, ∠A + ∠B + ∠C = 180°, [since, sum of all angles of a triangle is 180°], ⇒, ∠ A + 90° + 50° = 180°, ⇒, ∠A + 140 = 180, ⇒, ∠ A = 180° − 140° = 40°, ∠A or ∠OAB = 40°, Now, AT is the tangent to the circle at point A. So, OA is, perpendicular to AT., [from figure], ∴, ∠ OAT = 90°, ⇒, ∠ OAB + ∠ BAT = 90°, On putting ∠OAB = 40°, we get, ⇒, ∠ BAT = 90° − 40° = 50°, Hence, the value of ∠BAT is 50°., 17. (b) Clearly, ∠CAD = 90°, [angle in a semi-circle], So, in ΔACD, CD 2 = AC 2 + AD 2 = 36 + 64 = 100, [by Pythagoras theorem], ⇒, CD = 10 cm, Therefore,, OC = OD = OB = 5 cm [Q radius of a circle], Since,, ∠OBT = 90°, [angle between radius and tangent], So, in ΔOBT, OT 2 = OB 2 + BT 2, = 25 + 16 = 41, OT = 41 cm, ⇒, , [by Pythagoras theorem], , 18. (d) In ΔACB,, [angle in a semi-circle], ∠BCA = 90°, [by Pythagoras theorem], ∴, AB 2 = AC 2 + BC 2, ⇒, AB 2 = 42 + 32, ⇒, AB 2 = 16 + 9 cm, ⇒, AB 2 = 25 cm, ⇒, AB = 5 cm, 5, ⇒, OA = cm, 2, 19. (c) ∠OAT = 90°, [Q angle between radius and tangent], Now, ∠BOA = 100°, [angle subtended by an arc at, centre is twice the angle, subtended at remaining part of circle], ⇒ ∠ATO = 180° − ( ∠TOA + ∠OAT ), [angles property of a triangle], = 180° − ( 50° + 90° ), = 180° − 140° = 40°, 20. (a) Take a point Qʹ on circle and join PQʹ and QQʹ., Qʹ, Q, O, , º R, , 60, P, , A, , T, , Now, ∠OPQ = 90°−60° = 30°, ⇒ ∠OQP = 30°, ⇒, ⇒, , ⇒, are, , [Q ∠OPT = 90°], [angles opposite to equal, sides are equal], , ∠POQ = 120°, [angle sum property of a triangle], ∠PQʹ Q = 60°, [angle subtended by an arc at centre is twice the angle, subtended at remaining part of circle], [Q opposite angles, ∠PRQ = 120°, , supplementary in a cyclic quadrilateral PQQʹ R ], 21. (b) Given, AB ||PR, D, , A, , B, , O, , 70°, P, , Q, , R, , ∴, ∠ ABQ = ∠ BQR = 70° [alternate angles], Also, QD is perpendicular to AB and QD bisects AB., In ΔQDA and ΔQDB,, [each 90°], ∠QDA = ∠QDB, AD = BD, [common side], QD = QD, ∴, Δ ADQ ~ Δ BDQ, [by SAS similarity criterion], , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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65, , CBSE Term II Mathematics X (Standard), , Then,, [by CPCT] ...(i), ∠QAD = ∠ QBD, Also,, ∠ABQ = ∠BQR [alternate interior angle], [Q ∠BQR = 70°], ∴, ∠ABQ = 70°, Hence,, [from Eq. (i)], ∠QAB = 70°, Now, in Δ ABQ, ∠ A + ∠ B + ∠ Q = 180°, ⇒, ∠ Q = 180° − ( 70° + 70° ) = 40°, 22. (c) We have, PA = PB = 14 cm, A, 14, , cm, , C, , P, , E, , O, , D, B, , Also, CD is tangent at point E on the circle., So, CA and CE are tangents to the circle from point C., Therefore, CA = CE, similarly DB = DE, Now, perimeter of ΔPCD, = PC + CD + PD = PC + CE + ED + PD, = PC + CA + PD + DB, [Q CA = CE and DE = DB], = PA + PB = 14 + 14 = 28 cm, 23. (a) Here, AP = AQ, ⇒, , ∠AQP = ∠APQ = x (say), [Q angles opposite to equal sides of a triangle are equal], P, A, , O, Q, , In ΔAPQ, ∠PAQ = 180° − ( ∠APQ + ∠AQP ), = 180° − ( x + x ) = 180° − 2x, Q, OP ⊥ AP, ∴, ∠OPA = 90°, ⇒ ∠OPQ + ∠APQ = 90°, ⇒, ⇒, ⇒, ⇒, , 24. (a) In ΔOAB and ΔOAC, we have, ∠OBA = ∠OCA = 90°, OA = OA, and, OB = OC, So, by RHS congruence criterion,, ΔOBA ≅ ΔOCA, ⇒, ∠OAB = ∠OAC, 1, = × 120° = 60°, 2, In ΔOBA, we have, AB, cos 60° =, OA, , = 90°−50° = 40°, Similarly, ∠ORS = 30°, Now, ∠QSR = ∠OSR + ∠OSQ = ∠ORS + ∠OQS, [Q angle opposite to equal sides are equal], = 30° + 40°, = 70°, 1, 1, 27. (c) ∠ABQ = ∠AOQ = ( 58° ) = 29°, 2, 2, [angle subtended by an arc at the centre is twice the angle, subtended at remaining part of circle], and ∠BAT = 90°, [angle between radius and tangent], In ΔABT, we get,, ∠ATQ = 180° − (29°+90° ) = 61°, [angle sum property of a triangle], 28. (d) We know that, the opposite sides of a quadrilateral, circumscribing a circle subtend supplementary angles at the, centre of the circle., i.e. ∠AOB + ∠COD = 180°, ⇒, ∠ COD = 180° − ∠AOB, = 180° − 125° = 55°, 29. Here, in right angled ΔABC, AB = 3 m and BC = 4 m., ∴By Pythagoras theorem,, AC = ( AB )2 + (BC )2, = ( 3) 2 + ( 4 ) 2, , ∠OPQ + x = 90°, ∠OPQ = 90° − x, 2 ∠OPQ = 180°−2 x, ∠PAQ = 2 ∠OPQ, , 1 AB, =, 2 OA, ⇒, OA = 2 AB, 25. (c), ∠OPB = 30°, [Q angles opposite to equal sides are equal], and, ∠OPT = 90°, Now,, ∠PTA = 180°−( ∠OBP + ∠BPT ), [angle sum property of a ΔBPT], = 180° − ( 30° + 120° ), [Q ∠BPT = 90° + 30° = 120° ], = 180° − 150°, = 30°, 26. (d) ∠OQS = ∠OQL − ∠SQL [since, OQ ⊥ LP], ⇒, , = 9 + 16 = 25 = 5 m, [multiplying by 2], , [common], [radii of circle], , Also, AP = x m, (i) (c) AR = AP = x m, … (i), [since, length of tangents drawn from an, external point are equal], (ii) (b) BQ = BR = AB − AR = ( 3 − x ) m, [using Eq. (i)], (iii) (d) CQ = CP = AC − AP = ( 5 − x ) m, Also, CQ = BC − BQ = BC − BR, = 4 − ( 3 − x) = 1 + x, (iv) (b) Since, CQ = 5 − x = 1 + x, ⇒, 4 = 2x ⇒ x = 2, ∴, AR = AP = 2 m, BR = BQ = 1 m, and, CP = CQ = 3 m, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 66, A, , r, R, , O, , B, , Q, , 2×6, 3, ×, 3, 3, OQ = 4 3 cm, , ⇒, , P, , OQ =, , ⇒, C, , Also, OQ ⊥ BQ and OR ⊥ BR, ∴ BROQ is a square., (v) (a) Radius of the pit, OR = BR = 1 m, 30. Here, AS = 6 cm, BT = 4 cm, [Q radii of circles], , A, , E, , = 64 + 36, = 100 = 10 cm, (ii) (b) Again, by Pythagoras theorem, we have, BQ = TQ2 + BT 2 = 32 + 42, = 9 + 16 = 25 = 5 cm, (iii) (c) PK = PA + AK = 10 + 6 = 16 cm, (iv) (c) QY = BQ − BY = 5 − 4 = 1 cm, (v) (b) If two circles touch externally, then the number of, common tangents can be drawn is 2., Subjective Questions, 1. Given, PQ is a tangent, OP = 6 cm and ∠PQO = 60°, We know that, tangent at any point of a circle is, perpendicular to the radius through the point of contact., ∴, OP ⊥ PQ, , O, , 60°, P, , Now, in right angled ΔOPQ,, OP, tan 60° =, PQ, 6, 3=, ⇒, PQ, 6, ×, 3, , Q, , perpendicular ⎤, ⎡, Q tan θ =, ⎢⎣, ⎥⎦, base, [Q tan 60° = 3], 3, 3, , ⇒, , PQ =, , ⇒, , PQ = 2 3 cm and sin 60° =, , F, , C, , B, , EAF ||BC, , To prove, , = 82 + 62, , [rationalising], , Hence, length of a tangent PQ is 2 3 cm and a line OQ, is 4 3 cm., 2. Let EAF be tangent to the circumcircle of ΔABC., , (i) (c) Since, radius at point of contact is perpendicular to, tangent., ∴ By Pythagoras theorem, we have, PA = PS2 + AS2, , ⎡, 3⎤, ⎢Q sin 60°= 2 ⎥, ⎣, ⎦, , 3, 6, =, OQ, 2, , ⇒, , ∠EAB = ∠ABC, Here,, AB = AC, …(i), ⇒, ∠ACB = ∠ABC, [angle between tangent and its chord equal to angle made by, chord in the alternate segment], …(ii), ∴ Also,, ∠ EAB = ∠ BCA, From Eqs. (i) and (ii), we get, ∠ EAB = ∠ ABC, ⇒, EAF ||BC, 3. Since, AC is a diameter line, so angle in semi-circle makes, an angle 90°., ∴, ∠ ABC = 90°, In Δ ABC, ∠ CAB + ∠ ABC + ∠ ACB = 180°, [Q sum of all interior angles of any triangle is 180°], …(i), ⇒, ∠ CAB + ∠ ACB = 180° − 90° = 90°, Since, diameter of a circle is perpendicular to the tangent., i.e., CA ⊥ AT, ∴, ∠ CAT = 90°, …(ii), ⇒, ∠ CAB + ∠ BAT = 90°, From Eqs. (i) and (ii), we get, ∠ CAB + ∠ ACB = ∠ CAB + ∠ BAT, Hence proved., ⇒, ∠ ACB = ∠ BAT, 4. Given, AB is a diameter of the circle., A tangent is drawn from point A. Draw a chord CD parallel, to the tangent MAN., M, , A, , C, , O, E, , B, , [rationalising], OP, OQ, , ⎡, perpendicular ⎤, ⎢Q sin θ = hypotenuse ⎥, ⎦, ⎣, , N, , D, , So, CD is a chord of the circle and OA is a radius of the circle., ∠ MAO = 90°, [tangent at any point of a circle is perpendicular to the radius, through the point of contact], , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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67, , CBSE Term II Mathematics X (Standard), , ∠ CEO = ∠ MAO [corresponding angles], ∴, ∠CEO = 90°, Thus, OE bisects CD, [perpendicular from centre of circle to, the chord bisects the chord], Similarly, the diameter AB bisects all chords, which are, parallel to the tangent at the point A., 5. Given that, PQ is any line segment and S1 , S2 , S 3 , S 4 , ..., circles are touch a line segment PQ at a point A. Let the, centres of the circles S1 , S 2 , S 3 , S 4, ... be C1 , C 2 , C 3 , C 4,..., respectively., , C4, , P, , C3, S2, C2 S1, C1, , S3, , C2, , C1, O, , D, , 60°, A, , B, , C, , To prove Centres of these circles lie on the perpendicular, bisector of PQ., Now, joining each centre of the circles to the point A on the, line segment PQ by a line segment, i.e., C1A , C 2A , C 3A , C 4A ,... so on., We know that, if we draw a line from the centre of a circle to, its tangent line, then the line is always perpendicular to the, tangent line. But it not bisect the line segment PQ., [for S1], So,, C1A ⊥ PQ, [for S2], C 2A ⊥ PQ, [for S3], C 3A ⊥ PQ, [for S4], C 4A ⊥ PQ, ... so on., Since, each circle is passing through a point A. Therefore, all, the line segments C1A , C 2A , C 3A , C 4A,..., so on are coincident., So, centre of each circle lies on the perpendicular line of PQ, but they do not lie on the perpendicular bisector of PQ., Hence, a number of circles touch a given line segment PQ at, a point A, then their centres lie., 6. Let C1 and C 2 be the two circles having same centre O. AC is, a chord which touches the C1 at point D., , A, , O, , S4, , Q, , A, , ⇒, DO2 = 52 − 42 = 25 − 16 = 9, ⇒, DO = 3 cm, ∴ Radius of the inner circle OD = 3 cm, 7. Since, a chord AB subtends an angle of 60° at the centre of a, circle., , C, , Join OD., Also,, ∴, , OD ⊥ AC, AD = DC = 4 cm, [perpendicular line OD bisects the chord], In right angled ΔAOD,, OA 2 = AD 2 + DO2 [by Pythagoras theorem,, i.e. (hypotenuse)2 =(base)2 +(perpendicular)2], , i.e., ∠ AOB = 60°, As,, OA = OB = Radius of the circle, ∴, ∠ OAB = ∠ OBA = 60°, The tangent at points A and B is drawn, which intersects, at C., We know, OA ⊥ AC and OB ⊥ BC., ∴, ∠ OAC = 90° and ∠OBC = 90°, ⇒, ∠ OAB + ∠ BAC = 90°, and, ∠ OBA + ∠ABC = 90°, ⇒, ∠ BAC = 90° − 60° = 30°, and, ∠ ABC = 90° − 60° = 30°, In Δ ABC, ∠ BAC + ∠ CBA + ∠ ACB = 180°, [since, sum of all interior angles of a triangle is 180°], ⇒, ∠ ACB = 180° − ( 30° + 30° ) = 120°, 8. Two tangents PA and PB are drawn to a circle with centre O, from an external point P., A, , C, E, , O, , P, D, B, , Perimeter of ΔPCD = PC + CD + PD, = PC + CE + ED + PD, = PC + CA + DB + PD, [Q CE = CA, DE = DB], = PA + PB, [PA = PB tangents from, = 2PA = 2(10), external point to a circle are equal], = 20 cm, 9. Given Two tangents PQ and PR are drawn from an external, point P to a circle with centre O., R, O, , P, , Q, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 68, To prove Centre of a circle touching two intersecting lines, lies on the angle bisector of the lines., In ∠RPQ., Construction Join OR and OQ., In ΔPOR and ΔPOQ,, ∠ PRO = ∠ PQO = 90°, [tangent at any point of a circle is perpendicular, to the radius through the point of contact], [radii of same circle], OR = OQ, Since, OP is common., [by RHS], ∴, Δ PRO ≅ Δ PQO, Hence,, [by CPCT], ∠ RPO = ∠ QPO, Thus, O lies on angle bisecter of PR and PQ. Hence proved., 10. Two tangents BD and BC are drawn from an external point B., , Also,, , ∠ OAB + ∠ OCD = 180°, , ∴, , AB||CD, , Similarly, BD is a straight line., and, ∠ Oʹ BA = ∠ Oʹ DC = 90°, Also,, AC = BD, [radii of two circles are equal], In quadrilateral ABCD,, ∠ A = ∠ B = ∠ C = ∠ D = 90°, and, AC = BD, ABCD is a rectangle, Hence,, AB = CD, [opposite sides of rectangle are equal], 12. Given Common tangents AB and CD of two circles, intersecting at E., To prove, AB = CD, , C, 120°, , B, , A, , D, , O, E, , D, , BO = 2BC, , To prove, , Proof, , Given,, ∠ DBC = 120°, Join OC , OD and BO., Since, BC and BD are tangents., ∴, OC ⊥ BC and OD ⊥ BD, We know, OB is a angle bisector of ∠DBC., ∴, ∠ OBC = ∠ DBO = 60°, In right angled Δ OBC,, BC, cos 60° =, OB, 1 BC, =, ⇒, 2 OB, ⇒, OB = 2 BC, Also,, BC = BD, [tangents drawn from external point to circle are equal], ∴, OB = BC + BC, ⇒, OB = BC + BD, 11. Given AB and CD are tangents to two circles of equal radii., To prove, AB = CD, B, , A, , C1, , O, , Oʹ, , C, , C2, , D, , Construction Join Oʹ A , Oʹ C , OB and OD, Proof, , B, , C, , Now, ∠ OAB = 90°, , [tangent at any point of a circle is perpendicular to radius, through the point of contact], Thus, AC is a straight line., , EA = EC, ...(i), [the lengths of tangents drawn from an external, point to a circle are equal], EB = ED, ...(ii), On adding Eqs. (i) and (ii), we get, EA +EB = EC + ED, AB = CD, Hence proved., ⇒, 13. Let OP intersect AB at a point C., Clearly,, …(i), ∠APO = ∠BPO, [Q O lies on bisector of ∠APB], A, , P, , C O, B, , Now, in ΔACP and ΔBCP,, AP = BP, [Q length of tangents drawn from an external point, to a circle are equal], [common sides], PC = PC, and, [from Eq. (i)], ∠APO = ∠BPO, ∴, ΔACP ≅ ΔBCP [by SAS congruence rule], Then,, [by CPCT], AC = BC, and, [by CPCT], ∠ACP = ∠BCP, 1, = × 180° = 90°, 2, [Q AB is a straight line], Hence, OP is perpendicular bisector of AB. Hence proved., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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69, , CBSE Term II Mathematics X (Standard), , 14., , 1, 1, 5r 2, × OF × AC = × r × 5 =, cm, 2, 2, 2, ∴ ar (ΔABC ) = ar (ΔOAB ) + ar (ΔOBC ) + ar (ΔOAC ), 1, 4 r 3 r 5r, 1, 12 r, ⇒, AB × BC =, +, +, ⇒ ×3×4 =, 2, 2, 2, 2, 2, 2, r = 1 cm, ⇒, 17. Produce BD to meet the bigger circle at E. Join AE., Then,, [Q angle in semi-circle], ∠AEB = 90°, and ar (ΔOAC ) =, , P, , A, , O, , Q, , AP = AQ, [say], ∠APQ = ∠AQP = x, [Q angles opposite to equal sides are equal], In ΔAPQ,, ∠PAQ = 180° − ( ∠APQ + ∠AQP ), [angle sum property of a triangle], = 180° − ( x + x ) = 180° − 2x, Q, OP ⊥ AP, [Q radius is perpendicular to the tangent, at the point of contact], ∴, ∠OPA = 90°, ⇒, ∠OPQ + ∠APQ = 90°, ⇒, ∠OPQ + x = 90°, ⇒, ∠OPQ = 90° − x, Hence proved., ∴, ∠PAQ = 2 ∠OPQ, ⇒, , CR = CQ = BC − BQ, , 15., , = 38 − 25 = 13 cm, ∴, RD = CD − CR = 28 − 13 = 15 cm, Here, OR ⊥ RD and OS ⊥ DA., [Q tangent is perpendicular to the radius, through the point of contact], Also, ∠ADC = 90°, then fourth angle in quadrilateral ORDS, will be 90°. Thus, ORDS will be a rectangle., Q D is an external point of a circle., ∴, DR = DS, Also, opposites sides of rectangle are equal., ∴, RD = OR = OS = SD, Hence, quadrilateral DROS is a square., Radius = OR = RD = 15 cm, ∴, 16. Let D, E and F are the points, where the incircle touches the, sides AB, BC and CA, respectively. Join OA, OB and OC., A, , D, , F, , r, O, , r, , r, B, , In Δ ABC,, , C, , E, 2, , 2, , 2, , AC = AB + BC [by Pythagoras theorem], = 42 + 32 = 16 + 9 = 25, AC = 5 cm, ∴, [taking positive square root, as length cannot be negative], 1, 1, 4r 2, Now, ar (ΔOAB ) = × OD × AB = × r × 4 =, cm ,, 2, 2, 2, 1, 1, 3r 2, ar (ΔOBC ) = × OE × BC = × r × 3 =, cm, 2, 2, 2, , A, O, B, , Clearly,, , E, , D, , OD ⊥ BE, , [QBE is tangent to the smaller, circle at D and OD is its radius], , ∴, , BD = DE, [QBE is a chord of the bigger circle and OD ⊥ BE], Now, in ΔAEB, O and D are the mid-points of AB and BE,, respectively., Therefore, by mid-point theorem, we have, 1, OD = AE ⇒ AE = 2 × OD = 2 × 8 = 16 cm, 2, [Q OD = radius of smaller circle = 8 cm], In right angled ΔODB,, [by Pythagoras theorem], OB 2 = OD 2 + BD 2, ⇒, BD 2 = 169 − 64 = 105, [QBD = DE], BD = 105 cm = DE, ⇒, Now, in right angled ΔAED,, AD 2 = AE 2 + ED 2, , ⇒, , AD =, , 2, , (16) + ( 105 ), , [by Pythagoras theorem], 2, , = 256 + 105 = 361 = 19 cm, 18. We know that, tangents drawn from an exterior point to a, circle are equal in length., [say], ∴, AD = AF = x cm, [say], BD = BE = y cm, [say], CE = CF = z cm, Given,, AB = 8 cm, ⇒, AD + BD = 8 cm, ...(i), ⇒, x+y=8, BC = 10 cm, ⇒, BE + CE = 10 cm, ...(ii), ⇒, y + z = 10, and, CA = 12 cm, ⇒, CF + AF = 12 cm, ...(iii), ⇒, z + x = 12, On adding Eqs. (i), (ii) and (iii), we get, 2( x + y + z ) = 30, ...(iv), ⇒, x + y + z = 15, On subtracting Eq. (ii) from Eq. (iv), we get, x = 15 − 10 = 5, On subtracting Eq. (iii) from Eq. (iv), we get, y = 15 − 12 = 3, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 70, On subtracting Eq. (i) from Eq. (iv), we get, z = 15 − 8 = 7, ∴, AD = x cm = 5 cm,, BE = y cm = 3 cm, and, CF = z cm = 7 cm, Hence, the length of AD, BE and CE are 5 cm, 3 cm and, 7 cm, respectively., 19. A circle is inscribed in the ΔABC, which touches the BC , CA, and AB., A, , F, , B, , E, , D, , C, , Given,, BC = a , CA = b and AB = c, By using the property, tangents are drawn from an external, point to the circle are equal in length., ∴, BD = BF = x [say], DC = CE = y [say], and, [say], AE = AF = z, Now, BC + CA + AB = a + b + c, ⇒(BD + DC ) + ( CE + EA ) + ( AF + FB ) = a + b + c, ⇒, ( x + y ) + ( y + z) + ( z + x) = a + b + c, ⇒, 2 ( x + y + z ) = 2s, [Q 2s = a + b + c = perimeter of ΔABC], ⇒, s = x + y + z ⇒ x = s − ( y + z), ⇒, BD = s − b, [Qb = AE + EC = z + y ], Hence proved., 20., ∠OCA = 90° [angle between tangent and radius], Now,, ∠OCA = ∠OCD + ∠ACD, ⇒, ∠ACD = ∠OCA − ∠OCD, ⇒, ∠ACD = 90° − 44° = 46°, As,, AC = AD, [tangents drawn from an external point, are equal in length], So,, ∠ADC = ∠ACD = 46°, [Q angles opposite to the equal sides are equal], Also, ∠CAD + ∠ADC + ∠ACD = 180°, [angle sum property of a ΔACD], ⇒, ∠CAD = 180°− ( 46° + 46° ) = 88°, Again,, ∠COD = 180°− ∠CAD = 92 °, Further, ∠OBD = ∠ODB, [OB = OD radii of circle], In ΔOBD, use exterior angle theorem, exterior angle ∠COD = ∠OBD + ∠ODB, = ∠OBD + ∠OBD, [exterior angle theorem], ⇒, 2∠OBD = ∠COD, 1, ∠CBD = × 92 ° = 46°, ⇒, 2, Hence, ∠CAD = 88°, ∠ADC = 46°, ∠CBD = 46° and, ∠ACD = 46°, , 21. Given, hexagon ABCDEF circumscribe a circle., A, P, , Q, , F, , B, , U, , R, , E, , C, T, , S, D, , Since, tangents drawn from an external point to a circle are, equal in length, ∴ AQ = AP, BQ = BR, CR = CS, DS = DT,, ET = EU, FP = FU, So, AB + CD + EF = ( AQ + QB ) + ( CS + SD ) + (EU + UF ), = AP + BR + CR + DT + ET + FP, = ( AP + FP ) + (BR + CR ) + (DT + ET ), Hence proved., ⇒ AB + CD + EF = AF + BC + DE, 22. ∠ DAB = 90°, In ΔABD, ∠ DAB + ∠ ABD + ∠ ADB = 180°, ⇒, ∠ ADB = 180° − 140° = 40°, In ΔODC,, [radii of same circle], OD = OC, ⇒, ∠OCD = ∠CDO = 40°, [Q angles opposite to equal sides are equal], ∴, ∠DOC + ∠OCD + ∠ CDO = 180°, [Q sum of all angles in a triangle is 180°], ⇒, ∠ DOC = 100°, Since, AD is a straight line., ∴ ∠ DOC + ∠ COA = 180° ⇒ ∠ COA = 80°, 23. Let O be the centre of circle., Q, , S, , O, , 30°, , P, , A, R, , Join OQ and OR. Then,, OQ ⊥ PQ and OR ⊥PR, [Q tangent is perpendicular to the radius, at the point of contact], So,, ∠ROQ + ∠RPQ = 180°, [Q sum of all interior angles of, quadrilateral is 360°], ⇒, ∠ROQ = 150°, 1, But ∠RSQ = ∠ROQ, 2, 1, = × 150° = 75°, 2, Now, on extending QO to intersect RS at A, we get, ∠OQP = ∠QAS = 90° [alternate interior angle], [Q PQ||RS and ∠OQP = 90°], Therefore, from ΔQSA,, ∠SQA = 180° − 90° − 75° = 15°, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 72, From Eqs. (i) and (ii),, 2, , 2, , 2, , 6 − x = 9 − ( 9 − x), , ⇒, , 2, , ⇒, 36 − x 2 = 81 − ( 81 + x 2 − 18 x ), ⇒, 36 = 18x ⇒ x = 2, ∴, AM = x = 2, In right angled ΔABM,, [by Pythagoras theorem], AB 2 = BM 2 + AM 2, 62 = BM 2 + 2 2, ⇒, , BM 2 = 36 − 4 = 32 ⇒ BM = 4 2, , ∴, , BC = 2 BM = 2 × 4 2 = 8 2 cm, 1, ∴ Area of ΔABC = × Base × Height, 2, 1, = × BC × AM, 2, 1, = × 8 2 × 2 = 8 2 cm 2, 2, Hence, the required area of ΔABC is 8 2 cm 2., , 27. In the given figure, join AO, OC and Oʹ D , Oʹ B., Now, in ΔEOʹ D and ΔEOʹ B,, [radius], Oʹ D = Oʹ B, [common side], Oʹ E = Oʹ E, ED = EB, [since, tangents drawn from an external point to the circle, are equal in length], A, D, O, , Oʹ, , E, B, C, , [by SSS similarity rule], ∴, ΔEOʹ D ≅ ΔEOʹ B, [by CPCT], ⇒, ∠Oʹ ED = ∠Oʹ EB, …(i), Oʹ E is the angle bisector of ∠DEB., Similarly, OE is the angle bisector of ∠AEC., Now, in quadrilateral DEBOʹ,, ∠Oʹ DE = ∠Oʹ BE = 90°, [since, CED is a tangent to the circle and Oʹ D is the radius,, i.e. Oʹ D ⊥ CED], ⇒, ∠Oʹ DE + ∠Oʹ BE = 180°, ∴, ∠DEB + ∠DOʹ B = 180°, [since, DEBOʹ is cyclic quadrilateral] …(ii), Since, AB is a straight line., ∴, ∠AED + ∠DEB = 180°, [from Eq. (ii)], ⇒, ∠AED + 180° − ∠DOʹ B = 180°, …(iii), ⇒, ∠AED = ∠DOʹ B, Similarly,, ...(iv), ∠AED = ∠AOC, Again from Eq. (ii),, ∠DEB = 180° − ∠DOʹ B, Divided by 2 on both sides, we get, 1, 1, ∠DEB = 90° − ∠DOʹ B, 2, 2, , ∠DEOʹ = 90° −, , 1, ∠DOʹ B, 2, , …(v), , [since, Oʹ E is the angle bisector of ∠DEB i.e., 1, ∠DEB = ∠DEOʹ], 2, Similarly,, ∠AEC = 180° − ∠AOC, Divided by 2 on both sides, we get, 1, 1, ∠AEC = 90° − ∠AOC, 2, 2, 1, …(vi), ⇒, ∠AEO = 90° − ∠AOC, 2, [since, OE is the angle bisector of ∠AEC, 1, i.e. ∠AEC = ∠AEO], 2, Now,, 1, ⎛, ⎞, ∠AED + ∠DEOʹ + ∠AEO = ∠AED + ⎜ 90° − ∠DOʹ B⎟, ⎝, ⎠, 2, 1, ⎛, ⎞, + ⎜ 90° − ∠AOC⎟, ⎝, ⎠, 2, 1, = ∠AED + 180° − ( ∠DOʹ B + ∠AOC ), 2, 1, = ∠AED + 180° − ( ∠AED + ∠AED ), 2, [from Eqs. (iii) and (iv)], 1, = ∠AED + 180° − (2 × ∠AED ), 2, = ∠AED + 180° − ∠AED = 180°, ∴ ∠AEO + ∠AED + ∠DEOʹ = 180°, So, OEOʹ is straight line., Hence, O, E and Oʹ are collinear., Hence proved., 28. Given, OT = 13 cm and OP = 5 cm, Since, if we draw a line from the centre to the tangent of the, circle, then it is always perpendicular to the tangent i.e., OP ⊥ PT., P, A, , O, , E, , T, , B, Q, , In right angled ΔOPT ,, OT 2 = OP 2 + PT 2, , ⇒, , [by Pythagoras theorem,, (hypotenuse)2 = (base)2 + (perpendicular)2], 2, PT = (13)2 − ( 5)2, , = 169 − 25 = 144, ⇒, PT = 12 cm, Since, the length of pair of tangents from an external point T, is equal., ∴, QT = 12 cm, Now,, TA = PT − PA, ...(i), ⇒, TA = 12 − PA, and, TB = QT − QB, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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73, , CBSE Term II Mathematics X (Standard), , ...(ii), ⇒, TB = 12 − QB, Again, using the property,length of pair of tangents from an, external point is equal., ...(iii), ∴, PA = AE and QB = EB, ∴, OT = 13 cm, [Q OE = 5 cm = radius], ∴, ET = OT − OE, ⇒, ET = 13 − 5 ⇒ ET = 8 cm, Since, AB is a tangent and OE is the radius., ∴, OE ⊥ AB, ⇒, ∠OEA = 90°, [linear pair], ∴, ∠AET = 180° − ∠OEA, ⇒, ∠AET = 90°, Now, in right angled ΔAET,, ( AT )2 = ( AE )2 + (ET )2, ⇒, ⇒, , [by Pythagoras theorem], (PT − PA )2 = ( AE )2 + ( 8)2, [from Eq. (iii)], (12 − PA )2 = (PA )2 + ( 8)2, , Given, BD = FB = 9 cm, CD = CE = 3 cm, In ΔABC, AB 2 = AC 2 + BC 2, ⇒ ( AF + FB )2 = ( AE + EC )2 + (BD + CD )2, ⇒, , ( x + 9)2 = ( x + 3)2 + 12 2, , ⇒, , x 2 + 81 + 18x = x 2 + 9 + 6x + 144, , ⇒, 18x + 81 = 6x + 9 + 144, ⇒, 12 x = 72 ⇒ x = 6 cm, ∴, AB = 6 + 9 = 15 cm, (iii) (b) As we know that, tangents drawn from an external, point are equal in length. Therefore, AP = AS = 4 cm, ∴ DS = DR = 10 − 4 = 6 cm, And BP = BQ = 2 cm. So, CR = CQ = 5 − 2 = 3 cm, So, CD = DR + CR = 6 + 3 = 9 cm, (iv) (d) Here ∠OAP = 90°, A, , ⇒ 144 + (PA )2 − 24 ⋅ PA = (PA )2 + 64, ⇒, ∴, Similarly, Hence,, , 24 ⋅ PA = 80 ⇒ PA =, 10, cm, 3, 10, BE =, cm, 3, AE =, , AB = AE + EB =, , 10, cm, 3, , P, , [from Eq. (iii)], , 10 10 20, +, =, cm, 3, 3, 3, , 20, cm ., 3, 29. (i) (d) We have, AP = AQ, BP = BD, CQ = CD, … (i), [Q tangents drawn from an external point, are equal in length], Now, AB + BC + AC = 7 + 4 + 9 = 20 cm, ⇒ AB + BD + CD + AC = 20 cm, ⇒, AP + AQ = 20 cm, ⇒, 2 AP = 20 cm ⇒ AP = 10 cm, (ii) (c) Let AF = AE = x cm, [Q tangents drawn from an external point to, a circle are equal in length], Hence,the required length AB is, , O, , 50°, B, , In ΔAOP and ΔBOP,, [90° each], ∠OAP = ∠OBP, [radii of circle], OA = OB, PA = PB, [tangents drawn from an external point are equal], [by SAS similarity], ∴, ΔAOP ~ ΔBOP, [by CPCT], ∴, ∠APO = ∠OPB, = 50°, ∴ ∠BPA = 50° + 50° = 100°, (v) (c) For bigger circle, PA = PB, … (i), [Qtangents drawn from an external point are, equal in length] … (ii), Similarly, for smaller circle, PB = PC, From Eqs. (i) and (ii), we get, PA = PB = PC = 11 cm, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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Chapter Test, Multiple Choice Questions, , (iv) In PA and PB are two tangents, drawn to a circle, with centre O from P such that ∠PBA = 60°, then, ∠OAB is, , 1. Two concentric circles are of radii 10 cm and, 8 cm, then the length of the chord of the larger, circle, which touches the smaller circle is, (a) 6 cm, (c) 18 cm, , (a) 50°, , (b) 12 cm, (d) 9 cm, , (b) 25°, , from the centre O of a circle of radius 5 cm, the, pair of tangents PQ and PR to the circle is, drawn. Then, the area of the quadrilateral, [NCERT Exemplar], PQOR is, , B, O, 62°, , (b) 65 cm2, (d) 32.5 cm2, , P, , drawn to a circle of radius 3 cm, then the, length of each tangent is, [NCERT Exemplar], 3, 3 cm, 2, (c) 3 cm, , A, , (a) 18°, (c) 60°, , 3. If two tangents inclined at an angle 60° are, , 5. Two tangents PQ and PR are drawn from an, external point to a circle with centre O. Prove, that QORP is a cyclic quadrilateral., , (d) 3 3 cm, , Case Based MCQs, , 6. In figure, AB and CD are common tangents to, two circles of unequal radii. Prove, [NCERT, that Exemplar], AB = CD., , 4. For revision of chapter circles, a teacher, planned a game with some questions written, on the paper, which are to be answered by the, students. For each correct answer, a student, will get a prize. Some of the questions are given, below., , A, , Answer the questions to check your, knowledge., , C, , B, , D, , Long Answer Type Questions, , (i) In the given figure, x + y is, , 7. A is a point at a distance 13 cm from the centre, , O, , (a) 60°, (b) 90°, (c) 120°, (d) 145°, , O of a circle of radius 5 cm. AP and AQ are the, tangents to the circle at P and Q. If a tangent, BC is drawn at a point R lying on the minor arc, PQ to intersect AP at B and AQ at C, find the, perimeter of the ΔABC., [NCERT Exemplar], , x, y, C, , A, , (ii) In the given figure, PQ and PR are two tangents, to the circle, then ∠ROQ is, S, , 8. If a chord and a tangent intersect externally,, then the product of the lengths of the, segments of the chord is equal to the square of, the length of the tangent from the point of, contact to the point of intersection., , R, O, , 20°, Q, , (a) 30°, 160°, , (b) 60°, , (c) 105°, , P, , Answers, , (d), , (iii) In the adjoining figure, AB is a, chord of the circle and AOC is, its diameter such that, ∠ACB = 45°, then ∠BAT is, (a) 35°, (b) 45°, (c) 125°, (d) 110°, , C, , (b) 20°, (d) Can’t be determined, , Short Answer Type Questions, , (b) 6 cm, , (a), , (d) 130°, , (v) In the adjoining figure, if PC is the tangent at A, of the circle with ∠PAB = 62° and ∠AOB = 132°,, then ∠ABC is, , 2. From a point P, which is at a distance of 13 cm, , (a) 60 cm2, (c) 30 cm2, , (c) 30°, , C, , 1. (b) 2. (a) 3. (d), 4. (i) (b) (ii) (d) (iii) (b) (iv) (c) (v) (b), , 45°, O, , A, , 7. 24 cm, , B, T, , For Detailed Solutions, Scan the code, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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75, , CBSE Term II Mathematics X (Standard), , CHAPTER 04, , Constructions, In this Chapter..., !, , Division of a Line Segment Internally in the Given Ratio, , !, , Construction of a Tangent to a Circle at a Point that lies on it, , !, , Construction of Tangent to a Circle from a Point Outside the Circle, , !, , Construction of Tangents to a Circle When Angle, , Constructions 1, Division of a Line Segment Internally in the Given Ratio, , To divide a line segment AB (say) internally in the given ratio, m : n, where m and n are both positive integers, we use the, following steps, Step I Draw the given line segment AB and any ray AX,, making an acute angle with the line segment AB., This ray AX can be drawn above or below AB., Step II Mark m + n = p points, (i.e. A 1 , A 2 , . . . , A m ,. . . , A p ) on the ray AX, such, that AA 1 = A 1 A 2 = ... = A p − 1 A p, Step III Join BA p ., Step IV Through the point A m , draw a line parallel to A p B, (by making an angle equal to ∠AA p B at A m ) which, intersects the line segment AB at point C. Thus,, point C divides the line segment AB internally in the, ratio m : n, i.e. AC : CB = m : n., n, , X, , ), , +, (A m, Ap, , Am, A1, A, , Justification, Since, A m C||A p B, so use the basic proportionality theorem, in ΔABA p ., AA m, AC, Then,, …(i), =, A m A p CB, By using construction, the ratio is, AA m, m, =, A m A p ( p − m), , …(ii), , ∴From Eqs. (i) and (ii),, AC, m, =, CB ( p − m), Alternate Method, To divide a line segment in the given ratio m : n, where, m and n are both positive integers, we can also use the, following steps., Step I Draw the given line segment AB (say) and any ray, AX making an acute angle with the line, segment AB., Step II Draw another ray BY || AX by making, ∠ABY = ∠BAX., Step III Mark m points i.e. A1 , A 2 ,. . . , A m on AX, and n points i.e. B1 , B 2 , . . . , B n on BY such that, , A2, C, , B, , AA1 = A1 A 2 = . . . = A m − 1 A m, = BB1 = B1B 2 = . . . = B n − 1B n, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 76, Step IV Join A m B n which intersects line segment AB at the, point C., Now, C is the required point which divides line, segment AB internally in the ratio m : n., , By construction, we have, AA 3, 3, 3, =, =, A 3A 5 ( 5 − 3 ) 2, , X, Am – 1, , Am, , A1, B, , C, , Construction 2, , B1, , Construction of a Tangent to a Circle at a Point, that lies on it, , B2, Bn, , Bn – 1, , We can construct a tangent to a circle at a point that lies on it, by two cases which are given below, , Y, , Justification, Step V Use the condition of similarity of two triangles in, AA m AC, ...(i), ΔAA m C and ΔBB n C. Then,, =, BB n, BC, Step VI Write the ratio by using construction,, AA m m, =, BB n, n, AC m, Step VII Equating Eqs. (i) and (ii), we get, =, BC n, , ..(ii), , Example 1. Draw a line segment AB = 8 cm and divide it, internally in the ratio 3 : 2 and also justify it., Sol. Steps of Construction, (i) First, draw line segment, AB = 8 cm and draw a ray AX,, which makes an acute angle with line segment AB., (ii) Mark m + n = 3 + 2 = 5 points i.e. A1 , A 2 , A 3 , A 4 and A 5, on the ray AX such that, AA1 = A1A 2 = A 2A 3 = A 3A 4 = A 4A 5, (iii) Join BA 5., (iv) Through the point A 3 (Q m = 3 ), draw a line A 3C || A 5B, (by making an angle equal to ∠AA 5 B at A 3), which, intersects the line segment AB at C., X, A3, , A4, , A5, , A2, A1, A, , ...(ii), , On equating Eqs. (i) and (ii), we get, AC 3, =, BC 2, This shows that C divides AB internally in the ratio 3 : 2., , A2, , A, , So, in ΔABA 5, by basic proportionality theorem, we get, AA 3, AC, ...(i), =, A 3A 5 CB, , C, 8 cm, , B, , Thus, point C divides the line segment AB internally in the, ratio 3 : 2., Justification, Since, A 3C || A 5B., , Case I By using the centre of circle, To construct a tangent to a circle by using the centre, we use, the following steps., Step I Take a point O as centre and draw a circle of, given radius., Step II Take a point P on the circle, at which we want to, draw tangent., Step III Join OP, which is the radius of circle., Step IV Take OP as base and construct ∠OPT = 90 ° at P., Step V Draw a ray PT and produce TP to Tʹ to get the, required tangent TPTʹ., , O, , Tʹ, , T, P, , Example 2. Draw a circle of diameter AB = 5 cm with, centre O and then draw a tangent to the circle at, point A or B., Sol. Given, diameter of circle = AB = 5 cm and centre is O., 5, ∴ Radius = OA = OB = = 2 . 5 cm, 2, Steps of Construction, (i) Take a point O as centre and draw a circle of radius, 2.5 cm., (ii) Draw diameter AOB., (iii) Take OA as base and construct ∠OAT = 90° at A., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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77, , CBSE Term II Mathematics X (Standard), , (iv) Produce TA to Tʹ to get the required tangent TATʹ., Similarly, we can draw a tangent at point B or any other, point on the circle., T, , 2.5 cm, O, , B, , A, , Tʹ, , Case II Without using the centre of circle, To construct a tangent to a circle without using the, centre of circle, we use the following steps., Step I Draw a circle of given radius and take a point P, (at which we want to draw tangent) on the circle., Step II Draw any chord PQ through the given point P on the, circle., Step III Take a point R in either the major arc or minor arc, and join PR and QR., Step IV On taking PQ as base, construct ∠QPY equal to, , Construction 3, Construction of Tangents to a Circle from a Point, Outside the Circle, , If a point lies outside the circle, then there will be two, tangents to the circle from this point., Case I When centre of circle is known, If centre of circle is known, then to draw tangents from a, given external point, we use the following steps, Step I Draw a circle with centre O of given radius and take, a point P outside it., Step II Join OP and bisect it. Let its mid-point be M. Then,, MP = MO., Step III On taking M as centre and MO or MP as radius,, draw a dotted circle, which intersects the given, circle at points Q and Q ʹ (say)., Q, , P, , O, , M, , ∠PRQ and on the opposite side of R., Step V Draw a ray PY and produce YP upto X to get the, required tangent YPX., , Qʹ, , Step IV Join PQ and PQʹ. Thus, PQ and PQʹ are the required, , R, Q, , X, P, , Y, , Example 3. Draw a circle of radius 6 cm. Take a point P, on it. Without using the centre of the circle, draw a, tangent to the circle at point P., Sol. Given, radius of circle = 6 cm, Steps of Construction, (i) Draw a circle of radius 6 cm and take a point P on the, circle., (ii) Draw a chord PQ through the point P on the circle., (iii) Take a point R in the major arc and join PR and RQ., (iv) On taking PQ as base, construct ∠QPY equal to, ∠PRQ on the opposite side of the point R., (v) Produce YP to X. Then, YPX is the required tangent at, point P., , tangents drawn to the circle from the external point, P. Here, we observe that PQ = PQ ʹ., Justification, Join OQ. Then, ∠PQO = 90 °, since it is constructed in the, semi-circle of dotted circle. It shows that OQ ⊥ PQ. Also, OQ, is radius of given circle, so PQ has to be a tangent of given, circle. Similarly, PQʹ is also a tangent to the given circle., , Example 4. Draw a circle of radius 3.5 cm. From a point, P, 6 cm from its centre, draw two tangents of the, circle., Sol. Given, a circle of radius 3.5 cm whose centre is O (say) and, a point P, 6 cm away from its centre., A, , O, , R, , 6 cm, M, C, 3.5 cm, , P, , Q, X, , B, P, Y, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 78, Steps of Construction, (i) Draw a circle with O as centre and radius, OC = 3.5 cm. Take a point P such that OP = 6 cm., (ii) Draw the bisector of OP which intersect OP at M., (iii) Take M as centre and MO as radius, draw a dotted, circle. Let this circle cuts the given circle at A and B., (iv) Join PA and PB., Hence, PA and PB are the required tangents., , Case II When centre of circle is unknown, If centre of the circle is unknown, then to draw tangents of, the circle, by using the following steps, Step I Firstly, draw the circle and then draw two, non-parallel chords of the circle., Step II Draw the perpendicular bisectors of both chords, which intersect each other at a point, say O. Then,, this point O gives the centre of given circle. Now,, we further use the steps given in case I to draw, tangents., , Sol. Given, radius of circle = 2 cm and distance between point P, and centre = 6.5 cm, Steps of Construction, (i) Draw a circle of radius 2 cm with centre O., (ii) Take a point P outside it, such that its distance from, centre O is 6.5 cm., (iii) Consider O and P as centre and draw arcs of radius, more than half of OP on both sides of OP which, intersect each other at R and S. Join RS which bisects, OP at M. Then, MP = MO., (iv) Consider M as centre and MO as radius, draw a dotted, circle which intersects given circle at Q and Qʹ., (v) Join PQ and PQʹ., Hence, we get the required tangents drawn from point, P to the given circle., R, Q, , 1.1 cm, , Alternate Method, , P, , If centre of circle is unknown, then we can draw tangents, without finding centre of the circle. For this, we use the, following steps of construction., Step I Draw a circle of given radius and take a point P, outside it., D, , K, P, , O, , Qʹ, S, , Construction 4, Construction of Tangents to a Circle When Angle, between Them is Given, , T1, , C, , 2 cm, , M, 6.5 cm, , B, , A, T2, , Step II Through P, draw a line (i.e. secant) intersecting the, given circle at points A and B, respectively and, produce it to C in opposite direction of AB such that, AP = CP., Step III Now, bisect the segment CB at K. Then, take K as, centre and KB (or KC) as radius, draw a semi-circle., Step IV At point P, draw PD ⊥ CB which cuts the, semi-circle at D., Step V Take P as centre and PD as radius draw arcs to, intersect the given circle at points T1 and T2 ., Step VI Join PT1 and PT2 which are the required tangents., , Example 5. Draw a circle of radius 2 cm with centre O, and take a point P outside the circle such that, OP = 6.5 cm. From P, draw two tangents to the, circle., , Sometimes, angle between two tangents (or pair of tangents), is given and we have to draw these tangents. Then, we use, the following steps of construction., Step I First, draw the given circle with centre O and radius, r cm., A, O, r, α, , Q, , α, P, , R, , Step II Draw any diameter say AOQ of this circle., Step III Make given angle α at centre O with OQ (say) as, base which intersect the circle at point R (say) or, draw the radius OR meets the circle at R such that, ∠QOR = α., Step IV Now, draw perpendiculars to OA at A and to OR at, R, which intersect the tangents each other at a point, say P., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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79, , CBSE Term II Mathematics X (Standard), , Then, AP and RP are the required pair of tangents to given, circle, inclined at an angle α, i.e. angle between pair of, tangents is α., Justification, By construction, ∠OAP = 90 ° and OA is radius., So, PA is a tangent to the circle., Similarly, PR is a tangent to the circle., Also, ∠AOR = 180 ° − ∠QOR, [Q AOQ is a straight line], = 180 ° − α, Now, in quadrilateral AORP,, ∠APR + ∠PAO + ∠AOR + ∠PRO = 360 °, ⇒, ∠APR + 90 ° + 180 ° − α + 90 ° = 360 °, ⇒, ∠APR = α, , Example 6. Draw a pair of tangents to a circle of radius, 4 cm which are inclined to each other at an angle of, 30°., Sol. Given, a circle of radius 4 cm. We have to construct a pair of, tangents, which are inclined to each other at an angle of 30°., Steps of Construction, (i) Draw a circle with O as centre and radius 4 cm., (ii) Draw any diameter POQ of this circle., , (iii) Draw the radius OR meets the circle at R such that, ∠QOR = 30°., P, O, , 4 cm, 30°, , E, D, , Q, , 30°, N, , R, , (iv) Draw PD ⊥ PQ and RE ⊥ OR , which intersect each, other at point N. Then, NP and NR are the required, tangents to the given circle inclined to each other at an, angle of 30°., Justification, By construction, ∠OPN = 90° and OP is radius., ∴ PN is a tangent to the circle., Similarly, NR is a tangent to the circle., Now,, ∠POR = 180° − 30° = 150°, [QPOQ is a straight line and ∠QOR = 45°], In quadrilateral OPNR,, ∠OPN = 90° , ∠POR = 150° and ∠ORN = 90°, ∴, ∠PNR = 360° − ( 90° + 150° + 90° ) = 30°, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 80, , Chapter, Practice, PART 1, Objective Questions, ●, , Multiple Choice Questions, 1. To divide a line segment AB in ratio m : n, (m and n are positive integers), draw a ray AX to, that ∠BAX is an acute angle and the mark point on, ray AX at equal distances such that the minimum, number of these points is, (b) m + n, (d) mn, , (a) greater of m and n, (c) m + n − 1, , 2. To divide a line segment AB in the ratio 5 : 7, first, a ray AX is drawn, so that ∠ BAX is an acute angle, and then at equal distances points are marked on, the ray AX such that the minimum number of, these points is, (a) 8, , (b) 10, , (c) 11, , (d) 12, , 3. To divide a line segment AB in the ratio 3 : 5 first a, ray AX is drawn so that ∠BAX is an acute angle, and then at equal distances points are marked on, the ray AX such that the minimum number of, these points is, (a) 8, , (b) 9, , (c) 10, , (b) A 5, (d) A 7, , 5. The ratio of division of the line segment AB by the, point P from A in the following figure is [CBSE 2012], A4, , A5, , A3, A2, A1, A, , (a) 2 : 3, (c) 3 : 5, , P, , (b) 3 : 2, (d) 2 : 5, , (a) A12, (c) A10, , (b) A11, (d) A 9, , 7. To divide a line segment AB in the ratio 5 : 6, draw a, ray AX such that ∠ BAX is an acute angle, then draw, a ray BY parallel to AX and the points, A 1 , A 2 , A 3 ,. . . and B 1 , B 2 , B 3 ,. . . are located to, equal distances on ray AX and BY, respectively., Then, the points joined are, (a) A 5 and B 6, (c) A 4 and B 5, , B, , (b) A 6 and B 5, (d) A 5 and B 4, , 8. To divide a line segment AB in the ratio 6 : 7, a ray, AX is drawn first such that ∠BAX is an acute angle, and then points A 1 , A 2 , A 3 , … are located equal, distances on the ray AX and the point B is joined, with, (a) A12, , (b) A13, , (c)A10, , (d)A11, , 9. In the given figure, find the ratio, when P divides, AB internally., A3, , X, , A2, A1, , (d) 11, , 4. To divide a line segment AB in the ratio 4 : 5, first, a ray AX is drawn making ∠BAX an acute angle, and then points A 1 , A 2 , A 3 , . . at equal distances, are marked on the ray AX and the point B is joined, to, (a) A 4, (c) A 9, , 6. To divide a line segment AB in the ratio 4 : 7, a ray, AX is drawn first such that ∠ BAX is an acute angle, and then points A 1 , A 2 , A 3 ,. . . are located at equal, distances on the ray AX and the point B is joined to, , A, , Y, , (a) 3 : 2, , B, , P, , (b) 2 : 3, , B4, , B3, , B2, , B1, , (c) 4 : 3, , (d) 3 : 4, , 10. From the following ratios, a line segment cannot be, divided into A ratio., 1, 5, 2, 5, (c) A →, :, 5, 2, (a) A →, , 5:, , 1 1, :, 5 5, 1, (d) A → : 1, 5, (b) A →, , 11. To draw a pair of tangents to a circle, which are, inclined to each other at an angle of 60°, it is, required to draw tangents at end points of those two, radii of the circle, the angle between them should be, (a) 135°, , (b) 90°, , (c) 60°, , (d) 120°, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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81, , CBSE Term II Mathematics X (Standard), , 7. Draw a circle of radius 3.5 cm. Take a point P outside, the circle at a distance of 7 cm from the centre of the, circle and construct a pair of tangents to the circle, [CBSE 2020 (Standard)], from that point., 8. Draw a line segment AB of length 9 cm. Taking A as, centre, draw a circle of radius 5 cm and taking B as, centre, draw another circle of radius 3 cm. Construct, tangents to each circle from the centre of the other, [CBSE 2020 (Standard)], circle., 9. Draw a circle with the help of circular solid ring., Construct a pair of tangents from a point P outside the, circle. Also, justify the construction., , 12. A pair of tangents can be constructed from a, point P to a circle of radius 3.5 cm, situated at a, distance of 3 cm from the centre., (a) True, (b) False, (c) Can’t determined, (d) None of the above, , 13. A pair of tangents can be constructed to a circle, inclined at an angle of 170°., (a) True, (b) False, (c) Can’t determined, (d) None of the above, ●, , PART 2, Subjective Questions, ●, , Short Answer Type Questions, 1. Draw a line segment of length 7 cm. Find a point, P on it, which divides it in the ratio 3 : 5., 2. Draw a circle of diameter AB = 6 cm with centre, O and then draw a tangent to the circle at point A, or B., 3. Draw a circle of radius 5 cm. Take a point P on it., Without using the centre of the circle, draw a, tangent to the circle at point P., 4. Draw a circle of radius 6 cm and draw a tangent, to this circle, making an angle of 30° with a line, passing through the centre., 5. Draw a circle of radius 4 cm. From a point 6 cm, away from its centre, construct a pair of tangents, to the circle and measure their lengths., [CBSE 2019], , 6. Draw a circle of radius 1cm. From a point P,, 2.2 cm apart from the centre of the circle, draw, tangents to the circle., , Long Answer Type Questions, 10. Draw a circle of radius 4 cm. Construct a pair of, tangents to it, the angle between which is 60°., Also, justify the construction. Measure the distance, between the centre of the circle and the point of, intersection of tangents., 11. Construct a tangent to a circle of radius 1.8 cm from a, point on the concentric circle of radius 2.8 cm and, measure its length. Also, verify the measurement by, actual calculation., 12. Draw a circle of radius 2.8 cm. From an external point, P, draw tangents to the circle without using the centre, of the circle., 13. Draw a pair of tangents to a circle of radius 3 cm,, which are inclined to each other at an angle of 45°., 14. Let ABC be a right angled triangle, in which, AB = 6 cm, BC = 8 cm and ∠ B = 90°. BD is the, perpendicular from B on AC. The circle through, B , C and D is drawn. Construct the tangents from A to, this circle. Also, justify the construction., 15. Draw a circle of radius 3 cm. Take two points P and Q, on one of its extended diameter each at a distance of, 7 cm from its centre. Draw tangents to the circle from, [NCERT], these two points P and Q., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 82, , SOLUTIONS, Objective Questions, 1. (b) To divide a line segment in the ratio m : n , the maximum, number of the points to mark are m + n ., 2. (d) We know that, to divide a line segment AB in the ratio, m : n , first draw a ray AX, which makes an acute angle, ∠BAX, then marked m + n points at equal distance., Here, m = 5, n = 7, So, minimum number of these points = m + n = 5 + 7 = 12., 3. (a) Minimum number of points = 3 + 5 = 8, 4. (c) Here, 4 + 5 = 9 points are located at equal distances on, the ray AX, so B is joined to last point A 9 ., 5. (b) The ratio of division of the line segment AB by the point, P from A is AP : BP = 3 : 2 ., 6. (b) Here, minimum 4 + 7 = 11 points are located at equal, distances on the ray AX and then B is joined to last point, is A 11., 7. (a) Given, a line segment AB and we have to divide it in the, ratio 5 : 6., Y, , B, , 6, , B, , 5, , B, , 4, , B, , 3, , B, , 2, , B, , 1, , A, A1, , B, , C, , A2, , A3, , A4, , A5, , 9. (d) From given figure, it is clear that there are three points at, equal distances on AX and four points at equal distances on, BY. Here, P divides AB on joining A 3B 4. So, P divides AB, internally in the ratio 3 : 4., 10. (c) Since,, 1 1, 1, (a) 5 :, (b), :, =1:1, = 5:1, 5, 5 5, 2, 5, :, =2 2 : 5, 5, 2, , (d), , P, , O, , θ, , 60°, , R, , Q, , From figure, it is quadrilateral, ∠POQ + ∠ PRQ = 180°, [Q sum of opposite angles are 180°], 60° + θ = 180°, ∴, θ = 120, Hence, the required angle between them is 120°., 12. (b) False, since, the radius of the circle is 3.5 cm, i.e. r = 3. 5 cm and a point P is situated at a distance, of 3 cm from the centre i.e. d = 3 cm, We see that, r > d, i.e. a point P lies inside the circle. So, no tangent can be, drawn to a circle from a point lying inside it., 13. True, , X, , Steps of Construction, 1. Draw a ray AX, making an acute ∠BAX., 2. Draw a ray BY parallel to AX by making ∠ABY equal, to ∠BAX., 3. Now, locate the points A 1, A 2, A 3, A 4 and A 5 ( m = 5), on AX and B1 , B 2 , B 3 , B 4 , B 5 and B 6 ( n = 6 ) such that, all the points are at equal distance from each other., 4. Join B 6 A 5, which intersect AB at a point C., Then, AC : BC = 5 : 6, 8. (b) A 6 + 7 i.e. A13 is joined to the point B., , (c), , 11. (d) The angle between them should be 120° because in that, case the figure formed by the intersection point of pair of, tangent, the two end points of those two radii (at which, tangents are drawn) and the centre of the circle is a, quadrilateral., , 1, :1 = 1: 5, 5, , Since, (a), (b) and (d) are the ratio of both integers. So, it is, possible to divide a line segment into these points., Hence, option (c) is correct., , 170°, , If the angle between the pair of tangents is always greater, than 0 or less than 180°, then we can construct a pair of, tangents to a circle., Hence, we can draw a pair of tangents to a circle inclined at, an angle of 170°., Subjective Questions, 1. Steps of Construction, 1. Draw a line segment AB = 7 cm., , 2. Draw a ray AX, making an acute ∠BAX., 3. Along AX, mark 3 + 5 = 8 points, i.e. A 1 , A 2 , A 3 , A 4 , A 5 , A 6 , A 7 and A 8 such that, AA 1 = A 1 A 2 = A 2 A 3 = A 3 A 4, = A 4A 5 = A 5A 6 = A 6A 7 = A 7A 8, 4. Join A 8B., 5. From A 3 , draw A 3 C|| A 8 B, meeting AB at C., [by making an angle equal to ∠BA 8 A at A 3 ], Then, C is the point on AB, which divides it in the, ratio 3 : 5., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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83, , CBSE Term II Mathematics X (Standard), , 4. Steps of Construction, (i) Draw a circle with centre O and radius 6 cm., , Thus, AC : CB = 3 : 5, X, , P, , A8, , 30°, , A7, A6, , 60°, , A5, , O, , Q, , B, , A, 6 cm, , A4, A3, A2, A1, C, , A, , B, 7 cm, , 2. Given, diameter of circle = AB = 6 cm and centre is O., 6, ∴ Radius = OA = OB = = 3 cm, 2, Steps of Construction, (i) Take a point O as centre and draw a circle of radius 3 cm., (ii) Draw diameter AOB., (iii) Take OA as base and construct ∠OAT = 90° at A., (iv) Draw a ray AT and produce TA to Tʹ to get the required, tangent TATʹ., Similarly, we can draw a tangent at point B or any other, point on the circle., , (ii) Draw a radius OA and produce it to B., (iii) Construct an ∠AOP equal to the complement of 30°, i.e. equal to 60°., (iv) Draw a perpendicular to OP at P, which intersects OB at, point Q., Hence, PQ is the required tangent such that ∠OQP = 30°., 5. Given, a point Mʹ is at a distance of 6 cm from the centre of a, circle of radius 4 cm., Steps of Construction, (i) Draw a circle of radius 4 cm. Let centre of this circle is, O., (ii) Join OMʹ and bisect it. Let M be mid-point of OMʹ., , P, , T, Mʹ, , 3 cm, , B, , A, , O, , M, 4 cm, 6 cm, , O, , Q, , Tʹ, , 3. Given, radius of circle = 5 cm, Steps of Construction, (i) Draw a circle of radius 5 cm and take a point P on the circle., (ii) Draw a chord PQ through the point P on the circle., (iii) Take a point R in the major arc and join PR and RQ., (iv) On taking PQ as base, construct ∠QPY equal to ∠PRQ on, the opposite side of the point R., (v) Draw a ray PY and produce YP to X. Then, YPX is the, required tangent at point P., R, , (iii) Taking M as centre and MO as radius, draw a circle to, intersect circle (0, 4) at two points, P and Q., (iv) Join PMʹ and QMʹ. PMʹ and QMʹ are the required, tangents from Mʹ to circle C ( 0, 4)., The measure length of the tangents are 4.48 cm., 6. Given, radius of circle = 1 cm and distance between point P, and centre = 2.2 cm., Steps of Construction, (i) Draw a circle of radius 1 cm with centre O., (ii) Take a point P outside it such that its distance from, centre O is 2.2 cm., , Q, X, P, , (iii) Take O and P as centre and draw arcs of radius more, than half of OP on both sides of OP, which intersect, each other at R and S. Join RS, which bisects OP at M., Then, MP = MO., , Y, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 84, R, , M, , Q, , N, , 5 cm, P, , 1 cm, 1.1 cm M, 2.2 cm, , Q, , P, , 9. Steps of Construction, (i) First, draw a circle with the help of given circular solid, ring and then draw two non-parallel chords AB and CD., , Qʹ, , (iv) On taking M as centre and MO as radius, draw a dotted, circle, which intersects given circle at Q and Qʹ., (v) Join PQ and PQʹ. Thus, we get the required tangents, drawn from point P to the given circle., 7. Steps of Construction, (i) Draw a circle with O as centre and radius, OC = 3.5 cm. Take a point P such that OP = 7 cm., (ii) Draw the bisector of OP, which intersects OP at M., (iii) On taking M as centre and MO as radius, draw a dotted, circle. Let this circle cuts the given circle at A and B., (iv) Join PA and PB., Thus, PA and PB are the required tangents., , A, , 7 cm, C, 3.5 cm M, , m, , B, , O, , S, , O, , 3c, , O, 9 cm, , A, , P, , B, , 8. Given, a line segment AB = 9 cm, two circles with centres A, and B of radii 5 cm and 3 cm, respectively., We have to construct two tangents to each circle from the, centre of the other circle., Steps of Construction, (i) Draw a line segment AB = 9 cm., (ii) Draw a circle with centre A and radius 5 cm and another, circle with centre B and radius 3 cm., (iii) Now, bisect AB. Let O be the mid-point of AB., (iv) Take O as centre and AO as radius and draw a dotted, circle, which intersects the two given circles at N , Q, M, and P., (v) Join AN , AQ, BM and BP. These are the required tangents, to each circle from the centre of the other circle., , A, , T, , D, P, , M, , O, , Tʹ, , B, , C, , (ii) Draw perpendicular bisectors of AB and CD, which, intersect each other at point O. Then, O is the centre of, the circle., (iii) Now, take a point P outside the circle and join OP., (iv) Draw bisector of OP. Let its mid-point be M., (v) On taking M as centre and MP as radius, draw a dotted, circle which intersect the given circle at T and Tʹ., (vi) Join PT and PTʹ., Then, PT and PTʹ are the required pair of tangents drawn to, the circle from P., Justification, Join OT., Then, ∠PTO = 90°, [angle in semi-circle of dotted circle], This shows that OT ⊥ PT., Also, OT is radius of given circle, so PT has to be a tangent of, given circle. Similarly, PTʹ is also a tangent of given circle., 10. Steps of Construction, (i) Take a point O on the plane of the paper and draw a circle, with centre O and radius OA = 4 cm., (ii) At O construct radii OA and OB such that ∠AOB equal to, 120° i.e. supplement of the angle between the tangents., (iii) Draw perpendiculars to OA and OB at A and B,, respectively. Suppose these perpendiculars intersect at P., Then, PA and PB are required tangents., A, 4 cm, P, , 60°, , 120°, , O, , B, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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85, , CBSE Term II Mathematics X (Standard), , The distance between the centre of the circle and the point, of intersection of tangents is 8 cm., Justification, In quadrilateral OAPB, we have, ∠OAP = ∠OBP = 90°, and, ∠AOB = 120°, ∴ ∠OAP + ∠OBP + ∠AOB + ∠APB = 360°, ⇒, 90° + 90° + 120° + ∠APB = 360°, ∴, ∠APB = 360° − ( 90° + 90° + 120° ), = 360° − 300° = 60°, 11. Given, two concentric circles of radii 2.8 cm and 1.8 cm with, common centre say, O., A, C1 C2, P, , M, , O, , B, , Steps of Construction, (i) Draw two circles with common centre O and radii, 2.8 cm and 1.8 cm, respectively., (ii) Take a point P on the outer circle and join OP., (iii) Draw bisector of OP. Let mid-point of OP be M., (iv) Taking M as centre and PM as radius, draw a dotted circle,, which intersects the inner circle at two points say A and B., (v) Join AP and BP. Then, AP and BP are required tangents., On measuring the lengths, we get PA = PB = 2.14 cm, Calculation, Join OA. Then, OA = 1.8 cm, [radius of inner circle C1], OP = 2.8 cm, [radius of outer circle C 2], and, ∠PAO = 90°, [Q angle in semi-circle of constructed circle], So, in ΔPAO, by Pythagoras theorem,, OP 2 = OA 2 + AP 2, ⇒, , (2.8)2 = (1.8)2 + AP 2, , ⇒, , 7.84 = 3.24 + AP 2, , ⇒, , AP 2 = 7.84 − 3.24 = 4.6, , AP = 2.14 cm, [taking positive square root, as length cannot be negative], ⇒, PA = PB = 2.14 cm, ⇒, , Hence, the length of tangents is 2.14 cm., 12. Given, a circle of radius 2.8 cm and we have to draw, tangents without using the centre., Steps of Construction, (i) First, draw a circle of radius 2.8 cm and take a point P, outside the circle., (ii) Through P, draw a secant PAB, which intersects the, circle at A and B and extend it to C in opposite, direction of AB such that PC = PA., , D, M, , O, , C, , A, , B, , P, N, , (iii) Now, bisect BC and take its mid-point as O. Draw a, semi-circle with centre O and radius OB (or OC )., (iv) Draw PD ⊥ BC, which intersects the semi-circle at D., (v) With centre P and radius PD draw two arcs, which, intersects the given circle at points M and N., (vi) Join PM and PN. Thus, PM and PN are the required, tangents to the given circle., 13. Given, a circle of radius 3 cm. We have to construct a pair of, tangents, which are inclined to each other at an angle of 45°., Steps of Construction, (i) Draw a circle with O as centre and radius 3 cm., (ii) Draw any diameter POQ of this circle., (iii) Draw the radius OR meets the circle at R such that, ∠QOR = 45°., P, O, , 3 cm, 45°, , Q, , 45°, , E, , N, , R, , D, , (iv) Draw PD ⊥ PQ and RE ⊥ OR ,, which intersects each other at point N., Then, NP and NR are the required tangents to the given, circle inclined to each other at an angle of 45°., Justification, By construction, ∠OPN = 90° and OP is radius., ∴ PN is a tangent to the circle., Similarly, NR is a tangent to the circle., Now,, ∠POR = 180° − 45° = 135°, [QPOQ is a straight line and ∠QOR = 45°], In quadrilateral OPNR,, ∠OPN = 90° , ∠POR = 135°, and, ∠ORN = 90°, ∴, ∠PNR = 360° − ( 90° + 135° + 90° ) = 45°, 14. Given, ABC is a right angled triangle, in which AB = 6 cm,, BC = 8 cm, ∠B = 90° and BD is perpendicular to AC., Then, ∠ADB = ∠CDB = 90°, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 86, Steps of Construction, (i) Draw the line segments AB = 6 cm and BC = 8 cm, perpendicular to each other. Join AC. Thus, ΔABC is, the given right angled triangle., (ii) Draw perpendicular bisector of BC, which meets BC, at O., (iii) With O as centre and OB as radius draw a circle,, which intersects AC at D, then ∠BDC = 90°. Thus,, BD is perpendicular to AC., (iv) With A as centre and AB as radius draw an arc,, cutting the circle at M., (v) Join AM. Thus, AB and AM are required tangents., X, A, D, , 6 cm, B, , M, , O, , C, , 8 cm, , Justification, Since ΔABC is right angled triangle with ∠ABC = 90°, ∴ BO ⊥ AB., Also, BO is the radius of circle. So, AB has to be tangent of, the circle. Similarly, AM is also a tangent to the circle., 15. Given, two points P and Q on the extended diameter of a, circle with radius 3 cm such that OP = OQ = 7 cm, , We have to construct the tangents to the circle from the, given points P and Q., Pʹ, , M, 3 cm, P, , O, , Q, , F, , E, 7 cm, , 7 cm, N, , Qʹ, , Steps of Construction, (i) Draw a circle of radius 3 cm with centre at O., (ii) Produce its diameter on both sides and take points P and Q, on it such that OP = OQ = 7 cm, (iii) Draw bisector of OP and OQ, which intersect OP and OQ at, E and F, respectively., (iv) Now, take E as centre and OE as radius, draw a dotted circle, which intersects the given circle at two points M, N. Again,, take F as centre and OF as radius, draw another dotted circle, which intersects the given circle at two points Pʹ and Qʹ., (v) Join PM , PN , QPʹ and QQʹ. These are the required tangents, from P and Q to the given circle., Justification, Join OM and ON. The ∠OMP is the angle that lies in the, semi-circle of the dotted circle with centre E. Therefore,, ∠OMP = 90° ⇒ OM ⊥PM, Since, OM is radius of the circle. So, MP has to be a tangent to, the circle. Similarly, PN, QPʹ and QQʹ are also tangents to the, given circle., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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Chapter Test, Multiple Choice Questions, , these two radii of the circle, the angle between, two radii is, , 1. To divide a line segment AB in the ratio 2 : 5,, first a ray AX is drawn, so that ∠BAX is an acute, angle and then at equal distances, then the, number of points located on the ray AX is, [CBSE 2011], (a) 7, (c) 2, , (b) 10, (d) 5, , (a) 105°, (b) 70°, (c) 125°, (d) 135°, , Short Answer Type Questions, , 5. Draw a line segment AB = 6. 5 cm and divide it, internally in the ratio 3 : 5., , 2. To divide a line segment AB in the ratio 7 : 5,, first a ray AX is drawn, so that ∠BAX is acute, angle and then at equal distance points are, marked. Then, the minimum number of these, points is, (a) 5, (c) 7, , 6. Draw two tangents at the end points of the, diameter of a circle of radius 3.5 cm. Are these, tangents parallel?, Long Answer Type Questions, , (b) 35, (d) 12, , 7. Draw two concentric circles of radii 3 cm and, 5 cm. Taking a point on outer circle construct, the pair of tangents to the other. Measure the, length of a tangent and verify it by actual, calculation., , 3. By geometrical construction, it is possible to, divide a line segment in the ratio 3 :, (a) True, , (b) False, , (c) Can’t determined, , (d) None of these, , 1, 3, , ., , 8. Draw a line segment AB of length 7 cm. Taking, , 4. To draw a pair of tangents to a circle, which are, inclined to each other at an angle of 55°, it is, required to draw tangents at the end points of, , A as centre, draw a circle of radius 3 cm and, taking B as centre, draw another circle of radius, 2 cm. Construct tangents to each circle from, the centre of the other circle., , Answers, 1. (a), , 2. (d), , 3. (a), , 4. (c), , 6. These are parallel., , 7. 4 cm, , For Detailed Solutions, Scan the code, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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88, , CBSE Term II Mathematics X (Standard), , CHAPTER 05, , Applications of, Trigonometry, In this Chapter..., !, , Line of Sight & Horizontal Line, , !, , Angle of Elevation, , !, , Angle of Depression, , Some Important Points, , Line of Sight, The line of sight is the line drawn from the eye of an observer to, the point where the object is viewed by the observer., , Horizontal Line, , (i) The angle of elevation of a point P as seen from a, point O is always equal to the angle of depression of, O as seen from P., , The line which goes parallel from eye to ground, is called, horizontal line., , Angle of Elevation, The angle of elevation of an object viewed, is the angle formed by, the line of sight with the horizontal, when it is above the horizontal, level, i.e. the case when we raise our head to look at the object., P(object), of, , O, , sig, , ht, , e, Lin, Angle of, elevation, A, Horizontal line, , Eye, , Angle of Depression, The angle of depression of an object, viewed, is the angle formed by the line, of sight with the horizontal, when it is, below the horizontal level, i.e. the case, when we lower our head to look at the, object., , O, , Horizontal line, A, Angle of, depression, Lin, e, of, sig, ht, P(object), , O, , Angle of, depression, ht, sig, f, o, e, Lin Angle of, elevation, Horizontal line, , P, , A, , (ii) The angles of elevation and depression are always, acute angles., (iii) If the observer moves towards the perpendicular, line (tower/building), then angle of elevation, increases and if the observer moves away from the, perpendicular line (tower/building), then angle of, elevation decreases., (iv) If the height of tower is doubled and the distance, between the observer and foot of the tower is also, doubled, then the angle of elevation remains same., (v) If the angle of elevation of Sun, above a tower, decreases, then the length of shadow of a tower, increases and vice-versa., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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89, , CBSE Term II Mathematics X (Standard), , Solved Examples, Example 1. In figure, a tightly stretched rope of length, 20 m is tied from the top of a vertical pole to the, ground. Find the height of the pole, if the angle, made by the rope with the ground is 30°., [CBSE 2020 (Standard)], B, , In right angled ΔACB,, , ⇒, ⇒, , 20 m, , 30º, A, , C, , AB, BC, 1, h, =, 3 30, 30, 3, h=, ×, 3, 3, [multiply numerator and denominator by 3], 30 × 3, =, 3, = 10 3 m, tan 30° =, , Hence, height of the tower is 10 3 m., , Sol. Let AB = h be the height of the pole., , Example 3. The ratio of the length of a vertical rod and, , Given, length of rope, BC = 20 m, , the length of its shadow is 1 : 3. Find the angle of, elevation of the Sun at that moment?, , B, , [CBSE 2020 (Standard)], Sol. Let AB be the vertical rod and BC be its shadow and θ be, the angle of elevation of the Sun., , 20 m, , hm, , 30º, A, , C, , A, , Perpendicular, Hypotenuse, , sin 30° =, , 1 h, =, 2 20, , ∴, , !, , 20, = 10 m, 2, Hence, height of the pole is 10 m., ⇒, , Vertical rod, , In right angled ΔACB,, , h=, , B, , Shadow, , C, , We have, AB : BC =1 : 3, , Example 2. In figure, the angle of elevation of the top, of a tower from a point C on the ground, which is, 30 m away from the foot of the tower, is 30°. Find, the height of the tower., [CBSE 2020 (Standard)], A, , Let AB = x, then BC = x 3, AB, In ΔABC, tan θ =, BC, x, 1, tan θ =, =, ⇒, x 3, 3, ⇒, tan θ = tan 30°, ⇒, θ = 30°, , Example 4. A vertical tower stands on a horizontal, 30°, B, , C, , 30 m, , Sol. Let height of a tower be AB = h m, A, , h, , B, , 30°, 30 m, , C, , plane and is surmounted by a vertical flag-staff of, height 6 m. At a point on the plane, the angle of, elevation of the bottom and top of the flag-staff are, 30° and 45°, respectively. Find the height of the, tower. (take, 3 = 1.73), [CBSE 2020 (Standard)], Sol. Let BC = h be the height of the tower, CD = 6 m be the, height of the flag-staff and A is any point on the ground., Consider, AB = x m., Given, the angle of elevation from point A to the points C, and D are ∠CAB = 30° and ∠DAB = 45°., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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90, , CBSE Term II Mathematics X (Standard), , Also, let AB = x m., In right angled ΔABD,, , D, 6m, , Perpendicular BD, =, Base, AB, BC + CD, 3=, [Q tan 60° = 3 ], x, h + 1. 6, 3=, x, 3x = h + 1 . 6, , tan 60° =, , C, , ⇒, , h, A, , 30° 45°, , ⇒, xm, , B, , In right angled ΔABC,, BC, tan 30° =, AB, 1, h, =, ⇒, 3 x, x= 3hm, ⇒, Now, in right angled ΔABD,, BD, tan 45° =, AB, 6+ h, ⇒, 1=, x, ⇒, x=6+ h, ⇒, 3h = 6 + h, ⇒, , …(i), , In right angled ΔCBA,, BC, AB, h, 1=, ⇒, x, ⇒, x=h, On putting x = h in Eq. (i), we get, h = 3 h − 1. 6, tan 45° =, , …(i), , [QBD = BC + CD = 6 + h ], , ⇒, , …(ii), [from Eq. (i)], , ⇒, , h( 3 − 1) = 6, , [Q tan 45° = 1], , h ( 3 − 1) = 1. 6, h=, , 1. 6, ×, ( 3 − 1), , 3+1, 3+1, , [rationalising], , 1 . 6( 3 + 1), [Q ( a + b )( a − b ) = a 2 − b 2 ], ( 3 ) 2 − (1 ) 2, 1 .6, =, ( 3 + 1), 2, = 0. 8 (1.73 + 1), = 0.8 (2.73)= 2.184 m, Hence, the height of the pedestal is 2.184 m., =, , 6, 3 +1, h=, ×, ( 3 − 1), 3+ 1, , ⇒, , h = 3x − 1. 6, , ⇒, , (rationalisation), , 6( 3 + 1), [Q( a − b )( a + b ) = a 2 − b 2], ( 3 ) 2 − (1 ) 2, 6(1.73 + 1) 6 × 2.73, =, =, 3 −1, 2, , =, , Example 6. From a point on the ground, the angles of, , = 3 × 2.73 = 8.19 m, Hence, height of tower is 8.19 m., , Example 5. A statue 1.6 m tall, stands on the top of a, pedestal. From a point on the ground, the angle of, elevation of the top of the statue is 60° and from the, same point the angle of elevation of the top of the, pedestal is 45°. Find the height of the pedestal., (use 3 = 1.73), Sol. Let BC = h m be the height of the pedestal and CD = 1.6 m, be the length of the statue, which is standing on the, pedestal., , elevation of the bottom and the top of a transmission, tower fixed at the top of a 20 m high building are, 45° and 60°, respectively. Find the height of the, tower. (use 3 = 1.73 ), Sol. Let AB = 20 m be the height of the building and BC = h m be, the height of transmission tower. The angles of elevation, from a ground point D to the points B and C are, ∠ADB = 45° and ∠ADC = 60°, C, hm, , D, , B, , C, , 20 m, 45° 60°, , h, A, , xm, , B, , Again, let point A be a fixed point on the ground such that, the angles of elevation of the top of the statue and bottom of, the statue (i.e. top of the pedestal) are, ∠DAB = 60° and ∠CAB = 45°., , A, , D, , In right angled ΔADB,, AB, AD, 20, 1=, AD, AD = 20 m, , tan 45° =, ⇒, ⇒, , …(i), , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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91, , CBSE Term II Mathematics X (Standard), , Now, in right angled ΔADC,, AC, tan 60° =, AD, AB + BC, ⇒, 3=, 20, ⇒, 20 3 = 20 + h, ⇒, , ⇒, , Hence, height of the pole is 25 3 m and distances of the, point from the poles are 25 m and 75 m., [Q from Eq. (i)], , = 20 × 0.73 = 14. 60 m, Hence, height of the transmission tower is 14.6 m., , Sol. Let AB = 50 m, CD = h be the height of the tower and, building. Then,, , Example 7. Two poles of equal heights are standing, opposite to each other on either side of the road,, which is 100 m wide. From a point between them, on the road, the angles of elevation of the top of the, poles are 60° and 30°, respectively. Find the height, of the poles and the distance of the point from the, [CBSE 2020 (Standard)], poles., Sol. Let AB = 100 m be the width of the road. On both sides of, the road, poles AE = BD = h m are standing. Let C be any, point on AB such that from point C, angles of elevation are, ∠BCD = 60° and ∠ACE = 30°, D, , A, , 30º, (100 – x) m C, 100 m, , 60º, xm, , B, D, , 50 m, hm, 30º, , A, , 60º, , C, , ∠CAD = 30° and ∠ACB = 60°, In right angled ΔACD ,, 1, h, CD, =, tan 30° =, ⇒, AC, 3 AC, ⇒, AC = 3 h, , …(i), , In right angled ΔCAB,, , hm, , h, , Example 8. The angle of elevation of the top of a, building from the foot of a tower is 30° and the, angle of elevation of the top of a tower from the foot, of the building is 60°. If the tower is 50 m high,, then find the height of the building., , h = 20( 3 − 1) = 20(1.73 − 1), , E, , h = 25 3 m, , AB, AC, 50, 3=, ⇒, AC, 50, 3=, ⇒, 3h, 50, h=, = 16.67 m, ⇒, 3, Hence, the height of the building is 16.67 m., tan 60° =, , B, , Let BC = x m , then AC = AB − BC = (100 − x ) m, In right angled ΔCAE,, Perpendicular, tan 30° =, Base, 1, 1, h, AE, ∴, =, =, ⇒, 3 AC, 3 (100 − x ), (100 − x ), h=, ⇒, 3, h 3 = 100 − x, ⇒, and in right angled ΔBCD,, BD, tan 60° =, BC, h, ⇒, 3=, x, ⇒, h = 3x, Put h = 3x in Eq. (i), we get, , Example 9. From the top of a 7 m high building the, …(i), , …(ii), , angle of elevation of the top of a tower is 60° and, the angle of depression of its foot is 45°. Determine, [CBSE 2020 (Standard)], the height of the tower., Sol. Let AB = 7 m be the height of the building and EC be the, height of tower., A is the point from where elevation of tower is 60° and the, angle of depression of its foot is 45°., Here, EC = DE + CD, Also, CD = AB = 7 m, and BC = AD, , 3x × 3 = 100 − x, 3x + x = 100 ⇒ 4x = 100, 100, x=, = 25 m, 4, ∴ BC = 25 m and AC = 100 − x = 100 − 25 = 75 m, Put x = 25 in Eq. (i), we get, , [from Eq. (i)], , E, , ⇒, , ⇒, , ⇒, , h 3 = 100 − 25, h=, , 75, 3 75 3, ×, =, 3, 3, 3, , 60°, 45°, , D, , 7m, , ⇒, , A, , B, , 45° C, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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92, , CBSE Term II Mathematics X (Standard), Cʹ B + BM, PM, 10 + h + 10, [from Eq. (i)], 3=, ⇒, 3h, ⇒, 3h = 20 + h, ⇒, 2 h = 20, ⇒, h = 10 m, Now, the height of the cloud from the surface of lake, =BC, = BM + h, = 10 + 10 = 20 m, , Now, in right angled ΔABC,, , ⇒, , tan 45° =, , AB, BC, , 1=, , 7, BC, , ⇒, ⇒, , BC = 7 m, , Also, in right angled ΔADE,, tan 60° =, , DE, AD, , DE, 7, DE = 7 3 m, , ⇒, , 3=, , ⇒, , [Q AD = BC ], , 3=, , Example 11. From a point on a bridge across a river,, the angles of depression of the banks on opposite, sides of the river are 30° and 45°, respectively. If, the bridge is at a height of 30 m from sea level, then, find the width of the river. (use 3 = 1 . 73), , ∴ Height of the tower,, EC = DE + CD, = (7 3 + 7 ) m, = 7( 3 + 1) m, Hence, height of the tower is 7( 3 + 1) m., , Example 10. If the angle of elevation of a cloud from a, point 10 m above a lake is 30° and the angle of, depression of its reflection in the lake is 60°, find, the height of the cloud from the surface of lake., , Sol. Let A be a point on the bridge and points B and D are on the, opposite side of the banks. Then, angles of depression from, point A to the opposite banks are, ∠EAB = 30° and ∠FAD = 45°, ⇒, ∠CBA = 30° and ∠CDA = 45°, [Q alternate angles are equal], A, , E, , 30º, , [CBSE 2020 (Standard)], Sol. Let AB be the surface of the lake and P be the point of, observation such that AP = 10 m. Let C be the position of the, cloud and Cʹ be the reflection in the lake, then CB = Cʹ B., , 30 m, , C, , 30º, B, (River bank), , h, P, , 30°, , M, , 60°, , ⇒, ⇒, , (10 +h)m, , Let CM = h , then CB = 10 + h, ⇒, Cʹ B = 10 + h, In right angled ΔCMP,, tan 30° =, ⇒, ⇒, In right angled ΔPMCʹ,, tan 60° =, , Cʹ M, PM, , AC, BC, 1, 30, =, 3 BC, , BC = 30 3 m, , and in right angled ΔACD,, AC, tan 45° =, CD, AC, 1=, ⇒, 30, ⇒, AC = 30 m, Hence, width of the river is, BD = BC + CD, = 30 3 + 30, , Cʹ, , CM, PM, 1, h, =, 3 PM, PM = 3h, , D, (Opposite, river bank), , tan 30° =, , B, , A, , 45º, C, , In right angled ΔACB,, , 10 m, , 10 m, , F, Bridge, , 45º, , ...(i), , = 30 × 1.73 + 30, = 51.9 + 30, = 81.9, Hence, width of the river is 81.9 m., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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93, , CBSE Term II Mathematics X (Standard), , Example 12. From the top of a 7 m building, the angle, of elevation of the top of a cable tower is 60° and, the angle of depression of its foot is 45°. Determine, the height of the tower. (use 3 = 1.73)., [CBSE 2020 (Standard)], Sol. Let AB = 7 m be the height of the building and DE = h m, be the height of cable tower., E, , B, , 60º, , C, , 45º, , hm, , 7m, , 45º, , A, , xm, , Given, speed of plane is 720 km/h and time of flight is 10 s., Also, given ∠AOC = 60° and ∠BOD = 30°, In right angled ΔOCA,, OC, cot 60° =, AC, 1, OC, ⇒, =, h, 3, h, OC =, m, ⇒, 3, In right angled ΔODB,, OD, cot 30° =, BD, OD, 3=, ⇒, h, OD = h 3, ⇒, Now, CD = OD − OC, h, 2h, =h 3−, =, m, 3, 3, , D, , Then, ∠CBE = 60°, and ∠CBD = 45° ⇒ ∠ADB = 45°, (alternate angle), Let distance between two towers be, AD = BC = x m, CE, In right angled ΔBCE, tan 60° =, BC, CE, CE, … (i), ⇒, 3=, ⇒x =, x, 3, and in right angled ΔADB,, AB, 7, tan 45° =, ⇒1 = ⇒ x = 7 m, x, AD, Put x = 7 in Eq. (i), we get, CE, 7=, ⇒ CE = 7 3 m, 3, Now, height of cable tower,, h = DC + CE, = 7 + 7 3 = 7 (1 + 3 ), = 7(1 + 1.73) = 7 × 2 . 73 = 19.11 m, , Example 13. The angle of elevation of an aeroplane, from point O on the ground is 60°. After a flight of, 10 s, on the same height, the angle of elevation, from point O becomes 30°. If the aeroplane is flying, at the speed of 720 km/h, find the constant height at, which the aeroplane is flying. [CBSE 2020 (Standard)], , Thus, distance covered by aeroplane in 10 s is, Distance, Time, 2h, 5, 720 ×, = 3, 18 10, , Q Speed of aeroplane =, , ∴, , ⇒, h = 1000 3 m, Hence, height at which the aeroplane is flying is 1000 3 m., , Example 14. A straight highway leads to the foot of a, tower. A man standing at the top of the tower, observes a car at an angle of depression of 30°,, which is approaching the foot of the tower with a, uniform speed. After covering a distance of 50 m,, the angle of depression of the car becomes 60°., Find the height of the tower. (use 3 = 1.73)., [CBSE 2020 (Standard)], Sol. Let AB = h m be the height of the tower. Let C be the initial, position of the car and D be the final position of the car, when, it covers a distance, CD = 50 m ., E, , B, 60°, , D, , h, h, , 60º, O, , 30º, C, , D, , X, , 30°, , h, 60°, , 30°, , B, , 5, ⎡, ⎤, Q 1 km =, m /s, ⎥⎦, 18, ⎣⎢, , 40 × 5 × 10 3 = 2 h, , ⇒, , Sol. Let OX be the horizontal ground, A and B be the two, positions of the plane and O be the points of observation., Let height of an aeroplane from A to the ground is, AC = BD = h m, A, , 2h, m., 3, , 50 m, , C, , A, x, , Given, the angle of depressions from point B to the points C, and D are, ∠EBC = 60° and ∠EBD = 30°, ⇒, ∠BCA = 60° and ∠BDA = 30° [alternate angles], , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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94, , CBSE Term II Mathematics X (Standard), , A, , AB, tan 60° =, AC, h, ⇒, 3=, x, h, ⇒, x=, m, 3, And in right angled ΔDAB,, AB, tan 30° =, AD, 1, h, ⇒, =, 3 50 + x, ⇒, ⇒, , …(i), , 2 h = 50 3, , ⇒, , h = 25 3, , F, 60º, , 45º, x, , E, , 60º, C, , …(ii), , From Eqs. (i) and (ii), we have, h, = h 3 − 50, 3, h = 3h − 50 3, ⇒, ⇒, , B, , 45º, , hm, , 50 + x = h 3, x = h 3 − 50, , 60 "3 m, , In right angled ΔCAB,, , ⇒, h = 25 × 1 .73, ⇒, h = 43. 25 m, Hence, height of the tower is 43.25 m., , Example 15. The angles of depression of the top and, bottom of a tower as seen from the top of a 60 3 m, high cliff are 45° and 60°, respectively. Find the, height of the tower. (use 3 = 1.73), Sol. Let AC = 60 3 m be the height of the cliff, DE = h m be the, height of the tower and distance between tower and cliff be, CD = BE = x m., , x, , D, , Given, the angle of depressions from point A are, ∠ FAE = 45° and ∠FAD = 60°, ⇒, ∠BEA = 45° and ∠CDA = 60° [alternate angles], In right angled ΔAEB,, AB, AB, tan 45° =, ⇒ 1=, BE, x, …(i), ⇒, x = AB, and in right angled ΔADC,, AC, 60 3, 60 3, tan 60° =, ⇒ x=, ⇒ 3=, x, CD, 3, …(ii), ⇒, x = 60 m, ∴ From Eqs. (i) and (ii), we get, AB = x = 60 m, Now height of cliff,, h = AC − AB, = 60 3 − 60, = 60( 3 − 1), = 60( 1.73 − 1), = 60 × 0.73 = 43.8 m, Hence, height of the cliff is 43.8 m., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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95, , CBSE Term II Mathematics X (Standard), , Chapter, Practice, PART 1, Objective Questions, ●, , Multiple Choice Questions, 1. The angle of elevation of the Sun when the shadow, of a pole h m high is 3 h m long is, (a) 0°, , (b) 30°, , (c) 45°, , (d) 60°, , 2. If a pole 6 m high casts a shadow 2 3 m long on the, ground, then the Sun’s elevation is, (a) 60°, (c) 30°, , (b) 45°, (d) 90°, , 3. If 300 3 m high tower makes an angle of elevation, at a point on ground which is 300 m away from its, foot, then the angle of elevation is, (a) 0°, (c) 45°, , (b) 30°, (d) 60°, , 4. From the top of a 60 m high tower, the angle of, depression of a point on the ground is 30°. The, distance of the point from the foot of tower is, (b) 60 3 m, (d) 30 3 m, , (a) 180 m, (c) 150 m, , 5. The figure shows the observation of point C from, point A. The angle of depression from A is, [CBSE 2013], A, , D, , 4m, , C, , (a) 30°, , (b) 45°, , 4√3 m, , (d) 90°, , 6. A circus artist is climbing a 20 m long rope, which, is tightly stretched and tied from the top of a, vertical pole to the ground, then the height of pole,, if the angle made by the rope with the ground level, is 30°, is, (a) 5 m, (c) 15 m, , (b) 10 m, (d) 20 m, , [NCERT Exemplar], (a) 10 m, (c) 18 m, , (b) 16 m, (d) 19 m, , 8. A ramp for disabled people in a hospital have slope, 30°. If the height of the ramp be 1 m, then the, length of ramp is, (a) 2 m, (c) 2 3 m, , (b) 0.5 m, (d) 1 m, , 9. A kite is flying at a height of 80 m above the, ground. The string attached to the kite is, temporarily tied to a point on the ground. The, inclination of the string with ground is 60°, then the, length of the string is, (a) 62.37 m, (c) 52.57 m, , (b) 92.37 m, (d) 72.57 m, , 10. The length of a string between a kite and a point on, the ground is 85 m. If the string makes an angle θ, 15, with the ground level such that tan θ = , then the, 8, height of kite is, (a) 75 m, (c) 226 m, , (b) 78.05 m, (d) None of these, , 11. A tower stands near an airport. The angle of, elevation θ of the tower from a point on the ground, is such that its tangent is 5/12. The height of the, tower, if the distance of the observer from the tower, [CBSE 2015], is 120 m is, (a) 40 m, (c) 60 m, , B, , (c) 60°, , 7. A ladder, leaning against a wall, makes an angle of, 60° with the horizontal. If the foot of the ladder is, 9.5 m away from the wall. The length of the ladder is, , (b) 50 m, (d) 70 m, , 12. The top of two poles of height 20 m and, 14 m are connected by a wire. If the wire makes an, angle of 30° with the horizontal, then the length of, the wire is, (a) 12 m, , (b) 10 m, , (c) 8 m, , (d) 6 m, , 13. An observer, 1.5 m tall is 20.5 m away from a tower, 22 m high, then the angle of elevation of the top of, the tower from the eye of the observer is, (a) 30°, , (b) 45°, , (c) 60°, , (d) 90°, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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96, , CBSE Term II Mathematics X (Standard), , 14. The angle of elevation of the top of the tower from a, point, which is 40 m away from the base of the, tower in the horizontal level, is 45°. Find the height, of the tower., (a) 70 m, (c) 40 m, , (b) 60 m, (d) 30 m, , 15. The angle of elevation of the top of a building, 150 m high, from a point on the ground is 45°. The, distance of the point from foot of the building is, (a) 120 m, (c) 140 m, , (b) 130 m, (d) 150 m, , 16. The angle of depression of the car parked on the, road from the top of a 150 m high tower is 30°. The, [CBSE 2014], distance of the car from the tower is, (a) 150 m, , (b) 75 m, , (c) 150 3 m, , (d), , 150, m, 3, , 17. From a point on the ground, the angles of elevation, of the bottom and the top of a transmission tower, fixed at the top of a 20 m high building are 45° and, 60° respectively, then the height of the tower is, (a) 14.64 m, (c) 38.64 m, , (a), , h, m, 3, , (b) h 3 m, , (c), , h, m, 2, , (d) h m, , (iii) Value of h is, (a) 2, , (b) 3( 3 + 1), , (c) 4, , (d) 3( 3 − 1), , (iv) Height of the Parachute from the ground is, (a) 4 m, , (b) 3( 4 − 3 ), , (c) 8 m, , (d) 3( 4 + 3 ), , (v) If the Parachute is moving towards the building,, then both angles of elevation will, (a) remain same, (c) decreases, , (b) increases, (d) Can’t be determined, , 20. A cyclist is climbing through a 20 m long rope, which is highly stretched and tied from the top of a, vertical pole to the ground as shown below, , (b) 28.64 m, (d) 19.64 m, , 18. A bridge on a river makes an angle of 45° with its, edge. If the length along the bridge from one edge, to the other is 150 m, then the width of the river is, (a) 107.75 m, (c) 75 m, ●, , (ii) Value of DF is equal to, , (b) 105 m, (d) 106.05 m, , Case Based MCQs, 19. There are two balcony in a house. First balcony is, at a height of 3 m above the ground and other, balcony is 6 m vertically above the lower balcony., Ankit and Radha are sitting inside the two balcony, at points G and F, respectively. At any instant, the, angles of elevation of a Parachute from these, balcony are observed to be 60° and 45° as shown, below, E, F, 6m, G, , 45°, 60°, , hm, D, C, , 3m, A, , B, , Based on the above information, answer the, following questions., (i) Who is more closer to the Parachute., (a) Ankit, (b) Radha, (c) Both are at equal distance, (d) Can’t be determined, , Based on the above information, answer the, following questions., (i) The height of the pole, if angle made by rope with, the ground level is 60°, is, (a) 15 m, 10, (c), m, 3, , (b) 10 3 m, 15, (d), m, 2, , (ii) If the angle made by the rope with the ground, level is 60°, then the distance between artist and, pole at ground level is, 10, m, 2, (c) 10 m, (a), , (b) 10 2 m, (d) 10 3 m, , (iii) If the angle made by the rope with the ground, level is 45°. The height of the pole is, (a) 2.5 m, (c) 7.5 m, , (b) 10 m, (d) 10 2 m, , (iv) If the angle made by the rope with the ground, level is 45° and 3 m rope is broken, then the height, of the pole is, 17, m, 2, (c) 14 m, (a), , (b) 7 m, (d) 7 2 m, , (v) Which mathematical concept is used here?, (a) Similar triangles, (b) Pythagoras theorem, (c) Application of trigonometry, (d) None of the above, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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97, , CBSE Term II Mathematics X (Standard), , respectively. If the distance between the peaks of, two mountains is 1937 km, and the satellite is, vertically above the mid-point of the distance, between the two mountains. (use 3 = 1.73), , 21. A group of students of class X visited India Gate on, an educational trip. The teacher and students had, interest in history as well. The teacher narrated that, India Gate, official name Delhi Memorial,, originally called All-India War Memorial,, monumental sandstone arch in New Delhi,, dedicated to the troops of British India who died in, wars fought between 1914 and 1919.The teacher, also said that India Gate, which is located at the, eastern end of the Rajpath (formerly called the, Kingsway), is about 138 feet (42 m) in height., , F, A, , G, P, , H, Q, , B, , D, C, Nanda devi, , I, , R, S, Mullayanagiri, , (i) The distance of the satellite from the top of Nanda, Devi is, (a) 1139.4 km, (c) 1937 km, , (b) 1119.65 km, (d) 1025.36 km, , (ii) The distance of the satellite from the top of, Mullayanagiri is, (a) 1139.4 km, (c) 1937 km, , (i) What is the angle of elevation if they are standing, at a distance of 42 m away from the monument?, (a) 30°, (c) 60°, , (iii) The distance of the satellite from the ground is, (a) 1139.4 km, (c) 1937 km, , (b) 45°, (d) 0°, , (a) 30°, (c) 60°, , (b) 20.12 m, (d) 24.24 m, , (a) 20 3 m, (c), , 15, m, 3, , (d) 15 3 m, , (iv) The ratio of the length of a rod and its shadow is, 1 : 1. The angle of elevation of the Sun is, (a) 30°, (c) 60°, , (b) 45°, (d) 90°, , (v) The angle formed by the line of sight with the, horizontal when the object viewed is below the, horizontal level is, (a) corresponding angle, (b) angle of elevation, (c) angle of depression, (d) complete angle, , 22. A Satellite flying at height h is watching the top of, the two tallest mountains in Uttarakhand and, Karnataka ,they being Nanda Devi (height 7,816m), and Mullayanagiri (height 1,930 m). The angles of, depression from the satellite to the top of Nanda, Devi and Mullayanagiri are 30° and 60°,, , (b) 45°, (d) 0°, , (v) If a mile stone very far away, makes 45° to the top, of Mullayanagiri mountain. Hence, find the, distance of this mile stone from the mountain., , (iii) If the altitude of the Sun is at 60°, then the height, of the vertical tower that will cast a shadow of, length 20 m is, 20, m, (b), 3, , (b) 567.64 km, (d) 1025.36 km, , (iv) What is the angle of elevation, if a man is standing, at a distance of 7816 m from Nanda Devi?, , (ii) They want to see the tower at an angle of 60°. So,, they want to know the distance where they should, stand and hence find the distance., (a) 25.24 m, (c) 42 m, , (b) 577.52 km, (d) 1025.36 km, , (a) 1118.327 m, (c) 1930 m, , (b) 566.976 m, (d) 1025.36 m, , PART 2, Subjective Questions, ●, , Short Answer Type Questions, 1. If the height of a tower and the distance of the, point of observation from its foot, both are, increased by 10%, then the angle of elevation of its, top remains unchanged. Explain., 2. A straight tree is broken due to thunderstorm. The, broken part is bent in such a way that the peak of, the tree touches the ground at an angle of 60° at a, distance of 2 3 m. Find the whole height of the tree., 3. Determine the height of a mountain, if the, elevation of its top at an unknown distance from the, base is 30° and at a distance 10 km farther off from, the mountain, along the same line, the angle of, elevation is 15°. (take tan 15° = 0.27), , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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98, , CBSE Term II Mathematics X (Standard), , 4. There is a flag staff on a tower of height 20 m. At a, point on the ground, the angles of elevation of the, foot and top of the flag are 45° and 60°, respectively., Find the height of the flag staff., 5. If the length of the shadow of a tower is increasing,, then the angle of elevation of the Sun is also, increasing. Why or why not?, 6. A window in a building is at a height of 10 m from, the ground. The angle of depression of a point P on, the ground from the window is 30°. The angle of, elevation of the top of the building from the point P, is 60°. Find the height of the building. [CBSE 2007], 7. A player sitting on the top of a tower of height 20 m, observes the angle of depression of a ball lying on, the ground as 60°. Find the distance between the, foot of the tower and the ball., 8. If two towers of height x m and y m subtend angles, of 30° and 60°, respectively at the centre of the line, [CBSE 2015], joining their feet, then find x : y., 9. If a man standing on a platform 3 m above the, surface of a lake observes a cloud and its reflection, in the lake, then the angle of elevation of the cloud, is equal to the angle of depression of its reflection., State true or false. Justify., 10. From the top of a hill, the angles of depression of, two consecutive kilometre stones due East are, found to be 30° and 45°. Find the height of the hill., [CBSE 2015], , 11. The shadow of a tower is 30 m long, when the Sun’s, angle of elevation is 30°. What is the length of the, shadow, when Sun’s elevation is 60°?, 12. Two ships are there in the sea on either side of a, light house in such a way that the ships and the base, of the light house are in the same straight line., The angle of depression of two ships as observed, from the top of the light house are 60° and 45°., If the height of the light house is 200 m, then find, [CBSE 2014], the distance between the two ships., 13. From the top of a tower of height 50 m, the angles, of depression of the top and bottom of a pole are 30°, and 45°, respectively. Find, [CBSE 2015], (i) how far the pole is from the bottom of the tower., (ii) the height of the pole. [ take, 3 = 1.732 ], , 14. A man standing on the deck of a ship, which is 10 m, above the water level. He observes that the angle of, elevation of the top of a hill is 60° and the angle of, depression of the base of the hill is 30°. Calculate, the distance of the hill from the ship and height of, [CBSE 2016], the hill., , 15. The angle of elevation of an aeroplane from a point, on the ground is 45°. After flying for 15 s, the angle, of elevation changes to 30°. If the aeroplane is, flying at a constant height of 2500 m, then find the, [CBSE 2013], average speed of the aeroplane., 16. The shadow of a flag staff is three times as long as the, shadow of the flag staff, when the Sun rays meet the, ground at an angle of 60°. Find the angle between the, Sun rays and the ground at the time of longer shadow., 17. An aeroplane, when flying at a height of 4000 m, from the ground, passes vertically above another, aeroplane at an instant when the angles of elevation, of two planes from the same point on the ground, are 60° and 45°, respectively. Find the vertical, distance between the aeroplanes at that instant., 18. There is a small island in the middle of a 100 m, wide river and a tall tree stands on the island. P and, Q are points directly opposite to each other on two, banks and in line with the tree. If the angles of, elevation of the top of the tree from P and Q are, respectively 30° and 45°, then find the height of the, tree. [take, 3 = 1.732], ●, , Long Answer Type Questions, 19. An aeroplane is at an altitude of 1200 m. If two, ships are sailing towards it in the same direction., The angles of depression of the ships as observed, from the aeroplane are 60° and 30°, respectively., Find the distance between both ships., 20. The angles of depression of two consecutive kilometre, stones on the road on right and left of an aeroplane, are 60° and 45°, respectively as observed from the, aeroplane. Find the height of the aeroplane., 21. The angle of elevation of the top of a tower at a, distance of 120 m from a point A on the ground, is 45°. If the angle of elevation of the top of a flag, staff fixed at the top of the tower, at A is 60°, then, find the height of the flag staff. [use, 3 =1.73], [CBSE 2014], , 22. A balloon is connected to an electric pole. It is, inclined at 60° to the horizontal by a cable of length, 215 m. Determine the height of the balloon from, the ground. Also, find the height of the balloon, if, the angle of inclination is changed from 60° to 30°., [CBSE 2015], , 23. A man in a boat rowing away from a light house 100, m high takes 2 min to change the angle of elevation of, the light house from 60° to 45°. Find the speed of, boat., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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99, , CBSE Term II Mathematics X (Standard), , 27. The angle of elevation of the top of a tower 30 m, high from the foot of another tower in the same, plane is 60° and the angle of elevation of the top of, the second tower from the foot of the first tower is, 30°. Find the distance between the two towers and, also the height of the tower., 28. From the top of a tower h m high, angles of, depression of two objects, which are in line with the, foot of the tower are α and β (β > α). Find the, distance between the two objects., 29. A ladder against a vertical wall at an inclination α to, the horizontal. Its foot is pulled away from the wall, through a distance p, so that its upper end slides a, distance q down the wall and then the ladder makes, an angle β with the horizontal. Show that, p cos β − cos α, ., =, q sin α − sin β, 30. The angle of elevation of the top of a vertical tower, from a point on the ground is 60°. From another, point 10 m vertically above the first, its angle of, elevation is 45°. Find the height of the tower., 31. A window of a house is h m above the ground. From, the window, the angles of elevation and depression of, the top and the bottom of another house situated on, , the opposite side of the lane are found to be α and β,, respectively. Prove that the height of the other house, is h(1 + tan α cot β) m., , 32. The lower window of a house is at a height of 2 m, above the ground and its upper window is 4 m, vertically above the lower window. At any instant the, angles of elevation of a balloon from these windows, are observed to be 60° and 30°, respectively. Find, the height of the balloon above the ground., ●, , Case Based Questions, 33. A girl 8 m tall spots a parrot sitting on the top of a, building of height 58 m from the ground. The angle, of elevation of the parrot from the eyes of girl at, any instant is 60°. The parrot flies away, horizontally in such a way that it remained at a, constant height from the ground. After 8 s, the, angle of elevation of the parrot from the same point, is 30°., , 58 m, , 24. The angle of elevation of the top of a tower from, certain point is 30°. If the observer moves 20 m, towards the tower, the angle of elevation of the top, increases by 15°. Find the height of the tower., 25. The shadow of a tower standing on a level plane is, found to be 50 m longer when Sun’s elevation is 30°, than when it is 60°. Find the height of the tower., 26. A vertical tower stands on a horizontal plane and is, surmounted by a vertical flag staff of height h. At a, point on the plane, the angles of elevation of the, bottom and the top of the flag staff are α and β, respectively. Prove that the height of the tower is, ⎛ h tan α ⎞, ⎜, ⎟., ⎝ tan β − tan α ⎠, , 60°, 30°, , 8m, , Based on the above information, answer the, following questions. (Take 3 = 1.73), (i) Find the distance of first position of the parrot, from the eyes of the girl., (ii) If the distance between the position of parrot, increases, then the angle of elevation decreases., Justify with girl., (iii) Find the distance between the girl and the building., (iv) How much distance covers parrot covers?, (v) Find the speed of the parrot in 8s., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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100, , CBSE Term II Mathematics X (Standard), , SOLUTIONS, Objective Questions, 1. (b) Let the angle of elevation of the Sun is θ., , 4. (b) Let the distance of the foot of tower from the point be, QR = x m., P, , A, , 30°, 60 m, , h, B, , C, , √3h, , Given, height of pole = h, AC, h, Now, in ΔABC, tan θ =, =, BC, 3h, 1, tan θ =, = tan 30°, ⇒, 3, ⇒, θ = 30°, Hence, the angle of elevation of the Sun is 30°., 2. (a) Let BC = 6 m be the height of the pole and AB = 2 3 m, be the length of the shadow on the ground., Let the Sun makes an angle θ on the ground., C, , θ, 2√3 m, , B, , BC, AB, 6, 3, 3, [by rationalising], tan θ =, =, ⋅, ⇒, 2 3, 3, 3, 3 3, ⇒, tan θ =, = 3 = tan 60°[Q tan 60°= 3 ], 3, ∴, θ = 60°, Hence, the Sun’s elevation is 60°., 3. (d) Let AB be the tower whose height is 300 3 m,, i.e. AB = 300 3 m. Again, let C be the point at a distance of, 300 m from the foot of the tower, i.e. AC = 300 m., Here, the angle of elevation is unknown, so let it be θ., Since, here base and perpendicular are given., Now, in ΔBAC, tan θ =, , B, , 300√3 m, , A, , θ, 300 m, , R, , x, , and height PR = 60 m and ∠PQR = 30°, PR, 60, In ΔPRQ, tan 30° =, =, QR, x, 1, 60, ⇒, =, ⇒ x = 60 3 m, x, 3, 5. (a) In right angled ΔABC, ∠B = 90°, D, , A, θ, 4m, , C, , 4√3m, , Sun, , 6m, A, , 30°, , Q, , θ, , Let, ∠DAC = θ, Then, ∠ACB = ∠DAC = θ, [alternate angles], Perpendicular AB, 4, 1, Now,, tan θ =, =, =, =, = tan 30°, Base, BC 4 3, 3, ∴ The angle of depression from A is 30°., 6. (b) Let AB be the vertical pole and CA be the 20 m long rope, such that its one end A is tied from the top of the vertical pole, AB and the other end C is tied to a point C on the ground., A, , 20, , m, , 30°, C, , B, , In ΔABC, we have, AB, AC, 1 AB, ⇒, =, 2 20, ⇒, AB = 10 m, Hence, the height of the pole is 10 m., 7. (d) Let the length of ladder AB = h m., sin 30° =, , C, , So, in right angled ΔBAC,, Perpendicular AB 300 3, tan θ =, =, =, Base, AC, 300, ⇒, tan θ = 3 = tan 60°, ∴, θ = 60°, Hence, the required angle of elevation is 60°., , B, , B, , h, , 60°, A, , 9.5 m, , C, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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101, , CBSE Term II Mathematics X (Standard), , ⇒, , sin θ =, , 15, 17, , m, , B, , 85, , AC = 9. 5 m, ∠BAC = 60°, 1 9.5, AC, In ΔABC, cos 60° =, ⇒ =, AB, 2, h, ⇒, h = 2 × 9.5 = 19 m, Hence, length of ladder is 19 m., 8. (a) Let XZ be the length and YZ be the height of the ramp., Here,, , Z, θ, A, , 1m, 30°, X, , Y, , Then, ∠ZXY = 30° and YZ = 1 m, In right angled ΔXYZ,, Perpendicular YZ, sin 30° =, =, Hypotenuse, XZ, 1, 1, =, ⇒ XZ = 2 m, ⇒, 2 XZ, Hence, the length of ramp is 2 m., 9. (b) Let C be the position of the kite and AC be the length of, the string which makes an angle of 60° with the ground. The, height of the kite from the ground is BC = 80 m., , 11. (b) Let BC = h m be the height of the tower and A be the, point on the ground such that, ∠BAC = θ., C, , hm, , C, , 80 m, 60°, A, , B, , In right angled ΔABC,, sin 60° =, , Perpendicular BC, =, Hypotenuse, AC, , 3, 80, =, AC, 2, 80 × 2, 3, [by rationalising], AC =, ×, ⇒, 3, 3, 160 3 160 × 1.732, =, =, = 92. 37 m, 3, 3, Hence, the length of the string is 92.37 m., 10. (a) Given, length of the string of the kite,, AB = 85 m, 15, and, tan θ =, 8, 8, cotθ =, ⇒, 15, 2, Q, cosec θ − 1 = cot 2 θ, 64, ∴, cosec2θ − 1 =, 225, 64 289, cosec2θ = 1 +, =, ⇒, 225 225, 289 17, ⇒, cosec θ =, =, 225 15, ⇒, , C, , BC, In ΔABC,, sin θ =, AB, 15 BC, ⇒, =, 17 85, ⇒, BC = 75 m, Hence, height of kite is 75 m, , θ, , A, , 120 m, , B, , Then, AB = 120 m, In right angled ΔABC,, Perpendicular BC, h, tan θ =, =, =, Base, AB 120, 5, 5, h, ⎡, ⎤, ⇒, =, Q tan θ =, , given, ⎢⎣, ⎥⎦, 12, 12 120, ⇒, , h = 50 m, , Hence, height of the tower is 50 m., 12. (a) Here, CD = 20 m, , [height of big pole], , AB = 14 m, , [height of small pole], D, 6m, , B, , 14 m, , A, , ∴, ⇒, ⇒, , 30°, x, , E, , 20 m, , 14 m, , C, , DE = CD − CE, DE = CD − AB, DE = 20 − 14 = 6 m, DE, In ΔBDE,, sin 30° =, BD, 1, 6, =, ⇒ BD = 12 m, ⇒, 2 BD, ∴ Length of wire = 12 m, , [Q AB = CE ], , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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102, , CBSE Term II Mathematics X (Standard), , 13. (b) Let BE = 22 m be the height of the tower and, AD = 1. 5 m be the height of the observer. The point D be, the observer’s eye. Draw DC || AB., , θ, , D, 1.5 m, A, , 1=, , A, , 22 m, , 20.5 m, , E, , 150, ⇒ BC = 150 m, BC, Hence, the distance of the point from foot of the building is, 150 m., 16. (c) Let AB =150 m be the height of the tower and angle of, depression is ∠DAC = 30°., ⇒, , D, , 30º, 150 m, , C, , 20.5 m, , 30º, , B, , 20.5 m, , Then, AB = 20. 5 m = DC, and EC = BE − BC = BE − AD, = 22 − 1. 5 = 20. 5 m, [Q BC = AD ], Let θ be the angle of elevation made by observer’s eye with, the top of the tower i.e. ∠EDC = θ., In right angled ΔDCE,, Perpendicular CE 20. 5, tan θ =, =, =, Base, DC 20. 5, ⇒, tan θ = 1, ⇒, tan θ = tan 45°, ⇒, θ = 45°, 14. (c) In the figure, let AB be a tower which has height h m., The angle of elevation from point C at a distance of 40 m, from point B is ∠ACB = 45° and BC = 40 m., , C, , B, , Then, ∠ACB = ∠DAC = 30°, [alternate angles], Now, in right angled ΔABC,, Perpendicular AB, tan 30° =, =, Base, BC, 1, 150, =, ⇒, 3 BC, ⇒, BC = 150 3 m, Hence, the distance of the car from the tower is 150 3 m., 17. (a) Let the height of the building be, BC = 20 m, and height of the tower be CD = x m, D, xm, C, , A, , 20 m, , hm, 60°, , C, , A, , 45°, 40 m, , B, , Then, in right angled ΔABC,, Perpendicular AB, tan 45° =, =, Base, BC, h, h, ⇒ 1=, [Q tan 45° = 1], ⇒ tan 45° =, 40, 40, ∴, h = 40 m, Hence, the height of tower is 40 m., 15. (d) Let AB = 150 m be the height of building and C be a, point on the ground such that ∠ACB = 45°., A, , 150 m, , C, , 45°, B, , ym, , B, , Let the point A be at a distance y m, from the foot of the building., Now, in ΔABC,, BC, = tan 45° = 1, AB, 20, =1, ⇒, y, ⇒, y = 20 m, i.e., AB = 20 m, BD, Now, in ΔABD,, = tan 60° = 3, AB, BD, ⇒, = 3, AB, 20 + x, ⇒, = 3, 20, 20 + x = 20 3, ⇒, ⇒, , In right angled ΔABC,, Perpendicular AB, tan 45° =, =, Base, BC, , 45°, , x = 20 3 − 20, = 20( 3 − 1), , ⇒, , = 20(1.732 − 1), x = 20 × 0.732 = 14. 64 m, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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103, , CBSE Term II Mathematics X (Standard), , 18. (d) Let BC be the width of the river and A, B be the ends of, river such that AB = 150 m = Length of the bridge, [given], and ∠BAC = 45°., B, , 20, , m, , edge of river, River, edge of river, , C, , Perpendicular, Hypotenuse, 1, BC BC, =, =, 2 AB 150, 150, 2, BC =, ×, 2, 2, 150, =, 2 = 75 2, 2, = 75 × 1.414, , sin 45° =, ⇒, ∴, , [Qsin 45° =, , 1, ], 2, , [by rationalising], , [Q 2 = 1.414], , = 106.05 m (approx.), Hence, width of the river is 106.05 m., 19. (i) (b) Radha is more closer to the Parachute because the, angles of elevation of Parachute from these balcony are, observed to be 45°. So, Radha is more closer to the, Parachute than Ankit., DE, (ii) (d) In ΔDEF, ∠D = 90°, ∠DFE = 45°, tan 45° =, DF, DE, 1=, ⇒, DF, ⇒, DE = DF = h m, (iii) (b) In ΔEGC, ∠EGC = 60°, ∠C = 90°, CE, tan 60° =, CG, CD + DE, h+6, 3=, ⇒ 3=, [Q CG = DF ], CG, DF, h+6, ⇒, 3=, h, 6, 6, 3 =1+, ⇒ ( 3 − 1) =, ⇒, h, h, 6, 3+1, [by rationalising], ∴, h=, ×, ( 3 − 1), 3+1, =, , 6( 3 + 1), ( 3 ) 2 − (1 ) 2, , =, , 6 ( 3 + 1) 6 ( 3 + 1), =, = 3 ( 3 + 1) m, 3−1, 2, , [Q ( a + b )( a − b ) = a 2 − b 2 ], , = 9 + 3 ( 3 + 1), =9+ 3 3 + 3, , (iv) (a) Length of rope = 20 − 3 = 17 m, AB, sin 45° =, AC, 1, AB, 17, m, =, ⇔ AB =, 17, 2, 2, , (v) (c) If the Parachute is moving towards the building, then, both angles of elevation will decreases., , [∠C = 45°], , A, , 45°, , C, , B, , (v) (c) in this, mathematical concept trigonometric ratio is, used here, which is application of trigonometry., 21. (i) (b) Let AB be the monument of height 42 m and C is the, point where they are standing, such that BC = 42 m., A, , 42 m, , C, , θ, 42 m, , B, , Now, in ΔABC,, AB, BC, 42, tan θ =, =1, 42, tan θ = 1, ⇒, tan θ =, , = 12 + 3 3, 3) m, , B, , Then, AC = 20 m, ∠C = 60°, ∠B = 90°, AB, In ΔABC,, sin 60° =, AC, 3 AB, =, 2, 20, AB = 10 3 m, BC, 1 BC, (ii) (c) cos 60° =, ⇒ =, AC, 2 20, BC = 10 m, AB, (iii) (d) sin 45° =, [ ∠C = 45° ], AC, 1, 20, AB, =, ⇒, = AB, 20, 2, 2, 20, 2, [by rationalising], AB =, ×, ⇒, 2, 2, = 10 2 m, , (iv) (d) Height of the Parachute from the ground is BE, then, BE = BC + CD + DE, BE = 3 + 6 + 3( 3 + 1), , = 3( 4 +, , 60°, , C, , In right angled ΔACB,, , m, , 45°, A, , A, , 17, , River, , 20. (i) (b) Let in ΔABC, AC will be rope and AB be a vertical, pole., , ⇒, ⇒, , θ = 45°, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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104, , CBSE Term II Mathematics X (Standard), , (ii) (d) In ΔABC,, , 22. As it is given that satellite is the mid-point of the two, mountain hills i.e. I is the mid-point of DS., , A, , F, 30°, , 60°, , 42 m, , A, C, , 60°, , A, , hm, 60°, , B, , 20 m, , AB, h, ⇒ h = 20 3 m, ⇒ 3=, 20, BC, (iv) (b) Let h and x be the height and length of shadow of the, vertical tower., A, , hm, θ, , B, xm, , Now, in ΔABC,, AB, h, ⇒ tan θ =, BC, x, ⇒, tan θ = 1, [Q h : x = 1 : 1], ⇒, θ = 45°, (v) (c) The angle of depression of the object viewed, is the, angle formed by the line of sight with the horizontal,, when it is below the horizontal level., tan θ =, , D, , P, 1930 m, , 968500 m, , 968500 m, , S, , I, 1937000 m, , (i) (b) We have, AG = DI = 968500 m, Now, in ΔFAG,, AG, cos 30° =, AF, 3 968500, ⇒, =, AF, 2, 968500 × 2 1937000, AF =, =, ⇒, 1.73, 3, = 1119653.18 m, = 1119. 65 km, (ii) (c) We have, HP = IS = 968500 m, HP, FP, 1 968500, =, FP, 2, FP = 968500 × 2 = 1937000 m = 1937 km, 1, FG, FG, (iii) (b) In ΔFAG, tan 30° =, =, ⇒, AG, 3 968500, 968500, FG =, = 559826. 59 m, ⇒, 3, = 559. 82 km, ∴Height of satellite from ground = FI = FG + GI, = 559. 82 + 7. 816, [Q GI = AD = 7816 m = 7. 816 km ], = 567. 64 km, (iv) (b) Let E be the position of man., cos 60° =, , A, , D, , Horizontal line, O, , 60°, , H, , 7816 m, , Now, in ΔFHP,, , Now, in ΔABC, tan 60° =, , C, , G, , B, , AB, tan 60° =, BC, 42, 3=, BC, 42 42, 3, [by rationalising], BC =, =, ×, 3, 3, 3, 42 3, =, = 14 3, 3, = 14 × 1.732 = 24.24 m, (iii) (a) Let AB = h be the height of the tower., , C, , 30°, , Angle of depression, ne, Li, of, sig, ht, , P (object), , θ, 7816 m, , E, , Then, DE = 7816 m, AD 7816, In ΔADE, tan θ =, =, =1, DE 7816, [Q height of mountain AD = 7816 m], ∴, θ = 45°, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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105, , CBSE Term II Mathematics X (Standard), , (v) (c) Let T be the point where mile stone is kept., , 2. Let AB be the tree whose part AC breaks and touches the, ground at D., Then,, [given], BD = 2 3 m, , P, , T, , 45°, , A, , S, , So, In ΔPST, tan 45° =, ⇒, , AC = CD, , and, , 1930 m, , PS, TS, , C, , 1930, 1=, ⇒ TS = 1930 m, TS, , Subjective Questions, 1. Case I Let the height of a tower be h and the distance of the, point of observation from its foot be x., In ΔABC,, AC h, tan θ1 =, =, BC x, ⎛ h⎞, ...(i), θ1 = tan −1 ⎜ ⎟, ⇒, ⎝ x⎠, , h, θ1, , C, , x, , Case II Now, the height of the tower increased by 10%, 10 11 h, = h + 10% of h = h + h ×, =, 100, 10, and the distance of the point of observation from its foot, = x + 10% of x, 10, 11x, =x+x ×, =, 100, 10, P, 11h, 10, θ2, 11x, 10, , Q, , ⇒, , tan θ 2 =, , 1, ⎤, ⎡, Q cos 60° = and BD = 2 3 m, ⎥⎦, ⎢⎣, 2, [Q 3 = 1.732], , ∴, AC = CD = 6.928 m, Again, in right angled ΔCBD,, BC, tan 60° =, BD, BC, [Q tan 60° = 3 and BD = 2 3 m], ⇒, 3=, 2 3, ⇒, BC = 3 × 2 3 = 6 m, AB = AC + BC, = 6.928 + 6 = 12.928 m (approx.), Hence, whole height of the tree is 12.928 m., , Now,, , 3. Let AB = h km be the height of the mountain. Let C be a, point at a distance of x km from the base of the mountain, such that ∠ACB = 30° and let D be a point at a distance of, 10 km from C along the same line. Then, ∠ADB = 15°, and AD = AC + DC = ( x + 10) km, B, , R, h km, , ⎛ 11h ⎞, ⎜, ⎟, ⎝ 10 ⎠, PR, In ΔPQR, tan θ 2 =, =, QR ⎛ 11x ⎞, ⎜, ⎟, ⎝ 10 ⎠, ⇒, , In right angled ΔCBD,, BD, cos 60° =, CD, 1 2 3, ⇒, =, 2 CD, , D, , ⇒ CD = 2 × 2 3 = 4 3, = 4 × 1.732 = 6.928 m, , A, , B, , 60°, 2√3 m, , B, , D, , From Eqs. (i) and (ii),, θ1 = θ 2, Hence, the required angle of elevation of its top remains, unchanged., , C, , A, , In right angled ΔBAC,, , h, x, , ⎛ h⎞, θ 2 = tan −1 ⎜ ⎟, ⎝ x⎠, , 30°, , 15°, , AB, AC, 1, h, =, 3 x, x=h 3, , tan 30° =, ...(ii), , ⇒, ⇒, , 1 ⎤, ⎡, Q tan 30° =, ⎢⎣, 3 ⎥⎦, ...(i), , In right angled ΔBAD,, tan 15° =, , AB, AD, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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106, , CBSE Term II Mathematics X (Standard), , ⇒, , 0.27 =, , h, x + 10, , ⇒, 0.27 ( x + 10) = h, On putting x = 3h from Eq. (i) in Eq. (ii), we get, 0.27( 3h + 10) = h, ⇒, ⇒, ⇒, , ⇒, , [ given , tan 15° = 0.27 ], ...(ii), , 0.27 × 3h + 0.27 × 10 = h, h(1 − 0.27 × 3 ) = 0.27 × 10, h(1 − 0.27 × 1.732 ) = 2 .7, , [Q, , 3 = 1.732], , h (1 − 0. 47 ) = 2.7, 0. 53h = 2.7, 2.7, h=, = 5.09 ≈ 5 km, ⇒, 0.53, Hence, the height of mountain is 5 km., 4. Let AB be the tower and AC be the flag staff on the tower., Let D be a point on the ground such that the angles of, elevation of foot A and top C of the flag staff are 45° and 60°,, respectively., , ∴, θ = 60°, II. A same hight of tower casts a shadow 4m long from, preceding shadow, when the Sun’s elevation is 30°., AB, In ΔAPB,, tan θ =, PB, AB, =, PC + CB, , 60°, , 45°, , B, , D, , Then, we have AB = 20 m , ∠ADB = 45° and ∠CDB = 60°, In right angled ΔABD,, AB, Perpendicular ⎤, ⎡, tan 45°=, Q tan θ =, ⎢⎣, ⎥⎦, BD, Base, 20, 1=, ⇒, BD, ⇒, BD = 20 m, [Q tan 45° = 1], and in right angled ΔCBD,, BC, BC, tan 60°=, [Q tan 60° = 3 ], ⇒ 3=, 20, BD, ⇒, BC = 20 3 = 20 × 1.732, [Q 3 = 1.732], = 34. 64 m (approx.), Now,, AC = BC − AB = 34.64 − 20, = 14.64 m (approx.), Hence, the height of the flag staff is 14.64 m., 5. To understand the fact of this question, consider the, following example, I. A tower 2 3 m high casts a shadow 2 m long on the, ground, when the Sun’s elevation is 60°., BC 2 3, In ΔACB,, tan θ =, =, AB, 2, C, , 2√3 m, , 2 3 2 3, =, 4+2, 6, A, , 2√3 m, θ, 4m, , P, , 60°, C, , 2m, , B, , 3, 3, 3, ⋅, =, 3, 3 3 3, 1, ⇒, tan θ =, = tan 30°, 3, ∴, θ = 30°, Hence, we conclude from above two examples that if the, length of the shadow of a tower is increasing, then the, angle of elevation of the Sun is decreasing., 6. Let QS be the building and R be the position of window., ⇒, , A, , A, , tan θ =, , ⇒, , ⇒, ⇒, , C, , tan θ = 3 = tan 60°, , tan θ =, , Given, height of the window, QR = 10 m, ∠QPR = ∠XRP = 30° [alternate angles], and, ∠SPQ = 60°, S, , X, P, , 30°, 60°, 30°, , R (Window), 10 m, Q, , In right angled ΔPQR ,, QR, tan 30° =, PQ, ⇒, ⇒, , 1, 10, =, 3 PQ, PQ = 10 3 m, , 1 ⎤, ⎡, Q tan 30°=, ⎢⎣, 3 ⎥⎦, ...(i), , In right angled ΔPQS,, QS, tan 60° =, PQ, QS, 3=, [Q tan 60° = 3 and from Eq. (i)], ⇒, 10 3, ⇒, , QS = 10 × 3 = 30 m, , Hence, height of the building is 30 m., , θ, 2m, , B, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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107, , CBSE Term II Mathematics X (Standard), , 7. Let AB = 20 m be the height of tower and let the ball lying, on the ground at point C., Given, angle of depression,, [alternate angles], ∠TAC = 60° = ∠ACB, , So, angle of depression is different in the lake from the angle, of elevation of the cloud above the surface of a lake., C, hm, , A, , T, 60°, , θ1, θ2, , P, , M, 3m, , Q, , 20 m, , O, , h, , 60°, C, , B, , In right angled ΔABC,, AB, tan 60° =, BC, 20, 3=, ⇒, BC, 20, 20, ⇒, BC =, =, = 11.55 m, 3 1.732, Hence, the distance between the foot of the tower and the, ball is 11.55 m., 8. Let AB be the tower of height x m , and CD be the tower of, height y m., D, B, ym, , xm, 60°, , 30°, A, , a, , E, , a, , C, , Let E be the mid-point of the line AC. Then, ∠AEB = 30°, and ∠CED = 60°., Also, AE = EC = a m (let), In right angled ΔBAE,, AB x, tan 30° =, =, AE a, 1, x, a, …(i), ⇒, = ⇒x =, 3 a, 3, and in right angled ΔDCE,, DC y, tan 60°=, =, CE a, y, …(ii), ⇒, 3 = ⇒ y = 3a, a, [Q Eq. (i) divide by Eq. (ii)], a, 1, 1, x, ∴, = 3 =, =, y, 3a, 3× 3 3, Hence, x : y = 1 : 3, 9. From figure, we observe that, a man standing on a platform, at point P, 3 m above the surface of a lake observes a cloud at, point C. Let the height of the cloud from the surface of the, platform is h and angle of elevation of the cloud is θ1., Now at same point P, a man observes a cloud reflection in, the lake at this time the height of reflection of cloud in lake, is ( h + 3) because in lake platform height is also added to, reflection of cloud., , θ, , R, , CM, h, =, PM PM, tan θ1, 1, …(i), ⇒, =, h, PM, RM OR + OM h + 3, In ΔRPM,, =, =, tan θ 2 =, PM, PM, PM, tan θ 2, 1, …(ii), =, ⇒, h + 3 PM, From Eqs. (i) and (ii),, tan θ1 tan θ 2, ⎛ h + 3⎞, =, ⇒ tan θ 2 = ⎜, ⎟ tan θ1, ⎝ h ⎠, h, h+3, , In ΔMPC,, , tan θ1 =, , So,, θ1 ≠ θ 2, Hence, it is a false statement., 10. Let AB = h km be the height of the hill and C, D be two, consecutive stones such that CD = 1 km ., Let BC be x km, then BD = BC + CD = ( x + 1) km, A, , 30°, X, 45°, , h km, , B, , 45°, 30°, x km C 1 km, (x+1) km, , D, , Now,, ∠ ADB = ∠XAD = 30°, and, ∠ ACB = ∠XAC = 45°, In right angled ΔABC,, Perpendicular AB, tan 45° =, =, Base, BC, h, 1 = ⇒x = h, ⇒, x, AB, Now, in right angled ΔABD, tan 30° =, BD, 1, h, =, ⇒, 3 x+1, 1, h, =, ⇒, 3 h+1, , [alternate angles], [alternate angles], , …(i), , 1 ⎤, ⎡, Q tan 30°=, ⎢⎣, 3 ⎥⎦, [from Eq. (i)], , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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108, , CBSE Term II Mathematics X (Standard), , ⇒, ⇒, , In right angled ΔBMP,, , h + 1 = 3h, h( 3 − 1) = 1, , ⇒, , h=, , PM, BM, 200, 1=, [Q tan 45° = 1], ⇒, y, ⇒, y = 200 m, Now, distance between the two ships = AB = x + y, = 115. 47 + 200, = 315.47 m, 13. Let distance of the pole, say AE, from the bottom of the tower,, say BD, be x m and let the height of the pole, AE = y m, tan 45° =, , 3 + 1 ⎛ 3 + 1⎞, =⎜, ⎟ km, 3+1 ⎝ 2 ⎠, , 1, ×, 3 −1, , 3+1, km., 2, 11. Let AB be the tower, BC be the shadow of tower, when angle, of elevation of Sun is 30° and BD be the shadow of tower,, when angle of elevation of Sun is 60°., Then, we have, BC = 30 m, ∠ACB = 30° and ∠ADB = 60°, Now, let AB = h m and BD = x m, Hence, height of the hill is, , Now, draw EC ||AB., D, , A, 50 m, , hm, 30°, , E, , D, , xm, , 30 m, , B, , 45º, A, , P, 45°, , 60°, , 45°, ym, , In right angled ΔAMP,, , M, , ⇒, , ⇒, ⇒, ⇒, , A, , Perpendicular PM, =, Base, AM, 200, 3=, [Q tan 60° = 3 ], x, 200, 200, m=, x=, m = 115. 47 m, 1.732, 3, , tan 60° =, ⇒, , 60°, xm, , xm, , B, , Then, ∠DEC = 30°, ∠DAB = 45°, and, DC = DB − BC = DB − AE, [Q BC = AE ], ⇒, DC = ( 50 − y ) m, (i) In right angled ΔABD,, Perpendicular BD, tan 45° =, =, Base, AB, 50, …(i), 1=, ⇒ x = 50 m, ⇒, x, ∴ The pole is 50 m away from the foot of the tower., (ii) In right angled ΔECD,, Perpendicular DC, tan 30° =, =, Base, EC, 1 ⎤, 1, 50 − y, ⎡, tan 30° =, =, ⇒, ⎢⎣, x, 3, 3 ⎥⎦, 1, 50 − y, [Q x = 50 m from Eq. (i)], ⇒, =, 50, 3, , Y, , 200 m, B, , C, , ym, , Clearly, in ΔABC, we have, Perpendicular AB h, tan 30° =, =, =, Base, BC 30, 1, h, 30, =, ⇒h =, ⇒, 3 30, 3, 30, 3 30 3, = 10 3 m, ⇒, ×, =, 3, 3, 3, AB h, Also, in ΔABD, tan 60° =, =, BD x, 10 3, 3=, ⇒ x = 10 m, ⇒, x, Hence, length of shadow is 10 m, when angle of elevation is, 60°., 12. Let PM be the light house of height 200 m and let A and B be, two ships on either sides of light house such that the angles, of depression of A and B are 60° and 45°, respectively., Let, AM = x m and BM = y m, Then, ∠XPB = ∠MBP = 45°, [alternate angles], and, [alternate angles], ∠YPA = ∠MAP = 60°, X, , xm, , ym, , 60°, , C, , 30º, , 3 ( 50 − y ) = 50, 50, 50 − y =, 3, 1 ⎞, ⎛, y = 50 ⎜1 −, ⎟, ⎝, 3⎠, 1 ⎞, ⎛, = 50⎜1 −, ⎟, ⎝ 1.732 ⎠, , = 50(1 − 0. 57737 ), = 50 × 0. 4226, = 21.13 m, ∴ Height of the pole = 21.13 m, 14. Let a man is standing on the deck of a ship at point A such, that AB = 10 m and let CD be the hill., Then, ∠EAD = 60°, and, [alternate angles], ∠CAE = ∠BCA = 30°, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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109, , CBSE Term II Mathematics X (Standard), , 16. Let AB be the flag staff of height h units and AC = x units be, length of its shadow, when the Sun rays meet the ground at, an angle of 60°., Also, let θ be the angle between the Sun rays and the, ground, when the length of the shadow of the flag staff is, AD = 3x units., , BC = x m = AE and DE = h m, , Let, , D, , hm, , B, , 60°, 30°, , A, 10 m, , E, , h, , 10 m, 30°, xm, , B, , C, , In right angled ΔAED,, Perpendicular DE, tan 60° =, =, =, Base, EA, h, 3 = ⇒h = 3x, ⇒, x, In right angled ΔABC,, AB, 1, 10, ⇒, tan 30° =, =, BC, x, 3, ⇒, , x = 10 3 m, , ⇒, , h = 10 3 × 3 = 30 m, , h, x, , Perpendicular AB, =, Base, AC, h, 3=, [Q tan 60° = 3 ], x, ...(i), h = 3x, , tan 60° =, ⇒, , 1 ⎤, ⎡, Q tan 30° =, ⎢⎣, 3 ⎥⎦, , B, , D, , Here, AC = BD = 2500 m , ∠AOC = 45°, and, ∠BOD = 30°, In right angled ΔOCA,, Base, OC, cot 45°=, =, perpendicular AC, OC, 1=, ⇒, AC, ⇒, OC = AC = 2500 m, In right angled ΔODB,, OD, OD, cot 30° =, ⇒ 3=, 2500, BD, ⇒, OD = 2500 3 m, Now, CD = OD − OC = 2500 3 − 2500, , ⇒, , Now, in right angled ΔDAB,, AB, AB, tan θ =, =, [Q AD = DC + CA ], AD DC + CA, h, h, tan θ =, =, ⇒, 2 x + x 3x, 3x, [from Eq. (i)], =, 3x, 1, 1 ⎤, ⎡, =, = tan 30°, Q tan 30° =, 3, 3 ⎦⎥, ⎣⎢, ∴, θ = 30°, Hence, the angle between the sun rays and the ground at the, time of longer shadow is 30°., 17. Let P and Q be the positions of two aeroplanes, where P is, vertically above Q and OP = 4000 m., , 45°, C, , A, , x, , In right angled ΔCAB,, , 2500 m, , O, , C, , 2x, , ∴The height of hill, CD = h + 10 = 30 + 10 = 40 m, Hence, The distance of the hill from the ship is 10 3 m and, height of the hill is 40 m., 15. Let OX be the horizontal ground; A and B be the two, positions of the plane and O be the point of observation., A, , 60°, , θ, , D, , X, , P, , Q 4000 m, , A, , [Q cot 45° = 1], , = 2500( 3 − 1) = 2500(1.732 − 1), = 2500 × 0.732 = 1830 m, Thus, distance covered by plane in 15 s is 1830 m., 1830 60 × 60, ∴Speed of plane =, ×, = 439.2 km/h, 15, 1000, , 45°, , 60°, O, , Here, ∠PAO = 60° and ∠QAO = 45°, Now, in right angled ΔAOP,, Perpendicular OP, tan 60° =, =, Base, AO, 4000, 3=, ⇒, AO, 4000, ⇒, AO =, 3, In right angled ΔAOQ,, OQ, OQ, ⇒ 1=, tan 45° =, OA, OA, ⇒, OA = OQ, , [Q tan 60° = 3 ], ...(i), , [∴ tan 45° = 1], ...(ii), , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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110, , CBSE Term II Mathematics X (Standard), , From Eqs. (i) and (ii), we get, 4000, m, OQ =, 3, ∴vertical distance between the aeroplanes, 4000, 3, 1 ⎞, 1 ⎞, ⎛, ⎛, = 4000 ⎜1 −, ⎟, ⎟ = 4000⎜1 −, ⎝ 1.732 ⎠, ⎝, 3⎠, = 4000 (1 − 0. 577 ), = 4000 × 0. 423 = 1692 m, 18. Let OA be the tree of height h m., Given, PQ = 100 m and angles of elevation are ∠APO = 30°, and ∠OQA = 45°., = PQ = OP − OQ = 4000 −, , A, , hm, P, , 45°, , 30°, 100 m, , ⇒, , Q, , O, , Perpendicular OA, =, Base, OP, 1, h, 1 ⎤, ⎡, =, Q tan 30° =, ⎢⎣, 3 OP, 3 ⎥⎦, ...(i), OP = 3h, , In right angled ΔPOA, tan 30° =, ⇒, ⇒, , We have,, AB = 1200 m, Let, AC = x m and CD = y m., In right angled ΔBAC, we have, Perpendicular AB, tan 60° =, =, Base, AC, 1200, [Q tan 60° = 3], 3=, ⇒, x, 1200, 3, [rationalising], ⇒, x=, ×, 3, 3, 1200 3, ...(i), ⇒, x=, = 400 3 m, 3, In right angled ΔBAD, we have, AB, AB, [Q AD = DC + CA], tan 30° =, =, AD DC + CA, 1, 1200, =, ⇒, 3 x +y, , ⇒, , y = 1200 3 − x, , On putting the value of x from Eq. (i) in Eq. (ii), we get, y = 1200 3 − 400 3, = 800 3, = 800 × 1.732, , OA, OQ, h, 1=, OQ, , tan 45° =, , [Q tan 45° = 1], , ⇒, OQ = h, On adding Eqs. (i) and (ii), we get, OP + OQ = 3 h + h, , ...(ii), , Hence, the distance between both ships is 1385. 6 m., 20. Let A be the aeroplane and AD be its height. Again, let B, and C be two consecutive kilometre stones on the road on, the left and right of plane A and the angles of depression of, C and B from plane A are 60° and 45°, respectively., , ⇒, , PQ = ( 3 + 1)h, , 100 = ( 3 + 1)h [QPQ = 100 m, given], , 100, 3 −1, [by rationalising], ×, 3+1, 3 −1, 100 (1.732 − 1), =, 2, = 50 × 0.732 = 36. 6 m, Hecne, height of the tree is 36.6 m., 19. Let the aeroplane be at B and two ships be at C and D such, that their angles of depression from B are 60° and 30°,, respectively. Then, the angles of elevation of B from D and C, are 30° and 60°, respectively., B, 30°, , 1200 m, 60°, , 30°, D, , ym, , C, , A, xm, , 45°, , 60°, , Q, , [Q OP + OQ = PQ], , h=, , 60°, , A, , P, , ⇒, ⇒, , [Q 3 = 1.732], , = 1385. 6 m, , Now, in right angled ΔQOA,, , ⇒, , 1 ⎤, ⎡, Q tan 30° =, ⎢⎣, 3 ⎥⎦, …(ii), , x + y = 1200 3, , 45°, B, , x km, 1 km, , D, , 60°, C, (1 – x) km, , [alternate angles], ∠ABC = ∠PAB = 45°, [alternate angles], ∠ACB = ∠QAC = 60°, BC = 1 km, BD = x km , then, …(i), DC = BC − BD = (1 − x ) km, In right angled ΔADB,, Perpendicular AD, tan 45° =, =, Base, BD, AD, 1=, [Q tan 45° = 1], ⇒, x, ⇒, AD = x, and in right angled ΔADC,, AD, tan 60° =, DC, Then,, and, Also,, Let, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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111, , CBSE Term II Mathematics X (Standard), x, 1−x, [Q tan 60° = 3 and from Eq. (i)], , ⇒, , 3=, , ⇒, , 22. Let D be the position of the balloon, when it is inclined at, angle of 60° and AB be the height of the pole., D (Balloon), , 3 − 3x = x, , ⇒, , 3 = 3x + x, , ⇒, , C, , ( 3 + 1) x = 3, , ⇒, , x=, , B, , 3, =, 3+1, , 3, ×, 3+1, , 3 −1, 3 −1, , 60°, , [by rationalising], =, , 3− 3, ( 3 ) 2 − (1 ) 2, [Q ( a + b )( a − b ) = a 2 − b 2], , 3 − 3 3 − 1.732, [Q 3 = 1.732], =, 2, 2, 1.268, =, = 0.634 km, 2, Hence, the height of the aeroplane is 0.634 km., 21. Let height of the tower, BC = h m and height of the flagstaff, CD = H m ., …(i), ∴, BD = BC + CD = ( h + H )m, Given, AB =120 m, ∠CAB = 45° and ∠DAB = 60°, =, , Hm, , ⇒, , hm, 45° 60°, B, , In right angled ΔABC, we get, BC, tan 45° =, AB, h, 1=, ⇒, 120, ⇒, h = 120 m, , Perpendicular ⎤, ⎡, Q tan θ =, ⎢⎣, ⎥⎦, Base, , Now, in right angled ΔABD, we get tan 60° =, ⇒, , 3=, , [Q tan 45° = 1], …(ii), BD, AB, , h +H, …(iii), 120, [Q tan 60° = 3 and from Eq. (i)], , From Eqs. (ii) and (iii),, 120 + H, 120, 120 3 = 120 + H, 3=, , ⇒, ⇒, , ⎡, 3⎤, ⎢Q sin 60°= 2 ⎥, ⎣, ⎦, , 3 AD, =, 2, 215, , ⇒, , C, (h+H) m, , 120 m, , A, , Given, length of cable, DE = 215 m, In right angled ΔEAD,, Perpendicular AD, sin 60° =, =, Hypotenuse, ED, , 215 3, m, 2, Hence , initial height of the balloon from the ground is, 215 3, m., 2, Again, in right angled ΔEAD,, Base, AE AE, cos 60° =, =, =, Hypotenuse DE 215, , D, , A, , 30°, , E, , ⇒, , AD =, , 1 AE, 1⎤, ⎡, =, Q cos 60° =, ⎢⎣, 2 215, 2 ⎥⎦, 215, …(i), AE =, m, ⇒, 2, Now, the angle of inclination is changed, say ∠CEA = 30°., In right angled ΔEAC,, Perpendicular AC, tan 30° =, =, Base, EA, 1, AC × 2, ⇒, =, 215, 3, , ⇒, , 2 3AC = 215, , 215, m, 2 3, 23. Let the height of the light house AB be 100 m. C and D be, the positions of man when angle of elevation changes from, 60° to 45°, respectively. The man has covered a distance CD, in 2 min., Distance, CD, Speed =, ...(i), Q, ⇒ Speed =, Time, 2, In right angled ΔABC,, ⇒, , AC =, , A, , H = 120 ( 3 − 1), , = 120(1.732 − 1), = 120 × 0.732 = 87. 84 m, Hence, height of flag staff is 87.84 m., , 100 m, Light house, D, , 45°, , 60°, C, , B, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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112, , CBSE Term II Mathematics X (Standard), Perpendicular AB, =, Base, BC, 100, 3=, ⇒, BC, 100 3, m, ⇒, BC =, 3, In right angled ΔABD,, AB, tan 45° =, BD, 100, 1=, ⇒, BD, ⇒, BD = 100 m, , ⇒, , tan 60° =, , ∴, , h=, , 20, ⋅, 3 −1, , ⇒, , =, , 20 ( 3 + 1), 3−1, , ...(ii), , 3+1, 3+1, , [Q tan 45° = 1], , 20 ( 3 + 1), 2, = 10 ( 3 + 1) m, , ⇒, , Hence, the required height of tower is 10 ( 3 + 1) m., 25. Let the height of the tower be h and RQ = x m, , 100 3, 3, , PR = 50 m, ∠SPQ = 30° , ∠SRQ = 60°, SQ, Now, in ΔSRQ, tan 60° =, RQ, h, h, ⇒, 3=, ⇒ x=, x, 3, SQ, SQ, h, and in ΔSPQ, tan 30° =, =, =, PQ PR + RQ 50 + x, Given that,, and, , 50, CD, =, =, ( 3 − 3 ) m/min, 3, 2, 2, 24. Let the height of the tower be h., Also,, SR = x m, ∠PSR = θ, Given that,, QS = 20 m, and, ∠PQR = 30°, , S, , P, , 30°, 20 m, , θ, S, , xm, , P, , R, , ⇒, , PR h, tan θ =, =, SR x, h, tan θ =, x, h, x=, tan θ, , ⇒, ...(i), , ⇒, ⇒, , PR, PR, =, QR QS + SR, h, tan 30° =, 20 + x, h, h, 20 + x =, =, tan 30° 1 / 3, tan 30° =, , ⇒, ⇒, ⇒, , 20 + x = h 3, h, 20 +, =h 3, tan θ, , 60°, 30°, xm, 50 m R, , Q, , 3 ⋅ h = 50 + x, h, 3 ⋅ h = 50 +, 3, 1, ( 3−, ) h = 50, 3, ( 3 − 1), h = 50, 3, 50 3, h=, 2, h = 25 3 m, , ⇒, , Now, in ΔPQR,, , ⇒, , Sun, , 1, h, =, 3 50 + x, , Now, in ΔPSR,, , ⇒, , …(i), , hm, , h, Q, , [by rationalisation], , =, , ⎛ 3 − 3⎞, 100 ⎜, ⎟, ⎝ 3 ⎠, , and Speed =, , h ( 3 − 1) = 20, , ⇒, [Q tan 60° = 3 ], , CD = BD − BC = 100 −, , Now,, , 20 = h 3 − h, , ∴, , [from Eq. (i)], , Hence, the required height of tower is 25 3 m., 26. Let the height of the tower be H and OR = x, [from Eq. (i)] …(ii), , Since, after moving 20 m towards the tower the angle of, elevation of the top increases by 15°., i.e., ∠PSR = θ = ∠PQR + 15°, ⇒, θ = 30° + 15 = 45°, h, ∴From Eq. (i) 20 +, =h 3, tan 45°, h, ⇒, 20 + = h 3, 1, , Given that, height of flag staff = h = FP and ∠PRO = α,, ∠FRO = β, F, flag staff h, P, , H, β, R, , α, x, , O, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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115, , CBSE Term II Mathematics X (Standard), , ⇒, 2H = 16, ⇒, H=8, So, the required height is 8 m., Hence, the required height of the balloon above the ground, is 8 m., 33. (i) Distance of first position of parrot from the eyes of girl, = AC, C, , E, , 58 m, , 32. Let the height of the balloon above the ground is H, and, OP = W2R = W1Q = x, Given that, height of lower window from above the ground, = W2P = 2 m = OR, Height of upper window from above the lower window, = W1 W2 = 4 m = QR, ∴, BQ = OB − ( QR + RO), =H − ( 4 + 2 ) =H − 6, and, ∠BW1 Q = 30°, ⇒, ∠BW2R = 60°, Balloon, , 60°, D, , (H – 6), , B, , w1, , 30°, xm, , Upper, window, w2, P, , G, , x, R, 2m, O, , Now, in ΔBW2R ,, tan 60°=, ⇒, ⇒, , BR BQ + QR, =, W2R, x, , (H − 6 ) + 4, x, H −2, x=, 3, 3=, , and in ΔBW1Q,, tan 30° =, ⇒, , tan 30° =, , ⇒, From Eqs. (i) and (ii),, , (H − 2 ), 3, 3 (H − 6 ) = H − 2, 3H − 18 = H − 2, 3 (H − 6 ) =, , ⇒, , ...(i), , BQ, W1Q, , H −6 1, =, x, 3, x = 3 (H − 6 ), , A, 8m, F, , BC, AC, CH − BH, ⇒, AC =, sin 60°, 58 − 8 100, m, =, =, 3/2, 3, (ii) If the distance increases, then the angle of elevation, decreases., (iii) Distance between girl and building = AB, Now, in ΔABC,, BC, 50, m, tan 60° =, ⇒ 3 AB = 50 ⇒ AB =, AB, 3, DE, (iv) In ΔAED, tan 30° =, AD, AD = 3 BC = 50 3 m, ⇒, sin 60° =, , 4m, , xm, , H, , 30°, , In ΔABC,, , Q, Hm, , 60°, Lower, window, , B, , ...(ii), , [QED = BC = 58 − 8 = 50], Now, distance between two position of parrot = EC, = BD = AD − AB, 50 ⎞, ⎛, = ⎜ 50 3 −, ⎟m, ⎝, 3⎠, 50( 3 − 1) 100, =, = 57. 80 m, 1.73, 1.73, Distance covered, (v) Speed of parrot =, Time taken, ⎛ 57. 80⎞, =⎜, ⎟ m/s = 7.225 m/s, ⎝ 8 ⎠, =, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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Chapter Test, Multiple Choice Questions, , Based on the above information, answer the following, questions., , 1. A circus artist is climbing from the ground along a, rope stretched from the top of a vertical pole and tied, at the ground. The height of the pole is 12 m and the, angle made by the rope with ground level is 30°. The, distance covered by the artist in climbing to the top of, the pole is, (a) 12 m, (c) 24 m, , (i) Measure of ∠ACD is equal to, (a) 30°, (c) 60°, , (ii) If ∠YAB = 45°, then ∠ABD is also 45°, Why?, (a) vertically opposite angles, (b) alternate interior angles, (c) alternate exterior angles, (d) corresponding angles, , (b) 6 m, (d) 32 m, , 2. A ladder 15 m long just reaches the top of a vertical, , (iii) Length of CD is equal to, , wall. If the ladder makes an angle of 60° with the wall,, then the height of the wall is, , (a) 90 m, (c) 50 / 3 m, , 15, (b), m, 2, (d) 25 m, , (a) 30 m, (c) 15 m, , (a) 50 m, (c) 100 2 m, , The length of string from the kite to the ground is, 60 m. Assuming that there is no slack in the string,, then the angle of elevation of the kite at the ground is, (b) 45°, (d) None of these, , connected by a wire. If the wire makes an angle of 45°, with the horizontal, then find the length of the wire., , (v) Length of AC is equal to, (a) 100 / 3 m, (c) 50 m, , (b) 100 3 m, (d) 100 m, , Short Answer Type Questions, the height of the tower is doubled, then the angle of, elevation of its top will also be doubled. State true or, false. Explain., , 8. A peacock is sitting on the top of a tree. It observes a, , (b) 3 m, (d) 6 2 m, , serpent on the ground making an angle of depression, of 30°. The peacock catches the serpent in 12 s with the, speed of 300 m/min. What is the height of the tree?, [CBSE 2015], , 5. An observer 3.5 m tall is 38.5 m away from a tower, 42 m high. The angle of elevation of the top of the, tower from the eye of the observer is, , 9. The angles of elevation and depression of the top and, , (a) 30°, (b) 90°, (c) 45°, (d) 60°, , bottom of a light house from the top of a 60 m high, building are 30° and 60°, respectively. Find the, difference between the heights of the light house and, building., , Case Based MCQs, , 10. As observed from the top of a 100 m high light house, , 6. A boy is standing on the top of mountain. He, observed that boat P and boat Q are approaching, towards mountain from opposite directions. He finds, that angle of depression of boat P is 60° and angle of, depression of boat Q is 45°. He also knows that height, of the mountain is 50 m., X, , A, 45°, , 60°, , Y, , 50 m, P, , (b) 100 m, (d) 100 3 m, , 7. The angle of elevation of the top of a tower is 30°. If, , 4. The tops of two poles of height 30 m and 24 m are, (a) 14 m, (c) 4 m, , (b) 50 3 m, (d) 100 m, , (iv) Length of BD is equal to, , 3. A kite is flying at a height of 30 m from the ground., , (a) 30°, (c) 60°, , (b) 45°, (d) 90°, , Q, , C, , D, , B, , from the sea-level, the angles of depression of two, ships are 30° and 45°. If one ship is exactly behind the, other on the same side of the light house, find the, distance between the two ships., [CBSE 2018], Long Answer Type Questions, , 11. Two ships are sailing in the sea on either side of the, light house. The angles of depression of two ships as, observed from the top of the light house are 60° and, 45°, respectively. If the distance between the ships is, ⎛ 3 + 1⎞, 100 ⎜, ⎟ m, then find the height of the light house., 3 ⎠, ⎝, , Answers, 1. (c), 7. False, , 2. (b), , 3. (a), , 8. 30 m, , 4. (d), , 5. (c), , 9. 20 m, , 6. (i) (c) (ii) (b) (iii) (c) (iv) (a) (v) (a), , 10. 100 ( 3 − 1 ) m, , For Detailed Solutions, Scan the code, , 11. 100 m, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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117, , CBSE Term II Mathematics X (Standard), , CHAPTER 06, , Surface Areas, and Volumes, In this Chapter..., !, , Solid Figures, , !, , Surface Area, , !, , Volume, , !, , Combination of Two Figures, , !, , Conversion of Solid from One Shape to Another, , Solid Figures, , Different Types of Solid Figures, , The objects having definite shape, size and occupies a fixed, amount of space in three dimensions are called solids such as, cube, cuboid, cylinder, cone, sphere and hemisphere, etc., , 1. Cuboid, , A cuboid is a solid figure having 6 rectangular faces. Let its, length = l units, breadth = b units and height = h units., , Surface Area (SA), Surface area of a solid body is the area of all of its surfaces, together and it is always measured in square unit., e.g. A cube has 6 surfaces and each surface is in a square, shape. Therefore, its surface area will be 6 a 2 sq units, where, a 2 is the area of each surface of the cube., , Volume, Space occupied by an object/solid body is called the volume, of that particular object/solid. Volume is always measured in, cube unit., e.g. Suppose, a cube has edge of length a units. Volume of a, cube is equal to the product of area of base and height of, a cube i.e. a 2 × a = a 3 cu units., , b, , h, l, , Then,, (i) Total surface area of cuboid (TSA), = 2 ( lb + bh + hl) sq units, (ii) Lateral surface area of cuboid = 2( l + b )h sq units, or Lateral surface area = Area of the 4 vertical faces, (iii) Diagonal of the cuboid = l 2 + b 2 + h 2 units, (iv) Volume of cuboid = l × b × h cu units, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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118, , CBSE Term II Mathematics X (Standard), , 2. Cube, Cube is a special case of cuboid which has 6 equal square faces., , a, , a, a, , Let its length = breadth = height = a units, ∴ Each edge of cube = a units, Then,, (i) Total surface area (TSA) of a cube, = 6 × (Edge ) 2 = 6 a 2 sq units, (ii) Lateral surface area of cube = 4 × (Edge ) 2 = 4 a 2 sq units, (iii) Diagonal of a cube = 3 × Edge = 3 a units, , Then,, (i) Curved surface area (CSA), = CSA of outer cylinder + CSA of inner cylinder, = 2 πRh + 2 πrh, = 2π(R + r )h sq units, (ii) Total surface area (TSA), = CSA of hollow cylinder + Area of both ends, = 2π ( R + r )h + 2π ( R 2 − r 2 ), = 2 π(R + r )h + 2 π(R + r ) (R − r ), = 2π(R + r ) [ h + R − r ] sq units, (iii) Total outer surface area = 2 πRh + 2 π(R 2 − r 2 ) sq units, (iv) Volume of hollow cylinder, = Volume of outer cylinder, − Volume of inner cylinder, = πR 2 h − πr 2 h, = π(R 2 − r 2 )h cu units, , (iv) Volume of a cube = (Edge ) 3 = a 3 cu units, 5. Sphere, , 3. Right Circular Cylinder, , Cylinder is a solid figure obtained by revolving the rectangle,, say ABCD, about its one side, say BC. Let base radius of right, circular cylinder be r units and its height be h units. Then,, , A sphere is a solid generated by the revolution of a, semi-circle about its diameter. Let radius of sphere be r, units., A, , B, , A, , r, r, , h, , O, , r, , D, , B, , C, , (i) Curved surface area (CSA), = Circumference of the base × Height = 2πrh sq units, (ii) Total surface area (TSA), = Curved surface area (CSA) + Area of two ends, = 2 πrh + 2 πr 2 = 2 πr( h + r ) sq units, (iii) Volume of the cylinder = Area of base × Height, = πr 2 h cu units, 4. Right Circular Hollow Cylinder, , Let R units and r units be the external and internal radii of the, hollow cylinder, respectively and h units be its height., , Then,, (i) Surface area (SA) of sphere = 4 πr 2 sq units, 4, (ii) Volume of sphere = πr 3 cu units, 3, 6. Spherical Shell, , If R and r are respectively the outer and inner radii of a, spherical shell, then, (i) Outer surface area = 4 πR 2 sq units, (ii) Inner surface area = 4 πr 2 sq units, 4, (iii) Volume of a hollow sphere = π(R 3 − r 3 ) cu units, 3, , R, B, r, A, , h, , O, , R, , r, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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119, , CBSE Term II Mathematics X (Standard), , 7. Hemisphere, , A plane passing through the centre, cuts the sphere in two, equal parts, each part is called a hemisphere. Let radius of, hemisphere be r units. Then,, r, , e.g. A combined solid is formed by joining hemisphere and, right circular cone., , O, r, , (i) Curved surface area (CSA) of hemisphere = 2 πr 2 sq units, (ii) Total surface area (TSA) of hemisphere, = CSA of hemisphere + Area of one end, = 2 πr 2 + πr 2 = 3 πr 2 sq units., 2, (iii) Volume of hemisphere = πr 3 cu units, 3, 8. Right Circular Cone, , A right circular cone is a solid generated by the revolution of, a right angled triangle about one of its sides containing the, right angle as axis as shown in figure. Let height of a right, circular cone be h units and its radius be r units. Then,, (i) Slant height of the cone,, l = AC = r 2 + h 2 units, (ii) Curved surface area (CSA) of cone = πrl sq units, (iii) Total surface area (TSA) of a cone, = Curved surface area (CSA) + Area of the base, = πrl + πr 2 = πr( l + r ) sq units, A, , h, , B, , (iv) Volume of cone =, , l, , r O, , C, , 1 2, πr h cu units, 3, , Combination of Two Solids, Sometimes, we have to find the curved surface area and, volume of a solid, which is a combination of two solids. Then,, for finding the surface area, we add the curved surface areas, of individual solids and for finding the volume of this solid,, we add the volumes of individual solids., , (i) Surface area of combined solid figure, = CSA of cone + CSA of hemisphere, (ii) Volume of combined solid figure, = Volume of cone + Volume of hemisphere, While calculating the surface area, we have not added the, surface areas of the two individual solids, rather we have, added curved surface area because some part of the surface, area disappeared in the process of joining them. But this will, not be in the case, when we calculate the volume., , Conversion of Solid from One Shape to, Another, Sometimes, we need to convert solid figure of one shape to, another. When we come across objects which are converted, from one shape to another or when a liquid which is, originally filled in one container of a particular shape is, poured into another container of a different shape or size, the, volume remains same. e.g., (i) If a solid metallic sphere is melted and recast into more, than one spherical balls, then volume of metallic sphere, = Sum of volumes of all spherical balls., (ii) If the Earth taken out by digging a well and spreading it, uniformly around the well to form an embankment in the, shape of a cylindrical shell from its original shape of right, circular cylinder, then volume of embankment, = Volume of Earth taken out by digging a well., , Important Results or Formulae, If a solid of one shape is converted into solid (or solids) of, another shape, then, (i) Volume of the solid to be converted = Total volume of the, solids into which the given solid is to be converted, (ii) Number of solids of a given shape in which a given solid, is to be converted, Volume of the solid to be converted, =, Volume of one converted solid, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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120, , CBSE Term II Mathematics X (Standard), , Solved Examples, Example 1. Three metallic solid cubes whose edges are, 3 cm, 4 cm and 5 cm are melted and formed into a, single cube. Find the edge of the cube so formed., , We know that,, Volume of cylinder = πr 2h, , Sol. Given, edges of three solid cubes are 3 cm, 4 cm and 5 cm,, respectively., Volume of first cube = ( 3)3 = 27 cm 3, ∴, , 10cm, , [Q volume of cube = (side) 3], Volume of second cube = ( 4)3 = 64 cm 3, , 240 cm, , Sum of volume of three cubes = (27 + 64 + 125), = 216 cm 3, , Let the edge of the resulting cube = R cm, Then, volume of the resulting cube, R 3 = 216 ⇒ R = 6 cm, , Example 2. The volume of a right circular cylinder with, 1, its height equal to the radius is 25 cm 3 . Find the, 7, 22, height of the cylinder. (Use π = ), 7, , Sol. Let h and r be the height and radius of right circular, cylinder, respectively., Given, height of cylinder = Radius of cylinder, i.e., h=r, Q Volume of cylinder = πr 2h, 1 22, 1, ∴, 25 =, × h 2 × h [Q h = r and V = 25 , given], 7 7, 7, 176 22, =, × h3, ⇒, 7, 7, ⇒, h3 = 8, ⇒, , 13 cm, , volume of third cube = ( 5)3 = 125 cm 3, , and, ∴, , 66 cm, , 3, , 3, , h =2 ⇒ h =2, , [taking cube root], , Hence, height of cylinder is 2 cm., , Example 3. An iron pole consists of a cylinder of height, 240 cm and base diameter 26 cm, which is, surmounted by another cylinder of height 66 cm, and radius 10 cm. Find the mass of the pole given, that 1 cm3 of iron has approximately 8 g mass., [take, π = 3.14], Sol. Here, solid iron pole is a combination of two cylinders., For first cylinder,, Height = 240 cm, Base diameter = 26 cm, 26, cm = 13 cm, ∴ Base radius =, 2, For second cylinder,, Height = 66 cm, Radius = 10 cm, , ∴ Total volume of iron pole = Volume of first cylinder, + Volume of second cylinder, = π (13)2 × 240 + π (10)2 × 66, = π [169 × 240 + 100 × 66], = 3.14 [ 40560 + 6600], = 3.14 × 47160, = 148082.4 cm 3, Hence, total mass of the iron pole, = 148082.4 × 8 g = 1184659.2 g, [given, 1 cm 3 ≈ 8 g], 1184659.2, =, kg, 1000, 1, ⎡, ⎤, Q 1g =, kg, = 1184.66 kg, ⎢⎣, 1000 ⎥⎦, , Example 4. A spherical metal ball of radius 8 cm is, melted to make 8 smaller identical balls. The radius, of each new ball is ……… cm., Sol. Let radius of larger sphere be R = 8 cm, and radius of smaller sphere be r cm, Let number of smaller sphere be n = 8, According to the given condition,, Volume of larger sphere = n × volume of smaller sphere, 4, 4, ∴, πR 3 = n × πr 3, 3, 3, ∴, ( 8) 3 = 8 × r 3, ⇒, , r 3 = 82 ⇒ r 3 = 64 = ( 4)3, , ⇒, r = 4 cm, Hence, radius of new ball is 4 cm., , [taking cube root], , Example 5. A solid is in the shape of a cone mounted, on a hemisphere of same base radius. If the curved, surface areas of the hemispherical part and the, conical part are equal, then find the ratio of the, radius and the height of the conical part., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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121, , CBSE Term II Mathematics X (Standard), , Sol. Let radius, height and slant height of a cone are r, h and l,, respectively. Then, radius of hemisphere will be r., l, , h, r, , Example 7. In figure, a tent is in the shape of a cylinder, surmounted by a conical top. The cylindrical part is, 2.1 m high and conical part has slant height, 2.8 m. Both the parts have same radius 2 m., Find the area of the canvas used to make the tent., 22 ⎤, ⎡, ⎢⎣Use π = 7 ⎥⎦, , Now, curved surface area of cone C1 = πrl, and curved surface area of hemisphere, C 2 = 2 πr 2, , 2.8 m, , According to the question,, , C1 = C 2, πrl = 2 πr 2 ⇒ l = 2 r, , ∴, , Also, l = r 2 + h 2, (2 r ) = r 2 + h 2, , ⇒, , [Q from Eq. (i)], , (2 r ) 2 = ( r 2 + h 2 ) 2 ⇒ 4 r 2 = r 2 + h 2, 2, , 2, , 3r = h ⇒ ( 3r ) = h, , ⇒, , 2m, , Sol. Given radius of conical and cylindrical part is r = 2 m ., , On squaring both sides, we get, 2, , 2.1 m, , …(i), , 2, , Slant height of cone is l = 2. 8 m, And height of cylinder is h = 2.1 m, , Taking square root both sides, we get, r, 1, =, 3r = h ⇒, h, 3, Hence, the ratio of the radius and height of the conical part, is 1 : 3., , 2.8 m, , 2.1 m, , Example 6. A solid is in the shape of a hemisphere, surmounted by a cone. If the radius of hemisphere, and base radius of cone is 7 cm and height of cone, is 3.5 cm, find the volume of the solid., 22 ⎤, ⎡, ⎢⎣ take, π = 7 ⎥⎦, Sol. Given, radius of hemisphere and cone is r = 7 cm., And height of cone ( h ) = 3.5 cm, A, 3.5 cm, B, , 7 cm, , C, , 7 cm, , 1 22, 1 2, × (7 )2 × 3.5, πr h = ×, 3 7, 3, = 179.67 cm 3, 2 22, 2, and Volume of hemisphere , V2 = πr 3 = ×, × (7 ) 3, 3 7, 3, = 718.67 cm 3, , Now, volume of cone V1 =, , ∴ The volume of solid figure, = Volume of cone + Volume of hemisphere, = V1 + V2, = 179.67 + 718. 67 = 898.34 cm 3, Hence, volume of solid shape is 898.34 cm 3., , 2m, , ∴The area of the canvas used, to make the tent = curve, surface area of cone + curve surface area of cylinder, = πrl + 2 πrh, 22, 22, =, × 2 × 2. 8 + 2 ×, × 2 × 2.1, 7, 7, = 17. 6 + 26. 4 = 44 cm 2, Hence, the area of the canvas used to make the tent is, 44 cm 2., , Example 8. From a solid right circular cylinder of, height 14 cm and base radius 6 cm, a right circular, cone of same height and same base radius is, removed. Find the volume of the remaining solid., Sol. Given radius and height of cylinder are, r = 6 cm and h = 14 cm, , 14 cm, , 6 cm, , Also, radius and height of cone will be, r1 = 6 cm and h1 = 14 cm, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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122, , CBSE Term II Mathematics X (Standard), , Now, volume of cylinder,, V1 = πr12h, 22, =, × ( 6)2 × 14 = 1584 cm 3, 7, 1, Volume of cone, V2 = πr12h1, 3, 1 22, = ×, × ( 6)2 × 14 = 528 cm 3, 3 7, ∴Volume of remaining solid, = Volume of cylinder − Volume of cone, = V1 − V2, = 1584 − 528 = 1056 cm 3, , Example 10. Two cones with same base radius 8 cm and, height 15 cm are joined together along their bases., Find the surface area of the shape so formed., Sol. If two cones with same base and height are joined together, along their bases, then the shape so formed is look like as, figure shown., 8 cm, 16 cm, , 15 cm, 8 cm, , Hence, volume of the remaining solid is 1056 cm 3., , Example 9. An ice-cream cone full of ice-cream having, radius 5 cm and height 10 cm as shown in figure, 5 cm, , 10 cm, , 30 cm, , Given that, radius of cone, r = 8 cm and height of cone,, h = 15 cm, So, surface area of the shape so formed, = Curved area of first cone, + Curved surface area of second cone, = 2 ⋅ Surface area of cone [since, both cones are identical], = 2 × πrl = 2 × π × r × r 2 + h 2, 2 × 22 × 8 ×, 22, × 8 × ( 8)2 + (15)2 =, 7, 7, 44 × 8 × 289 44 × 8 × 17, =, =, 7, 7, 5984, 2, =, = 854. 85 cm, 7, = 855 cm 2 (approx.), =2 ×, , Calculate the volume of ice-cream, provided that its, 1, part is left unfilled with ice-cream., 6, Sol. Given, ice-cream cone is the combination of a hemisphere, and a cone., Also ,, radius of hemisphere = 5 cm, 2, 2 22, Volume of hemisphere = πr 3 = ×, × ( 5) 3, ∴, 3, 3 7, 5500, =, = 261. 90 cm 3, 21, Now,, radius of the cone = 5 cm, and, height of the cone = 10 − 5 = 5 cm, 1, Volume of the cone = πr 2h, ∴, 3, 1 22, = ×, × ( 5) 2 × 5, 3 7, 2750, =, = 130. 95 cm 3, 21, Now, total volume of ice-cream cone, = 261.90 + 130.95 = 392.85 cm 3, 1, Since, part is left unfilled with ice-cream., 6, 1, ∴Required volume of ice-cream = 392.85 − 392.85 ×, 6, = 392.85 − 65.475, = 327.4 cm 3, , 64 + 225, , Hence, the surface area of shape so formed is 855 cm 2., , Example 11. The barrel of a fountain pen, cylindrical in, shape, is 7 cm long and 0.5 cm in diameter. A full, barrel of ink in the pen can be used for writing 275, words on an average. How many words would be, written using a bottle of ink containing one-fourth, [CBSE 2015, 14], of a litre?, Sol. Given, height of cylindrical pen = 7 cm, Diameter 0. 5, cm, Radius =, =, 2, 2, ∴Volume of barrel of a fountain pen = πr 2h, 2, , 22, 22 ⎛ 0. 5⎞, ×⎜, cm 3, ⎟ ×7=, ⎝ 2 ⎠, 16, 7, It is given that, a pen can write 275 words by using the ink, 22, cm 3., 16, Volume of ink = 275 words, ∴, 22, cm 3 = 275 words, ⇒, 16, 275 × 16 1, 1, × 1000 cm 3 =, × × 1000 = 50000, ⇒, 4, 22, 4, 1, [Q he will use L of ink to write words], 4, 1, Hence, the pen can write 50000 words by L of ink., 4, =, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , Example 12. 500 persons are taking a dip into a, cuboidal pond which is 80 m long and 50 m broad., What is the rise of water level in the pond, if the, average displacement of the water by a person is, 0.04 m 3 ?, Sol. Let the rise of water level in the pond be h m when 500, persons are taking a dip into a cuboidal pond., h, Water, , Given that,, Length of the cuboidal pond = 80 m, Breadth of the cuboidal pond = 50 m, Now, volume for the rise of water level in the pond, = Length × Breadth × Height, = 80 × 50 × h, = 4000 h m 3, and the average displacement of the water by a person, = 0. 04 m 3, So, the average displacement of the water by 500 persons, = 500 × 0. 04 m 3, Now, by given condition,, Volume for the rise of water level in the pond = Average, displacement of the water by 500 persons, 4000 h = 500 × 0. 04, ⇒, 500 × 0. 04, 20, 1, m, ∴, h=, =, =, 4000, 4000 200, = 0. 005 m, = 0. 005 × 100 cm, [Q 1 m = 100 cm ], = 0. 5 cm, Hence, the required rise of water level in the pond is 0.5 cm., , Example 13. A small terrace at a hockey ground, comprises of 10 steps each of which 20 m long and, 1, built of solid concrete. Each step has a rise of m, 4, 1, and a tread of m. Calculate the total volume of, 2, concrete required to build the terrace., , 20, , m, , 1m, 2, , 1m 1, m 3m, 4, 4, 2, , 123, Sol. It is clear, from the figure, length = 20 m, 1, and width = m of each step., 2, 1, and height of Ist step which is in the bottom = m, 4, 1 1, ∴Height of second step = 2 × = m, 4 2, 1 3, Height of third step = 3 × = m, 4 4, M, M, 1 10, Height of tenth step = 10 × = m, 4 4, Total volume of the concrete used, 1 1, 1 10, 1 2, 1 3, = 20 × × + 20 × × + 20 × × + ... + 20 × ×, 2 4, 2, 4, 2 4, 2 4, [Q volume of cuboid = l × b × h], 1 1, = 20 × × [1 + 2 + 3 + ... + 10], 2 4, 1 1 10 × 11, n ( n + 1) ⎤, ⎡, = 20 × × ×, Q 1 + 2 + ... + n =, ⎢⎣, ⎥⎦, 2 4, 2, 2, 3, = 137.5 m, , Example 14. A wall 24 m long, 0.4 m thick and 6 m, high is constructed with the bricks each of, dimensions 25 cm × 16 cm × 10 cm. If the mortar, 1, th of the volume of the wall, then find, occupies, 10, the number of bricks used in constructing the wall., Sol. Given that, a wall is constructed with the help of bricks and, mortar., ∴Number of bricks, ⎛1, ⎞, (Volume of wall) − ⎜ th volume of wall ⎟, ⎝ 10, ⎠, ...(i), =, Volume of a brick, Also, given that, Length of a wall ( l ) = 24 m ,, Thickness of a wall ( b ) = 0. 4 m,, Height of a wall ( h ) = 6 m, So, volume of a wall constructed with the bricks = l × b × h, = 24 × 0. 4 × 6, 24 × 4 × 6 3, =, m, 10, 1 24 × 4 × 6, 1, Now,, th volume of a wall =, ×, 10, 10, 10, 24 × 4 × 6 3, m, =, 102, 25, and Length of a brick ( l1 ) = 25 cm =, m, 100, 16, Breadth of a brick ( b1 ) = 16 cm =, m, 100, 10, m, Height of a brick ( h1 ) = 10 cm =, 100, So, volume of a brick = l1 × b1 × h1, 25, 16, 10 25 × 16 3, =, ×, ×, =, m, 100 100 100, 105, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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124, , CBSE Term II Mathematics X (Standard), , From Eq. (i),, ⎛ 24 × 4 × 6 24 × 4 × 6⎞, −, ⎜, ⎟, ⎝, 10, 100 ⎠, Number of bricks =, ⎛ 25 × 16⎞, ⎜, ⎟, ⎝ 105 ⎠, 24 × 4 × 6, 10 5, ×9×, 100, 25 × 16, 24 × 4 × 6 × 9 × 1000, =, 25 × 16, =, , Since, the water is flowing at the rate of 5 km/h., Therefore, length of the water flow in x h, = 5x km = 5000x m, [Q 1 km = 1000 m], We have, diameter of cylindrical pipe = 14 cm, 14, 7, m, ∴ Radius of cylindrical pipe, r =, = 7 cm =, 2, 100, Volume of the water flowing through the cylindrical, pipe in, 2, , = 24 × 6 × 9 × 10 = 12960, Hence, the required number of bricks used in constructing, the wall is 12960., , Example 15. Water is flowing at the rate of 5 km/h, through a pipe of diameter 14 cm into a rectangular, tank which is 50 m long and 44 m wide. Determine the, time in which the level of the water in the tank will rise, by 7 cm., 14 cm, , 7 cm, 44 m, , x h = πr 2h =, , 22 ⎛ 7 ⎞, ×⎜, ⎟ × 5000x, ⎝ 100⎠, 7, , = 77 x m 3, Also, volume of the water that falls into the tank in x h, = l×b×h, 7, = 50 × 44 ×, 100, = 154 m 3, ⎤, ⎡, 7, ⎢Q l = 50 m, b = 44 m and h = radius = 100 m ⎥, ⎦, ⎣, Q Volume of the water flowing through the cylindrical pipe, in x h = Volume of water that falls in the tank in x h, ⇒, 77 x = 154, ⇒, x =2, Hence, the level of water in the tank will rise by 7 cm in 2 h., , 50 m, , Sol. Suppose, the level of the water in the tank will rise by 7 cm, in x h., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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125, , CBSE Term II Mathematics X (Standard), , Chapter, Practice, PART 1, Objective Questions, ●, , Multiple Choice Questions, 1. Three cubes each of side 5 cm are joined end to, end, then the surface area of the resulting solid is, (a) 250 cm 2, (c) 350 cm 2, , (b) 180 cm 2, (d) None of these, , 2. A solid ball is exactly fitted inside the cubical box of, side a. The volume of the ball is, 1, (a) πa 3, 6, 1, (c) πa 3, 3, , 4, (b) πa 3, 3, (d) None of these, , 3. A cubical icecream brick of edge 22 cm is to be, distributed among some children by filling, icecream cones of radius 2 cm and height 7 cm upto, its brim. How many children will get icecream, cones?, (a) 163, (c) 363, , (b) 263, (d) 463, , 4. A right circular cylinder of radius r cm and height h, cm ( where, h > 2 r ) just encloses a sphere of, diameter, (a) r cm, (c) h cm, , 7. A cylindrical pencil sharpened at one edge is the, combination of, (a) a cone and a cylinder, (b) cube and a cylinder, (c) a hemisphere and a cylinder, (d) two cylinders, , 8. A surahi is the combination of, (a) a sphere and a cylinder, (b) a hemisphere and a cylinder, (c) two hemispheres, (d) a cylinder and a cone, , 9. Two cones have their heights in the ratio 1 : 3 and, radii in the ratio 3 : 1, then the ratio of their, volumes is, (a) 1 : 3, (c) 2 : 3, , (b) 3 : 1, (d) 3 : 2, , 10. The shape of a gilli, in the gilli-danda game (see, figure) is a combination of, , (a) two cylinders, (b) a cone and a cylinder, (c) two cones and a cylinder, (d) two cylinders and a cone, , 11. A plumbline (sahul) is the combination of, (see figure), , (b) 2r cm, (d) 2h cm, , 5. If two solid hemispheres of same base radius r are, joined together along their bases, then curved, surface area of this new solid is, (a) 4πr 2, (c) 3πr 2, , (b) 6πr 2, (d) 8πr 2, , 6. A solid cylinder of radius r and height h is placed, over other cylinder of same height and radius., The total surface area of the shape so formed is, (a) 4πr( h 2 + r 2 ), (b) 4πr[ h + r ], (c) 4π ( h 2 + r 2 ), (d) None of the above, , (a) a cone and a cylinder, (c) cube and a cylinder, , (b) a hemisphere and a cone, (d) sphere and cylinder, , 12. A solid cone of radius r and height h is placed over, a solid cylinder having same base radius and height, as that of a cone. The total surface area of the, combined solid is, (b) πr 2( l + 2 h ), , (a) πrl + 2 πrh, 2, , 2, , (c) πr [ r + h + 2 h + r ], , (d) None of these, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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126, , CBSE Term II Mathematics X (Standard), , 13. The capacity of a cylindrical vessel with a, hemispherical portion raised upward at the bottom, πr 2, as shown in the figure is, [ 3h − 2 r ]., 3, , 20. A wooden article was made by scooping out a, hemisphere from each end of a solid cylinder, as, shown in figure. If the height of the cylinder is, 10 cm and its base is of radius 3.5 cm. Find the total, surface area of the article., , r cm, , h cm, , 1, (a) πr 2[2 h − 3r ], 3, 1 2, (c) πr [ 3h − 2 r ], 3, , 2, (b) πr 2[ 3h − 2 r ], 3, , [CBSE 2018], (a) 374 cm 2, (c) 475 cm 2, , (d) None of these, , 21. A heap of rice is in the form of a cone of base, diameter 24 m and height 3.5 m. Find the volume of, the rice. How much canvas as cloth is required to, [CBSE 2018], just cover the heap?, , 14. The diameter of a sphere is 6 cm. It is melted and, drawn into a wire of diameter 2 mm., The length of the wire is, (a) 12 m, (c) 36 m, , (b) 18 m, (d) 66 m, , (a) 105.5 m 2, (c) 173.5 m 2, , 15. During conversion of a solid from one shape to, another, the volume of the new shape will, (a) increase, (c) remain unaltered, , 16. From a solid circular cylinder with height 10 cm, and radius of the base 6 cm, a right circular cone of, the same height and same base is removed, then the, volume of remaining solid is, (b) 330 π cm 3, , 3, , 3, , (c) 240 π cm, , (d) 440 π cm, , (a) 2.5 m, (c) 3 m, , (a) 11100, ●, , 17. A 20 m deep well, with diameter 7 m is dug and the, earth from digging is evently spread out to form a, platform 22 m by 14 m. The height of the platform, is, (b) 3.5 m, (d) 2 m, , (b) 471.42 m 2, (d) None of these, , 22. A mason constructs a wall of dimensions, 270 cm × 300 cm × 350 cm with the bricks each of, size 22.5 cm × 11.25 cm × 8.75 cm and it is assumed, 1, that space is covered by the mortar. Then, the, 8, number of bricks used to construct the wall is, , (b) decrease, (d) be doubled, , (a) 280 π cm 3, , (b) 370 cm 2, (d) None of these, , (b) 11200, , (c) 11000, , (d) 11300, , Case Based MCQs, 23. To make the learning process more interesting, creative and innovative Shavya’s class teacher, brings clay in the classroom, to teach the topic., Surface Areas and Volumes. With clay, she forms a, cylinder of radius 4 cm and height 18 cm. Then, she, moulds the cylinder into a sphere and ask some, question to students., , 18. If the radius of the base of a right circular cylinder, is halved, keeping the height same, then find the, ratio of the volume of the cylinder thus obtained to, [CBSE 2009], the volume of original cylinder., (a), , 1, 3, , (b), , 1, 4, , (c), , 1, 2, , (d), , 1, 5, , 19. Marbles of diameter 1.4 cm are dropped into a, cylindrical beaker of diameter 7 cm containing, some water. The water level rises by 5.6 cm. When, marble dropped into the beaker, then the number of, marble is, (a) 150, (c) 175, , (b) 160, (d) 235, , (i) The radius of the sphere so formed is, (a) 4 cm, (c) 7 cm, , (b) 6 cm, (d) 8 cm, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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127, , CBSE Term II Mathematics X (Standard), , (ii) The volume of the sphere so formed is, (a) 905.14 cm3, (c) 1296.5 cm3, , (b) 903.27 cm3, (d) 1156.63 cm3, , (iii) Find the ratio of the volume of sphere to the, volume of cylinder., (a) 2 : 1, (c) 1 : 1, , (b) 1 : 2, (d) 3 : 1, , (iv) Total surface area of the cylinder is, (a) 553.14 cm2, (c) 625 cm2, , (b) 751.52 cm2, (d) 785.38 cm2, , (v) During the conversion of a solid from one shape to, another the volume of new shape will, (a) be increase, (c) remain unaltered, , 25. The Great Stupa at Sanchi is one of the oldest stone, structures in India, and an important monument of, Indian Architecture. It was originally, commissioned by the emperor Ashoka in the 3rd, century BCE. Its nucleus was a simple, hemispherical brick structure built over the relics, of the Buddha. It is a perfect example of, combination of solid figures. A big hemispherical, dome with a cuboidal structure mounted on it., 22, (take π = ), [CBSE Question Bank], 7, , (b) be decrease, (d) be double, , 24. Geeta and Meena have 10 and 6 CD respectively,, each of radius 4 cm and thickness 1 cm. They place, their CD one above the other to form solid, cylinders., Dome, Toranas, , Chattra, Harmika, , Balustrade, , Stairs, , Based on the above information, answer the, following questions., (i) Curved surface area of the cylinder made by Geeta, is, (a) 308.17 cm2, (c) 154 cm2, , (b) 132 cm2, (d) 251.42 cm2, , (ii) The ratio of curved surface area of the cylinder, made by Geeta and Meena is, (a) 3 : 5, (c) 5 : 3, , (b) 3 : 2, (d) 5 : 7, , (iii) The volume of the cylinder made by Meena is, (a) 301.44 cm3, (c) 132 cm3, , (b) 144 cm3, (d) 208.42 cm3, , (iv) The ratio of the volume of the cylinders made by, Geeta and Meena is, (a) 1 : 2, (c) 3 : 5, , (b) 2 : 5, (d) 5 : 3, , (v) When two CD Cassette are shifted from Geeta, cylinder to Meena’s cylinder, then, (a) Volume of two cylinder become equal, (b) Volume of Geeta’s cylinder > Volume of Meena’s, cylinder, (c) Volume of Meena’s cylinder > Volume of Geeta’s, cylinder, (d) None of the above, , (i) Calculate the volume of the hemispherical dome if, the height of the dome is 21 m., (a) 19404 cu m, (c) 15000 cu m, , (b) 2000 cu m, (d) 19000 cu m, , (ii) The formula to find the volume of sphere is, 2, (a) π r 3, 3, (c) 4π r 2, , 4, (b) πr 3, 3, (d) 2 πr 2, , (iii) The cloth require to cover the hemispherical dome, if the radius of its base is 14m is, (a) 1222 sq m, (c) 1200 sq m, , (b) 1232 sq m, (d) 1400 sq m, , (iv) The total surface area of the combined figure, i.e. hemispherical dome with radius 14 m and, cuboidal shaped top with dimensions 8 m × 6 m, × 4 m is, (a)1200 sq m, (c) 1392 sq m, , (b) 1232 sq m, (d) 1932 sq m, , (v) The volume of the cuboidal shaped top is with, dimensions mentioned in question (iv)., (a) 182.45 m 3, (c) 292 m 3, , (b) 282.45 m 3, (d) 192 m 3, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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128, , CBSE Term II Mathematics X (Standard), , PART 2, Subjective Questions, ●, , Short Answer Type Questions, 1. Two identical cubes each of volume 64 cm 3 are, joined together end to end. What is the surface area, of the resulting cuboid?, 2. How many shots each having diameter 3 cm can be, made from a cuboidal lead solid of dimensions, 9 cm × 11 cm × 12 cm?, 3. 16 glass spheres each of radius 2 cm are packed into, a cuboidal box of internal dimensions 16 cm × 8 cm, × 8 cm and then the box is filled with water. Find, the volume of water filled in the box., 4. If a solid piece of iron in the form of a cuboid of, dimensions 49 cm × 33 cm × 24 cm, is moulded to, form a solid sphere. Then, find radius of the sphere., 5. If volumes of two spheres are in the ratio 64 : 27,, then find the ratio of their surface areas., 6. The decorative block shown in the following figure, is made of two solids, a cube and a hemisphere., The base of the block is a cube with edge 6 cm and, the hemisphere fixed on the top has a diameter of, 2.1 cm, then find the total surface area of the, block and find the total area to be painted., 22 ⎤, ⎡, ⎢⎣ take, π = 7 ⎥⎦, 2.1 cm, , 6 cm, 6 cm, , 6 cm, , 7. From a solid cube of side 7 cm, a conical cavity of, height 7 cm and radius 3 cm is hollowed out. Find, the volume of the remaining solid., 8. A hemispherical bowl of internal radius 9 cm is full, of liquid. The liquid is to be filled into cylindrical, shaped bottles each of radius 1.5 cm and height, 4 cm. How many bottles are needed to empty the, bowl?, , 9. A building is in the form of a cylinder surmounted, by a hemispherical dome (see the figure). The base, 2, diameter of the dome is equal to of the total, 3, height of the building. Find the height of the, 1, building, if it contains 67 m 3 of air., 21, r, , r, H, , h, , 10. Twelve solid spheres of the same size are made by, melting a solid metallic cylinder of base diameter, 2 cm and height 16 cm. Find the diameter of each, sphere., 11. A solid is composed of a cylinder with, hemispherical ends. If the whole length of the solid, is 104 cm and the radius of each hemispherical end, is 7 cm, then find the cost of polishing its surface at, the rate of ` 2 per dm 2 ., 22 ⎤, ⎡, ⎢⎣ take, π = 7 ⎥⎦, 12. A solid metallic hemisphere of radius 8 cm is, melted and recasted into a right circular cone of, base radius 6 cm. Determine the height of the, cone., 13. A rectangular water tank of base 11 m × 6 m, contains water upto a height of 5 m. If the water in, the tank is transferred to a cylindrical tank of, radius 3.5 m, find the height of the water level in, the tank., 14. A copper rod of diameter 1 cm and length 8 cm is, drawn into a wire of length 8 m of uniform, thickness. Find the thickness of the wire., 15. The rain water from a roof of dimensions 22 m, × 20 m drains into a cylindrical vessel having, diameter of base 2 m and height 3.5 m. If the rain, water collected from the roof just fill the, cylindrical vessel, then find the rainfall (in cm)., 16. A cylindrical bucket of height 32 cm and base, radius 18 cm is filled with sand. This bucket is, emptied on the ground and a conical heap of sand, is formed. If the height of the conical heap is 24, cm, find the radius and slant height of the heap., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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129, , CBSE Term II Mathematics X (Standard), , 17. The barrel of a fountain pen, cylindrical in shape, is, 7 cm long and 5 mm in diameter. A full barrel of ink, in the pin is used up on writing 3300 words on an, average. How many words can be written in a bottle, of ink containing one-fifth of a litre?, 18. Water flows at the rate of 10 m min −1 through a, cylindrical pipe 5 mm in diameter. How long would it, take to fill a conical vessel whose diameter at the, base is 40 cm and depth 24 cm?, 19. Water flows through a cylindrical pipe, whose inner, radius is 1 cm, at the rate of 80 cms −1 in an empty, cylindrical tank, the radius of whose base is 40 cm., What is the rise of water level in tank in half an hour?, 20. A factory manufactures 120000 pencils daily. The, pencils are cylindrical in shape each of length 25 cm, and circumference of base as 1.5 cm. Determine the, cost of colouring the curved surfaces of the pencils, manufactured in one day at ` 0.05 per dm 2 ., 21. A well of diameter 10 m is dug 14 m deep. The Earth, taken out of it is spread evenly all around to a width, of 5 m to form an embankment. Find the height of, embankment., 22. Marbles of diameter 1.4 cm are dropped into a, cylindrical beaker of diameter 7 cm containing some, water. Find the number of marbles that should be, dropped into the beaker, so that the water level rises, by 5.6 cm., , 4 cm, 24 cm, , 4 cm, , 6 cm, , Long Answer Type Questions, 23. If a hollow cube of internal edge 22 cm is filled with, spherical marbles of diameter 0.5 cm and it is, 1, assumed that space of the cube remains unfilled., 8, Then, the number of marbles that the cube can, accomodate is, 24. A solid iron cuboidal block of dimensions, 4.4 m × 2.6 m × 1 m is recast into a hollow cylindrical, pipe of internal radius 30 cm and thickness 5 cm., Find the length of the pipe., 25. A building is in the form of a cylinder surmounted by, 19 3, m, a hemispherical vaulted dome and contains 41, 21, of air. If the internal diameter of dome is equal to its, total height above the floor, find the height of the, building?, , Base of cylinder, Base of cone, , The base of the conical portion has a diameter of, 6 cm, while the base diameter of the cylindrical, portion is 4 cm. If the conical portion is to be, painted orange and the cylindrical portion yellow,, then find the area of the rocket painted with each, of these colours. [take, π = 3.14], , 30. Two solid cones A and B are placed in a cylindrical, tube as shown in the figure. The ratio of their, capacities is 2 : 1. Find the heights and capacities, of cones. Also, find the volume of the remaining, portion of the cylinder., 21 cm, , 6 cm, , ●, , 26. A medicine-capsule is in the shape of a cylinder of, diameter 0.5 cm with two hemispheres stuck to, each of its ends. The length of entire capsule is, 2 cm. The capacity of the capsule is, 27. A rocket is in the form of a right circular cylinder, closed at the lower end and surmounted by a cone, with the same radius as that of the cylinder. The, diameter and height of the cylinder are 6 cm and, 12 cm, respectively. If the slant height of the, conical portion is 5 cm, then find the total surface, area and volume of the rocket. [use π = 3.14], 28. A solid toy is in the form of a hemisphere, surmounted by a right circular cone. The height, of the cone is 3 cm and the diameter of the base is, 4 cm. Determine the volume of the solid toy. If a, right circular cylinder circumscribes the toy, then, find the difference of the volumes of the cylinder, and the toy., [take, π = 3 . 14], 29. A wooden toy rocket is in the shape of a cone, mounted on a cylinder, as shown in figure. The, height of the entire rocket is 24 cm, while the, height of the conical part is 4 cm., , A, , B, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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130, 31. How many spherical lead shots of diameter 4 cm, can be made out of a solid cube of lead whose edge, measures 44 cm., 32. A metallic spherical shell of internal and external, diameters 4 cm and 8 cm, respectively is melted, and recast into the form of a cone of base diameter, 8 cm. Find the height of the cone., 33. How many spherical lead shots each of diameter, 4.2 cm can be obtained from a solid rectangular, lead piece with dimensions 66 cm, 42 cm and, 21 cm?, 34. Find the number of metallic circular disc with, 1.5 cm base diameter and of height 0.2 cm to be, melted to form a right circular cylinder of height, 10 cm and diameter 4.5 cm., 35. A heap of rice is in the form of a cone of diameter, 9 m and height 3.5 m. Find the volume of the rice., How much canvas cloth is required to just cover, heap?, 36. How many cubic centimetres of iron is required to, construct an open box whose external dimensions, are 36 cm, 25 cm and 16.5 cm provided the, thickness of the iron is 1.5 cm. If one cubic, centimetre of iron weights 7.5 g, then find the, weight of the box., 37. Water is flowing at the rate of 15 kmh −1 through a, pipe of diameter 14 cm into a cuboidal pond which, is 50 m long and 44 m wide. In what time will the, level of water in pond rise by 21 cm?, ●, , Case Base Questions, 38. Mathematics teacher of a school took her 10th, standard students to show Ram Mandir. It was a, part of their Educational trip. The teacher had, interest in history as well. She narrated the facts of, Ram Mandir to students., Ram mandir is a Hindu temple that is being built in, Ayodhya, which is in Uttar Pradesh. The temple, construction is being supervised by the Shri Ram, Janmabhoomi Teerth Kshetra., Then the teacher said in this monuments one can, find combination of solid figures. She pointed that, there are cubical bases and in centre cylinder with, the cone shape structure on the top is constructed., , CBSE Term II Mathematics X (Standard), , (i) Ram Mandir is constructed in the form of the, cubical base of 30 cm × 20 cm × 10 cm, then find, the area covered., (ii) If the radius of the cylinder is 7 cm and Height of, the cylinder is 60 cm and the radius of the cone is, similar to that of cylinder and Height of the cone is, 24 cm, then the ratio of curved surface area of, cylinder to curved surface area of the cone., (iii) Given structure in based on the concept of, (a) Area and perimeter, (b) Surface area and volume, (c) Both (a) and (b), (d) None of the above, , 39. Adventure camps are the perfect place for the, children to practice decision making for themselves, without parents and teachers guiding their every, move. Some students of a school reached for, adventure at Sakleshpur. At the camp, the waiters, served some students with a welcome drink in a, cylindrical glass and some students in a, hemispherical cup whose dimensions are shown, below., , d=7 cm, h=10.5 cm, , d=7 cm, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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131, , CBSE Term II Mathematics X (Standard), , After that they went for a jungle trek. The jungle trek, was enjoyable but tiring. As dusk fell, it was time to take, shelter. Each group of four students was given a canvas, of area 551m2. Each group had to make a conical tent to, accommodate all the four students. Assuming that all the, stitching and wasting incurred while cutting, would, amount to 1 m2, the students put the tents. The radius of, [CBSE Question Bank], the tent is 7 m., , Area = 551 m2, r=7m, , (i) Find the volume of cylindrical cup., (ii) Find the volume of hemispherical cup., (iii) Find the height of the conical tent prepared to, accommodate four students., , 40. On a Sunday, your Parents took you to a fair., You could see lot of toys displayed, and you, wanted them to buy a RUBIK’s cube and, strawberry ice-cream for you., Observe the figures and answer the questions., [CBSE Question Bank], , (i) Find the length of the diagonal if each edge, measures 6cm., (ii) Find volume of the solid figure if the length, of the edge is 7cm., (iii) What is the curved surface area of, hemisphere (ice-cream) if the base radius is 7, cm?, (iv) Find the slant height of a cone if the radius is, 7 cm and the height is 24 cm, (v) Find the total surface area of cone with, hemispherical ice cream., , SOLUTIONS, Objective Questions, 1. (c) Here, on joining three cubes, we get a cuboid whose, length, l = 5 + 5 + 5 = 15 cm, breadth, b = 5 cm and, height, h = 5 cm, 5 cm, , 5 cm, , 5 cm, 5 cm, , I, , II, , III, 5 cm, , ∴ Required surface area of the resulting solid, = Surface area of new cuboid, = 2 ( lb + bh + hl ) = 2 (15 × 5 + 5 × 5 + 5 × 15), = 2 (75 + 25 + 75) = 2 (175) = 350 cm 2, 2. (a) Because solid ball is exactly fitted inside the cubical box of side, a. So, a is the diameter for the solid ball., a, Radius of the ball =, ∴, 2, 3, , So,, , volume of the ball =, , 4 ⎛ a⎞, 1, π ⎜ ⎟ = πa 3, 3 ⎝2⎠, 6, , 3. (c) Given, volume of brick = 22 3 cm 3, 1, ∴Volume of 1 cone = πr 2h, 3, 1 22, 22 × 4, = ×, ×2 ×2 ×7=, 3 7, 3, Let number of cones = n, 4, Then, n × 22 × = 22 × 22 × 22, 3, 22 × 22 × 3, n=, ⇒, 4, ∴, , n = 121 × 3 = 363, , 4. (b) Because the sphere encloses in the cylinder,, therefore the diameter of sphere is equal to diameter, of cylinder which is 2r cm., 5. (a) Because curved surface area of a hemisphere is, 2 πr 2 and here, we join two solid hemispheres along, their bases of radius r, from which we get a solid, sphere., Hence, the curved surface area of new solid, = 2 π r 2 + 2 π r 2 = 4π r 2, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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132, , CBSE Term II Mathematics X (Standard), , 6. (d) Since, the total surface area of cylinder of radius r and, height h = 2 πrh + 2 πr 2, When one cylinder is placed over the other cylinder of same, height and radius,, then height of the new cylinder = 2 h, and radius of the new cylinder = r, ∴Total surface area of the new cylinder = 2 πr(2 h ) + 2 πr 2, = 4πrh + 2 πr 2, = 2 πr(2 h + r ), 7. (a) Because the shape of sharpened pencil is, , =, , +, , = Cylinder + Cone, , 8. (a) Because the shape of surahi is, , =, , +, , = Sphere + Cylinder, , 9. (b) Let the radii of two cones are r1 , r2 and their heights are, h1 and h 2., r, 3, h, 1, Given, 1 = and 1 =, r2 1, h2 3, 1 2, 2, πr h, V1 3 1 1 ⎛ r1 ⎞ ⎛ h1 ⎞, Now, the ratio of their volumes,, =⎜ ⎟ ⎜ ⎟, =, ⎝ r2 ⎠ ⎝ h 2 ⎠, V2 1 πr 2h, 2 2, 3, 2, , 1 3, ⎛ 3⎞, ⎛ 1⎞, = ⎜ ⎟ ×⎜ ⎟ = 9 × = = 3:1, ⎝ 1⎠, ⎝ 3⎠, 3 1, Hence, the ratio of their volumes is 3 : 1., 10. (c), , =, , +, , +, , = Cone + Cylinder + Cone, = Two cones and a cylinder, 11. (b), , +, , =, , = Hemisphere + Cone, , 12. (c) We know that, total surface area of a cone of radius, r, and height, h = Curved surface Area + area of base, = πrl + πr 2, where,, , l = h 2 + r2, , and total surface area of a cylinder of base radius, r and, height, h, = Curved surface area + Area of both base, = 2 πrh + 2 πr 2, Here, when we placed a cone over a cylinder, then one base, is common for both., So, total surface area of the combined solid, = πrl + 2 πrh + πr 2, = πr [ l + 2 h + r ], = πr ⎡ r 2 + h 2 + 2 h + r ⎤, ⎣, ⎦, 13. (c) We know that, capacity of cylindrical vessel = πr 2h cm 3, 2, and capacity of hemisphere = πr 3 cm, 3, From the figure, capacity of the cylindrical vessel, 1, 2, = πr 2h − πr 3 = πr 2 [ 3h − 2 r ], 3, 3, 14. (c) We have, diameter of metallic sphere = 6 cm, ∴Radius of metallic sphere, r1 = 3 cm, Also, diameter of cross-section of cylindrical wire = 0.2 cm, ∴Radius of cross-sections of cylindrical wire, r2 = 0.1 cm, Let the length of the wire be h cm., Since, metallic sphere is converted into a cylindrical shaped, wire of length h cm., ∴Volume of the metal used in wire = Volume of the sphere, 4, πr22h = πr13, ⇒, 3, 2, , 4, ⎛ 1⎞, π × ⎜ ⎟ × h = × π × 27, ⎝ 10⎠, 3, 1, π×, × h = 36π, ⇒, 100, 36π × 100, h=, ⇒, π, = 3600 cm = 36 m, [Q 1m = 100 cm ], 15. (c) During conversion of a solid from one shape to another,, the volume of the new shape will remain unaltered., 16. (c) Volume of the remaining solid, = Volume of the cylinder − Volume of the cone, 1, ⎫, ⎧, = ⎨π × 62 × 10 − × π × 62 × 10⎬, 3, ⎭, ⎩, = ( 360π − 120π ) = 240π cm 3, 7, 17. (a)Q Radius of the well = m = 3. 5 m, 2, 22, × ( 3. 5)2 × 20, ∴Volume of the earth dug out =, 7, 22, =, × 3. 5 × 3. 5 × 20, 7, = 770 m 3, ⇒, , Area of platform = (22 × 14) m 2 = 308 m 2, 770, Height =, = 2.5 m, ∴, 308, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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133, , CBSE Term II Mathematics X (Standard), , 18. (b) Let the radius of original right circular cylinder be ‘r’ cm., and height be ‘h’ cm. Then,, Volume of original cylinder V1 = ( πr 2h ) cm 3, Volume of circular cylinder, when radius is halved, 2, πr 2h, ⎛ r⎞, cm 3, V2 = π ⎜ ⎟ h cm 3 =, ⎝ 2⎠, 4, Thus,, , 1, 1, V2 πr 2h, =, ×, = ., 4, V1, π r 2h 4, , 4, 19. (a) So, volume of one spherial marble = π (0.7)3, 3, 4, ⎡, ⎤, Q volume of sphere = πr 3, ⎢⎣, ⎥⎦, 3, 1.372, 3, =, π cm, 3, ∴ Volume of the raised water in beaker = π (3.5)2 × 5.6, [Q volume of cylinder = πr 2h ], = 68.6 π cm 3, Now, required number of marbles, Volume of the raised water in beaker, =, Volume of one spherical marble, 68.6 π, =, × 3 = 150 marbles, 1.372 π, 20. (a) Given, wooden article is a combination of a cylinder and, two hemispheres., 3.5 cm, , 1 2, πr h, 3, d 24, ⎡, ⎤, Qr = =, = 12 m, ⎢⎣, ⎥⎦, 2 2, , Now, volume of rice = volume of cone =, 1 22, ×, × (12 )2 × 3. 5, 3 7, 22 × 144 × 3. 5, =, 21, 11088, =, = 528 m3, 21, =, , Now, slant height l = h 2 + r 2, = ( 3. 5)2 + (12 )2 = 12 . 25 + 144, = 156.25 = 12 . 5 m, ∴The canvas required to cover the heap = πrl, 22, 3300, =, × 12 × 12 . 5 =, = 471. 42 m2, 7, 7, 22. (b) Volume of the wall = 270 × 300 × 350 = 28350000 cm 3, [Q volume of cuboid = length × breadth × height], 1, Since, space of wall is covered by mortar., 8, So, remaining space of wall = Volume of wall, − Volume of mortar, 1, = 28350000 − 28350000 ×, 8, = 28350000 − 3543750, = 24806250 cm 3, Now, volume of one brick = 22 . 5 × 11.25 × 8.75, = 2214. 844 cm 3, , 10 cm, , Here, height of the cylinder, h = 10 cm, Q Radius of base of the cylinder, = Radius of hemisphere, r = 3.5 cm, Now, required TSA of the wooden article, = 2 × CSA of one hemisphere + CSA of cylinder, = 2 × (2 π r 2 ) + 2 π r h, = 2 π r (2 r + h ), 22, =2 ×, × 3.5 × (2 × 3.5 + 10), 7, 22, =, × 7 × ( 7 + 10) = 22 × 17 = 374 cm 2, 7, 21. (b) Given diameter d = 24 m, A, , 3.5m, , B, 24 m, , C, , [Q volume of cuboid = length × breadth × height], 24806250, ∴ Required number of bricks =, = 11200 (approx.), 2214. 844, Hence, the number of bricks used to construct the wall is, 11200., 23. (i) (b) Since, volume of sphere = Volume of cylinder, 4, ⇒ πR 3 = πr 2h , where R , r are the radii of sphere and, 3, cylinder, respectively., 3, ⇒ R 3 = r2 × h ×, 4, 4 × 4 × 18 × 3, 3, ⇒ R =, = 8 × 27 ⇒ R 3 = (2 × 3)3, 4, ∴, R = 6 cm, 4, 4 22, (ii) (a) Volume of sphere = πR 3 = ×, ×6×6×6, 3, 3 7, = 905.14 cm3, (iii) (c) Since the volume of sphere is equal to volume of, cylinder, then the ratio of volume of the sphere to the, volume of cylinder = 1 : 1, (iv) (a) Total surface area of cylinder = 2πr( r + h ), 22, 22, =2 ×, × 4( 4 + 18) = 2 ×, × 4 × 22, 7, 7, 2, = 553.14 cm, (v) (c) During the conversion of a solid from one shape to, another the volume of new shape will remain unaltered., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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134, , CBSE Term II Mathematics X (Standard), , 24. We have radius of each CD cassette = 4 cm, and thickness of each cassette = 1 cm, So, height of cylindrical made by Geeta, h1 = 10 × 1 = 10 cm, and height of cylindrical made by Meena, h 2 = 6 × 1 = 6 cm, (i) (d) Curved surface area of cylinder made by Geeta, 22, = 2 πrh = 2 ×, × 4 × 10 = 251. 42 cm2, 7, ⎡Curved surface area of ⎤, ⎢cylinder made by Geeta ⎥, ⎦, (ii) (c)∴Required ratio = ⎣, ⎡Curved surface area of ⎤, ⎢cylinder made by Meena ⎥, ⎦, ⎣, 2 πrh1 h1 10, =, =, =, = 5: 3, 2 πrh 2 h 2, 6, (iii) (a) Volume of cylinder made by Meena, 22, = πr 2h 2 =, ×4×4×6, 7, = 301. 44 cm3, (iv) (d) Required ratio, Volume of the cylinder made by Geeta, =, Volume of the cylinder made by Meena, 2, , =, , πr h1 h1, =, = 5: 3, πr 2h 2 h 2, , (v) (a) When two CD Cassette are shifted from Geeta’s, cylinder to Meena’s cylinder, then length of both, cylinders become equal., So, volume of both cylinders become equal., 25. (i) (a) As, we know that hemisphere is a type of solid in, which radius is the height. So, radius = 21 m, 2, ∴Required volume = πr 3, 3, 2 22, = ×, × 21 × 21 × 21, 3 7, = 19404 cu m, 4, (ii) (b) Volume of sphere = πr 3, 3, (iii) (b) Given, radius ( r ) = 14 m, ∴Curved surface area of hemisphere dome = 2 πr 2, 22, =2×, × 14 × 14, 7, = 1232 sq m, (iv) (c) Here, radius of hemispherical dome ( r ) = 14 m, Surface area of dome = 2 πr 2, 22, =2×, × 14 × 14 = 1232 m 2, 7, and CSA of cuboidal shaped top = 2 × h( l + b ) + lb, = 2 × 4( 8 + 6) + 8 × 6, = 8(14) + 48, = 112 + 48 = 160 m 2, ∴Total surface area = 1232 + 160 = 1392 m 2, (v) (d) Volume of cuboidal shape = lbh, = 8 × 6 × 4 = 192 m 3, , Subjective Questions, 1. Let the length of a side of a cube = a cm, a, a, , a, , 2a, , Given, volume of the cube, a 3 = 64 cm 3 ⇒ a = 4 cm, On joining two cubes, we get a cuboid whose, length, l = 2a cm, breadth, b = a cm, and, height, h = a cm, Now, surface area of the resulting cuboid, = 2 ( lb + bh + hl ), = 2 (2 a ⋅ a + a ⋅ a + a ⋅ 2 a ), = 2 (2 a 2 + a 2 + 2 a 2 ) = 2 ( 5 a 2 ), = 10 a 2 = 10 ( 4)2 = 160 cm 2, 2. Given, dimensions of cuboidal = 9 cm × 11 cm × 12 cm, ∴Volume of cuboidal = 9 × 11 × 12 = 1188 cm 3, and diameter of shot = 3 cm, 3, = 1. 5 cm, 2, 4, 4 22, Volume of shot = πr 3 = ×, × (1.5)3, 3, 3 7, 297, =, = 14.143 cm 3, 21, 1188, = 84 (approx.), ∴Required number of shots =, 14.143, 3. Given, dimensions of the cuboidal = 16 cm × 8 cm × 8 cm, ∴ Radius of shot, r =, , ∴Volume of the cuboidal = 16 × 8 × 8 = 1024 cm 3, Also, given radius of one glass sphere = 2 cm, 4 3 4 22, πr = ×, × (2 ) 3, 3, 3, 7, 704, =, = 33.523 cm 3, 21, Now, volume of 16 glass spheres = 16 × 33.523 = 536.37 cm 3, ∴ Required volume of water = Volume of cuboidal, − Volume of 16 glass spheres, = 1024 − 536.37, = 487.6 cm 3, 4. Given, dimensions of the cuboid = 49 cm × 33 cm × 24 cm, ∴ Volume of the cuboid = 49 × 33 × 24 = 38808 cm 3, ∴Volume of one glass sphere =, , [Q volume of cuboid = length × breadth × height], Let the radius of the sphere is r, then, 4, Volume of the sphere = πr 3, 3, 4, [Q volume of the sphere = π × (radius)3], 3, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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135, , CBSE Term II Mathematics X (Standard), , According to the question,, Volume of the sphere = Volume of the cuboid, 4 3, πr = 38808, ⇒, 3, 22 3, 4×, r = 38808 × 3, ⇒, 7, 38808 × 3 × 7, ⇒, r3 =, = 441 × 21, 4 × 22, ⇒, , Clearly, the total area to be painted = Total surface area of, the decorative block − Area of base of cube, = 219. 465 − 62 = 219. 465 − 36 = 183. 465 cm 2, 7. Given that, side of a solid cube ( a ) = 7 cm, Height of conical cavity i.e. cone, h = 7 cm, , r 3 = 21 × 21 × 21, , ∴, r = 21 cm, Hence, the radius of the sphere is 21 cm., 5. Let the radii of the two spheres are r1 and r2, respectively., 4, …(i), ∴ Volume of the sphere of radius, r1 = V1 = πr13, 3, 4, [Q volume of sphere = π (radius)3], 3, 4, and volume of the sphere of radius, r2 = V2 =, π r23 …(ii), 3, 4 3, πr1, 64, Given, ratio of volumes = V1 : V2 = 64 : 27 ⇒ 3, =, 4 3 27, πr2, 3, [using Eqs. (i) and (ii)], r13 64, r1 4, …(iii), =, =, ⇒, ⇒, r2 3, r23 27, Now, ratio of surface area =, , 4πr12, 4πr22, , [Q surface area of a sphere = 4π (radius) 2], r2, = 12, r2, 2, , 2, ⎛r⎞, 16, ⎛ 4⎞, = ⎜ 1⎟ = ⎜ ⎟ =, ⎝ 3⎠, ⎝ r2 ⎠, 9, , [using Eq. (iii)], , Hence, the required ratio of their surface area is 16 : 9., 6. Here, the decorative block is a combination of a cube and a, hemisphere., For cubical portion,, Each edge = 6 cm, For hemispherical portion,, Diameter = 2.1 cm, 2.1, cm, Radius, r =, ∴, 2, Now, total surface area of the cube, = 6 × (Edge) 2= 6 × 6 × 6 = 216 cm 2, Here, the part of the cube where the hemisphere is, attached, is not included in the surface area., So, the total surface area of the decorative block, = Total surface area of cube − Area of base of hemisphere, + Curved surface area of hemisphere, = 216 − πr 2 + 2 πr 2 = 216 + πr 2, 22 2 . 1 2 . 1, = 216 +, ×, ×, 7, 2, 2, = 216 + 3. 465 = 219. 465 cm 2, , 7 cm, , 3 cm, , Since, the height of conical cavity and the side of cube is, equal that means the conical cavity fit vertically in the cube., Radius of conical cavity i.e. cone, r = 3 cm, Diameter = 2 × r = 2 × 3 = 6 cm, ⇒, Since, the diameter is less than the side of a cube that means, the base of a conical cavity is not fit inhorizontal face of, cube., Now, volume of cube = (side) 3= a 3 = (7 )3 = 343 cm 3, 1, and volume of conical cavity i.e. cone = π × r 2 × h, 3, 1 22, = ×, × 3× 3×7, 3 7, = 66 cm 3, ∴Volume of remaining solid = Volume of cube, − Volume of conical cavity, = 343 − 66 = 277 cm 3, Hence, the required volume of solid is 277 cm 3., 8. Given, radius of hemispherical bowl, r = 9 cm, and radius of cylindrical bottles, R = 1. 5 cm and height,, h = 4 cm, ∴Number of required cylindrical bottles, Volume of hemispherical bowl, =, Volume of one cylindrical bottle, 2 3, πr, = 3 2, πR h, 2, ×π ×9×9×9, = 54, = 3, π × 1. 5 × 1. 5 × 4, 9. Let r be the radius of the hemispherical dome, and total height of building be H m., It is given that diameter of dome, 2, = × Total height of the building, 3, 1, ⇒, r= Hm, 3, Let h m be the height of the cylinder., 1, 2, h = H − r = H − H = Hm, ∴, 3, 3, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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136, , CBSE Term II Mathematics X (Standard), , Volume of the air inside the building = Volume of air, inside the dome + Volume of air inside the cylinder, 2, = πr 3 + πr 2h, 3, ⎡2 ⎛ 1 ⎞ 3 ⎛ 1 ⎞ 2 ⎛ 2 ⎞ ⎤, = π ⎢ ⎜ H⎟ + ⎜ H⎟ ⎜ H⎟ ⎥, ⎝3 ⎠ ⎝3 ⎠, ⎝, ⎠, ⎣3 3, ⎦, 3, ⎡ 2H, 2 3⎤, = π⎢, + H ⎥, ⎣ 81 27 ⎦, 8, =, πH 3 m 3, 81, 1, Given, volume of the air inside the building = 67 m 3, 21, 8, 1408, πH 3 =, ∴, 81, 21, 1408 81, 3, ⇒, ×, H =, 21, 8π, 1408 × 81 × 7, 3, ⇒, H =, 21 × 8 × 22, ⇒, , H 3 = 216 ⇒ H = 6 m, , 10. Given, diameter of the cylinder = 2 cm, ∴Radius = 1 cm and height of the cylinder = 16 cm, [Q diameter = 2 × radius], ∴Volume of the cylinder = π × (1)2 × 16 = 16 π cm 3, [Q volume of cylinder= π × (radius)2 × height], Now, let the radius of solid sphere = r cm, 4, Then, its volume = πr 3 cm 3, 3, [Q volume of sphere =, , 4, × π × (radius)3], 3, , According to the question,, Volume of the twelve solid sphere = Volume of cylinder, 4, ⇒, 12 × πr 3 = 16 π, 3, ⇒, r 3 = 1 ⇒ r = 1 cm, ∴ Diameter of each sphere, d = 2 r = 2 × 1 = 2 cm, Hence, the required diameter of each sphere is 2 cm., 11. Given, whole length of the solid = 104 cm, and the radius of each hemisphere = 7 cm, Therefore, the length of the cylindrical part of the solid, = (104 − 2 × 7 ) = 90 cm, For hemispherical portion,, Radius, r = 7 cm, For cylindrical portion,, Radius, r = 7 cm, Height, h = 90 cm, ∴ Total surface area of the solid, = 2 × Curved surface area of a hemisphere, + Curved surface area of cylindrical part, = 2 [2 πr 2 ] + 2 πrh, = 2 × [2 π (7 )2 ] + 2 π (7 ) ( 90), , 22, 22, ⎤, ⎡, × (7 ) ( 90), =2 × 2 ×, × (7 ) 2 + 2 ×, 7, 7, ⎦⎥, ⎣⎢, = 4 × 22 × 7 + 2 × 22 × 90 = 22 [28 + 180], = 4576 cm 2, 7 cm, 7 cm, , 104 cm, , 90 cm, , 7 cm, 7 cm, , Then, the cost of polishing at the rate of ` 2 per dm 2, 4576 × 2, [Q1 dm 2 = 100 cm 2 ], = ` 91.52, =`, 100, 12. Let height of the cone be h., Given, radius of the base of the cone = 6 cm, 1, 1, Volume of circular cone = πr 2h = π ( 6)2 h, ∴, 3, 3, 36 π h, =, = 12 π h cm 3, 3, Also, given radius of the hemisphere = 8 cm, 2, 2, ∴Volume of the hemisphere = π r 3 = π ( 8)3, 3, 3, 512 × 2 π, =, cm 3, 3, According to the question,, Volume of the cone = Volume of the hemisphere, 512 × 2 π, 12 πh =, ⇒, 3, 512 × 2 π 256, ∴, h=, =, = 28. 44 cm, 12 × 3 π, 9, 13. Given, dimensions of base of rectangular tank = 11 m × 6 m, and height of water = 5 m, Volume of the water in rectangular tank = 11 × 6 × 5, = 330 m 3, Also, given radius of the cylindrical tank = 3.5 m, Let height of water level in cylindrical tank be h., Then, volume of the water in cylindrical tank = πr 2h, = π (3.5)2 × h, 22, × 3.5 × 3.5 × h = 11. 0 × 3. 5 × h = 38. 5 h m 3, 7, According to the question,, 330 = 38. 5 h, [since, volume of water is same in both tanks], 330 3300, h=, =, ∴, 38. 5 385, =, , ∴, = 8. 57 m or 8.6 m, Hence, the height of water level in cylindrical tank is 8.6 m., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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137, , CBSE Term II Mathematics X (Standard), , 14. Here, a rod of cylindrical shape is converted into a wire of, cylindrical shape., For rod,, 1, Diameter = 1 cm ⇒ Radius, r1 = cm and length,, 2, h = 8 cm, 2, ⎛ 1⎞, ∴ Volume of the rod = πr12h = π × ⎜ ⎟ × 8 = 2 π cm 3, ⎝ 2⎠, For wire,, Length = 8 m = 800 cm, [Q1m = 100 cm ], Let r be the radius (in cm) of cross-section of the wire, then, Volume of wire = π × r 2 × 800 cm 3, Since, the rod is converted into wire, so, Volume of wire = Volume of rod, 1, 1, cm, ⇒, π × r 2 × 800 = 2 π ⇒ r 2 =, ⇒r =, 400, 20, 2, 1, Then, diameter =, cm, =, 20 10, Hence, the diameter of the cross-section, i.e. the thickness, 1, cm, i.e. 1 mm [Q1cm = 10 mm ], of the wire is, 10, 15. Given, length of roof = 22 m and breadth of roof = 20 m, Let the rainfall be a cm., a, 22 a 3, m, =, 100, 5, Also, we have radius of base of the cylindrical vessel, r = 1 m, and height of the cylindrical vessel, h = 3.5 m, ∴Volume of water in the cylindrical vessel when it is just full, 7⎞, ⎛ 22, = πr 2h = ⎜, × 1 × 1 × ⎟ = 11 m 3, ⎝7, 2⎠, ∴Volume of water on the roof = 22 × 20 ×, , Now, volume of water on the roof = Volume of water in, the vessel, 22 a, ⇒, = 11, 5, 11 × 5, ∴, a=, = 2.5 cm, 22, [Q volume of cylinder=π × (radius) 2 × height], Hence, the rainfall is 2.5 cm., 16. Given, radius of the base of the bucket = 18 cm, Height of the bucket = 32 cm, So, volume of the sand in cylindrical bucket = π r 2h, = π (18)2 × 32 = 10368 π, Also, given height of the conical heap ( h ) = 24 cm, Let radius of heap be r cm., 1, Then, volume of the sand in the heap = π r 2 h, 3, 1 2, 2, = πr × 24 = 8 π r, 3, According to the question,, Volume of the sand in cylindrical bucket = Volume of the, sand in conical heap, ⇒, 10368 π = 8πr 2, ⇒, , 10368 = 8 r 2, , 10368, = 1296 ⇒ r = 36 cm, 8, Again, let the slant height of the conical heap = l, Now,, l 2 = h 2 + r 2 = (24)2 + ( 36)2, ⇒, , r2 =, , = 576 + 1296 = 1872, ∴, l = 43. 267 cm, Hence, radius of conical heap of sand = 36 cm, and, slant height of conical heap = 43. 267 cm, 17. Given, length of the barrel of a fountain pen = 7 cm, 5, 1, 1, ⎡, ⎤, Q 1 mm = cm, cm = cm, ⎥⎦, ⎢⎣, 10, 10, 2, 1, ∴ Radius of the barrel =, = 0.25 cm, 2 ×2, and diameter = 5 mm =, , Volume of the barrel = πr 2h [since, its shape is cylindrical], 22, =, × (0.25)2 × 7, 7, = 22 × 0.0625 = 1.375 cm 3, 1, Also, given volume of ink in the bottle = of litre, 5, 1, 3, = × 1000 cm = 200 cm 3, 5, Now, 1.375 cm 3 ink is used for writing number of words, = 3300, 3300, 3, ∴1 cm ink is used for writing number of words =, 1.375, ∴200 cm 3 ink is used for writing number of words, 3300, =, × 200 = 480000, 1.375, 18. Given, speed of water flow = 10 m min −1 = 1000 cm/min, and diameter of the pipe = 5 mm =, ∴, , Radius of the pipe =, , 5, 1, ⎡, ⎤, cm Q 1 mm = cm, ⎢⎣, ⎥⎦, 10, 10, , 5, = 0.25 cm, 10 × 2, , ∴ Area of the face of pipe, 22, = πr 2 =, × (0.25)2 = 0.1964 cm 2, 7, Also, given diameter of the conical vessel = 40 cm, 40, = 20 cm, ∴Radius of the conical vessel =, 2, and depth of the conical vessel = 24 cm, 1, 1 22, × (20)2 × 24, ∴Volume of conical vessel = πr 2h = ×, 3, 3 7, 211200, =, = 10057.14 cm 3, 21, Volume of the conical vessel, ∴Required time =, Area of the face of pipe × Speed of water, 10057.14, =, 0.1964 × 1000, 20, × 60 s, = 51.20 min = 51 min, 100, = 51 min 12 s, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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138, , CBSE Term II Mathematics X (Standard), , 19. Given, radius of tank, r1 = 40 cm, Let height of water level in tank in half an hour = h1, Also, given internal radius of cylindrical pipe, r2 = 1 cm, and speed of water = 80 cm/s i.e. in 1s water flow = 80 cm, ∴ In 30 (min) water flow = 80 × 60 × 30 = 144000 cm, According to the question,, Volume of water in cylindrical tank = Volume of water flow, from the circular pipe in half an hour, ⇒, π r12 h1 = π r22 h 2, ⇒, , 40 × 40 × h1 = 1 × 1 × 144000, 144000, = 90 cm, 40 × 40, , ∴, , h1 =, , Hence, the level of water in cylindrical tank rises 90 cm in, half an hour., 20. Given, pencils are cylindrical in shape., Length of one pencil = 25 cm, and circumference of base, 2πr = 1.5 cm, 1.5 × 7, ⇒, r=, = 0.2386 cm, 22 × 2, Now, curved surface area of one pencil = 2πrh, 22, =2 ×, × 0. 2386 × 25, 7, 262. 46, =, = 37.49 cm 2, 7, 37.49, 1, ⎡, ⎤, =, Q 1 cm =, dm 2, dm, ⎢⎣, ⎥⎦, 100, 10, = 0.375 dm 2, ∴Curved surface area of 120000 pencils = 0.375 × 120000, = 45000 dm 2, 2, Now, cost of colouring 1 dm curved surface of the pencils, manufactured in one day = ` 0.05, ∴Cost of colouring 45000 dm 2 curved surface, = 45000 × 0. 05 = ` 2250, 21. Here, a well is dug and Earth taken out of it is used to form, an embankment., Given, Diameter of well = 10 m, 10, = 5m, ∴ Radius =, 2, Also, depth = 14 m, ∴Volume of Earth taken out on digging the well, 22, = πr 2h =, × ( 5)2 × 14 = 1100 m 3, 7, 10 m, 5m, , 5m, , 14 m, , h, , The embankment is in the form of cylindrical shell, so area, of embankment = π(R 2 − r 2 ) = π(102 − 52 ), 22, × 75 m 2, = π (100 − 25) =, 7, Since, Earth taken out from well is used to form, embankment., ∴ Volume of embankment, = Volume of Earth taken out on digging the well, ⇒ Area of embankment × Height of embankment, = Volume of Earth dugout, Volume of Earth dugout, ⇒ Height of embankment =, Area of the embankment, 1100, =, = 4. 67 m, 22, × 75, 7, 22. Given, diameter of a marble = 1.4 cm, 1.4, Radius of marble ( r ) =, ∴, = 0.7 cm, 2, 4, 4, So,, volume of one marble = πr 3 = π ( 0.7 )3, 3, 3, 4, 1.372, = π × 0. 343 =, π cm 3, 3, 3, Also, given diameter of beaker = 7 cm, 7, ∴ Radius of beaker = = 3.5 cm, 2, Height of water level raised = 5.6 cm, ∴Volume of the raised water in beaker, = π (3.5)2 × 5.6 = 68.6π cm 3, Now, required number of marbles, Volume of the raised water in beaker, =, Volume of one spherical marble, 68.6 π, =, × 3 = 150, 1.372 π, 23. Given, edge of the cube = 22 cm, ∴, , Volume of the cube = (22 )3 = 10648 cm 3, [Q volume of cube = (side)3], , Also, given diameter of marble = 0.5 cm, 0. 5, = 0. 25 cm, 2, [Q diameter =2 × radius], 4, 4 22, Volume of one marble = πr 3 = ×, × (0.25)3, 3, 3 7, 4, [Q volume of sphere = × π × (radius)3], 3, 1. 375, 3, =, = 0. 0655 cm, 21, 1, Filled space of cube = Volume of the cube −, 8, × Volume of cube, 1, = 10648 − 10648 ×, 8, 7, = 10648 × = 9317 cm 3, 8, , ∴ Radius of a marble, r =, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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139, , CBSE Term II Mathematics X (Standard), , ∴ Required number of marbles, Total space filled by marbles in a cube, =, Volume of one marble, 9317, =, = 142244 (approx.), 0.0655, Hence, the number of marbles that the cube can accomodate, is 142244., 24. Given that, a solid iron cuboidal block is recast into a hollow, cylindrical pipe., Length of cuboidal pipe ( l ) = 4. 4 m, Breadth of cuboidal pipe ( b ) = 2. 6 m and height of cuboidal, pipe ( h ) = 1 m, So, volume of a solid iron cuboidal block = l ⋅ b ⋅ h, = 4. 4 × 2. 6 × 1 = 11. 44 m 3, Also, internal radius of hollow cylindrical pipe ( ri ) = 30 cm, = 0. 3 m, and thickness of hollow cylindrical pipe = 5 cm = 0. 05 m, So, external radius of hollow cylindrical pipe, ( re ) = ri + Thickness, = 0. 3 + 0. 05 = 0. 35 m, ∴Volume of hollow cylindrical pipe, = Volume of cylindrical pipe with external radius, − Volume of cylindrical pipe with internal radius, = πre2h1 − πri2h1 = π ( re2 − ri2 ) h1, 22, =, [( 0. 35)2 − ( 0. 3)2 ] ⋅ h1, 7, 22, 22, =, [ 0. 1225 − 0. 09] ⋅ h1 = [ 0. 0325] ⋅ h1, 7, 7, = 0.715 × h1 / 7, where, h1 be the length of the hollow cylindrical pipe., Now, by given condition,, Volume of solid iron cuboidal block = Volume of hollow, cylindrical pipe, ⇒, 11. 44 = 0.715 × h / 7, 11. 44 × 7, ∴, h=, = 112 m, 0.715, Hence, required length of pipe is 112 m., 25. Let total height of the building = Internal diameter of the, dome = 2r m, r, r, , r, 2r, r, , 2r, =rm, 2, Height of cylinder = 2r − r = r m, Volume of the cylinder = π r 2 ( r ) = πr 3 m 3, ∴, 2, and volume of hemispherical dome on cylinder = πr 3 m 3, 3, , ∴, , Radius of building (or dome) =, , ∴ Total volume of the building = Volume of the cylinder, + Volume of hemispherical dome, ⎛ 3 2 3⎞ 3 5 3 3, = ⎜ πr + πr ⎟ m = πr m, ⎝, ⎠, 3, 3, According to the question,, Volume of the building = Volume of the air, 5 3, 19, ⇒, πr = 41, 3, 21, 5 3 880, πr =, ⇒, 3, 21, 880 × 7 × 3, r3 =, ⇒, 21 × 22 × 5, 40 × 21, =, =8, 21 × 5, 3, ⇒, r = 8 ⇒ r =2m, ∴ Height of the building = 2 r = 2 × 2 = 4 m, 26. Given, diameter of cylinder = Diameter of hemisphere, = 0.5 cm, [since, both hemispheres are attach with cylinder], ∴ Radius of cylinder ( r ) = radius of hemisphere, 0.5, (r) =, = 0.25 cm, 2, [Q diameter = 2 × radius], , 0.25, , 0.5 cm, , 0.25, , 2 cm, , and total length of capsule = 2 cm, ∴Length of cylindrical part of capsule,, h = Length of capsule − Radius of both hemispheres, = 2 − (0.25 + 0.25) = 1.5 cm, Now, capacity of capsule = Volume of cylindrical part, + 2 × Volume of hemisphere, 2, 2, = πr h + 2 × πr 3, 3, [Q volume of cylinder = π × (radius)2, 2, × height and volume of hemisphere = π (radius)3], 3, 22, 4, 2, 3, =, [(0.25) × 1.5 + × (0.25) ], 7, 3, 22, [0.09375 + 0.0208], =, 7, 22, =, × 0.11455 = 0.36 cm 3, 7, Hence, the capacity of capsule is 0.36 cm 3., 27. Since, rocket is the combination of a right circular cylinder, and a cone., Given, diameter of the cylinder = 6 cm, 6, ∴ Radius of the cylinder = = 3 cm, 2, and height of the cylinder = 12 cm, ∴ Volume of the cylinder = πr 2h = 3.14 × ( 3)2 × 12, = 339.12 cm 3, and curved surface area = 2πrh, = 2 × 3.14 × 3 × 12 = 226.08, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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140, , CBSE Term II Mathematics X (Standard), , 2 3 1 2, πr + πr h, 3, 3, 1, ⎡2, ⎤, =, × 3.14 × (2 )3 + × 3.14 × (2 )2 × 3, ⎢⎣ 3, ⎥⎦, 3, , A, , =, , 5 cm, h, , = 16.75 + 12 . 56 = 29.31 cm 3, O, , C, 12 cm, , B, , 3 cm, , Hence, volume of the solid toy is 29.31 cm 3., Now, let the right circular cylinder EFGH circumscribe the, given solid toy. Then, radius of the base of the right circular, cylinder = HP = BO = 2 cm and its height,, EH = AP = AO + OP = 3 + 2 = 5 cm, So, volume of the cylinder = π r 2 h, , = 3.14 × 2 2 × 5, E, , D, 6 cm, , Now, in right angled ΔAOC,, h = 52 − 32, = 25 − 9 = 16 = 4, ∴Height of the cone, h = 4 cm, and radius of the cone, r = 3 cm, Now, volume of the cone, 1, 1, = πr 2h = × 3.14 × ( 3)2 × 4, 3, 3, 113.04, =, = 37.68 cm 3, 3, and curved surface area = πrl = 3.14 × 3 × 5 = 47.1, Hence, total volume of the rocket, = 339.12 + 37.68 = 376.8 cm 3, and total surface area of the rocket, = CSA of cone + CSA of cylinder + Area of base of cylinder, = 47.1 + 226. 08 + 28. 26, = 301. 44 cm 2, 28. Here, given solid toy is a combination of a right circular, cone and a hemisphere., Let BPC be the hemisphere and ABC be the cone standing, on the base of the hemisphere as shown in the figure., For conical portion, height, h = 3 cm, Diameter, d = 4 cm, E, , B, , H, , A, , O, , P, , F, , C, , G, , 4, cm = 2 cm, 2, For hemispherical portion,, Radius, r = 2 cm [Q radii of hemisphere and cone are same], So, volume of the solid toy, = Volume of hemisphere + Volume of cone, , ∴Radius, r =, , = 62.8 cm 3, Now, required difference of the volume of the cylinder and, the solid toy = Volume of cylinder − Volume of solid toy, = 62.8 − 29.31 = 33.49 cm 3, 29. Here, the given wooden toy rocket is combination of a cone, and a cylinder., For conical portion,, Diameter = 6 cm, 6, Radius, r1 = cm = 3 cm, ∴, 2, Height, h1 = 4 cm, Then, slant height, l = ( 3)2 + 42, , [Q l = r 2 + h 2 ], , = 9 + 16 = 25 = 5 cm, For cylindrical portion,, Diameter = 4 cm, 4, Radius, r2 = cm = 2 cm, ∴, 2, Height, h 2 = Total height of rocket − Height of cone, = 24 − 4 [Q total height of rocket = 24 cm], = 20 cm, Here, we have to find the area of the rocket painted with, orange and yellow colours separately., Since, radius of base of cone is larger than radius of base of, cylinder and cone is mounted on cylinder., ∴ Area to be painted orange = Curved surface area of cone, + Area of base of cone − Area of base of cylinder, [Q area of base of cylinder is common, in area of base of cone], = πr1l + πr12 − πr22, = 3.14 × 3 × 5 + 3.14 × (3)2 − 3.14 × (2)2, = 3.14 [15 + 9 −4], = 3.14 × 20 = 62.8 cm 2, Now, area to be painted yellow, = Curved surface area of cylinder, + Area of base of the cylinder = 2 πr2h 2 + π r22, = 2 × 3.14 × 2 × 20 + 3.14 × (2)2, = 3.14 [80 + 4], = 3.14 × 84 = 263.76 cm 2, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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141, , CBSE Term II Mathematics X (Standard), , 30. Let volume of cone A be 2 V and volume of cone B be V., Again, let height of the cone A = h1 cm, then height of cone, B = (21 − h1 ) cm, , Now, since edge of a solid cube ( a ) = 44 cm, So, volume of a solid cube = ( a )3, = ( 44)3, = 44 × 44 × 44 cm 3, , A, , B, , h1, , 6 cm, , 6 cm, , 21 cm, , 21 – h1, , Given, diameter of the cone = 6 cm, 6, Radius of the cone = = 3 cm, ∴, 2, 1, 1, Now, volume of the cone, A = 2 V = πr 2h = π( 3)2 h1, 3, 3, 1, 3, …(i), V = π 9 h1 = h1π, ⇒, 6, 2, and volume of the cone,, 1, …(ii), B = V = π ( 3)2 (21 − h1 ) = 3π (21 − h1 ), 3, From Eqs. (i) and (ii),, 3, h1π = 3π (21 − h1 ), 2, ⇒, h1 = 2 (21 − h1 ), ⇒, 3h1 = 42, 42, = 14 cm, ⇒, h1 =, 3, ∴ Height of cone, B = 21 − h1 = 21 − 14 = 7 cm, 22, Now, volume of the cone, A = 3 × 14 ×, = 132 cm 3, 7, [using Eq. (i)], 1 22, and volume of the cone, B = ×, × 9 × 7 = 66 cm 3, 3 7, [using Eq. (ii)], 22 2, 2, Now, volume of the cylinder = πr h = ( 3) × 21 = 594 cm 3, 7, ∴ Required volume of the remaining portion, = Volume of the cylinder, − (Volume of cone A + Volume of cone B), = 594 − (132 + 66) = 396 cm 3, 31. Given that, lots of spherical lead shots made out of a solid, cube of lead., ∴ Number of spherical lead shots, Volume of a solid cube of lead, ...(i), =, Volume of a spherical lead shot, Given that, diameter of a spherical lead shot i.e. sphere = 4 cm, 4, Radius of a spherical lead shot (r) =, ⇒, 2, [Q diameter = 2 × radius], r = 2 cm, So, volume of a spherical lead shot i.e. sphere, 4, = π r3, 3, 4 22, 4 × 22 × 8, cm 3, = ×, × (2 ) 3 =, 3 7, 21, , From Eq. (i),, Number of spherical lead shots =, , 44 × 44 × 44, × 21, 4 × 22 × 8, , = 11 × 21 × 11, = 121 × 21, = 2541, Hence, the required number of spherical lead shots is 2541., 32. Given, internal diameter of spherical shell = 4 cm, and external diameter of shell = 8 cm, 4, cm = 2 cm, 2, [Q diameter = 2 × radius], , ∴ Internal radius of spherical shell, r1 =, , and external radius of shell, r2 =, , 8, = 4 cm, 2, [Q diameter = 2 × radius], , 8 cm, 4 cm, , Spherical shell, , 4, π [ r23 − r13 ], 3, 4, [Q volume of the spherical shell = π, 3, {(external radius)3 − (internal radius)3}], 4, = π ( 43 − 2 3 ), 3, 4, = π ( 64 − 8), 3, 224, =, π cm 3, 3, Let height of the cone = h cm, Diameter of the base of cone = 8 cm, 8, ∴ Radius of the base of cone = = 4 cm, 2, [Q diameter = 2 × radius], According to the question,, Volume of cone = Volume of spherical shell, 1, 224, π ( 4) 2 h =, π, ⇒, 3, 3, 224, ⇒, h=, = 14 cm, 16, 1, [Q volume of cone = × π × (radius)2 × (height)], 3, Hence, the height of the cone is 14 cm., Now, volume of the spherical shell =, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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142, , CBSE Term II Mathematics X (Standard), , 33. Given that, lots of spherical lead shots made from a solid, rectangular lead piece., ∴ Number of spherical lead shots, Volume of solid rectangular lead piece, ...(i), =, Volume of a spherical lead shot, Also, given that diameter of a spherical lead shot, i.e. sphere = 4. 2 cm, 4.2, ∴ Radius of a spherical lead shot, r =, = 2.1 cm, 2, 1, ⎡, ⎤, Q radius = diameter, ⎢⎣, ⎥⎦, 2, So, volume of a spherical lead shot i.e. sphere, 4, = π r3, 3, 4 22, = ×, × (2.1)3, 3 7, 4 22, = ×, × 2.1 × 2.1 × 2.1, 3 7, 4 × 22 × 21 × 21 × 21, =, 3 × 7 × 1000, Now, length of rectangular lead piece, l = 66 cm, Breadth of rectangular lead piece, b = 42 cm, Height of rectangular lead piece, h = 21 cm, ∴Volume of a solid rectangular lead piece i.e. cuboid, =l×b×h, = 66 × 42 × 21, From Eq. (i),, Number of spherical lead shots, 66 × 42 × 21, =, × 3 × 7 × 1000, 4 × 22 × 21 × 21 × 21, 3 × 22 × 21 × 2 × 21 × 21 × 1000, =, 4 × 22 × 21 × 21 × 21, = 3 × 2 × 250, = 6 × 250 = 1500, Hence, the required number of spherical lead shots is 1500., 34. Given that, lots of metallic circular disc to be melted to form, a right circular cylinder. Here, a circular disc work as a, circular cylinder., Base diameter of metallic circular disc = 1. 5 cm, 1. 5, cm, ∴ Radius of metallic circular disc =, 2, [Q diameter = 2 × radius], and height of metallic circular disc i.e. = 0.2 cm, ∴Volume of a circular disc = π × (Radius)2 × Height, , ⇒, ∴, , 4. 5, cm, 2, Volume of right circular cylinder = π r 2h, , Radius of a right circular cylinder ( r ) =, , 2, , ⎛ 4.5⎞, = π ⎜ ⎟ × 10, ⎝ 2 ⎠, π, × 4. 5 × 4. 5 × 10, 4, ∴ Number of metallic circular disc, Volume of a right circular cylinder, =, Volume of a metallic circular disc, π, × 4. 5 × 4. 5 × 10, = 4, π, × 1. 5 × 1. 5 × 0. 2, 4, 3 × 3 × 10 900, =, =, = 450, 0. 2, 2, =, , Hence, the required number of metallic circular disc is 450., 35. Given that, a heap of rice is in the form of a cone., Height of a heap of rice i.e. cone ( h ) = 3. 5 m, and diameter of a heap of rice i.e. cone = 9 m, 9, Radius of a heap of rice i.e. cone ( r ) = m, 2, 1, So, volume of rice = π × r 2h, 3, 1 22 9 9, = ×, × × × 3. 5, 3 7 2 2, 6237, =, = 74. 25 m 3, 84, Now, canvas cloth required to just cover heap of rice, = Surface area of a heap of rice, = πrl, 22, =, × r × r2 + h 2, 7, 2, , =, , 22 9, ⎛ 9⎞, × × ⎜ ⎟ + ( 3.5)2, ⎝ 2⎠, 7 2, , 11 × 9, 81, ×, + 12. 25, 7, 4, 99, 130 99, =, ×, =, × 32 . 5, 7, 4, 7, = 14.142 × 5. 7 = 80. 61 m 2, =, , Hence, 80.61 m 2 canvas cloth is required to just cover heap., 36. Let the length(l), breadth (b) and height (h) be the external, dimension of an open box and thickness be x., , 2, , ⎛ 1. 5⎞, =π×⎜, ⎟ × 0. 2, ⎝ 2 ⎠, π, × 1. 5 × 1. 5 × 0. 2, 4, Now, height of a right circular cylinder ( h ) = 10 cm, and diameter of a right circular cylinder = 4. 5 cm, =, , x, , x, x, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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143, , CBSE Term II Mathematics X (Standard), , and speed of water flowing through the pipe = (15 × 1000), = 15000 mh −1, , Given that,, external length of an open box ( l ) = 36 cm, external breadth of an open box ( b ) = 25 cm, and external height of an open box ( h ) = 16.5 cm, ∴ External volume of an open box = lbh, = 36 × 25 × 16. 5 = 14850 cm 3, , Now, volume of water flow in 1 h = π R 2H, 7, 7, ⎛ 22, ⎞, =⎜, ×, ×, × 15000⎟, ⎝ 7 100 100, ⎠, = 231 m 3, , Since, the thickness of the iron ( x ) = 1. 5 cm, So, internal length of an open box ( l1 ) = l − 2 x, = 36 − 2 × 1. 5, = 36 − 3 = 33 cm, Therefore, internal breadth of an open box ( b 2 ) = b − 2 x, = 25 − 2 × 1. 5, = 25 − 3 = 22 cm, and internal height of an open box ( h 2 ) = ( h − x ), = 16. 5 − 1. 5 = 15 cm, So, internal volume of an open box, = ( l − 2 x) ⋅ ( b − 2 x) ⋅ ( h − x), = 33 × 22 × 15, = 10890 cm 3, , Since, 231 m 3 of water falls in the pond in 1 h., 1, h., So, 1 m 3 water falls in the pond in, 231, ⎛ 1, ⎞, Also, 462 m 3 of water falls in the pond in ⎜, × 462⎟ h = 2 h, ⎝ 231, ⎠, , Therefore, required iron to construct an open box, = External volume of an open box, − Internal volume of an open box, = 14850 − 10890 = 3960 cm 3, 3, , Hence, required iron to construct an open box is 3960 cm ., 7. 5, kg, Given that, 1 cm 3 of iron weights = 7 . 5 g =, 1000, = 0. 0075 kg, ∴ 3960 cm 3 of iron weights = 3960 × 0. 0075 = 29.7 kg, 37. Given, length of the pond = 50 m and width of the pond, = 44 m, 21, m, 100, ∴ Volume of water in the pond = l × b × h, 21 ⎞, ⎛, = ⎜ 50 × 44 ×, ⎟, ⎝, 100⎠, Depth required of water = 21 cm =, , = 462 m 3, 7, m, Also, given radius of the pipe = 7 cm =, 100, 14 cm, Pipe, 44 m, , 21 m, 100, , Tank, , Hence, the required time is 2 h., 38. (i) Length of the cuboid (L ) = 30 cm, Breadth of the cuboid (B ) =20 cm, Height of the cuboid (H ) = 10 cm, Then,, Area covered by the cuboid, = L × B + 2H (L + B ), = 30 × 20 + 2 × 10( 30 + 20), = 600 + 1000, = 1600 cm2, (ii) We know that,, C.S.A of the cylinder = 2 πRH1, 22, C.S.A of the cylinder = 2 ×, × 7 × 60, 7, = 2 × 22 × 60, = 2640 cm2, In cone, height is given and radius of the cone is equal to, radius of the cylinder,, Q, L2 = H 22 + R 2, ⇒, , L2 = 7 × 7 + 24 × 24, , ⇒, , L2 = 49 + 576 = 625, , ∴, , L = 25 cm, 22, C.S.A. of cone = πrl =, × 7 × 25 = 550 cm2, 7, C.S. A of the cylinder 2640, ⇒, =, = 24 : 5, C.S. A. of the cone, 550, (iii) (b) Given structure is based on the concept of surface, area and volume., 7, 21, cm, 39. (i) Given, r = cm, h = 10. 5 cm =, 2, 2, Volume of cylindrical cup = πr 2h, 22 7 7 21, =, × × ×, 7 2 2 2, 11 × 7 × 21, =, 4, = 404.25 cm 3, , 50 m, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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144, , CBSE Term II Mathematics X (Standard), , (ii) Given, r =, , 7, cm, 2, , (ii) Given edge of cube = 7 cm, ∴Volume of cube = 7 × 7 × 7 = 343 cm 3, 2 22 7 7 7, ×, × × ×, 3 7 2 2 2, = 89. 83 cm 3, , ∴ Volume of hemispherical cup =, , (iii) Area of canvas provided = 551 m 2, Area of remained after westage = 551 − 1 = 550 m 2, So, area of conical tent = πrl, Here, r = 7 m, 22, × 7 × l = 550 ⇒ l = 25 m, ∴ πrl = 550 ⇒, 7, Now, h = l 2 − r 2 = 625 − 49, = 576 = 24 m, 40. (i) Given, edge of cube = 6 cm, ∴Diagonal of cube = 3 × edge of cube, = 3 × 6 = 6 3 cm, , (iii) Given, radius ( r ) = 7 cm, ∴Curved surface area of hemisphere = 2 πr 2, 22, =2×, ×7 ×7, 7, = 44 × 7 = 308 cm 2, (iv) Given, radius ( r ) = 7 cm and height ( h ) = 24 cm, Slant height ( l ) = ?, ∴ l 2 = r 2 + h 2 = (7 )2 + (24)2, = 49 + 576 = 625, ⇒ l = 625 = 25 cm, (v) TSA of cone = πrl + 2 πr 2, 22, 22, =, × 7 × 25 + 2 ×, ×7 ×7, 7, 7, = 550 + 308 = 858 cm 2, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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Chapter Test, Multiple Choice Questions, , 1. How many cubes of side 2 cm can be made from a, solid cube of side 10 cm?, (a) 100, , (b) 125, , (c) 175, , (d) 200, , 2. 2 cubes, each of volume 125 cm 3 , are joined end to, end. Find the surface area of the resulting cuboid., (a) 100 cm 2, , (b) 200 cm 2, , (c) 225 cm 2, , (d) 250 cm 2, , 3. The radius of a sphere (in cm) whose volume is, (a) 3, , (b) 3 3, , (c) 3, , 1/ 3, , (d) 3, , 4. A solid spherical ball fits exactly inside the cubical box, of side 2a. The volume of the ball is, 16, πa 3, 3, 32 3, (c), πa, 3, , (iv) External curved surface area of the cylinder, is, (a) 2876 cm 2, (c) 4224 cm 2, , (b) 1250 cm 2, (d) 3824 cm 2, , (a) 6232 cm 3, (b) 7248 cm 3, (c) 5380 cm 3, (d) 9856 cm 3, , 7. A hemispherical depression is cut out from one face of, , 5. A cone and a cylinder have the same radii but the, height of the cone is 3 times that of the cylinder, then, the ratio of their volumes., (a) 1 : 2, (c) 1 : 1, , (b) 2200 cm 2, (d) 2400 cm 2, , Short Answer Type Questions, , 1 3, πa, 6, 4 3, (d) πa, 3, (b), , (a), , (a) 2250 cm 2, (c) 1800 cm 2, , (v) Volume of conical cavity is, , 12π cm 3 , is, , 2/ 3, , (iii) The curved surface area of the conical cavity, so formed is, , (b) 2 : 1, (d) None of these, , Case Based MCQs, , 6. One day Aakash was going home from market saw a, carpenter working on wood. He found that he is, carving out a cone of same height and same diameter, from a cylinder. The height of the cylinder is 48 cm and, base radius is 14 cm. While watching this, some, questions came into Aakash’s mind. Help Aakash to, find the answer of the following questions., , a cuboidal block of side 7 cm such that the diameter of, the hemisphere is equal to the edge of the cube. Find, the surface area of the remaining solid., , 8. The capacity of a cylindrical glass tumbler is 125.6 cm 3 ., If the radius of the glass tumbler is 2 cm, then find its, height. (Use π = 3.14), , 9. In given figure, a solid toy is in the form of a, hemisphere surmounted by a right circular cone. The, height of the cone is 2 cm and the diameter of the, base is 4 cm. Determine the volume of the toy., 22 ⎤, ⎡, take π =, 7 ⎦⎥, ⎣⎢, 2 cm, , 14 cm, , 2 cm, 48 cm, , Long Answer Type Questions, , 10. From a solid cylinder whose height is 2.4 cm and, (i) After carving out cone from the cylinder,, (a) Volume of the cylindrical wood will decrease, (b) Height of the cylindrical wood will increase, (c) Volume of cylindrical wood will increase, (d) Radius of the cylindrical wood will decrease, , (ii) Find the slant height of the conical cavity so, formed., (a) 28 cm, (c) 40 cm, , (b) 60 cm, (d) 50 cm, , diameter 1.4 cm, a conical cavity of the same height, and same diameter is hollowed out. Find the total, surface area of the remaining solid to the nearest cm 2 ., , 11. A well of diameter 3 m is dug 14 m deep. The earth, taken out of it has been spread evenly all around it in, the shape of a circular ring of width 4 m to form a, platform. Find the height of the platform., 22 ⎤, ⎡, take π =, ⎢⎣, 7 ⎥⎦, , Answers, 1. (b), , 2. (d), , 3. (c), , 4. (d), , 7. 332.465 cm 2 8. 10 cm, , 5. (c), , 6. (i) (a) (ii) (d) (iii) (b) (iv) (c) (v) (d), , 9. 25.12 cm 3, , 10. 18 cm 2, , 11. 1.125 m, , For Detailed Solutions, Scan the code, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 146, , CHAPTER 07, , Statistics, In this Chapter..., !, , Mean of Grouped Data, , !, , Mode of Grouped Data, , !, , Median of Grouped Data, , Arithmetic Mean or Mean or Average, The arithmetic mean of a set of observations is obtained by, dividing the sum of the values of all observations by the total, number of observations., Thus, the mean of n observations x1 , x 2 , x 3 , K , x n , is, defined as, n, , ∑ xi, x1 + x 2 + x 3 + … + x n i = 1, Mean ( x ) =, =, n, n, where, the Greek letter ‘Σ’ (sigma) means ‘Summation.’, Let x1 , x 2 , . . . , x n be n observations with respective, frequencies f 1 , f 2 , . . . , f n . This means observation x1 occurs, f 1 times, x 2 occurs f 2 times and so on., n, , ∑ f i xi, , ∴, , Mean ( x ) =, , i =1, n, , ∑ fi, , =, , In general, for the ith class interval, we have frequency f i, corresponding to the class mark x i . The sum of the values in, the last column gives us Σf i x i , so the mean x of the given data, is given by, Σf x, x= i i., Σf i, 2. Assumed Mean Method, , The cases, in which numerical values of x i and f i are large and, computation of product of x i and f i becomes tedious and time, consuming, assumed mean method is used. In this method,, first of all, one among x i ’s is chosen as the assumed mean, denoted by ‘a’. After that, the difference di between a and, each of the x i ’ s, i.e. d i = x i − a is calculated., Then, arithmetic mean is given by, , Σ f i xi, Σ fi, , i =1, , Method of Calculating, Mean of Grouped Data, 1. Direct Method, , In this method, we find the class marks of each class, interval. These class marks would serve as the, representative of whole class and are represented by x i ., , x = a+, where,, , Σf i d i, Σf i, , d i = xi − a, , Mode, , The observation, which occurs most frequently among the, given observations, i.e. the value of the observation having, maximum frequency is called mode. e.g. Mode of the numbers, 2, 3, 4, 4, 6, 6, 6, 6, 7 and 9 is 6 because it is repeated, maximum number of times, i.e. 4 times., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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147, , CBSE Term II Mathematics X (Standard), , Modal Class, , Case II If n is even, then, , In a grouped frequency distribution, it is not possible to, determine the mode by looking at the frequencies. So, here, we first locate a class with the maximum frequency. This class, is called modal class. e.g., Class interval, Number of students, , 0-10, 2, , 10-20 20-30 30-40 40-50 50-60, 9, , 14, , 20, , 22, , 8, , Here, the highest frequency is of the class 40-50, which is 22., Hence, the modal class is 40-50., , Mode of Grouped Data, In grouped data, mode is a value that lies in the modal class, and it is given by the formula,, ⎧ f1 − f 0 ⎫, Mode = l + ⎨, ⎬×h, ⎩2f 1 − f 0 − f 2 ⎭, where, l = lower limit of the modal class, h = size of the class intervals, (assuming all class sizes to be equal), f 1 = frequency of the modal class, f 0 = frequency of the class preceding the modal class, f 2 = frequency of the class succeeding the modal class, , Median, Median is defined as the middle-most or the central, observation, when the observations are arranged either in, ascending or descending order of their magnitudes., Median divides the arranged series into two equal parts, i.e., 50% of the observations lie below the median and the, remaining are above the median., Let n be the total number of observations and suppose that, they are arranged in ascending or descending order., Median of the data depends on the number of, observations (n)., Case I If n is odd, then, ⎛ n + 1⎞, Median = Value of ⎜, ⎟ th observation, ⎝ 2 ⎠, e.g. If five girls of different heights are made to, stand in a row, in descending order of their heights,, then the height of the third girl from either end is, median height., Since, n = 5 is odd., 5 +1, ⎛ n + 1⎞, ∴ Median = ⎜, ⎟ th observation =, ⎝ 2 ⎠, 2, 6, = = 3rd observation, 2, , ⎛n ⎞, ⎛ n⎞, Median = Mean of value of ⎜ ⎟ th and ⎜ + 1⎟ th, ⎠, ⎝2, ⎝ 2⎠, observations, ⎡⎛ n ⎞, 1, ⎛n ⎞ ⎤, = × Value of ⎢⎜ ⎟ th + ⎜ + 1⎟ th⎥, ⎝, ⎠, ⎝2, ⎠ ⎦, 2, 2, ⎣, observations, e.g. If six girls of different heights are made to stand, in a row, in descending order of their heights, then, the mean height of third and fourth girl from either, end is the median height., Since, n = 6 is even., n 6, So, = = 3rd observation, 2 2, 6 +2, ⎛n ⎞ 6, and ⎜ + 1⎟ = + 1 =, = 4th observation, ⎝2 ⎠ 2, 2, ∴ Median = Mean of 3rd and 4th observations, , Cumulative Frequency, The frequency of an observation in a data refers to how many, times that observation occur in the data. Cumulative, frequency of a class is defined as the sum of all frequencies, upto the given class., Less than type and more than type. Formation of these two, distributions can be understood with the help of following, example., e.g. Consider a grouped frequency distribution of marks, obtained out of 100, by 58 students, in a certain examination,, as follows:, Marks, , Number of students, , 0-10, , 5, , 10-20, , 7, , 20-30, , 4, , 30-40, , 2, , 40-50, , 3, , 50-60, , 6, , 60-70, , 7, , 70-80, , 9, , 80-90, 90-100, , 8, 7, , Cumulative frequency distribution of the less than type, Here, the number of students who have scored marks less, than 10 are 5. The number of students who have scored, marks less than 20 includes the number of students who have, scored marks from 0-10 as well as the number of students, who have scored marks from 10-20., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 148, Thus, the total number of students with marks less than 20 is, 5 + 7 , i.e. 12. So, the cumulative frequency of the class 10-20, is 12., Similarly, on computing the cumulative frequencies of the, other classes, which is shown in the table., Marks obtained, , Number of students, (cumulative frequency), , Less than 10, , 5, , Less than 20, , 5 + 7 = 12, , Less than 30, , 12 + 4 = 16, , Less than 40, , 16 + 2 = 18, , Less than 50, , 18 + 3 = 21, , Less than 60, , 21 + 6 = 27, , Less than 70, , 27 + 7 = 34, , Less than 80, , 34 + 9 = 43, , Less than 90, , 43 + 8 = 51, , Less than 100, , 51 + 7 = 58, , Cumulative frequency distribution of the more than type, For this type of distribution, we make the table for the, number of students with scores, more than or equal to 0,, more than or equal to 10, more than or equal to 20 and so on., From the example, we observed that all 58 students have, scored marks more than or equal to 0., There are 5 students scoring marks in the interval 0-10, it, shows that there are 58 − 5 = 53 students getting more than or, equal to 10 marks. In the same manner, the number of, students scoring 20 marks or above = 53 − 7 = 46 students,, and so on., Marks obtained, , Number of students, (cumulative frequency), , More than or equal to 0, , 58, , More than or equal to 10, , 58 − 5 = 53, , More than or equal to 20, , 53 − 7 = 46, , More than or equal to 30, , 46 − 4 = 42, , More than or equal to 40, , 42 − 2 = 40, , More than or equal to 50, , 40 − 3 = 37, , More than or equal to 60, , 37 − 6 = 31, , More than or equal to 70, , 31 − 7 = 24, , More than or equal to 80, , 24 − 9 = 15, , More than or equal to 90, , 15 − 8 = 7, , Median for Discrete Series, A series having observations x1 , x 2 , x 3 , K , x n with respective, frequencies f 1 , f 2 , f 3 K , f n is known as discrete series., Method to Find the Median of the Discrete Series, , Firstly, we arrange the data in the ascending or descending, order of x i , then we find the cumulative frequencies of all the, observations., , Let n be the total number of observations (sum of, frequencies), then median of the data depends on the number, of observations ( n )., ⎛ n + 1⎞, If n is odd, then Median = Value of ⎜, ⎟ th observation., ⎝ 2 ⎠, If n is even, then, ⎛n ⎞, ⎛ n⎞, Median = Mean of ⎜ ⎟ th and ⎜ + 1⎟ th observations, ⎝2, ⎝ 2⎠, ⎠, =, , ⎡⎛ n ⎞, 1, ⎛n ⎞ ⎤, × Value of ⎢⎜ ⎟ th + ⎜ + 1⎟ th⎥ observations, ⎠ ⎦, ⎝2, ⎠, ⎝, 2, ⎣ 2, , Here, for the value of observation, first look at the cumulative, frequency just greater than (and nearest to) the position of, required observations. Then, determine the corresponding, value of the observation., , Median for Grouped Data, In a grouped data, we may not find the middle observation by, looking at the cumulative frequencies, since the middle, observation will be some value in a class interval, so it is, necessary to find the value inside a class that divides the, whole distribution into two halves., For this, we find the cumulative frequencies of all the classes, n, and then determine , where n = number of observations., 2, Now, locate the class whose cumulative frequency is greater, n, than (i.e. nearest to) and this class is called median class., 2, After finding the median class, use the following formula for, calculating the median., ⎫, ⎧N, − cf, ⎪, ⎪2, Median = l + ⎨, ⎬×h, ⎪ f ⎪, ⎭, ⎩, where, l = lower limit of median class, N = sum of frequencies, cf = cumulative frequency of the class preceding, the median class, f = frequency of the median class, h = class width (assuming class sizes to be equal), , Relationship among Mean,, Median and Mode, There is an empirical relationship among the three measures, of central tendency, which is given by, Mode = 3(Median) − 2(Mean), 3 (Median) − Mode, or, Mean =, 2, Mode + 2 (Mean), or, Median =, 3, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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149, , CBSE Term II Mathematics X (Standard), , Solved Examples, Example 1. Find the mean of the following data., , Sol. The table for given data is, , x, , 10, , 30, , 50, , 70, , 89, , Class, , Frequency (fi ), , Mid-value ( x i ), , fi x i, , f, , 7, , 8, , 10, , 15, , 10, , 2-4, , 6, , 3, , 18, , 4-6, , 8, , 5, , 40, , Sol. Table for the given data is, , xi, , fi, , fi x i, , 6-8, , 15, , 7, , 105, , 10, , 7, , 70, , 8-10, , p, , 9, , 9p, , 30, , 8, , 240, , 10-12, , 8, , 11, , 88, , 50, , 10, , 500, , 12-14, , 4, , 13, , 70, , 15, , 1050, , 89, , 10, , 890, , Total, , Σ fi = 50, , Σ fi x i = 2750, , ∴, , Σfi = p + 41, Given, mean = 7.5, Σfi x i, 9p + 303, = 7. 5, = 7. 5 ⇒, Σfi, p + 41, , Σfi = 50 and Σfi x i = 2750, , Here,, , ∴, , Σfi x i, Σfi, , Mean ( x ) =, , ⇒, ⇒, ⇒, , 9p + 303 = 7.5p + 307. 5, 9p − 7.5p = 307.5 − 303, 1. 5p = 4. 5, 4.5, p=, =3, ⇒, 1.5, Hence, value of p is 3., , 2750, = 55, 50, Hence, mean of the given data is 55., =, , Example 2. Calculate the mean of the scores of 20, , Example 4. The weights of tea in 70 packets are shown, , students in a mathematics test, 10-20 20-30 30-40 40-50 50-60, , Marks, , 2, , Number of students, , 52, Σ fi x i, = 9p + 303, , 4, , 7, , 6, , in the following table, Weight (in gm), , 1, , Sol. We first, find the class marks x i of each class and then, proceed as follows, , Number of packets, , 200-201, , 13, , 201-202, , 27, , 202-203, , 18, , Marks, , Class marks ( x i ), , Frequency ( fi ), , fi xi, , 203-204, , 10, , 10-20, 20-30, 30-40, 40-50, 50-60, , 15, 25, 35, 45, 55, , 2, 4, 7, 6, 1, Σ fi = 20, , 30, 100, 245, 270, 55, Σ fi x i = 700, , 204-205, , 1, , 205-206, , 1, , Therefore, mean ( x ) =, , Σfi x i 700, =, = 35, Σfi, 20, , Hence, the mean of scores of 20 students in mathematics test, is 35., , Example 3. Find the value of p, if the mean of the, following distribution is 7.5., Classes, , 2-4, , 4-6, , 6-8, , 8-10, , 10-12, , 12-14, , Frequency (fi ), , 6, , 8, , 15, , p, , 8, , 4, , Find the mean weight of packets., Sol. First, we find the class marks of the given data as follows., Weight, (in gm), , Number of, Packets ( fi ), , Class, Deviation, marks ( x i ) ( di = x i − a ), , fi di, , 200-201, , 13, , 200.5, , −3, , − 39, , 201-202, , 27, , 201.5, , −2, , − 54, , 202-203, , 18, , 202.5, , −1, , − 18, , 203-204, , 10, , a = 203. 5, , 0, , 0, , 204-205, , 1, , 204.5, , 1, , 1, , 1, , 205.5, , 2, , 205-206, , N = ∑ fi = 70, , 2, ∑ fi di = − 108, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 150, Here, assume mean ( a ) = 203. 5, ∑ fi di, Mean ( x ) = a +, ∴, ∑ fi, 108, = 203.5 −, 70, = 203.5 − 1.54, = 201.96, Hence, the required mean weight is 201.96 gm., , Example 5. The following distribution gives cumulative, frequencies of ‘more than type’:, Marks obtained, (More than or equal to), , 5, , 10, , 15, , Sol. Here, the number of students who have scored marks less, than 10 are 10. The number of students who have scored, marks less than 20 includes the number of students who, have scored marks from 0-10 as well as the number of, students who have scored marks from 10-20., Thus, the total number of students with marks less than 20 is, 10 + 8, i.e. 18. So, the cumulative frequency of the class, 10-20 is 18., Similarly, on computing the cumulative frequencies of the, other classes, i.e. the number of students with marks less, than 30, less than 40, … less than 100, we get the, distribution which is called the cumulative frequency, distribution of the less than type., , 20, Marks obtained, , Number of students, (cumulative frequency), , Less than 10, , 10, , Change the above data into a continuous grouped, frequency distribution., [CBSE 2015], , Less than 20, , 10 + 8 = 18, , Less than 30, , 18 + 7 = 25, , Sol. Given, distribution is the more than type distribution., Here, we observe that, all 30 students have obtained marks, more than or equal to 5. Further, since 23 students have, obtained score more than or equal to 10. So, 30 − 23 = 7, students lie in the class 5-10. Similarly, we can find the other, classes and their corresponding frequencies. Now, we, construct the continuous grouped frequency distribution as, , Less than 40, , 25 + 4 = 29, , Less than 50, , 29 + 6 = 35, , Less than 60, , 35 + 8 = 43, , Number of students, (cumulative frequency), , 30, , 23, , 8, , 2, , Less than 70, , 43 + 5 = 48, , Less than 80, , 48 + 9 = 57, , Less than 90, , 57 + 5 = 62, , Less than 100, , 62 + 8 = 80, , Class (Marks obtained), , Number of students, , 5-10, , 30 – 23 = 7, , 10-15, , 23 – 8 = 15, , 15-20, , 8–2=6, , Example 7. In a class of 72 students, marks obtained by, , More than or equal to 20, , 2, , the students in a class test (out of 10) are given, below:, , Example 6. Consider a grouped frequency distribution, of marks obtained out of 100, by 70 students in a, certain examination, as follows:, Marks, , Number of students, , 0-10, , 10, , Here, 10, 20, 30,…, 100 are the upper limits of the, respective class intervals., , Marks obtained, (Out of 10), , 1, , 2, , 3, , 4, , 6, , 7, , 9, , 10, , Number of students, , 3, , 5, , 12, , 18, , 23, , 8, , 2, , 1, , Find the mode of the data., , 10-20, , 8, , Sol. The mode of the given data is 6 as it has the maximum, frequency, i.e. 23 among all the observations., , 20-30, , 7, , Example 8. The weight of coffee in 70 packets are, , 30-40, , 4, , 40-50, , 6, , 50-60, , 8, , 200-201, , 12, , 60-70, , 5, , 201-202, , 26, , 70-80, , 9, , 202-203, , 20, , 80-90, , 5, , 203-204, , 9, , 90-100, , 8, , 204-205, , 2, , 205-206, , 1, , Form the cumulative frequency distribution of less, than type., , shown in the following table, Weight (in gm), , Number of packets, , Determine the modal weight., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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151, , CBSE Term II Mathematics X (Standard), , Sol. In the given data, the highest frequency is 26, which lies in, the interval 201-202, Here, l = 201, f1 = 26, f0 = 12 , f2 = 20 and h = 1, ⎞, ⎛ f −f, ∴ Mode = l + ⎜ 1 0 ⎟ × h, ⎝ 2 f1 − f0 − f2 ⎠, ⎞, ⎛, 26 − 12, = 201 + ⎜, ⎟ ×1, ⎝ 2 × 26 − 12 − 20⎠, , Sol. In a given data, the highest frequency is 41, which lies in the, interval 10000-15000., Here, l = 10000, f1 = 41, f0 = 26,, f2 = 16 and h = 5000, ⎞, ⎛ f −f, ∴ Mode = l + ⎜ 1 0 ⎟ × h, ⎝ 2 f1 − f0 − f2 ⎠, ⎞, ⎛, 41 − 26, = 10000 + ⎜, ⎟ × 5000, ⎝ 2 × 41 − 26 − 16⎠, , ⎛ 14 ⎞, 14, = 201 + ⎜, ⎟ = 201 +, ⎝ 52 − 32 ⎠, 20, , ⎛ 15 ⎞, = 10000 + ⎜, ⎟ × 5000, ⎝ 82 − 42 ⎠, , = 201 + 0.7 = 201.7 gm, Hence, the modal weight is 201.7 gm., , ⎛ 15 ⎞, = 10000 + ⎜ ⎟ × 5000, ⎝ 40⎠, , Example 9. Find the mode of the following distribution, Marks, , 0-10, , Number of, students, , 4, , 10-20 20-30 30-40 40-50 50-60, 6, , 7, , 12, , 5, , 6, , Sol. Given, distribution table is, Marks, , Number of students, , 0-10, , 4, , 10-20, , 6, , 20-30, , 7 ( f0 ), , 30-40, , 12 ( f1 ), , 40-50, , 5 ( f2 ), , 50-60, , 6, , = 10000 + 15 × 125, = 10000 + 1875, = ` 11875, Hence, the modal income is ` 11875., , Example 11. Find the median of the following data., Marks obtained, , 20, , 29, , 28, , 42, , 19, , 35, , 51, , Number of students, , 3, , 4, , 5, , 7, , 9, , 2, , 3, , Sol. Let us arrange the data in ascending order of x i and make a, cumulative frequency table., Marks, obtained (x i ), , Number of, students (fi ), , Cumulative, frequency (cf ), , 19, , 9, , 9, , The highest frequency in the given distribution is 12, whose, corresponding class is 30 - 40., Thus, 30-40 is the required modal class., Here, l = 30, f1 = 12 , f0 = 7, f2 = 5 and h = 10, f1 − f0, ∴ Mode = l +, ×h, 2 f1 − f0 − f2, 12 − 7, × 10, = 30 +, 2 × 12 − 7 − 5, 50, 50, = 30 +, = 30 +, = 30 + 4.17 = 34.17, 24 − 12, 12, , Here, n = 33 (odd), , Hence, mode of the given distribution is 34.17., , ∴, , Example 10. The monthly income of 100 families are, given as below, Income (in `), , Number of families, , 0-5000, , 8, , 20, , 3, , 9 + 3 = 12, , 28, , 5, , 12 + 5 = 17, , 29, , 4, , 17 + 4 = 21, , 35, , 2, , 21 + 2 = 23, , 42, , 7, , 23 + 7 = 30, , 51, , 3, , 30 + 3 = 33, , ⎛ n + 1⎞, Median = Value of ⎜, ⎟ th observation, ⎝ 2 ⎠, ⎛ 33 + 1⎞, = Value of ⎜, ⎟ th observation, ⎝ 2 ⎠, , = Value of 17th observation, Corresponding value of 17th observation of cumulative, frequency in x i is 28. Hence, median is 28., , 5000-10000, , 26, , 10000-15000, , 41, , 15000-20000, , 16, , 20000-25000, , 3, , 25000-30000, , 3, , 30000-35000, , 2, , Number of letters, , 0-5, , 5-10, , 35000-40000, , 1, , Number of surnames, , 20, , 60, , Calculate the modal income., , Example 12. 200 surnames were randomly picked up, from a local telephone directory and the frequency, distribution of the number of letters in English, alphabets in the surnames was obtained as follows:, 10-15 15-20 20-25, 80, , 32, , 8, , Find the median of the above data., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 152, Sol. The cumulative frequency table of given data is, Number of, letters, , Number of, surnames ( fi ), , Cumulative, frequency (cf), , 0-5, , 20, , 20, , 5-10, , 60, , 20 + 60 = 80 ( cf ), , 10-15, , 80 ( = f ), , 80 + 80 = 160, , 15-20, , 32, , 160 + 32 = 192, , 20-25, , 8, , 192 + 8 = 200, , Total, , N = 200, , ( 45 − 40 − p), , 20, 5−p, ⎛ 5 − p⎞, ⇒ 50 = 50 + ⎜, ⎟ ⇒0 =, ⎝ 2 ⎠, 2, ∴, Also,, ⇒, ⇒, ∴, , × 10, [Median = 50], , p=5, 78 + p + q = 90, 78 + 5 + q = 90, q = 90 − 83, q =7, , [given], , Example 14. The median of the following data is 525., , Since, the cumulative frequency just greater than 100 is 160, and the corresponding class interval is 10-15., N 200, ∴ N = 200; ∴, =, = 100, 2, 2, Here, l = 10, cf = 80, h = 5 and f = 80, ⎫, ⎧N, − cf, ⎪2, ⎪, ⎧100 − 80 ⎫, Now, median = l + ⎨, ⎬ × h = 10 + ⎨, ⎬×5, ⎩ 80 ⎭, ⎪ f ⎪, ⎩, ⎭, ⎛ 20 ⎞, = 10 + ⎜ ⎟ × 5 = 10 + 1.25 = 11.25, ⎝ 80⎠, , Find the values of x and y, if total frequency is 100., Class, , 0- 100- 200- 300- 400- 500- 600- 700- 800- 900100 200 300 400 500 600 700 800 900 1000, , Frequency 2, , 5, , 12, , x, , 17, , 20, , 9, , y, , 7, , 4, , Sol. Given, frequency table is, Class, , Frequency ( f1 ), , Cumulative Frequency ( cf ), , 0-100, , 2, , 2, , 100-200, , 5, , 7, , 200-300, , x, , 7+x, , 300-400, , 12, , 19 + x, , 400-500, , 17, , 36 + x ( cf ), , Frequency, , 500-600, , 20 ( f ), , 56 + x, , 20-30, , p, , 600-700, , y, , 56 + x + y, , 30-40, , 15, , 700-800, , 9, , 65 + x + y, , 40-50, , 25, , 800-900, , 7, , 72 + x + y, , 50-60, , 20, , 900-1000, , 4, , 76 + x + y, , 60-70, , q, , 70-80, , 8, , 80-90, , 10, , Example 13. The median of the following data is 50., Find the values of p and q, if the sum of all the, frequencies is 90., Marks, , Sol., Marks, , Frequency ( fi ), , Cumulative frequency ( cf ), , 20-30, 30-40, 40-50, 50-60, 60-70, 70-80, 80-90, , p, 15, 25, 20 ( = f ), q, 8, 10, , p, 15 + p, 40 + p = cf, 60 + p, 60 + p + q, 68 + p + q, 78 + p + q, , N = 90, N 90, =, = 45, ∴, 2, 2, which lies in the interval 50-60., Here, l = 50, f = 20, cf = 40 + p and h = 10, ⎛N, ⎞, ⎜ − cf⎟, ⎝2, ⎠, Median = l +, Q, ×h, f, Given,, , = 50 +, , Given, total frequency is 100., ∴ 2 + 5 + x + 12 + 17 + 20 + y + 9 + 7 + 4 = 100, ⇒, 76 + x + y = 100, …(i), ⇒, x + y = 24, It is given that the median is 525., Clearly, 525 lies in the class 500-600. So, 500-600 is the, median class., Here,, l = 500, h = 100, f = 20 and cf = 36 + x, ∴, N = 100, N, − cf, ×h, Q Median = l + 2, f, 50 − 36 − x, 525 = 500 +, × 100, ⇒, 20, ⇒, 525 = 500 + (14 − x ) × 5, ⇒, 525 = 500 + 70 − 5x, ⇒, 5x = 570 − 525, 45, ⇒, 5x = 45 ⇒ x =, =9, 5, Put x = 9 in Eq. (i), we get, 9 + y = 24 ⇒ y = 24 − 9 = 15, Hence, x = 9 and y = 15, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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153, , CBSE Term II Mathematics X (Standard), , Chapter, Practice, 8. If the arithmetic mean of the following distribution, is 47, then the value of p is, , PART 1, Objective Questions, ●, , 1. Which of the following is a measure of central, tendency?, (b) Cumulative frequency, (d) Class-limit, , 2. While computing mean of grouped data, we assume, that the frequencies are, (a) evenly distributed over all the class, (b) centred at the class marks of the class, (c) centred at the upper limits of the class, (d) centred at the lower limits of the class, , (a) 10, , (b) 11, , (c) n x, , 1414.2, , 14.214.4, , 14.414.6, , 14.614.8, , 14.815, , Frequency, , 2, , 4, , 5, , 71, , 48, , 20, , (b) 71, , (b) −1, , (c) 1, , (d) 130, , 10. For the following distribution, Number of students, 3, , Below 20, , 12, , Below 30, , 27, , Below 40, , 57, , (d) 36, , Below 50, , 75, , n, , Below 60, , 80, , ∑ x i is, , (d), , x, n, , (d) 2, , Σf d, 7. In the formula x = a + i i , for finding the mean, Σf i, of grouped data d i ’s are deviation from a of, (a) lower limits of the class, (b) upper limits of the class, (c) mid-points of the class, (d) frequencies of the class marks, , (c) 82, , Marks, , 6. If x i ’s are the mid-points of the class intervals of, grouped data, f i ’s are the corresponding, frequencies and x is the mean, then Σ ( f i x i − x ) is, equal to, (a) 0, , (d) 12, , Below 10, , i =1, , (b) 2 x, , 5, , 13.814, , (a) 11, , 5. If x is the mean of x’s, then the value of, x, (a), 2, , (c) 13, , p, , Class, , 4. If the difference of mode and median of a data is 24,, then the difference of median and mean is, (c) 8, , 20, , 9. The times (in seconds) taken by 150 atheletes to run, a 110 m hurdle race are tabulated below, , (a) evenly distributed over all the class, (b) centred at the class marks of the class, (c) centred at the upper limits of the class, (d) centred at the lower limits of the class, , (b) 24, , 40-60 60-80 80-100, , 15, , The number of atheletes who completed the race in, less than 14.6 s is, , 3. While computing the mean of grouped data, we, assume that the frequencies are, , (a) 12, , 8, , Frequency, , Multiple Choice Questions, , (a) Frequency, (c) Mean, , 0-20 20-40, , Class interval, , The modal class is, (a) 10-20, , (b) 20-30, , (c) 30-40, , (d) 50-60, , 11. Consider the following distribution, Marks obtained, , Number of students, , More than or equal to 0, , 63, , More than or equal to 10, , 58, , More than or equal to 20, , 55, , More than or equal to 30, , 51, , More than or equal to 40, , 48, , More than or equal to 50, , 42, , The frequency of the class 30-40 is, (a) 3, , (b) 4, , (c) 48, , (d) 51, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 154, 12. For the following distribution, Marks, , Number of, students, , Marks, , Number of, students, , Below 10, , 3, , Below 40, , 57, , Below 20, , 12, , Below 50, , 75, , Below 30, , 28, , Below 60, , 80, , 18. The mean, mode and median of grouped data will, always be, (a) same, (b) different, (c) depends on the type of data, (d) None of the above, , 19. The mean and median of a distribution are 14 and, 15 respectively. The value of mode is, , The modal class is, (a) 0-20, , (b) 20-30, , (c) 30-40, , 13. A student noted the number of cars passing through, a spot on a road for 100 periods each of 3 min and, summarised in the table given below., Number of cars, 0-10, , 7, 14, , 20-30, , 13, , 30-40, , 12, , 40-50, , 20, , 50-60, , 11, , 60-70, , 15, , 70-80, , 8, , ●, , (b) 17, (d) 18, , Case Based Study, 20. Analysis of Water Consumption in a Society, An inspector in an enforcement squad of department, of water resources visit to a society of 100 families, and record their monthly consumption of water on, the basis of family members and wastage of water,, which is summarise in the following table., , Frequency, , 10-20, , (a) 16, (c) 13, , (d) 50-60, , Monthly Consumption 0- 10- 20- 30- 40- 5010 20 30 40 50 60, (in kWh), Number of Families, , 10, , x, , 25 30, , y, , Total, , 10, , 100, , Then, the mode of the data is, (a) 34.7, , (b) 44.7, , (c) 54.7, , (d) 64.7, , 14. Mode of the following grouped frequency, distribution is, Class, , 3-6, , 6-9, , Frequency, , 2, , 5, , (a) 13.6, , 9-12 12-15 15-18 18-21 21-24, 10, , (b) 15.6, , 23, , 21, , 12, , (c) 14.6, , 3, , (d) 16.6, , 15. If the number of runs scored by 11 players of a, cricket team of India are 5, 19, 42, 11, 50, 30, 21, 0,, 52, 36, 27, then median is, (a) 30, , (b) 32, , (c) 36, , (d) 27, , 16. Consider the following frequency distribution, Class, , 0-5, , 6-11, , 12-17, , 18-23, , 24-29, , Frequency, , 13, , 10, , 15, , 8, , 11, , The upper limit of the median class is, (a) 17, , (b) 17.5, , (c) 18, , (d) 18.5, , (a) 50, (c) 25, , 17. Consider the following frequency distribution, Class, , 6585, , 85105, , Frequency, , 4, , 5, , 105- 125- 145- 165- 185125 145 165 185 205, 13, , 20, , 14, , 7, , The difference of the upper limit of the median, class and the lower limit of the modal class is, (a) 0, , (b) 19, , (c) 20, , (d) 38, , Based on the above information, answer the, following questions., (i) The value of x + y is, , 4, , (b) 42, (d) 200, , (ii) If the median of the above data is 32, then x is, equal to, (a) 10, (c) 9, , (b) 8, (d) None of these, , (iii) What will be the upper limit of the modal class?, (a) 40, (c) 65, , (b) 60, (d) 70, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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155, , CBSE Term II Mathematics X (Standard), , (iv) If A be the assumed mean, then A is always, , (iv) If a machine work for 10 h in a day, then, approximate time required to complete the work, for a machine is, , (a) > (Actual mean), (b) < (Actual Mean), (c) = (Actual Mean), (d) Can’t say, , (a) 3 days, (c) 5 days, , (v) The measure of central tendency is, , (v) The class mark of the modal class is, (a) 25, (c) 30, , (a) Mean, (c) Mode, , (b) 35, (d) 45, , 21. As the demand for the products grew a, manufacturing company decided to purchase more, machines. For which they want to know the mean, time required to complete the work for a worker., The following table shows the frequency, distribution of the time required for each machine, to complete a work., Time (in hours), , 15-19 20-24 25-29 30-34 35-39, , Number of machines, , 20, , 35, , 32, , 28, , (b) 22, (d) 32, , (ii) If x i ’s denotes the class mark and f i ’s denotes the, corresponding frequencies for the given data, then, the value of Σx i f i equals to, (a) 3600, (b) 3205, (c) 3670, (d) 3795, , (iii) The mean time required to complete the work for a, worker is, (a) 27.10 h, (b) 23 h, (c) 24 h, (d) None of the above, , (b) Median, (d) All of these, , 22. Direct income in India was drastically impacted, due to the COVID-19 lockdown. Most of the, companies decided to bring down the salaries of the, employees upto 50%., The following table shows the salaries (in percent), received by 50 employees during lockdown., Salaries received (in %), Number of employees, , 50-60 60-70 70-80 80-90, 18, , 12, , 16, , 4, , 25, , Based on the above information, answer the, following questions., (i) The class mark of the modal class 30-34 is, (a) 17, (c) 27, , (b) 4 days, (d) 6 days, , Based on the above information, answer the, following questions., (i) Total number of persons whose salary is reduced, by more than 20% is, (a) 40, , (b) 46, , (c) 30, , (d) 22, , (ii) Total number of persons whose salary is reduced, by atmost 40% is, (a) 32, (c) 46, , (b) 40, (d) 18, , (iii) The modal class is, (a) 50-60, (c) 70-80, , (b) 60-70, (d) 80-90, , (iv) The median class of the given data is, (a) 50-60, (c) 70-80, , (b) 60-70, (d) 80-90, , (v) The empirical relationship among mean, median, and mode is, (a) 3 Median = Mode + 2 Mean, (b) 3 Median = Mode −2 Mean, (c) Median = 3 Mode −2 Mean, (d) Median = 3 Mode + 2 Mean, , ER, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 156, , Find the mean mileage., The manufacturer claimed that the mileage of the, model was 16 kmL −1 . Do you agree with this, claim?, , PART 2, Subjective Questions, ●, , 9. An NGO working for welfare of cancer patients,, maintained its records as follows:, , Short Answer Type Questions, 1. Find the class marks of the class 15-35 and, class 45-60., 2. What is the arithmetic mean of first n natural, numbers?, 3. Calculate the mean of the following data., Class, , 4-7, , 8-11, , 12-15, , 16-19, , 5, , 4, , 9, , 10, , Frequency, , Age of patients (in years), Number of patients, , Frequency, , 3-5, , 5-7, , 7-10, , 9, , 22, , 27, , 17, , 22-24, , 25-27, , 28-30, , Number of days, , 1, , 3, , 4, , 9, , 13, , Find the mean number of pages written per day., , 6. The mean of the following data is 14. Find the value, of k., 20, , 25, , f, , 7, , k, , 8, , 4, , 5, , 7. The mean of the following frequency distribution is, 18. The frequency f in the class interval 19-21 is, missing. Determine f ., Class interval, Frequency, , 11-13 13-15 15-17 17-19 19-21 21-23 23-25, 3, , 6, , 9, , 13, , 5, , f, , 45, , 120, , 50, , 30, , 75, , 25-30 30-35 35-40, 20, , Frequency, , 19-21, , 15, , 315, , 10-15 15-20 20-25 25-30 30-35 35-40, , Class, , 16-18, , 10, , 35, , 20, , 35, , 15, , 11. Find the mode of the following distribution., , Number of pages, written per day, , 5, , 60-80, , [CBSE 2016], , Class, , 5. The following table gives the number of pages, written by Sarika for completing her own book for, 30 days., , x, , 40-60, , Find mode., , Frequency, , 1-3, , 20-40, , 10. Find the mode of the following distribution., , 4. Find the mean of the distribution., Class, , 0-20, , 4, , 8. The mileage (kmL −1 ) of 50 cars of the same model, was tested by a manufacturer and details are, tabulated as given below., Mileage (kmL −1 ), , 10-12, , 12-14, , 14-16, , 16-18, , Number of cars, , 7, , 12, , 18, , 13, , 36, , 40-45, , 45-50, , 50-55, , 40, , 28, , 14, , 53, , 12. Compute the mode for the following frequency, distribution., Size of items, (in cm), , 0-4, , 4-8, , Frequency, , 5, , 7, , 8-12 12-16 16-20, 9, , 17, , 20-24, , 24-28, , 10, , 6, , 12, , 13. Find the mode of the following frequency, distribution., Class, , 15-20, , 20-25, , 25-30, , 30-35, , 35-40, , 40-45, , Frequency, , 3, , 8, , 9, , 10, , 3, , 2, , 14. The set of data given below shows the ages of, participants in a certain summer camp. Draw a, cumulative frequency table for the data., Age (in years), , 10, , 11, , 12, , 13, , 14, , 15, , Frequency, , 3, , 18, , 13, , 12, , 7, , 27, , 15. Given below is a cumulative frequency distribution, showing the marks secured by 50 students of a, class, Marks, Number of, students, , Below 20 Below 40 Below 60 Below 80, , 17, , 22, , 29, , 37, , Below, 100, 50, , Form the frequency distribution table for the data., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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157, , CBSE Term II Mathematics X (Standard), , 16. The following table shows the cumulative, frequency distribution of marks of 800 students in, an examination., Marks, , 19. From the following distribution, find the median, , Number of students, , Class, , Frequency, , 500-600, , 36, , 600-700, , 32, , Below 10, , 10, , 700-800, , 32, , Below 20, , 50, , 800-900, , 20, , Below 30, , 130, , 900-100, , 30, , Below 40, , 270, , Below 50, , 440, , Below 60, , 570, , Below 70, , 670, , Below 80, , 740, , Below 90, , 780, , Below 100, , 800, , 20. Size of agricultural holdings in a survey of 200, families is given in the following table, Size of agricultural, holdings (in hectare), , Number of, families, , Construct a frequency distribution table for the, data above., , 17. The following distribution of weights (in kg) of, 40 persons., Weight (in kg), , Number of persons, , 40-45, , 4, , 45-50, , 4, , 50-55, , 13, , 55-60, , 5, , 60-65, , 6, , 65-70, , 5, , 70-75, , 2, , 75-80, , 1, , 10, 15, , 10-15, , 30, , 15-20, , 80, , 20-25, , 40, , 25-30, , 20, , 30-35, , 5, , Compute median and modal class of the holdings., 21. If median = 137 units and mean = 137.05 units, then, find the mode., ●, , Long Answer Type Questions, 22. The weights (in kg) of 50 wrestlers are recorded in, the following table., Number of, wrestlers, , Weight (in kg), , Construct a cumulative frequency distribution (of, the less than type) table for the data above., 18. Form the frequency distribution table from the, following data, Marks (Out of 90), , 0-5, 5-10, , Number of candidates, , 100-110, , 4, , 110-120, , 14, , 120-130, , 21, , 130-140, , 8, , 140-150, , 3, , Find the mean weight of the wrestlers., , 23. If mode of the following series is 54, then find the, value of f ., , More than or equal to 80, , 4, , More than or equal to 70, , 6, , More than or equal to 60, , 11, , More than or equal to 50, , 17, , Class, interval, , 0-15, , More than or equal to 40, , 23, , Frequency, , 3, , More than or equal to 30, , 27, , More than or equal to 20, , 30, , More than or equal to 10, , 32, , More than or equal to 0, , 34, , 15-30 30-45 45-60 60-75 75-90, 5, , f, , 16, , 12, , 7, , Find the modal class in which the given mode lies, and find the value of f by using the formula,, ⎧ f1 − f 0 ⎫, Mode = l + ⎨, ⎬×h., ⎩2 f 1 − f 0 − f 2 ⎭, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 158, 24. Find the mode of the following distribution., , 29. Weekly income of 600 families is tabulated below, , Classes, , 0-20, , 20-40, , 40-60, , 60-80, , 80-100, , Frequency, , 10, , 8, , 12, , 16, , 4, , Weekly income (in `) Number of families, , 25. The following are the ages of 300 patients getting, medical treatment in a hospital on a particular day, Age (in years) 10-20 20-30 30-40 40-50 50-60 60-70, Number of, patients, , 60, , 42, , 55, , 70, , 53, , 20, , Form, (i) less than type cumulative frequency distribution., (ii) more than type cumulative frequency distribution., , 26. Find the unknown entries a, b, c, d, e and f in the, following distribution of heights of students in a, class, , 0-1000, , 250, , 1000-2000, , 190, , 2000-3000, , 100, , 3000-4000, , 40, , 4000-5000, , 15, , 5000-6000, , 5, , Total, , 600, , Compute the median income., , 30. A survey regarding the heights (in cm) of 51 boys of, Class X of a school was conducted and the following, data was obtained:, Heights (in cm), , Number of boys, , Less than 140, , 4, , Frequency, , Cumulative, frequency, , Less than 145, , 11, , Less than 150, , 29, , 150-155, , 12, , a, , Less than 155, , 40, , 155-160, , b, , 25, , Less than 160, , 46, , 160-165, , 10, , c, , Less than 165, , 51, , 165-170, , d, , 43, , 170-175, , e, , 48, , 175-180, , 2, , f, , Total, , 50, , Height (in cm), , Find the median height., , 31. Find the missing frequencies in the, following frequency distribution table, if N = 100, CBSE 2019, and median is 32., Marks obtained, , 27. The maximum bowling speeds (in km/h) of, 33 players at a cricket coaching centre are given as, follows, Speed (in km/h), , Number of, players, , 85-100, , 11, , 100-115, , 9, , 115-130, , 8, , 130-145, , 5, , Number of, students, , 0-10, , 10-20, , 20-30, , 30-40, , 40-50, , 50-60, , Total, , 10, , ?, , 25, , 30, , ?, , 10, , 100, , 32. The table below shows the salaries of 280 persons., Salary (in ` thousand), , Calculate the median bowling speed., , 28. Obtain the median for the following frequency, distribution., x, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , y, , 8, , 10, , 11, , 16, , 20, , 25, , 15, , 9, , 6, , Number of persons, , 5-10, , 49, , 10-15, , 133, , 15-20, , 63, , 20-25, , 15, , 25-30, , 6, , 30-35, , 7, , 35-40, , 4, , 40-45, , 2, , 45-50, , 1, , Calculate (i) median of the data,, (ii) mode of the data., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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159, , CBSE Term II Mathematics X (Standard), , ●, , (i) Estimate the mean time taken by a student to, finish the race., , Case Based Questions, 33. The men’s 200 m race event at the 2020, Tokyo Olympic took place 3 and 4 August., A stopwatch was used to find the time that it took a, group of Athletes to run 200 m., , Time, (in seconds), 0-20, , 8, , Time (in seconds), , 0-20, , 20-40, , 10, , Number of Students, , 8, , 40-60, , 13, , 20-40 40-60 60-80 80-100, 10, , 13, , 6, , 3, , Number of Students, , 60-80, , 6, , 80-100, , 3, , (ii) What is the sum of lower limits of median class and, modal class., (iii) How many students finished the race within, 1 min?, , SOLUTIONS, Objective Questions, 1. (c) Mean is the measure of central tendency., 2. (b) In computing the mean of grouped data, the frequencies, are centred at the class marks of the class., 3. (b) While computing mean of grouped data, we assume that, the frequencies are centred at the class marks of the classes., 4. (a) We have, Mode − Median = 24, We know that, Mode = 3 Median − 2 Mean, ∴ Mode − Median = 2 Median − 2 Mean, ⇒, 24 = 2(Median − Mean), ⇒ Median − Mean = 12, , 8. (d) Let us construct the following table for finding the, arithmetic mean, Class, interval, 0-20, , 8, , 10, , 80, , 20-40, , 15, , 30, , 450, , 40-60, , 20, , 50, , 1000, , 60-80, , p, , 70, , 70p, , 5, , 90, , 450, , Total, , Σfi = 48 + p, , n, , x=, , Q, , Now,, , i =1, , n, , n, , ⇒, , ∑ xi = n x, i =1, , Σ fi x i, n, ∴ Σ ( fi x i − x ) = Σ fi x i − Σx = nx − nx, =0, 7. (c) We know that, di = x i − a, , 6. (a)Q, , ⇒, , x=, , [Q Σx = nx ], , i.e. di ’s are the deviation from a of mid-points of the classes., , f i xi, , 80-100, , 5. (c) We know that,, , ∑ xi, , Frequency ( fi ) Class mark, ( xi ), , Σfi x i = 1980 + 70p, , Σfi x i, Σfi, 1980 + 70p, =, 48 + p, 1980 + 70p, 47 =, 48 + p, x=, , ⇒ 2256 + 47 p = 1980 + 70p, ⇒, 276 = 23p, ⇒, p = 12, 9. (c) The number of atheletes who completed the race in less, than 14.6 = 2 + 4 + 5 + 71 = 82, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 160, 10. (c), , f1 − f0, ×h, 2 f1 − f0 − f2, 23 − 10, = 12 +, ×3, 46 − 10 − 21, 13, 13, = 12 +, × 3 = 12 +, = 12 + 2. 6 = 14. 6, 15, 5, 15. (d) Arranging the terms in ascending order,, 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52, Here n = 11 (odd), ⎛ n + 1⎞, Q Median = ⎜, ⎟ th, ⎝ 2 ⎠, Q, , Marks, , Number of students Cumulative frequency, , Below 10, , 3=3, , 3, , 10-20, , 12 − 3 = 9, , 12, , 20-30, , 27 − 12 = 15, , 27, , 30-40, , 57 − 27 = 30, , 57, , 40-50, , 75 − 57 = 18, , 75, , 50-60, , 80 − 75 = 5, , 80, , Here, we see that the highest frequency is 30, which lies in, the interval 30-40., 11. (a) Given, the distribution table, , Mode = l +, , ⎛ 11 + 1⎞, Median value = ⎜, ⎟ th = 6th value = 27, ⎝ 2 ⎠, 16. (b) Given, classes are not continuous, so we make, continuous by subtracting 0.5 from lower limit and adding, 0.5 to upper limit of each class., , Marks obtained, , Number of students, , 0-10, , 63 − 58 = 5, , 10-20, , 58 − 55 = 3, , Class, , 20-30, , 55 − 51 = 4, , −0.5-5.5, , 13, , 13, , 30-40, , 51 − 48 = 3, , 5.5-11.5, , 10, , 23, , 40-50, , 48 − 42 = 6, , 11.5-17.5, , 15, , 38, , 50, , 42, , 17.5-23.5, , 8, , 46, , 23.5-29.5, , 11, , 57, , Frequency of the modal class 30-40 is 3 from the above table., 12. (c) Let us first construct the following frequency distribution, table., Marks, , Number of students, , 0-10, , 3, , Frequency, , Cumulative frequency, , N 57, =, = 28. 5, which lies in the interval 11.5-17.5., 2, 2, Hence, the upper limit is 17.5., 17. (c), Here,, , 10-20, , 9, , Frequency, , Cumulative frequency, , 20-30, , 16, , 65-85, , 4, , 4, , 30-40, , 29, , 85-105, , 5, , 9, , 40-50, , 18, , 105-125, , 13, , 22, , 125-145, , 20, , 42, , 145-165, , 14, , 56, , 165-185, , 7, , 63, , 185-205, , 4, , 67, , 50-60, , 5, , Since, the maximum frequency is 29 and the class, corresponding to this frequency is 30-40., So, the modal class is 30-40., 13. (b) Here, modal class is 40-50. Since, it has maximum, frequency which is 20., ∴ l = 40, f1 = 20, f0 = 12, f2 = 11 and h = 10, ⎛ f −f, ⎞, Mode = l + ⎜ 1 0 ⎟ × h, Q, ⎝ 2 f1 − f0 − f2 ⎠, ⎛ 20 − 12 ⎞, = 40 + ⎜, ⎟ × 10, ⎝ 40 − 12 − 11⎠, 80, 17, = 40 + 4.7 = 44.7, 14. (c) We observe that the class 12-15 has maximum frequency., Therefore, this is the modal class., We have, l = 12, h = 3 , f1 = 23 , f0 = 10 and f2 = 21, = 40 +, , Class, , N 67, =, = 33. 5 which lies in the interval 125 -145., 2, 2, Hence, upper limit of median class is 145., Here, we see that the highest frequency is 20 which lies in, 125-145. Hence, the lower limit of modal class is 125., ∴ Required difference = Upper limit of median class, − Lower limit of modal class, = 145 − 125 = 20, 18. (c) No, the value of these three measures can be the same, it, depends on the type of data., 19. (b) Given, mean = 14 and median = 15, By using empirical relationship,, Mode = 3 Median −2 Mean = 3 × 15 − 2 × 14 = 45 − 28 = 17, Here,, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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161, , CBSE Term II Mathematics X (Standard), Case Based Study, , ⎡{50 − ( x + 35)}, ⎤, × 10 = 32, ⎢⎣, ⎥⎦, 30, (15 − x ), = 32, ⇒ 30 +, 3, ⇒, (15 − x ) = 6 ⇒ x = 9, Put x = 9 in Eq. (1), we get, y = 16, Hence, x = 9 and y = 16, (iii) (a) Since, the maximum frequency is 30, so the modal, class is 30-40., Hence, upper limit of the modal class is 40., (iv) (d) The value of assumed mean can be less, more or equal, than the actual mean., (v) (b) The modal class is 30-40., 30 + 40 70, =, = 35, ∴ Class mark =, 2, 2, 21. (i) (d) Class mark of class 30-34, 30 + 34, =, 2, 64, =, 2, = 32, (ii) (d) Let’s make the table, ⇒ 30 +, , Class, , 27.10, 10, = 2.710, ~, − 3 days, (d) Measure of central tendency are mean, median and, mode., (b)∴Required number of employees = 18 + 12 + 16, = 46, (a)∴Required number of employees = 12 + 16 + 4, = 32, (a) The maximum frequency is 18 and the corresponding, class is 50-60., Hence, modal class is 50-60., (b) Consider the table, , (iv) (a) Approximate time =, , 20. (i) (c) Given x and y are the frequencies of class intervals, 10-20 and 40-50, respectively. Then,, 10 + x + 25 + 30 + y + 10 = 100, …(1), ⇒, x + y = 25, (ii) (c) Median is 32, which lies in 30-40., So, the median class is 30-40., ∴ l = 30, h = 10, f = 30, N = 100, and cf = 10 + x + 25 = x + 35, ⎛N, ⎞, ⎜ − cf ⎟, 2, Now, median = l + ⎜, ⎟ ×h, ⎜ f ⎟, ⎝, ⎠, , Class marks Frequency ( fi ), ( xi ), , x i fi, , 15-19, , 17, , 20, , 340, , 20-24, , 22, , 35, , 770, , 25-29, , 27, , 32, , 864, , 30-34, , 32, , 28, , 896, , 35-39, , 37, Total, , Σx i fi = 3795, Σx f, (iii) (a) Mean time ( X ) = i i, Σfi, 3795, =, 140, = 27.10, , 25, , 925, , Σfi = 140, , Σx i fi = 3795, , (v), 22. (i), (ii), (iii), , (iv), , Cumulative, frequency ( cf ), , Salaries received, (in %), , Number of, employees ( fi ), , 50-60, , 18, , 18, , 60-70, , 12, , 18 + 12 = 30, , 70-80, , 16, , 30 + 16 = 46, , 80-90, , 4, , 46 + 4 = 50, , Total, , Σfi = 50, , N 50, =, = 25, 2, 2, The cumulative frequency more than 25 lies in 60-70., Hence, median class is 60-70., (v) (a) As we know, Mode = 3 Median − 2 Mean, ∴ 3 Median = Mode + 2 Mean, Here,, , Subjective Questions, Lower limit + Upper limit, 2, 15 + 35 50, =, = 25, ∴ Class mark of 15-35 is =, 2, 2, 45 + 60 105, Class mark of 45-60 is, =, = 52.5, 2, 2, Sum of all the observations, 2. Arithmetic mean =, Number of observations, 1+2 +K+ n, =, n, n, [2 × 1 + ( n − 1)1], = 2, n, [Q1 + 2 + K + n is an AP series whose first term is a = 1 and, common difference is d = 1. We know that, the, n, sum of n th term of an AP is Sn = [2 a + ( n − 1)d ]], 2, 2 + n −1 n + 1, =, =, 2, 2, 3. Since, given data is not continuous, so we subtract 0.5 from, the lower limit and add 0.5 in the upper limit of each class., Now, we first find the class mark x i of each class and then, proceed as follows, , 1. We know that, Class mark =, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 162, , Class, , Class marks, ( xi ), , Frequency (fi ), , f i xi, , 3.5-7.5, , 5.5, , 5, , 27.5, , 7.5-11.5, , 9.5, , 4, , 38, , 11.5-15.5, , 13.5, , 9, , 121.5, , 15.5-19.5, , 17.5, , 10, , 175, , Σ fi = 28, , Σ fi x i = 362, , Therefore, ( x ) mean =, , Σfi x i 362, =, = 12 . 93, Σfi, 28, , Given, mean = 14, Σfi x i, 10k + 360, = 14 ⇒, = 14, ∴, Σfi, k + 24, ⇒, ⇒, ⇒, ⇒, , 10k + 360 = 14( k + 24), 10k + 360 = 14k + 336, 14k − 10k = 360 − 336, 4k = 24, 24, ∴, k=, =6, 4, Hence, the value of k is 6., 7. Table of given data is, , Hence, mean of the given data is 12.93., , Class interval, , Frequency (fi ), , Mid-value, (x i ), , x i fi, , 11-13, , 3, , 12, , 36, , 4. We first, find the class mark x i of each class and then, proceed as follows., , Class, , Class marks, (xi ), , Frequency, (f i ), , f i xi, , 13-15, , 6, , 14, , 84, , 1-3, 3-5, 5-7, 7-10, , 2, 4, 6, 8.5, , 9, 22, 27, 17, Σ fi = 75, , 18, 88, 162, 144.5, Σ fi x i = 412.5, , 15-17, , 9, , 16, , 144, , 17-19, , 13, , 18, , 234, , 19-21, , f, , 20, , 20f, , 21-23, , 5, , 22, , 110, , 23-25, , 4, , 24, , 96, , Total, , Σfi = 40 + f, , Therefore, mean ( x ) =, , Σ fi x i 412.5, =, = 5.5, Σ fi, 75, , Σfi x i =, 704 + 20f, , Hence, mean of the given distribution is 5.5., 5. Since, given data is not continuous, so we subtract 0.5 from, the lower limit and add 0.5 in the upper limit of each class., , Class mark, , Mid-value (x i ) Number of days, (f i ), , f i xi, , Q, , Mean =, , ∴, , 18 =, , Σfi x i, Σfi, 704 + 20f, 40 + f, , [Q mean = 18, given], , 15.5-18.5, , 17, , 1, , 17, , 18.5-21.5, , 20, , 3, , 60, , ⇒, , 21.5-24.5, , 23, , 4, , 92, , 24.5-27.5, , 26, , 9, , 234, , ⇒, f=8, Hence, missing frequency in the given data is 8., , 27.5-30.5, , 29, , 13, , 377, , 30, , 780, , Total, Mean ( x ) =, , Q, , Σ fi x i 780, =, = 26, Σ fi, 30, , Hence, the mean of pages written per day is 26., 6. Table for the given data is, xi, , fi, , fixi, , 5, , 7, , 35, , 10, , k, , 10k, , 15, , 8, , 120, , 20, , 4, , 80, , 25, , 5, , 125, , Total, , Σ fi = k + 24, , Σ fi x i = 10 k + 360, , Here, Σfi = k + 24 and Σfi x i = 10k + 360, , ⇒ 720 + 18f = 704 + 20f, 16 = 2 f, , 8., , Mileage, (kmL −1 ), , Class marks Number of cars, (x i ), (f i ), , f i xi, , 10-12, , 11, , 7, , 77, , 12-14, , 13, , 12, , 156, , 14-16, , 15, , 18, , 270, , 16-18, , 17, , Total, , 13, , 221, , Σ fi = 50, , Σ fi x i = 724, , Σ fi = 50, Σ fi x i = 724, Σ f x 724, = 14. 48, Q Mean ( x ) = i i =, Σ fi, 50, Here,, and, , Hence, mean mileage is 14.48 kmL −1., No, the manufacturer is claiming mileage 1.52 kmL −1 more, than average mileage., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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163, , CBSE Term II Mathematics X (Standard), , 9., , Age of patients (in years), Number of patients, , 0-20, , 20-40, , 40-60, , 60-80, , 35( f0 ), , 315( f1 ), , 120( f2 ), , 50, , Here, maximum frequency is 315 and the class, corresponding to this frequency is 20-40. So, the modal, class is 20-40., ∴ l = 20, f1 = 315, f0 = 35, f2 = 120 and h = 20, ⎞, ⎛ f −f, Now, Mode = l + ⎜ 1 0 ⎟ × h, ⎝ 2 f1 − f0 − f2 ⎠, 315 − 35, ⎛, ⎞, = 20 + ⎜, ⎟ × 20, ⎝ 2 × 315 − 35 − 120⎠, ⎛ 280 ⎞, = 20 + ⎜, ⎟ × 20, ⎝ 630 − 155⎠, 280, = 20 +, × 20, 475, = 20 + 11.79 = 31.79, Hence, average age of maximum number of patients is 31.79., 10. Given, distribution table is, , In the given table, the highest frequency is 53 and, corresponding class of this frequency is 35-40., Thus, 35-40 is a modal class., Here, l = 35, f1 = 53, f0 = 36, f2 = 40 and h = 5, f1 − f0, Mode = l +, Q, ×h, 2 f1 − f0 − f2, 53 − 36, ×5, = 35 +, 2 × 53 − 36 − 40, 85, 17 × 5, = 35 +, = 35 +, 30, 106 − 76, = 35 + 2 . 83 = 37 . 83, Hence, mode of given data is 37.83., 12. Given frequency distribution table is, Size of items (in cm), , Frequency, , 0-4, , 5, , 4-8, , 7, , 8-12, , 9, , 12-16, , 17, , 16-20, , 12, , Class, , Frequency, , 20-24, , 10, , 10-15, , 45, , 24-28, , 6, , 15-20, , 30( f0 ), , 20-25, , 75( f1 ), , 25-30, , 20( f2 ), , 30-35, , 35, , 35-40, , 15, , The highest frequency in the given data is 75 and the, corresponding class is 20-25, which is a modal class., Here, l = 20, f1 = 75, f0 = 30, f2 = 20 and h = 5, f1 − f0, ×h, Q Mode = l +, 2 f1 − f0 − f2, 75 − 30, ×5, = 20 +, 2 × 75 − 30 − 20, 45 × 5, = 20 +, 150 − 50, 225, = 20 +, = 20 + 2 .25 = 22 . 25, 100, 11. Given, distribution table is, , (approx), , The maximum frequency in the given distribution table is, 17, which lies in the class interval 12-16., ∴Modal class = 12 -16, So, l = 12 , f1 = 17, f0 = 9, f2 = 12 and h = 4, ⎞, ⎛ f −f, Q Mode = l + ⎜ 1 0 ⎟ × h, ⎝ 2 f1 − f0 − f2 ⎠, 17 − 9, ⎛, ⎞, = 12 + ⎜, ⎟ ×4, ⎝ 2 × 17 − 9 − 12 ⎠, ⎛ 8 ⎞, = 12 + ⎜, ⎟ ×4, ⎝ 34 − 21⎠, 32, = 12 + 2 . 46 = 14. 46, 13, Hence, mode of given distribution is 14.46., 13. Given, frequency distribution table is, = 12 +, , Class, , Frequency (fi ), , 15-20, , 3, , 20-25, , 8, , Class, , Frequency, , 25-30, , 9, , 25-30, , 20, , 30-35, , 10, , 30-35, , 36, , 35-40, , 3, , 35-40, , 53, , 40-45, , 2, , 40-45, , 40, , 45-50, , 28, , 50-55, , 14, , The maximum frequency in the given distribution table is, 10, which lies in the class interval 30-35., ∴ Modal Class = 30-35, So, l = 30, f1 = 10, f0 = 9, f2 = 3 and h = 5, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 164, f1 − f0, ×h, 2 f1 − f0 − f2, 10 − 9, = 30 +, ×5, 2 × 10 − 9 − 3, 5, = 30 + = 30 + 0. 625 = 30. 625, 8, Hence, mode of given distribution is 30.625., 14. The cumulative frequency of first observation is the same as, its frequency, since there is no frequency before it., Now, the cumulative frequency table is, Q, , Mode = l +, , Age, (in years), , Frequency, (f i ), , Cumulative frequency, (cf), , 10, , 3, , 3, , 11, , 18, , 3 + 18 = 21, , 12, , 13, , 21 + 13 = 34, , 13, , 12, , 34 + 12 = 46, , 14, , 7, , 46 + 7 = 53, , 15, , 27, , 53 + 27 = 80, , 17. The cumulative distribution (less than type) table is shown, below, , Weight (in kg), , 0-20, , Less than 45, , 4, , Less than 50, , 4+ 4=8, , Less than 55, , 8 + 13 = 21, , Less than 60, , 21 + 5 = 26, , Less than 65, , 26 + 6 = 32, , Less than 70, , 32 + 5 = 37, , Less than 75, , 37 + 2 = 39, , Less than 80, , 39 + 1 = 40, , 18. Here, we observe that, all 34 students have scored marks, more than or equal to 0. Since, 32 students have scored, marks more than or equal to 10. So, 34 − 32 = 2 students lies, in the interval 0-10 and so on., Now, we construct the frequency distribution table., , Class interval, , 15. Here, we observe that, 17 students have scored marks below, 20 i.e. it lies between class interval 0-20 and 22 students, have scored marks below 40, so 22 − 17 = 5 students lies in, the class interval 20-40 continuting in the same manner, we, get the complete frequency distribution table for given data., , Marks, , Cumulative frequency (cf ), , Number of students, 17, , Number of candidates (f i ), , 0-10, , 34 − 32 = 2, , 10-20, , 32 − 30 = 2, , 20-30, , 30 − 27 = 3, , 30-40, , 27 − 23 = 4, , 40-50, , 23 − 17 = 6, , 50-60, , 17 − 11 = 6, , 60-70, , 11 − 6 = 5, , 20-40, , 22 − 17 = 5, , 70-80, , 6− 4 =2, , 40-60, , 29 − 22 = 7, , 80-90, , 4, , 60-80, , 37 − 29 = 8, , 80-100, , 50 − 37 = 13, , 16. Here, we observe that 10 students have scored marks below, 10 i.e. it lies between class interval 0-10. Similarly, 50, students have scored marks below 20. So, 50 − 10 = 40, students lies in the interval 10 - 20 and so on. The table of a, frequency distribution for the given data is, , Class interval, , Number of students (f i ), , 0-10, , 10, , 10-20, , 50 − 10 = 40, , 20-30, , 130 − 50 = 80, , 30-40, , 270 − 130 = 140, , 40-50, , 440 − 270 = 170, , 50-60, , 570 − 440 = 130, , 60-70, , 670 − 570 = 100, , 70-80, , 740 − 670 = 70, , 80-90, , 780 − 740 = 40, , 90-100, , 800 − 780 = 20, , 19. The cumulative frequency table for given distribution is, Cumulative Frequency, (cf), , Class, , Frequency (fi ), , 500-600, , 36, , 36, , 600-700, , 32, , 36 + 32 = 68, , 700-800, , 32 (f), , 68 + 32 = 100, , 800-900, , 20, , 100 + 20 = 120, , 900-1000, , 30, , 120 + 30 = 150, , N 150, =, = 75, which lies in the cumulative frequency, 2, 2, 100, whose corresponding class is 700-800. Thus, modal class, is 700-800., Here, l = 700, cf = 68, f = 32 and h = 100, N, − cf, Q Median = l + 2, ×h, f, 75 − 68, 700, = 700 +, × 100 = 700 +, 32, 32, = 700 + 21. 88 = 721. 88, Hence, median of the given distribution is 721.88., Here,, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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165, , CBSE Term II Mathematics X (Standard), , By assumed mean method,, , 20., , Size of agricultural, holdings (in hec), , Number of, families ( f i ), , Cumulative, frequency (cf ), , 0-5, , 10, , 10, , 5-10, , 15, , 25, , 10-15, , 30, , 55, , 15-20, , 80 (f), , 135, , 20-25, , 40, , 175, , 25-30, , 20, , 195, , 30-35, , 5, , 200, , ⎛ 45⎞, = 15 + ⎜ ⎟, ⎝ 16 ⎠, = 15 + 2. 81 = 17. 81 hec, II. In a given table 80 is the highest frequency., So, the modal class is 15-20., 21. Given, median = 137 units and mean= 137. 05 units., We know that,, Mode = 3(Median) − 2(Mean), = 3 (137 ) − 2(137. 05), = 411 − 274. 10, = 136.90, Hence, the value of mode is 136.90 units., 22. We first find the class mark x i , of each class and then, proceed as follows, , Number of, Deviations, Class, wrestlers marks ( x ) d i = x i − a,, i, (f i ), a = 125, , f i di, , 100-110, , 4, , 105, , −20, , −80, , 110-120, , 14, , 115, , −10, , −140, , 120-130, , 21, , a = 125, , 0, , 0, , 130-140, , 8, , 135, , 10, , 80, , 3, , 145, , 20, , 140-150, , N = Σ fi = 50, , 60, Σ fi di = − 80, , ∴Assumed mean ( a ) = 125,, Class width ( h ) = 10 and total observation (N ) = 50, , Σ fi di, , Σ fi, ( − 80), = 125 +, 50, = 125 − 1. 6 = 123. 4 kg, 23. Here, given mode is 54, which lies between 45-60., Therefore, the modal class is 45-60., ∴, l = 45, f1 = 16, f0 = f , f2 = 12 and h = 15, ⎞, ⎛ f −f, Mode = l + ⎜ 1 0 ⎟ × h, Q, ⎝ 2 f1 − f0 − f2 ⎠, ∴, , I. Here, N = 200, N 200, Now, =, = 100, which lies in the interval 15-20., 2, 2, Here, l = 15 , h = 5, f = 80 and cf = 55, ⎞, ⎛N, ⎜ − cf ⎟, 2, Median = l + ⎜, ∴, ⎟ ×h, ⎜ f ⎟, ⎠, ⎝, ⎛ 100 − 55⎞, = 15 + ⎜, ⎟ ×5, ⎝ 80 ⎠, , Weight, (in kg), , Mean ( x ) = a +, , 16 − f, ⎛, ⎞, 54 = 45 + ⎜, ⎟ × 15, ⎝ 2 × 16 − f − 12 ⎠, 16 − f, × 15, 20 − f, , ⇒, , 9=, , ⇒, , 9(20 − f ) = 15(16 − f ), , ⇒, , 180 − 9f = 240 − 15f, , ⇒, ⇒, , 6f = 240 − 180 = 60, f = 10, , Hence, required value of f is 10., 24. The given distribution table is, Class, , Frequency ( f ), , 0-20, , 10, , 20-40, , 8, , 40-60, , 12 ( f0 ), , 60-80, , 16 ( f1 ), , 80-100, , 4 ( f2 ), , The highest frequency in the given distribution table is 16,, whose corresponding class is 60-80. Thus, 60-80 is the modal, class of the given distribution., Here, l = 60, f1 = 16, f0 = 12, f2 = 4 and h = 20, f1 − f0, Q Mode = l +, ×h, 2 f1 − f0 − f2, 16 − 12, = 60 +, × 20, 2 × 16 − 12 − 4, 4 × 20, 80, = 60 +, = 60 +, 32 − 16, 16, = 60 + 5 = 65, Hence, mode of the given distribution is 65., 25. (i) We observe that the number of patients which take, medical treatment in a hospital on a particular day less, than 10 is 0. Similarly, less than 20 include the number of, patients which take medical treatment from 0-10 as well, as the number of patients which take medical treatment, from 10-20., So, the total number of patients less than 20 is, 0 + 60 = 60, we say that the cumulative frequency of the, class 10-20 is 60. Similarly, for other classes, which is, shown below the table., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 166, (ii) Also, we observe that all 300 patients which take medical, treatment more than or equal to 10. Since, there are 60, patients which take medical treatment in the interval, 10-20, this means that there are 300 − 60 = 240 patients, which take medical treatment more than or equal to 20., Continuing in the same manner, which is shown below, the table., , (i) Less than type, Age (in years), , Number of, students, , Speed (in km/h), , (ii) More than type, Age (in years), , Number of, students, , Less than 10, , 0, , More than or, equal to 10, , 300, , Less than 20, , 60, , More than or, equal to 20, , 240, , Less than 30, , 102, , More than or, equal to 30, , 198, , Less than 40, , 157, , More than or, equal to 40, , 143, , Less than 50, , 227, , More than or, equal to 50, , 73, , Less than 60, , 280, , More than or, equal to 60, , 20, , Less than 70, , 300, , 26., , Frequency, (f i ), , Cumulative, frequency, (given), , Cumulative, frequency, ( cf ), , 150-155, , 12, , a, , 12, , 155-160, , b, , 25, , 12 + b, , 160-165, , 10, , c, , 22 + b, , 165-170, , d, , 43, , 22 + b + d, , 170-175, , e, , 48, , 22 + b + d + e, , 175-180, , 2, , f, , 24 + b + d + e, , Total, , 50, , Height (in cm), , 27. First we construct the cumulative frequency table, , On comparing last two tables, we get, a = 12, ∴, 12 + b = 25, ⇒, b = 25 − 12 = 13, 22 + b = c, ⇒, c = 22 + 13 = 35, 22 + b + d = 43, ⇒, 22 + 13 + d = 43, ⇒, d = 43 − 35 = 8, 22 + b + d + e = 48, ⇒, 22 + 13 + 8 + e = 48, ⇒, e = 48 − 43 = 5, and, 24 + b + d + e = f, ⇒, 24 + 13 + 8 + 5 = f, ∴, f = 50, , Number of, players ( f i ), , Cumulative, frequency ( cf ), , 85-100, , 11, , 100-115, , 9 (f), , 11 + 9 = 20, , 11, , 115-130, , 8, , 20 + 8 = 28, , 130-145, , 5, , 28 + 5 = 33, , It is given that, N = 33, N 33, =, = 16. 5, ∴, 2, 2, So, the median class is 100-115., Here, l = 100, f = 9, cf = 11 and h = 15, ⎛N, ⎞, ⎜ − cf⎟, ⎝2, ⎠, Median = l +, ×h, Q, f, (16. 5 − 11), = 100 +, × 15, 9, 5. 5 × 15, = 100 +, 9, 82. 5, = 100 +, 9, = 100 + 9.17 = 109.17, Hence, the median bowling speed is 109.17 km/h., 28. Here, the given data is in ascending order of x i ., Cumulative frequency table for the given data is, , fi, , cf, , 8, , 8, , 2, , 10, , 18, , 3, , 11, , 29, , 4, , 16, , 45, , 5, , 20, , 65, , 6, , 25, , 90, , 7, , 15, , 105, , 8, , 9, , 114, , 9, , 6, , 120, , x, , i, 1, , Here, n = 120 (even), ∴Median =, =, , 1⎡, Value of, 2 ⎢⎣, , 1⎡, Value of, 2 ⎢⎣, , ⎧⎛ n ⎞, ⎛n, ⎞ ⎫⎤, ⎨⎜⎝ ⎟⎠ th + ⎜⎝ + 1⎟⎠ th ⎬⎥ observations, 2, 2, ⎭⎦, ⎩, , ⎧⎛ 120⎞, ⎛ 120, ⎞ ⎫⎤, + 1⎟ th ⎬⎥ observations, ⎟ th + ⎜, ⎨⎜⎝, ⎝ 2, ⎠ ⎭⎦, ⎩ 2 ⎠, , 1, [Value of 60th observation + Value of 61th observation], 2, Both 60th and 61th observations lie in the cumulative, frequency 65 and its corresponding value of x i is 5., 1, Median = ( 5 + 5) = 5, ∴, 2, =, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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167, , CBSE Term II Mathematics X (Standard), , 29. First we construct a cumulative frequency table., , Weekly income Number of families Cumulative frequency, (in `), (f i ), ( cf ), 0-1000, , 250, , 250, , 190 = f, , 250 + 190 = 440, , 2000-3000, , 100, , 440 + 100 = 540, , 3000-4000, , 40, , 540 + 40 = 580, , 4000-5000, , 15, , 580 + 15 = 595, , 5000-6000, , 5, , 595 + 5 = 600, , 1000-2000, = mid class, , It is given that, N = 600, N 600, =, = 300, ∴, 2, 2, Since, cumulative frequency 440 lies in the interval, 1000 - 2000., Here, l = 1000, f = 190, cf = 250 and h = 1000, ⎫, ⎧N, ⎨ − cf ⎬, 2, ⎭×h, ⎩, Median = l +, Q, f, ( 300 − 250), = 1000 +, × 1000, 190, 50, = 1000 +, × 1000, 190, 5000, = 1000 +, 19, = 1000 + 263.15, = 1263.15, Hence, the median income is ` 1263.15., 30. To calculate the median height, we need to convert the given, data in the continuous grouped frequency distribution., Given, distribution is of less than type and 140, 145, 150, …,, 165 gives the upper limits of the corresponding class, intervals. So, the classes should be below 140, 140-145,, 145-150, …, 160-165., Clearly, the frequency of class interval below 140 is 4, since, there are 4 boys with height less than 140. For the frequency, of class interval 140-145 subtract the number of boys having, height less than 140 from the number of boys having height, less than 145., Thus, the frequency of class interval 140 - 145 is 11 − 4 = 7., Similarly, we can calculate the frequencies of other class, intervals and get the following table, , Class interval Frequency ( f i ), , Cumulative, frequency (cf ), , Below 140, , 4, , 4, , 140-145, , 11 − 4 = 7, , 11, , 145-150, , 29 − 11 = 18 = f, , 29, , 150-155, , 40 − 29 = 11, , 40, , 155-160, , 46 − 40 = 6, , 46, , 160-165, , 51 − 46 = 5, , 51, , Here,, , N = 51, N 51, ∴, =, = 25.5, 2, 2, Since, the cumulative frequency just greater than 25.5 is 29, and the corresponding class interval is 145-150., ∴ Median class = 145-150, Now, l = 145, f = 18, cf = 11 and h = 5, ⎧N, ⎫, − cf, ⎪2, ⎪, ⎧25. 5 − 11 ⎫, ∴ Median = l + ⎨, ⎬×5, ⎬ × h = 145 + ⎨, ⎩ 18 ⎭, ⎪ f ⎪, ⎭, ⎩, 72.5, = 145 +, = 145 + 4. 03 = 149. 03, 18, Hence, the required median height is 149.03 cm., 31. Given, median = 32, and, N = Σf = 100, Let f1 and f2 be the frequencies of the class interval 10-20, and 40-50, respectively., Since, sum of frequencies = 100, ∴, 10 + f1 + 25 + 30 + f2 + 10 = 100, ⇒, f1 + f2 = 100 − 75 ⇒ f1 + f2 = 25, …(i), ⇒, f2 = 25 − f1, Now, the cumulative frequency table for given distribution is, , Cumulative, frequency ( cf ), , Class, interval, , Frequency ( f i ), , 0-10, , 10, , 10, , 10-20, , f1, , 10 + f1, , 20-30, , 25, , 35 + f1, , 30-40, , 30 ( f ), , 65 + f1, , 40-50, , f2, , 65 + f1 + f2, , 50-60, , 10, , 75 + f1 + f2, , Total, , N = f1 + f2 + 75, , N, = 50, 2, Given, median = 32, which belongs to the class 30-40., So, the median class is 30-40., Then, l = 30, h = 10, f = 30 and cf = 35 + f1, ⎫, ⎧N, − cf, ⎪, ⎪, Median = l + ⎨ 2, Q, ⎬×h, f, ⎪, ⎪, ⎩, ⎭, ⎧ 50 − 35 − f1 ⎫, ∴, 32 = 30 + ⎨, ⎬ × 10, 30, ⎭, ⎩, 15 − f1, ⇒, 32 − 30 =, 3, ⇒, 2 × 3 = 15 − f1, ⇒, f1 = 15 − 6 = 9, On putting the value of f1 in Eq. (i), we get, f2 = 25 − 9 = 16, Hence, the missing frequencies are f1 = 9 and f2 = 16 ., Here, N = 100 ⇒, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 168, , = ` 12.727 (in thousand), = 12.727 × 1000 = ` 12727, Hence, the median and modal salary are ` 13421 and, ` 12727, respectively., , 32. First, we construct a cumulative frequency table, , Salary (in `, thousand), , Number of, persons ( f i ), , Cumulative frequency, ( cf ), , 5-10, , 49 ( f0 ), , 49 ( cf ), , 10-15, , f1 = 133, , 133 + 49 = 182, , 15-20, , 63 ( f2 ), , 182 + 63 = 245, , 20-25, , 15, , 245 + 15 = 260, , 25-30, , 6, , 260 + 6 = 266, , 30-35, , 7, , 266 + 7 = 273, , 35-40, , 4, , 273 + 4 = 277, , 40-45, , 2, , 277 + 2 = 279, , 45-50, , 1, , 33. (i), , Time, Number of Class mark, (in seconds) students ( f i ), ( xi ), , 279 + 1 = 280, , 0-20, , 8, , 10, , 8 × 10 = 80, , 20-40, , 10, , 30, , 10 × 30 = 300, , 40-60, , 13, , 50, , 13 × 50 = 650, , 60-80, , 6, , 70, , 6 × 70 = 420, , 80-100, , 3, , 90, , 3 × 90 = 270, , N 280, =, = 140, 2, 2, (i) Here, median class is 10-15, because 140 lies in it., ∴ l = 10, f = 133, cf = 49 and h = 5, ⎛N, ⎞, ⎜ − cf⎟, ⎝2, ⎠, Q Median = l +, ×h, f, (140 − 49), = 10 +, ×5, 133, 91 × 5, = 10 +, 133, 455, = 10 +, = 10 + 3. 421, 133, = ` 13. 421 (in thousand), = 13. 421 × 1000, = ` 13421, (ii) Here, the highest frequency is 133, which lies in the, interval 10-15, called modal class., ∴l = 10, h = 5, f1 = 133, f0 = 49, and f2 = 63., ⎞, ⎛ f −f, ∴ Mode = l + ⎜ 1 0 ⎟ × h, ⎝ 2 f1 − f0 − f2 ⎠, ⎧, ⎫, 133 − 49, = 10 + ⎨, ⎬×5, ×, −, −, 2, 133, 49, 63, ⎩, ⎭, 84 × 5, = 10 +, 266 − 112, 84 × 5, = 10 +, = 10 + 2.727, 154, , Σfi x i = 1720, , Σfi = 40, , N = 280, Mean ( x ) =, , ∴, , f i xi, , Σfi x i 1720, =, = 43, Σfi, 40, , ∴ Mean time is 43s., (ii), , Time, (in seconds), , Number of, students ( f i ), , Cumulative, frequency ( cf ), , 0-20, , 8, , 8, , 20-40, , 10, , 8 + 10 = 18, , 40-60, , 13, , 18 + 13 = 31, , 60-80, , 6, , 31 + 6 = 37, , 3, , 37 + 3 = 40, , 80-100, , Σfi = 40, Modal class is a class having highest frequency., So, 40-60 is modal class, To find median class, we find cumulative frequency, N 40, =, = 20, 2, 2, ∴ 40-60 has cumulative frequency greater than 20., Thus, 40-60 is the median class., ∴Sum of lower limits of median class and modal class, = 40 + 40 = 80., (iii) Students finished the race within 1 min, = Students between 0-20 + Students between 20-40, + Students between 40-60, = 8 + 10 + 13 = 31, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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Chapter Test, Multiple Choice Questions, , (ii) The modal value of the given data is, (a) 150, (c) 145.6, , 1. A survey conducted by a group of students is, given as, , (iii) The median value of the given data is, , Family Size, , 1-3, , 3-5, , 5-7, , 7-9, , 9-11, , Number of families, , 7, , 8, , 2, , 2, , 1, , (a) 140, , (a) 6.8, (c) 5.4, , (b) Median, (d) All of these, , (a) 144 km/charge, (c) 165 km/charge, , mode for a distribution is, (a) Mode = Median − 2 mean, (b) Mode = 3 Median − 2 mean, (c) Mode = 2 Median − 3 mean, (d) Mode = Median − mean, , (b) 155 km/charge, (d) 175 km/charge, , Short Answer Type Questions, , 5. Find the median of the first ten prime numbers., 6. An aircraft has 120 passenger seats. The, , 3. For the following distribution, 0-5, , 5 -10, , 10 -15, , 15 - 20, , 20 - 25, , 10, , 15, , 12, , 20, , 9, , number of seats occupied during 100 flights is, given in the following table., 100-104 104-108 108-112 112-116 116-120, , Number of seats, , The sum of lower limits of the median class and, modal class is, (b) 25, , (d) 136.6, , (v) The manufacturer can claim that the, mileage for his car is, , 2. The relationship among mean, median and, , Frequency, , (c) 130, , (a) Mean, (c) Mode, , (b) 4.2, (d) None of these, , Class, , (b) 146.67, , (iv) Assumed mean method is useful in, determining the, , The mean of the data is, , (a) 15, , (b) 150.91, (d) 140.9, , (c) 30, , (d) 35, , 15, , Frequency, , 20, , 32, , 18, , 15, , Determine the mean number of seats occupied, over the flights., , Case Based MCQs, , 4. A Tesla car manufacturing industry wants to, declare the mileage of their electric cars. For, this, they recorded the mileage (km/charge) of, 100 cars of the same model. Details of which, are given in the following table., Mileage, (km/charge), , 100-120, , 120-140, , 140-160, , 160-180, , Number of Cars, , 14, , 24, , 36, , 26, , 7. The following distribution gives the daily, income of 50 workers of a factory:, 100-120 120-140 140-160 160-180 180-200, , Daily income (in `), , 12, , Number of workers, , 14, , 8, , 6, , 10, , Write the above distribution as ‘less than type’, cumulative frequency distribution. [CBSE 2015], , 8. Find the mode of the following frequency, distribution., , [CBSE 2019], , Class, , 0-10, , 10-20, , 20-30, , 30-40, , 40-50, , 50-60, , 60-70, , Frequency, , 8, , 10, , 10, , 16, , 12, , 6, , 7, , Long Answer Type Questions, Based on the above information, answer the, following questions., (i) The average mileage is, (a) 140 km/charge, (b) 150 km/charge, (c) 130 km/charge, (d) 144.8 km/charge, , 9. Find the mean of the following frequency, distribution using assumed mean method., Class, , 2-8, , 8-14, , 14-20, , 20-26, , 26-32, , Frequency, , 6, , 3, , 12, , 11, , 8, , Answers, 1. (b), 5. 12, , 2. (b) 3. (b) 4. (i) (d) (ii) (b) (iii) (b) (iv) (a) (v) (a), 6. 109.92, 8. 36,, 9. 18.8, , For Detailed Solutions, Scan the code, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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170, , CBSE Term II Mathematics X (Standard), , Mathematics (Standard), Class 10th (Term II), , Practice Paper 1, , *, , (Solved), Instructions, , Time : 2 Hr, Max. Marks : 40, , 1. The question paper contains three sections A, B and C., 2. Section A has 5 questions with 3 internal choices., 3. Section B has 4 questions with 3 internal choices., 4. Section C has 1 Case Based MCQs comprises of 5 MCQs., 5. There is no negative marking., , * As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this, paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised, not to consider the pattern of this paper as official, it is just for practice purpose., , Section A, , (3 Marks Each), , This section consists of 5 questions of Short Answer Type., , 1. Find the value of k for which the quadratic equation (3k + 1 )x 2 + 2( k + 1 )x + 1 = 0, has equal roots . Also find, these roots., , 2. Write the expression a n − a k for the AP a, a + d, a + 2 d, …, Hence, find the common difference of the AP for which 25th term is 10 more than the 23rd term., Or If two towers of heights x m and y m subtend angles of 45° and 60°, respectively at the centre of a line, joining their feet, then find the ratio of ( x + y): y., , 3. The length of common chord of two intersecting circles is 30 cm. If the diameters of these two circles are, 50 cm and 34 cm, then calculate the distance between their centres., Or Given, a line segment AB. Divide it in the ratio m : n by construction, where both m and n are positive, integers and let m = 4 and n = 3., , 4. From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the, volume of the remaining solid., , 5. The mode of the following series is 36. Find the missing (x ) frequency in it., Class interval, , 0-10, , 10-20, , 20-30, , 30-40, , 40-50, , 50-60, , 60-70, , Frequency, , 8, , 10, , x, , 16, , 12, , 6, , 7, , Or The 8th term of an AP is 17 and its 14th term is 29. Find its common difference., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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171, , CBSE Term II Mathematics X (Standard), , Section B, , (5 Marks Each), , This section consists of 4 questions of Long Answer Type., , 6. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure, its length. Also, verify the measurement by actual calculation., Or A decorative block as shown in figure is made of two solids, a cube and a hemisphere., 4.2 cm, , 5, , cm, , 5 cm, , 5 cm, , The base of the block is the cube with edge of 5 cm and the hemisphere attached on the top has a diameter of, 4.2 cm. If the block is to be painted, then find the total area to be painted. [take, π = 22 / 7], , 7. From the point, 36 m above the surface of a lake, the angle of elevation of a bird is observed to be 30° and angle, of depression of its image in the water of the lake is observed to be 60°. Find the actual height of the bird above, the surface of the lake., Or Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the, centre., 1, x − 1 2x + 1 5, 8. Find the solution of the equation, = , x ≠ − , 1 by factorisation method., +, 2x + 1 x − 1 2, 2, Or Find the median for the following frequency distribution., Height (in cm), , Frequency, , 160-162, , 15, , 163-165, , 117, , 166-168, , 136, , 169-171, , 118, , 172-174, , 14, , 9. A girl of height 120 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 4.8 m, above the ground, then find the length of her shadow after 6 s., , Section C, , (1 Mark Each), , This section consists of 1 Case Based comprises of 5 MCQs., , 10. In a mathematic class, a teacher explain the concept for determine the mean by defining the formula x =, , Σf i x i, ., Σf i, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 172, Further, a teacher give one example for explaining the above, concepts., The marks obtained by 30 students of class X of a certain, school in a mathematics paper consisting of 100 marks are, presented in table below, Class interval 10-25, Number of, Students, , 2, , 25-40, , 40-55, , 55-70, , 70-85, , 85-100, , 3, , 7, , 6, , 6, , 6, , (ii) Find the cumulative frequency value in the, interval (40-55)., (a) 5, (c) 2, , (b) 12, (d) 18, , (iii) Through cumulative frequency table, which, central measurement can be determined., (a) mean, (c) median, , (b) mode, (d) None of these, , (iv) Find the lower limit of the median class., (a) 55, (c) 70, , (i) Find the average marks obtained by the students., (a) 61, (b) 62, (c) 63, (d) 64, , (b) 40, (d) 25, , (e) Find the upper limit of modal class., (a) 40, (c) 70, , (b) 55, (d) 25, , Solutions, On subtracting Eq. (ii) from Eq. (i), we get, , 1. Given quadratic equation is, ( 3k + 1) x 2 + 2 ( k + 1 ) x + 1 = 0, , a n − a k = [ a + ( n − 1) d] − [ a + ( k − 1) d], , On comparing with ax 2 + bx + c = 0, we get, a = 3k + 1, b = 2 ( k + 1), c = 1, Since, the roots are equal, so b 2 − 4ac = 0, ∴, , [2 ( k + 1)] − 4( 3k + 1) (1) = 0, 4( k + 1)2 − 4( 3k + 1) = 0, , ⇒, , 4( k 2 + 2 k + 1) − 4( 3k + 1) = 0, , ⇒, , 4( k 2 + 2 k + 1 − 3k − 1) = 0, , ⇒, , k 2 + 2 k + 1 − 3k − 1 = 0, , ⇒, , k 2 + 2 k + 1 − 3k − 1 = 0, , ⇒, , k2 − k = 0, , a n − a k = a + ( n − 1) d − a − ( k − 1 ) d, a n − a k = ( n − 1 − k + 1) d, , ⇒, , a n − a k = (n − k) d, , Now, a n − a k = ( n − k ) d, , 2, , ⇒, , ⇒, ⇒, , ⇒ a 25 − a 23 = (25 − 23) d, ⇒, ⇒, , [put n = 25and k = 23], [Qa 25 − a 23 = 10, given], , 10 = 2 d, d=5, , Hence, the common difference is 5., Or, [Q 4 ≠ 0], , ⇒, k ( k − 1) = 0, ⇒, k = 0 or k = 1, We know that, if roots are equal, then roots will be the form of, −b −b, ., ,, 2a 2a, −b −b, ., Thus, roots are, ,, 2a 2a, − ( k + 1) − ( k + 1), ∴ Equal roots are, ,, ( 3k + 1) ( 3k + 1), − ( 0 + 1), − ( 0 + 1), When k = 0, equal roots are, and, 0+1, 0+1, i.e. − 1 and − 1., − (1 + 1 ), − (1 + 1 ), and, When k = 1, equal roots are, 3+1, 3+1, 1, 1, i.e. − and − ., 2, 2, 2. Given, first term = a and common difference = d, ∴, , a n = a + ( n − 1) d, , …(i), , and, , a k = a + ( k − 1) d, , …(ii), , Let AB = x m be the height of a tower and CD = y m be the, height of other tower and ∠AEB = 45° and ∠CED = 60°., D, B, y, x, A, , 45°, a, , 60°, E, , a, , C, , Let E be the point (centre) on the line AC., i.e. AE = EC = a m, In right angled ΔBAE,, AB, tan 45° =, AE, x, =1, ⇒, a, ⇒, x=a, Again, in right angled ΔDCE,, DC, tan 60° =, CE, y, 3=, ⇒, a, ⇒, , y = 3a, , perpendicular ⎤, ⎡, Q tan θ =, ⎥⎦, base, ⎣⎢, [Q tan 45°= 1], …(i), , [Q tan 60°= 3 ], …(i), , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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173, , CBSE Term II Mathematics X (Standard), , ∴, , (iv) Now, through the point A 4( m = 4), draw a line parallel, to A 7B by making an angle equal to ∠AA 7 B at A 4, intersecting AB at a point C., Then, AC : BC = 4 : 3, , ( x + y ): y = ( a + 3a ): a 3 = (1 + 3 ): 3, , Hence, the requred ratio ( x + y ): y is (1 + 3 ): 3., 3. Let, PQ be the length of the common chord of two, intersecting circles., ∴, , [given], , PQ = 30cm, , Diameters of two circles are 50 cm and 34 cm., Join AB., P, , Alternative Method, (i) Draw any ray AX making an acute angle with AB., (ii) Draw a ray BY parallel to AX by making, ∠ABY = ∠BAX., (iii) Locate the points A 1 , A 2 , A 3 , A 4 (for m = 4) on AX and, similarly B 1, B 2 , B 3 (for n = 3) on BY, such that, , A, , A A 1 = A 1A 2 = A 2A 3 = A 3A 4, = BB1 = B1B 2 = B 2B 3, , B, , O, , X, , Q, , A4, A3, , Since, AB bisects the common chord PQ perpendicularly., 1, 1, OP = OQ = × PQ = × 30 = 15 cm, ∴, 2, 2, 1, Radius, AP = × 50 = 25 cm, 2, 1, and radius, PB = × 34 = 17 cm, 2, , A2, A1, C, B1, B2, B3, , In right angled ΔAOP,, Y, , OA = ( AP )2 − ( OP )2, [by using Pythagoras theorem], 2, , = (25) − (15), , 2, , = 625 − 225 =, , B, , A, , (iv) Join A 4B 3. Let it intersects AB at point C., AC 4, =, BC 3, 4. Given, side of a solid cube, a = 7 cm, Height of conical cavity, i.e. cone, h = 7 cm, Radius of conical cavity, r = 3 cm, Then,, , 400 = 20 cm, , In right angled ΔPOB,, OB = (PB )2 − ( OP )2 = (17 )2 − (15)2, [by using Pythagoras theorem], = 289 − 225, = 64 = 8 cm, , 7 cm, , ∴ Distance between centres,, AB = OA + OB = 20 cm + 8 cm = 28 cm, 3 cm, , Or, Given A line segment AB, m = 4 and n = 3., , Now, volume of cube = a = (7 )3 = 343 cm 3, , Steps of construction, , and volume of conical cavity =, , (i) Draw any ray AX making an acute angle with AB., (ii) Locate 7 (i.e. m + n ) points A1 , A 2 , ...,A 7 on AX,, such that AA 1 = A 1A 2 = ... = A 6 A 7 ., (iii) Join BA 7., X, A7, A6, A5, A4, A3, A2, A1, A, , C, , B, , 3, , 1, π × r2 × h, 3, , 1 22, ×, × 3 × 3 × 7 = 66 cm 3, 3 7, According to the question,, Volume of remaining solid, = Volume of cube − Volume of conical cavity, = 343 − 66 = 277 cm 3, =, , Hence, the required volume of solid is 277 cm 3., 5. Since, the mode of the given series is 36, which lies in the, class 30-40., So, the modal class is 30-40., Then, l = 30, f1 = 16, f0 = x, f2 = 12, and, h = 10, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 174, Also,, , mode = 36, , ∴, , (PM )2 = ( OP )2 − ( OM )2, , Q, , ⎧, f1 − f0 ⎫, Mode = l + ⎨h ×, ⎬, 2 f1 − f0 − f2 ⎭, ⎩, , ⇒, , (PM )2 = ( 6)2 − ( 4)2 = 36 − 16 = 20, , ⎧, ⎫, 16 − x, 36 = 30 + ⎨10 ×, ⎬, ×, −, −, 2, 16, x, 12, ⎩, ⎭, 10 (16 − x ), 36 = 30 +, (20 − x ), , ∴, ⇒, ⇒, , 36 − 30 =, , 6 10(16 − x ), 10 (16 − x ), ⇒ =, 1, (20 − x ), (20 − x ), , ⇒, , 10 (16 − x ) = 6 (20 − x ), , ⇒, ⇒, ⇒, , 160 − 10 x = 120 − 6x, − 10x + 6x = 120 − 160, − 4x = − 40, − 40, x=, = 10, −4, , ∴, , ⇒, , Now, area to be painted on the cube, = Total surface area of cube, − Base area of hemisphere, 22 4.2 4.2, 2, = 150 − πr = 150 −, ×, ×, 7, 2, 2, 2, = 150 − 13.86 = 136.14 cm, Area to be painted on the hemisphere, = Curved surface area of hemisphere, 22 4.2 4.2, = 2 πr 2 = 2 ×, ×, ×, = 27.72 cm 2, 7, 2, 2, ∴ Total area to be painted, = Area to be painted on the cube, + Area to be painted on the hemisphere, = 136.14 + 27.72 = 163.86 cm 2, 7. Let QR be the surface of the lake and P be point above the, surface such that PQ = 36 m. Let B represents the bird and, Bʹ be its image in the lake., B, x, P, 36 m, , 30°, 60°, , Q, , S, 36 m, R, , (36+x) m, , Bʹ, , ∴ ∠BPS = 30° and ∠Bʹ PS = 60°, , M, , Also,, Mʹ, , PM = 20 = 4. 47 ≈ 4.5, , Hence, the length of tangent is 4.5 cm., Or, Given, edge of cube = 5 cm, and diameter of hemisphere = 4. 2 cm, 4.2, radius of hemisphere =, cm = 2.1 cm, 2, Clearly, total surface area of the cube, = 6 (Edge) 2, = 6 × 5 × 5 = 150 cm 2, , Hence, the missing frequency is 10., Or, Given,, ε 8 = 17 , ε14 = 29, Let be a first term a, ε n = a + ( n − 1 )d, 17 = 9 + ( 8 − 1)d, …(i), ⇒, 17 = a + 7 d, Similarly, 29 = a + 13d, …(ii), Subtract Eq. (i) from Eq. (ii),, a + 13d − a − 7 d = 29 − 17 = d = 2, 6. Given, two concentric circles of radii 4 cm and 6 cm with, common centre O., Here, we have to draw two tangents to inner circle C1 from a, point of outer circle C 2., Steps of construction, (i) Draw two concentric circles C1 and C 2 with common, centre O and radii 4 cm and 6 cm, respectively., (ii) Take any point P on outer circle C 2 and join OP., (iii) Now, bisect OP. Let Mʹ be the mid-point of OP., (iv) Taking Mʹ as centre and OMʹ as radius, draw a dotted, circle which cuts the inner circle C1 at two points M, and Pʹ., (v) Join PM and PPʹ. Thus, PM and PPʹ are required, tangents., , P, , [by Pythagoras theorem], , O, C1 C2, , Pʹ, , On measuring PM and PPʹ, we get, PM = PPʹ = 4. 5 cm, Calculation In right angled ΔOMP, ∠ PMO = 90°, , Let, ⇒, ∴, , SR = PQ = 36 m, BS = x, Bʹ R = BR = ( 36 + x ) m, Bʹ S = SR + BR, , = 36 + 36 + x = (72 + x ) m, In right angled ΔPSB,, BS, = tan 30°, PS, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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175, , CBSE Term II Mathematics X (Standard), , x, 1, =, PS, 3, PS = 3x, , ∴, ⇒, , 1 ⎤, ⎡, Q tan 30° =, 3 ⎥⎦, ⎣⎢, …(i), , 1, ( ∠1 + ∠1 + ∠ 3 + ∠ 3), 2, 1, = ( ∠1 + ∠2 + ∠ 3 + ∠ 4), 2, [from Eqs. (i) and (ii)], 1, = (180°) = 90°, 2, [Q QR is a straight line, therefore, ∠1 + ∠2 + ∠ 3 + ∠ 4 = 180°], =, , Again, in right angled ΔBʹ SP ,, Bʹ S, 72 + x, = tan 60° ⇒, = 3, PS, PS, [Q B’S = (72+x) m and tan 60° = 3 ], 72 + x, …(ii), PS =, ⇒, 3, From Eqs. (i) and (ii), we get, 72 + x, 3x =, 3, ⇒, , 3 ⋅ 3x = 72 + x, , ⇒, , 3x − x = 72 ⇒ 2 x = 72, 72, x=, ⇒ x = 36, 2, , ⇒, , y+, , BR = BS + SR = 36 + 36 = 72 m, Or, Let, AB and CD are two tangents to a circle and AB || CD., Tangent BD subtends ∠BOD at the centre., To prove ∠BOD = 90°, Construction Join OP, OQ and OR., Q, , B, , 2, O, , C, , R, , 3, P, D, , Proof Here, OP ⊥ BD, [Q a tangent at any point of a circle is perpendicular to the, radius through the point of contact], In right angled ΔOQB and Δ OPB,, [Q the lengths of tangents drawn from an, external point are equal], , Then,, , 2 y( y − 2 ) − 1 ( y − 2 ) = 0, (2 y − 1)( y − 2 ) = 0, 2 y − 1 = 0 or y − 2 = 0, 1, y = or y = 2, ⇒, 2, x −1, , we get, Put y =, 2x + 1, x −1 1, x −1, = or, =2, 2x + 1 2 2x + 1, , Or, The given series is in inclusive form. Converting it to, exclusive form and preparing the cumulative frequency table, is given below, , Class interval, , BQ = BP, , ∴, , ⇒, ⇒, ⇒, , ⇒, 2 x − 2 = 2 x + 1 or x − 1 = 4x + 2, ⇒, −2 = 1, which is not true., Consider, x − 1 = 4x + 2, ⇒, 3x = − 3 ⇒ x = − 1, , 1, 4, , 1 5, y2 + 1 5, = ⇒, = ⇒ 2 y 2 − 5y + 2 = 0, y, 2, y 2, , This is a quadratic equation., By using factorisation method,, 2 y 2 − 4y − y + 2 = 0, , ∴ Height of bird above surface of the lake,, , A, , Hence proved., x −1, 2x + 1 5, 8. We have,, +, =, x −1 2, 2x + 1, x −1, , then given equation becomes, Let y =, 2x + 1, , Frequency (fi ), , Cumulative, frequency, , 159.5-162.5, , 15, , 15, , 162.5-165.5, , 117, , 132, , OQ = OP, , [radii], , 165.5-168.5, , 136, , 268, , OB = OB, , [common], , 168.5-171.5, , 118, , 386, , [by SSS congruency], , 171.5-174.5, , 14, , 400, , Total, , N = Σ f i = 400, , ΔOQB ≅ Δ OPB, ∠1 = ∠ 2, , [by CPCT] …(i), , Similarly, in right angled ΔOPD and ΔORD,, ∠ 3 = ∠4, 1, ∴, ∠BOD = ∠1 + ∠ 3 = (2 ∠1 + 2 ∠ 3), 2, , Here,, …(ii), , Now,, , N = 400, N 400, =, = 200, 2, 2, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 176, The cumulative frequency just greater than 200 is 268 and the, corresponding class is 165.5-168.5., Thus, the median class is 165.5-168.5., ∴ l = 165.5, h = 3 and f = 136 and C = 132, N, ⎫, ⎧, − C. f, ⎪, ⎪, 2, ∴ Median = l + ⎨h ×, ⎬ (1/2), f, ⎪, ⎪, ⎩, ⎭, (200 − 132 ) ⎫, ⎧, = 165.5 + ⎨3 ×, ⎬, 136, ⎭, ⎩, 3 × 68, = 165.5 + 1.5 = 167, = 165.5 +, 136, Hence, the median height is 167 cm., 9. Let AB be the lamp-post, CD be the girl and D be the, position of girl after 6 s., Again, let DE = x m be the length of shadow of the girl., A, , 7.2, = 2. 4 m, 3, Hence, the length of her shadow after 6 s is 2.4 m., 10. (i) (b) Let us make the following table for the given data., ⇒, , 3x = 7.2 ⇒ x =, , Class, Interval, , Frequency, , 10-25, , 2, , 25-40, , 3, , 40-55, , 7, , 55-70, , 6, , 70-85, , 6, , 85-100, , 6, , Total, , Σfi = 30, , Class marks, (xi ), 10 + 25, = 17. 5, 2, 25 + 40, = 32. 5, 2, 40 + 55, = 47. 5, 2, 55 + 70, = 62. 5, 2, 70 + 85, = 77. 5, 2, 85 + 100, = 92. 5, 2, , f i xi, , Cumulative, frequency, ( cf ), , 35.0, , 2, , 97.5, , 5, , 332.5, , 12, , 375.0, , 18, , 465.0, , 24, , 555.0, , 30, , 4.8 m, , Σfi x i, , C, , B, , D, , x, , E, , Given, CD = 120 cm = 1.2 m, AB = 4. 8 m, and speed of the girl = 1.2 m/s, ∴ Distance of the girl from lamp-post after 6 s., BD = 1.2 × 6 = 7.2 m, [Q distance = speed × time], In ΔABE and ΔCDE,, [each 90°], ∠B = ∠D, [common angle], ∠E = ∠E, [by AA similarity criterion], ∴ ΔABE ~ ΔCDE, BE AB, …(i), ⇒, =, DE CD, On substituting all the values in Eq. (i), we get, 7.2 + x 4.8, =, [Q BE = BD + DE = 7.2 + x ], x, 1.2, 7.2 + x, ⇒, =4, x, ⇒, 7.2 + x = 4x, , = 1860. 0, , Here, Σfi = 30 and Σfi x i = 1860. 0, Σf x 1860. 0, = 62, Q Average, x = i i =, Σfi, 30, Hence, average marks obtained by student is 62., (ii) (b) The cumulative frequency value in the interval, 40-55 is 12., (iii) (c) Through cumulative frequency table, median can be, determined., N 30, (iv) (a) Here, =, = 15, which lies in the cumulative, 2, 2, frequency 18, whose corresponding frequency is 55-70., Hence, lower limit of the median class is 55., (v) (b) In the given data, the highest frequency is 7, whose, corresponding interval is 40-55., Hence, upper limit of the modal class is 55., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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177, , CBSE Term II Mathematics X (Standard), , Mathematics (Standard), Class 10th (Term II), , Practice Paper 2, , *, , (Unsolved), Instructions, , Time : 2 Hr, Max. Marks : 40, , 1. The question paper contains three sections A, B and C., 2. Section A has 5 questions with 3 internal choices., 3. Section B has 4 questions with 3 internal choices., 4. Section C has 1 Case Based MCQs comprises of 5 MCQs., 5. There is no negative marking., , * As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this, paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised, not to consider the pattern of this paper as official, it is just for practice purpose., , Section A, , (3 Marks Each), , This section consists of 5 questions of Short Answer Type., , 1. Find the roots of the quadratic equation 9x 2 − 9 (a + b )x + (2 a 2 + 5ab + 2 b 2 ) = 0., 2, 1, 5, 5, Or From the top of a 10 m high building, the angle of elevation of the top of a tower is 60° and the angle of, depression of its foot is 45°. Determine the height of the tower., , 2. Which term of the progression 19, 18 , 17 , ... is the first negative term?, , 3. In the given figure, AB is the diameter of a circle with centre O and QC is a tangent to the circle at C. If, ∠CAB = 30°, then find ∠CQA and ∠CBA., C, , A, , 30°, O, , B, , Q, , Or A copper wire 4 mm in diameter is evenly wound about a cylinder whose length is 24 cm and diameter 20, cm so as to cover the whole surface. Find the length and weight of the wire assuming the specific density to, be 8.88 gm/cm 3 ., , 4. Draw a circle of radius 6 cm. Take a point P on it. Without using the centre of the circle, draw a tangent to, the circle at point P., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 178, , 5. The mean of the following frequency table is 50 but the frequencies f 1 and f 2 in class intervals 20-40 and 60-80, are missing. Find the missing frequencies., Class interval, Frequency, , 0-20, , 20-40, , 40-60, , 60-80, , 80-100, , Total, , 17, , f1, , 32, , f2, , 19, , 120, , Or Find the volume area of the largest right circular cone that can be cut out of a cube whose edge is 10 cm., , Section B, , (5 Marks Each), , This section consists of 4 questions of Long Answer Type., , 6. A cone of maximum size is cut-out from a cube of edge 14 cm. Find the surface area of the remaining solid left, out after the cone is cut-out., Or The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 sec, the angle of, elevation changes to 30°. If the jet plane is flying at a constant height of 1500 3 m, find the speed of the jet, plane., , 7. If α and β are the zeroes of the quadratic polynomial f ( x ) = 3x 2 − 4x + 1, find a quadratic polynomial whose, zeroes are, , β2, α2, and ., α, β, , Or If m times the mth term of an AP is equal to n times its nth term, then show that ( m + n ) th term of the AP is, zero., , 8. Construct a tangent to a circle of radius 1.8 cm from a point on the concentric circle of radius 2.8 cm and, measure its length. Also, verify the measurement by actual calculation., , 9. In the given figure, PT is a tangent and PAB is a secant. If PT = 6 cm and AB = 5 cm, then find the length of PA., T, , O, P, , B, , A, , Or Find the mean, mode and median of the following data., Class, , 0-10, , 10-20, , 20-30, , 30-40, , 40-50, , 50-60, , 60-70, , Frequency, , 3, , 4, , 7, , 15, , 10, , 7, , 4, , Section C, , (1 Mark Each), , This section consists of 1 Case Based comprises of 5 MCQs., , 10. In one corner of the drawing room, a flower basket is kept inside the glass, lies on the table. The basket is, designed in such a way that every one pleases to see it., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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179, , CBSE Term II Mathematics X (Standard), , The shape of flower basket is hemisphere with radius 60 cm and upper shape is conical with height 120 cm from the bottom, surface., P, , E, , 180 cm, , 120 cm, , F, , O, , 60 cm, , D, 60 cm, Oʹ, , A, , B, , (i) Find the capacity of the glass., (a), , 14.256 3, m, 7, , (b), , 12.256 3, m, 7, , (c), , 142. 56 3, m, 7, , (d), , 14.256 3, m, 5, , (ii) Find the volume of the cone., (a) 0.54 m 3, , (b) 0.45 m 3, , (c) 0.25 m 3, , (d) 0.52 m 3, , (c) 0.102 m 2, , (d) 0.401 m 2, , (iii) Find the curve surface area of hemisphere., (a) 0.201 m 2, , (b) 0.104 m 2, , (iv) The volume of two combined figure is equal to the sum of, (a) two individual volumes, (c) volumes and curve surface area, , (b) two individual curve surface area, (d) None of these, , (v) If the cost of painting the glass outside is ` 1.20 per m 2 , find the total cost of painting the CSA of the glass., (a) ` 55, , (b) ` 55.02, , (c) ` 57, , (d) ` 57.02, , Answers, (2a + b ) (a + 2b ), ,, 3, 3, 5. f1 = 28 and f 2 = 24 or 261.9 cm 2, 1., , 2. 25th term or 10( 3 + )1 m, 6. 1022 + 154 5 cm or 720 km/h, , 1⎞, ⎛ 2 28, 7. k ⎜ x −, x + ⎟ where k is any non-zero real number, ⎝, 9, 3⎠, 9. 4 cm or (i) 37.4, (ii) 36.15, (iii) 37.3, , 3. ∠CQA = 30° and ∠CBA = 60° or 4.21 kg, , 2, , 8. 2.14 cm, , 10. (i) (a) (ii) (b) (iii) (c) (iv) (a) (v) (d), , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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181, , CBSE Term II Mathematics X (Standard), , 5. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14, cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel., Or Two pillars of equal height are on either sides of a road, which is 100 m wide. The angles of the top of the pillars, are 60° and 30° at a point on the road between the pillars. Find the position of the point between the pillars., Also, find the height of each pillar., , Section B, , (5 Marks Each), , This section consists of 4 questions of Long Answer Type., , 6. A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is 4 cm and, the diameter of the base is 8 cm. If a right circular cylinder circumscribes the solid. Find how much more space, it will cover?, Or The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 m towards the, tower, the angle of elevation of the top increases by 15°. Find the height of the tower., , 7. Construct a tangent to a circle of radius 1.8 cm from a point on the concentric circle of radius 2.8 cm and, measure its length. Also, verify the measurement by actual calculation., Or From an external point P, two tangents PA and PB are drawn to the circle with centre O. Prove that OP is the, perpendicular bisector of AB., , 8. Solve the following quadratic equation by factorisation method., 1 1 1, 1, = + + , a + b ≠ 0., a+b+x a b x, Or The angles of a triangle are in AP. If the greatest angle equals to the sum of the other two, then find the angles., Also, conclude that find these angles are multiple of which angle., , 9. The median of the distribution given below is 14.4. Find the values of x and y, if the total frequency is 20., Class interval, , 0-6, , 6-12, , 12-18, , 18-24, , 24-30, , Frequency, , 4, , x, , 5, , y, , 1, , Section C, , (1 Mark Each), , This section consists of 1 Case Based comprises of 5 MCQs., , 10. Suppose, there are two windows in a house. A window of the house is at a height of 1.5 m above the ground and, the other window is 3 m vertically above the lower window., C, h – 4.5 m, (Shyam), , B, , 30°, , D, , 3m, , 3m, , (RAM) A, , E, , 1.5 m, , 1.5 m, , P, , Q, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm

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CBSE Term II Mathematics X (Standard), , 182, , Anil and Sanjeev are sitting in the two windows. At an instant, the angles of elevation of a balloon from these, windows are observed as 45° and 30°, respectively., (i) Find the height of the balloon from the ground., (a) 6.8 m, (c) 9.4 m, , (b) 8.6 m, (d) 9.6 m, , (ii) Among Anil and Sanjeev, who is more closer to the balloon?, (b) Anil, (d) None of these, , (a) Sanjeev, (c) cannot say, , (iii) If the balloon is moving towards the building, then will both the angles of elevation remain same?, (a) cannot say, (c) No, , (b) Yes, (d) None of these, , (iv) If the height of any tower is double and the distance between the observer and foot of the tower is also doubled,, then the angle of elevation, (a) remain same, (c) become triple, , (b) become double, (d) None of these, , (v) Suppose a tower and a pole is standing on the ground. And the angle of elevation from bottom of pole is θ1 and, elevation from top of pole to the top of tower is θ 2 ., , θ1, , B, , θ1, , A, , Choose the correct option., (a) θ1 > θ 2, (c) θ1 < θ 2, , (b) θ1 = θ 2, (d) None of these, , Answers, 9, 1. −4,, 4, 6., , 2. 5 or 0.8( 3 + 1) m, , 128, π cm 3 or10( 3 + 1 ) m, 3, , 9. x = 4 and y = 6, , 3. 45°, , 7. 2.14 cm, , 4. 110, , 5. 572 cm 2 or 25 m, 43.3 m, , 8. x = −a or x = −b or ` 30°, 60° and 90° angles are the multiple of 30°, , 10. (i) (b) (ii) (a) (iii) (c) (iv) (a) (v) (a), , Visit https://telegram.me/booksforcbse for more books., , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterm