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CHAPTER-WISE PREVIOUS YEARS' QUESTIONS, , MATHEMATICS, HINTS & SOLUTIONS, Class X (CBSE)

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MATHEMATICS, Chapter - 1 : Real Numbers, 1., , 2, , This contradicts the fact that, , 3, , a, , , , b, 7., , Let assume the missing entries be a, b., b = 3 × 7 = 21, , [½], , a = 2 × b = 2 × 21 = 42, , [½], , 1044 = 216 × 4 + 180, 180 = 36 × 5 + 0, , 441, Given a rational number 5 7 2 ., 2 5 7, ,  HCF of 1260 and 7344 is 36., 8., , Since 0  r < 4, the possible remainders are, 0, 1, 2 and 3., [½],  a can be 4q or 4q + 1 or 4q + 2 or 4q + 3,, where q is the quotient., , [1], , [∵, 6., , 2  1.414 and, , , , 3  1.732], , 2 and, , 3 is, , Since a is odd, a cannot be 4q and 4q + 2., , [½], ,  Any odd integer is of the form 4q + 1 or, 4q + 3, where q is some integer., [½], , [½], , , , Let us assume that 5  3 2 is rational. Then, there exist co-prime positive integers a and b, such that, a, 53 2 , b, , , , a  5b, 3b, , [½], , 9., , Let 'a' be any positive integer and b = 3., We know a = bq + r, 0  r < b., Now, a = 3q + r, 0  r < 3., The possible remainder = 0, 1 or 2, , [½], , Case (i) a = 3q, a2 = 9q2, , a, 3 2  5, b, 2 , , [½], , a = 4q + r, , Smallest composite number is 4., Therefore, HCF is 2., , [½], , Let a be positive odd integer., Using division algorithm on a and b = 4, , Smallest prime number is 2., , 15 3, , 10 2, , [½], , The remainder has now become zero., , 441, 9, [½],  5 7, 5 7 2, 2 5 7, 2 5, Since, the denominator is in the form of 2m 5n., So, the rational number has terminating decimal, expansion., [½], , 1.5 , , [½], , 216 = 180 × 1 + 36, , [½], , Rational number lying between, , [½], , Since remainder  0, , [½], , , , 5., , Since 7344 > 1260, , 1260 = 1044 × 1 + 216, , = 19000, , 4., , , , 7344 = 1260 × 5 + 1044, , Given two numbers 100 and 190.,  HCF × LCM = 100 × 190, , 3., , [½], , Hence, 5  3 2 is an irrational number., , 7, , 2., , 2 is irrational., , So, our assumption is incorrect., , = 3 × (3q2), = 3m (where m = 3q2), [½], , 2 is irrational., a  5b, is rational]., [∵ a, b are integers, , 3b, [½], , [1], , Case (ii) a = 3q + 1, a2 = (3q + 1)2, = 9q2 + 6q + 1, = 3(3q2 + 2q) + 1, = 3m + 1 (where m = 3q2 + 2q), , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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2, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Case (iii) a = 3q + 2, , Common prime factor = 2, Least exponent = 2, , a2 = (3q + 2)2, ,  HCF = 22 = 4, , =, , 9q2, , =, , 3(3q2, , + 12q + 4, , To find the LCM, we list all prime factors of 404, and 96 and their greatest exponent as follows :, , + 4q + 1) + 1, , = 3m + 1 (where m = 3q2 + 4q + 1), From all the above cases it is clear that square, of any positive integer (as in this a2) is either of, the form 3m or 3m + 1., [1], 10. Let assume 3  2 is a rational number., , , 3 2 , , , , [1], , 3  2 is not a rational number., , [1], , (where p, q are co-prime integers and q  0), p, 3 5, q, , [1], , = 9696, , [1], , 5 is an irrational, , [1], , Therefore, HCF × LCM = Product of two, numbers., [1], 13. Let, , 2 be rational. Then, there exist positive, , and b are co-prime, b  0]., ,  2, , 2, , a, . [Where a, b, [½], , 2, , [½], , a2, b2, ,  2b2 = a2,  2 divides a2, ...(i), [½], , a2 = 4c2,  2b2 = 4c2,  b2 = 2c2,  2 divides b2, ,  Our assumption is wrong., 2  3 5 is an irrational number., , Product of two numbers = 404 × 96 = 38784, , Let a = 2c for some integer c., ,  Rational  irrational., , , , 1, ,  2 divides a, , 5, , 2q  p, Since,, is a rational number but we also, 3q, know, , 101, , a, 2,  ( 2)   , b, , p,  23 5  ,, q, , 2q  p, , 3q, , 1, , integers a and b such that, , 11. Let assume 2  3 5 is a rational number., , , , 3, , HCF × LCM = 9696 × 4 = 38784, ,  Irrational  rational, ,  2, , 5, , Now,, , p  3q, Since,, is a rational number but we know, q, 2 is an irrational., , , , 2, , [1], , p, 3, q, , p  3q, 2, q, , Greatest Exponent, , = 25 × 31 × 1011, , {p, q are co-prime integers and q  0}, 2, , Prime factors of, 404 and 96, ,  LCM = 25 × 31 × 1011, , p, q, , , , [1], , [1], , 12. Using the factor tree for the prime factorization of, 404 and 96, we have, 404 = 22 × 101 and 96 = 25 × 3, To find the HCF, we list common prime factors, and their smallest exponent in 404 and 96 as, under :, ,  2 divides b, , ...(ii), , [½], , From (i) and (ii), we get, 2 is common factor of both a and b., But this contradicts the fact that a and b have, no common factor other than 1., [½],  Our supposition is wrong., Hence,, , 2 is an irrational number., , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Mathematics, , 3, , Chapter - 2 : Polynomials, 1., , (x + a) is factor of the polynomial p(x) = 2x2 +, 2ax + 5x + 10.,  p(–a) = 0, 2(–a)2, , 6., , x3 – 4x2 – 3x + 12, , {By factor theorem}, , + 2a(–a) + 5(–a) + 10 = 0, , 2a2 – 2a2 – 5a + 10 = 0, , Sum of all the zeroes of polynomial = –(–4) = 4, , [½], ,   3 3 4, , [½], 7., , –2a + 2 = 0, 3., , 4., , [½], , Given  and  are the zeroes of quadratic, polynomial with  +  = 6 and  = 4., ,   4, , , , It is given that 2  3, zeros of f(x) =, , 2x4, , –, , , , +, , , , p(x) = x4 + x3 – 34x2 – 4x + 120, ,  ( x  2)2 , , Let assume other two zeroes be , ., , = x2 – 4x + 1, , , (x2, , + 3x – 1, ,  x  2  3 , ,  3, , 2, , [1], , 2, , 2, , x – 4 x + 1 2 x4 – 9 x 3 + 5 x 2 + 3 x – 1, ,   1  , , 4, , 3, ,   30, , 3, , 2, , 2x – 8 x + 2x, (–), (+), (–), ____________________, 3, 2, –x + 3x + 3 x – 1, , ...(i), , ..2.(–2) = 120, ...(ii), , 2, , –x + 4x – x, _ _ _ _ _ _ _ _(+), ____________, (+) (–), , [1], , 2, , –x + 4x – 1, , Substituting (i) in (ii), we get, , 2, , –x + 4x – 1, _ _ _ _ _ _ _ _(+), ____________, 0, ____________________, , (–1 – ) = –30, , (+) (–), ,  + 2 = 30, 2 +  – 30 = 0, , We have,, ,   = –6, 5, ,  f(x) = (x2 – 4x + 1)(2x2 – x – 1), ,  = 5, –6, Zeroes of the polynomial are –6, –2, 2, 5., +, , 3x2, , [1], , – 2x – 6 = 0, , Given two zeros are  2,, , Hence, other two zeros of f(x) are the zeros of, the polynomial 2x2 – x – 1., , 2, , 2x2 – x – 1 = 2x2 – 2x + x – 1, = 2x(x – 1) + 1(x – 1), , [1], , Let the third zero be x, , x, , , , [1], , We have,, , Sum of all zeros = –3, , , are two, , – 4x + 1) is a factor of f(x), , Product of zeroes = 120, , x3, , , , 2x – x – 1, ,  +  = –1, , , 5x2, , x   2  3 x   2  3 ,  x2 3, , =+, , , , and 2  3, , 9x3, , Quadratic polynomial = k[x2 – 6x + 4], where k, is real., [1], , Sum of all zeroes =  +  + 2 – 2, , 5., , [1], ,  Third zero is 4., ,  a(1)2 – 3(a – 1) – 1 = 0, a=1, , [1], , Say the third zero = , , If x = 1 is the zero of the polynomial,  p(x) = ax2 – 3(a – 1)x – 1, Then p(1) = 0, , 3,  3., , Given two zeroes are, , [½], , a=2, 2., , Given a polynomial, , = (2x + 1)(x – 1), , , , , , 2   2  3, , f (x)  x  2  3, ,  x  2  3  (2x  1)( x  1), , x = –3,  All zeroes will be 3,  2,, , 2, , [1], , Hence, the other two zeros are , , 1, and 1. [1], 2, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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4, 8., , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , For given polynomial, x2 – (k + 6)x + 2(2k – 1),, ,   , , [½], , c 4k  2, b,  k  6,   , a, 1, a, , ∵ Sum of zeroes , , [½], , 1,  k  6  (4k  2), 2, , Let the zeroes be  and ., So,     , , 1, , 2, ,  k + 6 = 2k – 1, , [1], ,  k = 7, , 1, (product of zeroes), 2, , So, the value of k is 7., , [1], , Chapter - 3 : Pair of Linear Equations in Two Variables, 1., , x + 2y – 8 = 0, , Putting x = 22 in equation (ii), , 2x + 4y – 16 = 0, , 22 – y = 14  22 – 14 = y, , For any pair of linear equations, ,  y=8, , a1x + b2y + c1 = 0, ,  x = 22 and y = 8, 4., , a2x + b2y + c2 = 0, If, , a1 b1 c1, ,  , then, a2 b2 c 2, , , , II, III, c, 3, 3c, , , 12, c, c, , a1 1 b1 2 c1, 8,  ,,  ,, , a2 2 b2 4 c 2 16, , (i) c2 = 12 × 3, , a1 b1 c1 1, , , , a2 b2 c 2 2, , (ii), , c = 0, 6, (iii) c2 = 12(c – 3), , [½], , c – 12c + 36 = 0, , [½], , (c – 6)2 = 0, , Upon solving we get, , c=6, , k 7, , Hence the value of c is 6., , [1], 5., , Since it is a rectangle, (AB) = (CD), , [½], , x + 3y = 6, 2x – 3y = 12, , ...(i), , [½], , (AD) = (BC), ...(ii), , [½], , Graph of x + 3y = 6 :, When x = 0, we have y = 2 and when y = 0, we, have x = 6., [½], Therefore, two points on the line are (0, 2) and, (6, 0)., [½], , Adding (i) and (ii), we get, 2x = 44, x = 22, , [From I and III], , 2, , 2, 3, 7, , , k  1 k  2 3k, , x – y = 14, , [From II and III], , c2 – 6c = 0, , For this pair of linear equations to have infinitely, many solutions, they need to be coincident [½], , x + y = 30, , 3 3c, , c, c, , [½], , –3c = 3c – c2, , (k – 1)x + (k + 2)y = 3k, , 3., , [From I and II], , c = ±6, , 2x + 3y = 7, , , , [½], , I, ,  Lines are coincident and will have infinite, solutions., [½], 2., , For infinitely many solutions, , a1 b1 c1, , , a2 b2 c 2, , [½], , There exists infinite solutions, Here, , [½], , [½], , The line x + 3y = 6 is represented in the given, graph., , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Graph of 2x – 3y = 12 :, , Mathematics, , 7., , When x = 0, we have y = –4 and when y = 0,, we have x = 6., [½], , Denominator = y, Given x + y = 2y – 3, , , The line 2x – 3y = 12 is shown in the graph., , From the next condition, , y, , 7, , 2x  y  1  0, , 6, 5, 3, , x, , –8 –7 –6 –5 –4 –3 –2 –1, O, –1, , 1, 1, , x+, , 3, –, 2x, , –3, , 6, , 4, =, 3y, , 5, , 12, , 6, , 7, , 8, , x, , , , (0 , –4), , 8., , –6, –8, y, , [½], , ax by, ,  ab, b, a, , ...(i), , ax – by = 2ab, , ...(ii), , [½], , ...(i), , [½], [½], , [½], , x2, , [½], , Substitute, x = 2 in (i), 2 + 3y = 8, 3y = 6, , [1], , y=2, , [½], ,  x=2, y=2, , [½], 9., , y = –a, Substituting y = –a in (i), , [½], , Let the present age of father be x years and the, sum of present ages of his two children be y, years., [½], According to question, , [½], , x = 3y,  x – 3y = 0, , a, x a, b, , [½], ...(i), , After 5 years,, , x=b,  x = b and y = –a, , 4,  3y  8, x, , [1], , 34,  17, x, , a, x  y  2a, b, , a, b, x  ( a )  a  b, b, a, , 4, 7, , 18,  12y  15, x, , 1, and subtract (i) from (ii), b, , ba, y ,   ab,  a , , Fraction , , 16,  12y  32, x, , The line x + 3y = 6 intersects y-axis at (0, 2) and, the line 2x – 3y = 12 intersects y-axis at (0, –4). [½], , ax by, ,  a  b, , b  a, ____________________, , [1], , 6,  4y  5, ...(ii), x, Multiplying 4 to (i) and 3 to (ii), , –7, , , , ...(ii), , y=7, , (6, 0), , –5, , Multiply (ii) with, , [1], , x=4, 3y, =, , 2, , –2, –4, , ...(i), , Solving (i) and (ii), , 4, 2, , xy 3 0, , x 1 1, , y 1 2, , 8, , 6., , Lets say numerator = x, , Hence, the two points on the line are (0, –4) and, (6, 0)., [½], , (0 , 2), , 5, , x + 5 = 2(y + 10), [½], ,  x – 2y = 15, , ...(ii), , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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6, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , On subtracting equation (i) from (ii), we get :, x  2y, x  3y, , , y, , , , , 11. Let AB be the pillar of height 9 meter. The, peacock is sitting at point A on the pillar and B, is the foot of the pillar. (AB = 9), , 15, 0, , [1], , ,  15, , Let C be the position of the snake which is at, 27 meters from B. (BC = 27 and ABC = 90°), As the speed of the snake and of the peacock, is same they will travel the same distance in, the same time, , On substituting the value of y = 15 in (i), we get :, x – 3 × 15 = 0,  x = 45, , [½], , Now take a point D on BC that is equidistant, from A and C (Please note that snake is moving, towards the pillar), [½], , Hence, the present age of father is 45 years., 10. Let the numerator of required fraction be x and, the denominator of required fraction be y (y  0), According to question;, , A, , [½], , x 2 1, , y, 3, ,  3x – 6 = y, , [½], ,  3x – y = 6, , y, , C, , ...(i), , D, , x, , 27 mt, , [½], B, , Hence by condition AD = DC = y(say), , and, , Take BD = x, , x, 1, , y 1 2, ,  2x = y – 1, , Now consider triangle ABD which is a right, angled triangle, , [½], ,  2x – y = –1, , Using Pythagoras theorem (AB2 + BD2 = AD2), , ...(ii), , 92 + x2 = y2, , On subtracting (ii) from (i), we get :, , 3x, 2x, , , 9 mt, , y, ,  y,  y, , x, , 81 =, ,  6,  1, , , , y2, , –, , x2, , [½], = (y – x)(y + x), , 81/(y + x) = (y – x), [1], , [½], , y + x = BC = 27, , 7, , Hence, 81/27 = (y – x) = 3, , On substituting x = 7 in (i), we get :, 3(7) – y = 6,  –y = 6 – 21,  y = 15, , [½], , Hence, the required fraction is, , [½], , x, 7, , ., y 15, , y–x=3, , ...(i), , y + x = 27, , ...(ii), , [½], , [½], , Adding (i) and (ii), gives 2y = 30 or y = 15, , [1], , x = 12, y = 5, , [1], , Thus the snake is caught at a distance of, x meters or 12 meters from the hole., [½], , Chapter - 4 : Quadratic Equations, 1., ,  For the given quadratic equation, , x2 + 6x + 9 = 0, x2 + 2.3x + (3)2 = 0, (x +, , 3)2, , , ,  (10)2  4 3 3, , [½], , =0, , = 100 – 36, ,  x = –3 is the solution of x2 + 6x + 9 = 0. [½], 2., , 2, , 3 3 x  10 x  3  0., , Discriminant for ax 2 + bx + c = 0 will be, b2 – 4ac., [½], ,  3 , , = 64, 3., , [½], , Answer (B), Given a quadratic equation, x2 – 3x – m(m + 3) = 0, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , , , x2, , – (m + 3)x + mx – m(m + 3) = 0, , , , [½], ,  –4k  –16, , (x – (m + 3))(x + m) = 0, ,  k4,  k has all real values  4, , [½], 8., , Answer (A), , 3x2, , Therefore, y = 1 will satisfy both the equations., , Let the roots be and, , a(1)2, , + a(1) + 3 = 0, Product of roots , ,  2a + 3 = 0, , [½], ,  ., , 3,  a, 2, 9., ,  1+1+b=0, , c, a, , 1 k, ,  3, , [½], ,  mx2 – 7mx + 49 = 0, , D  (7m )2  4m  49, , 3,  2  3, 2, , ab , , [½], , Given; mx(x – 7) + 49 = 0, ,  b = –2, , , 1, ., , ,  k = 3, , Also, (1)2 + (1) + b = 0, , 49 m 2  4m  49, , Given quadratic equation is,, , Here a = 3, b = –2k and c = 12., ,  b2 – 4ac = 0, ,  2, , 5p, , , , , , 20p2, , – 60p = 0, , The quadratic equation will have equal roots if, =0, , [½], ,  b2 – 4ac = 0, ,  4 p(15)  0, ,  20p(p – 3) = 0, , Putting the values of a, b and c we get, (2k)2 – 4(3)(12) = 0, , , 4k2, , – 144 = 0, ,  p = 3 or p = 0, , , , 4k2, , = 144, , But, p = 0 is not possible., ,  k2 , ,  p=3, ∵ x = 3 is one of the root of, , [1], , 10. Given quadratic equation is 3x2 – 2kx + 12 = 0, , For real equal roots, discriminant = 0, , , , [∵ m  0], , m=4, , Here, a  p, b  2 5 p, c  15, , 2, , [1], , 49m 2  4m  49  0, , [½], , px 2  2 5 px  15  0, , [1], , 144,  36, 4, Considering square root on both sides,, , [½], x2, , – 2kx – 6 = 0, , (3)2 – 2k(3) – 6 = 0, , k  36  6, , 9 – 6k – 6 = 0, , Therefore, the required values of k are 6 and –6. [1], , 3 – 6k = 0, , [½], , k, , 3 1, , 6 2, , [½], , x2 + 4x + k = 0, ∵ Roots of given equation are real,, D0, , 11., , 4 3x 2  5x  2 3  0, ,  4 3x 2  8x  3x  2 3  0, , 3 = 6k, , 7., , – 10x + k = 0, , ∵ Roots of given equation are reciprocal of each, other., ,  a+a+3=0, , 6., , [½], , It is given that 1 is a root of the equations, ay2 + ay + 3 = 0 and y2 + y + b = 0., , , 5., , 7, , –4×k0, , x(x – (m + 3)) + m(x – (m + 3)) = 0,  x = –m, m + 3, 4., , Mathematics, , (4)2, , [½], ,  4x, , , , , , 3x  2  3, , , ,  4x  3 , , , , x, , , , , , 3x  2  0, , [1], , , , 3x  2  0, , 3, 2, or x  , 4, 3, , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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8, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 12. Comparing the given equation with the standard, quadratic equation (ax2 + bx + c = 0), we get, a = 2, b = a and c = –a2, Using the quadratic formula, x , we get :, , 2, , b  b  4ac, ,, 2a, , Substituting p = 7 in p(x2 + x) + k = 0, we get, 7(x2 + x) + k = 0,  7x2 + 7x + k = 0, The roots of the equation are equal.,  Discriminant = b2 – 4ac = 0, Here, a = 7, b = 7, c = k, , a  a 2  4  2  (  a ) 2, x, 22, , , , [1], ,  (7)2 – 4(7)(k) = 0, , a  9a 2, 4, ,  49 – 28k = 0,  28k = 49, , a  3a, , 4, , , x, , b2 – 4ac = 0, , , , a  3a a, a  3a, ,  a, or, 4, 2, 4, , k, , 49 7, , 28 4, , 15. Quadratic equation px2 – 14x + 8 = 0, Also, one root is 6 times the other, , So, the solutions of the given quadratic equation, , a, or x  a., 2, 13. 4x2 + 4bx – (a2 – b2) = 0, are x , , [1], , Let say one root = x, Second root = 6x, From the equation : Sum of the roots  , , , ,  a2  b2 , x 2  bx  , 0, 4 , , , Product of roots , , , , b, a2  b2, x 2  2   x , 4, 2, , , , , , b, b, a2  b2  b , x 2  2   x    ,  , 4, 2, 2, 2, , 2, , [1], , 2, , [1], , 2, , x  6x , x, , 8, p, , 14, ., p, , 2, p, , [1], ,  6x 2 , , b, a2,   x   , 2, 4, , , 8, p, , 2, , 2, 8,  6  , p, p,  , , , , b, a, x , 2, 2, , , , x, , b a, , 2 2, , 64 8, , p, p2, , , , x, , b  a b  a, ,, 2, 2, , p=3, , [1], , 16. Let assume two numbers be x, y., , ab,  a  b  . [1], Hence, the roots are  ,  and , ,  2 ,  2 , , 14. Given –5 is a root of the quadratic equation, 2x2 + px – 15 = 0.,  –5 satisfies the given equation., , Given, x + y = 8  x = 8 – y, 1 1, 8,  , x y 15, , ...(i), [1], , xy, 8, 8, 8, , , , xy, 15, xy 15, ,  2(–5)2 + p(–5) – 15 = 0, ,  xy = 15, ,  50 – 5p – 15 = 0, , From (i) xy = y(8 – y) = 15, ,  35 – 5p = 0, ,  y2 – 8y + 15 = 0, , [1], , y = 3, 5  x = 5, 3, ,  5p = 35,  p=7, , 14, p, , [1], ,  The numbers are 3 and 5., , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 17., , x 2  3 5 x  10  0, ,  k(k – 5) + 3(k – 5) = 0, , For any quadratic equation, ,  (k – 5)(k + 3) = 0, , b  b 2  4ac, ax  bx  c  0  x , 2a, 2, , , 4x2, , x, , 3 5  45  40, 2, , x, , 3 5 5, 2, , [1], , – 4ax +, , b2), , =0, ,  (4x2 – 4ax + a2) – b2 = 0, , [1], ,  [(2x2) – 2.2x.a + a2] – b2 = 0,  [(2x – a)2] – b2 = 0, , 19., , ab, ab, ; x, 2, 2, , [1], , 3   3 x  2   2  3 x  2   0, 3x  2, , , , , , , , 3x  2, , , , 2, , 3x  2, , 20. (k +, , x, 4)x2, , 2, , 3, , [1], ,  3, , 2, , , , 6, 3, , [1], , + (k + 1)x + 1 = 0, , For equal roots, discriminant, D = 0, , ,  (k +, , – 4(k + 4) × 1 = 0, ,  k2 + 2k + 1 – 4k – 16 = 0,  k2 – 2k – 15 = 0, , , k2, , Now,, , 3 x 2  2 2x  2 3  0, , – 5k + 3k – 15 = 0, , D  b 2  4ac, ,  2 2 , , 2, , 4, ,  3  2 3 , ,  8  24  32  4 2, , x, , b  b 2  4ac, 2a, , , , x, , , , x, , , , x, , [1], , – 4ac = 0, 1)2, , 22. For the given equation,, , [1], , Using quadratic formula, we obtain, , a = k + 4, b + k + 1, c = 1, , b2, , + 2x – x – 2 = 0, ,  (x + 2) = 0 or (x – 1) = 0, , , , 2 3, , [1], , a  3, b  2 2, c  2 3, , 0, , , , x2, , Comparing this equation with ax2 + bx + c = 0,, we obtain, , , , 3x  2  0, , , , [1], , 3x  2  0, , , ,  x2 + x – 2 = 0, , Thus, the solution of the given equation is –2, and 1., [1], ,  3x  6x  6x  2  0, , , ,  –6x2 + 8x – 9x + 12 = 5x, ,  x = –2 or x = 1, , 2, , , , [1], ,  (x + 2)(x – 1) = 0, , 3x 2  2 6x  2  0, , , , 4  3x, 5, , x, 2x  3, ,  (4 – 3x)(2x + 3) = 5x, , , ,  [(2x – a) – b] = 0 or [(2x – a) + b] = 0, , x, , , ,  6x2 + 6x – 12 = 0, [1], ,  [(2x – a) – b][(2x – a) + b] = 0, , , , 4, 5, 3, ; x  0, , 3 , 2x  3, 2, x, 4, 5, 3 , x, 2x  3, , [1], , –, , Thus, for k = 5 or k = –3, the given quadratic, equation has equal roots., [1], 21. Given equation :, , [1], , x  5, 2 5, , (a2, , 9, ,  k = 5 or k = –3, ,  For the given equation, , 18., , Mathematics, , [1], , , , , ,  2 2  4 2, 2 3, 2 24 2, 2 24 2, or, 2 3, 2 3, 22 2, or, 3, , [1], , 2 2 2, 3, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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10, , 23., , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 25. Let the usual speed of the plane be x km/hr., , 3 2,  2, or x , 3, 3, , , , x, , , ,  2, x  3 2 or x , 3, , , , x  6 or x , ,  2, 3, , Time taken to cover 1500 km with usual, , 1500, hrs, x, Time taken to cover 1500 km with speed of, speed , , [1], , , , 1, 1, 2, , ,  x  1 x  2   x  2   x  3  3, , 2, ,  x 2  3 x  2  ( x  3) 3, , [1], , –, , 2x(x2, , [1], , x = –600 or x = 500, But speed can't be negative., Hence, usual speed 500 km/hr., , + 16x = 0, , [1], , Then, their areas are x 2 and y 2 and their, perimeters are 4x and 4y., , x2 – 4x – 2x + 8 = 0, , By the given condition :, , x(x – 4) – 2(x – 4) = 0, , x2 + y2 = 400, , (x – 4)(x – 2) = 0, , and 4x – 4y = 16, , x – 4 = 0 or x – 2 = 0, ,  4(x – y) = 16  x – y = 4, , x = 4 and x = 2, , [1], , 24. Given ad  bc for the equation (a2 + b2)x2 +, 2(ac + bd)x + (c2 + d 2) = 0., For this equation not to have real roots its, discriminant < 0., [1], ,  x=y+4, , ...(i), , ...(ii), , [1], , Substituting the value of x from (ii) in (i), we get :, (y + 4)2 + y2 = 400,  y2 + 16 + 8y + y2 = 400,  2y2 + 16 + 8y = 400,  y2 + 4y – 192 = 0, , D = 4(ac + bd)2 – 4(a2 + b2)(c2 + d 2), D = 4a2c2 + 4b2d2 + 8acbd – 4a2c2 – 4b2d2 –, 4b2c2 – 4a2d 2, [1], ,  y2 + 16y – 12y – 192 = 0,  y(y + 16) – 12(y + 16) = 0, , D = –4(a2d 2 + b2c2 – 2acbd), ,  (y + 16)(y – 12) = 0, , D = –4(ad – bc)2, ,  y = –16 or y = 12, , [1], , [1], , Since, y cannot be negative, y = 12., , Given ad  bc, , So, x = y + 4 = 12 + 4 = 16, ,  D<0, Quadratic equation has no real roots., , [1], , 26. Let the sides of the two squares be x cm and, y cm where x > y., , – 6x + 8) = 0, , x2 – 6x + 8 = 0, ,  x  100  x  1, 1500 , ,  x ( x  100)  2, , x2 + 100x – 300000 = 0, , 6x – 12 = 2x3 – 12x2 + 22x – 12, 12x2, , 1500, 1500, 1, , , x, x  100 2, , x2 + 100x – 300000 = 0, , 2, , 2x  4, 2, , x 3  6 x 2  11x  6 3, , 2x3, , [1], , 150000 × 2 = x(x + 100), , 2x  4, 2, , 2, 3, x  3 x  3 x  9 x  2x  6, 3, , 1500, hrs., x  100, , 1500, 1500, 1, , , x, x  100 2, , ( x  3)  ( x  1), 2, ,  x  1 x  2  ( x  3) 3, , x  3  x 1, , (x + 100) km/hr , , [1], , Thus, the sides of the two squares are 16 cm, and 12 cm., [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 27., , 1, 1 1 1, ,  , 2a  b  2 x 2a b 2 x, , (k + 4)x2 + (k + 1)x + 1 = 0, , 1, 1, 1 1, , , , 2a  b  2 x 2 x 2a b, , , , 2 x  2a  b  2 x b  2a, , 2 x (2a  b  2 x ), 2ab, , , , b  2a, 2a  b, , 2 x (2a  b  2 x ), 2ab, , , , 1, 1, , x (2a  b  2 x ) ab, , [1], , [1], ,  k2 – 5k + 3k – 15 = 0,  (k – 5) (k + 3) = 0,  k – 5 = 0 or k + 3 = 0,  k = 5 or –3, [1], ,  x + a = 0 or 2x + b = 0, , b, 2, , [1], , Difference between the natural numbers = 5, ...(i), , Difference of their reciprocals, , 1, (given), 10, , 1 1, 1,  , y x 10, , 5, 1, , xy 10, ,  xy = 50, , Thus, the values of k are 5 and –3.,  9x2 + 6x + 1 = 0,  (3x)2 + 2(3x) + 1 = 0,  (3x + 1)2 = 0, , , Given :,  x–y=5, , [For k = –3], ,  (x – 1)2 = 0,  x = 1, 1, , [1], , Thus, the equal roots of the given quadratic, , [1], , Putting the value of x from equation (i) in, equation (ii), we get, (y + 5) y = 50, , 1, equation is either 1 or  ., [1], 3, 30. Let l be the length of the longer side and b be, the length of the shorter side., , Given that the length of the diagonal of the, rectangular field is 16 metres more than the, shorter side., Thus, diagonal = 16 + b, Since longer side is 14 metres more than, shorter side, we have,, l = 14 + b, ,  y2 + 5y – 50 = 0, , Diagonal is the hypotenuse of the triangle., ,  y2 + 10y – 5y – 50 = 0, , [1], , Consider the following figure of the rectangular, field., ,  y(y + 10) – 5(y + 10) = 0,  (y – 5)(y + 10) = 0,  y = 5 or –10, , 1, 1, x , , 3, 3, ,  x2 – 2x + 1 = 0, , [1], , ...(ii), , [1], , For k = 5, (k + 4)x2 + (k + 1)x + 1 = 0, , 28. Let the two natural numbers be x and y such, that x > y., , , ,  k2 + 2k + 1 – 4k – 16 = 0,  k2 – 2k – 15 = 0, ,  (x + a)(2x + b) = 0, , xy, 1, , xy, 10, , [1], ,  (k + 1)2 – 4 × (k + 4) × 1 = 0, ,  2x(x + a) + b(x + a) = 0, , , , Since the given quadratic equation has equal, roots, its discriminant should be zero.,  D=0, ,  2x2 + 2ax + bx + ab = 0, , x  a, or x , , 11, , 29. Given quadratic equation :, , , , , , Mathematics, , D, , [1], , As y is a natural number, therefore y = 5, , Breadth, , C, Diagonal, , Other natural number = y + 5 = 5 + 5 = 10, Thus, the two natural numbers are 5 and 10. [1], , A, , Length, , B, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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12, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , By applying Pythagoras Theorem in ABD, we, have,, Diagonal2 = Length2 + Breadth2, , Multiply throughout by the L.C.M., we get, (x + 2)(x + 4) + 2(x + 1)(x + 4) = 4(x + 1), (x + 2), , [1], ,  (16 + b)2 = (14 + b)2 + b2, , (x + 4)(x + 2 + 2x + 2) = 4(x2 + 3x + 2), ,  256 + b2 + 32b = 196 + b2 + 28b + b2,  256 + 32b = 196 + 28b +, , (x + 4)(3x + 4) = 4x2 + 12x + 8, , b2, , 3x2 + 16x + 16 = 4x2 + 12x + 8, ,  60 + 32b = 28b + b2,  b2 – 4b – 60 = 0, ,  x2 – 4x – 8 = 0, , [1], , Now, a = 1, b = –4, c = –8, ,  b2 – 10b + 6b – 60 = 0,  b(b – 10) + 6(b – 10) = 0, , x, ,  (b + 6)(b – 10) = 0, , b  b 2  4ac 4  16  32, , 2a, 2, , ,  (b + 6) = 0 or (b – 10) = 0,  b = –6 or b = 10, , , , As breadth cannot be negative, breadth = 10 m, Thus, length of the rectangular field = 14 + 10, = 24 m., [1], , Speed of the motor boat (downstream) = 24 + s, [1], According to the given condition,, , 32, 32, , 1, 24  s 24  s, , [1], , 54( x  6)  63 x, 3, x ( x  6), , , , 1, 1 , 32 , ,  1,  24  s 24  s , , , ,  24  s  24  s , 32 ,  1,  576  s 2, , ,  54(x + 6) + 63x = 3x(x + 6),  54x + 324 + 63x = 3x2 + 18x,  117x + 324 = 3x2 + 18x, , , 3x2, , [1], , [1], ,  32 × 2s = 576 – s2, , – 117x – 324 + 18x = 0, ,  s2 + 64s – 576 = 0, ,  3x2 – 99x – 324 = 0, ,  (s + 72)(s – 8) = 0, ,  x2 – 33x – 108 = 0, , [1], ,  x2 – 36x + 3x – 108 = 0, ,  s = –72 or s = 8, ,  x(x – 36) + 3(x – 36) = 0, , Since, speed of the stream cannot be negative,, the speed of the stream is 8 km/h., [1], ,  (x + 3)(x – 36) = 0, , [1], ,  (x + 3) = 0 or (x – 36) = 0,  x = –3 or x = 36, Speed cannot be negative. Hence, initial speed, of the train is 36 km/hour., [1], 32., , [1], , Speed of the motor boat (upstream) = 24 – s, , Thus, we have,, , , , x  22 3, , [1], , Speed of the motor boat 24 km/h, , Distance,  time, Speed, , 54, 63, , 3, x x6, , 4  48 4  4 3, , 2, 2, , 33. Let the speed of the stream be s km/h., , 31. Let x be the first speed of the train., We know that,, , [1], , 1, 2, 4, , , x 1 x  2 x  4, L.C.M. of all the denominators is (x + 1)(x + 2), (x + 4), [1], , 34., , 1, 3, 5, 1, , , , x   1, ,  4, x  1 5x  1 x  4, 5, Take L.C.M. on the left hand side of equation, 5 x  1  3( x  1), 5, , ( x  1)(5 x  1), x4, , [1], , 8x2 + 4x + 32x + 16 = 25x2 + 5 + 5x + 25x, 17x2 – 6x – 11 = 0, , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 17x2, , – 17x + 11x – 11 = 0, , 1 , , 17x(x – 1) + 11(x – 1) = 0, (x – 1)(17x + 11) = 0, , x, , , , [1], [1], , 1, hrs, 13, Say taps are A, B and, , in 3, , B fills tank in (x + 3) hrs, Portion of tank filled by A (in 1 hr) , Portion of tank filled by B (in 1hr) , , [1], , 1, x, , 1, 1, 13, , , x x  3 40, , , , –324 + x2 = –48x, , 13, 40, [1], , – 41x – 120 = 0, ,  13x2 – 65x + 24x – 120 = 0, , 24,  x = 5 or, 13, [But negative value not be taken], B fills tank in 8 hrs, , [1], , [½], , x = –54 km/hr (not possible), , [½], , 37. Let x be the original average speed of the train, for 63 km., , , , [1], , Now, for upstream: speed = (18 – x) km/hr, , 63 x  378  72 x, 3, x ( x  6), , [½], ,  135x + 378 = 3x2 + 18x, , [½], ,  x2 – 39x – 126 = 0, , [½], ,  (x – 42)(x + 3) = 0, , [½], , x  42, , OR, , x  3, , [½], , 38. Let the time in which tap with longer and smaller, diameter can fill the tank separately be x hours, and y hours respectively., [½], According to the question, , [½], , 1 1, 8,  , x y 15, , and x = y – 2, [½], , 1, 1, 8,  , y  2 y 15, , [½], , , , y y 2, y 2  2y, , , , [½], , ...(ii), , [½], , [½], , 8, 15, ,  15(2y – 2) = 8(y2 – 2y), , 24 (18  x )  (18  x ), ,  30y – 30 = 8y2 – 16y, [½], , ...(i), , On substituting x = y – 2 from (ii) in (i), we get, , 24, 24, , 18  x 18  x, , (18)2  x 2, , [½], , Since, speed cannot be negative., , Given that,, , 1 , , [½], , , Distance ,  Time =, , Speed , , , Now, for downstream: speed = (18 + x) km/hr, , 1 , , 63, 72, , 3, x ( x  6), , Therefore x = 42 km/hr., , 36. Let the speed of stream be x km/ hr., , 24, 24, , 1, 18  x 18  x, , + 54x – 6x – 324 = 0, , x = –54 or x = 6, , , ,  A fills tank in 5 hrs, , 24 ,  Time taken  ,  hr,  18  x , , [½], , (x + 54)(x – 6) = 0, , , , + 39x, , 24 ,  Time taken  ,  hr,  18  x , , x2, , Total time taken to complete the journey is 3 hrs., , (x + 3 + x)40 = 13(x)(x + 3), 13x2, , [½], , 324  x 2, , Then, (x + 6) will be the new average speed for, remaining 72 km., [½], , 1, x3, , Portion of tank filled by A and B (both in 1hr) , , 13x2, , 13, , Therefore, speed of the stream = 6 km/hr., , A fills the tank by itself in x hrs, , 80x + 120 =, , 24[ 2 x ], , x2 + 48x – 324 = 0, , 11, , 1, 17, , 35. Two taps when run together fill the tank, , , , Mathematics, ,  8y2 – 46y + 30 = 0, , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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14, , Mathematics, , , , 4y2, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , – 20y – 3y + 15 = 0, ,  (4y – 3)(y – 5) = 0, , , y, , 3, ,y 5, 4, , [½], , , Substituting values of y in (ii), we get, , 3, 2, 4, 5, x, 4, 5,  x, 4,  time cannot ,  be negative , , , x, , x  52, , x, , 9(a  b )  3(a  b ), 18, , x, , 3a  3b  a  b 3a  3b  a  b, ,, 6, 6, , [1], , x, , 2a  b a  2b, ;, 3, 3, , [1], , 41. –5 is root of 2x2 + px – 15 = 0,  2(–5)2 + p(–5) – 15 = 0, , x 3, [½], , 10 – p – 3 = 0,  p=7, ,  7x2 + 7x + k = 0, , Given that y – x = 4  y = 4 + x, , ...(i), , 1 1 4,  , x y 21, , [1], [1], , yx, 4, , xy, 21, , [1], , [As we know p = 7] [1], ,  Discriminant = 0, D = 49 – 28k, , [1], , 28k = 49, , 39. Let assume the two numbers to be x, y (y > x), , , , [1], , p(x2 + x) + k = 0 has equal roots., , Hence, the time taken by tap with longer, diameter is 3 hours and the time taken by tap, with smaller diameter is 5 hours, in order to fill, the tank separately., [½], , , , [1], , k, , 7, 4, , [1], , 42. Let the required three integers be (x – 1), x and, (x + 1)., [1], Now, (x – 1)2 + [x.(x + 1)] = 46, (x2 – 2x + 1) + [x2 + x] = 46, , 4, 4, , xy 21, , [1], , [1], , 2x2 – x – 45 = 0, 2x2 – 10x + 9x – 45 = 0, ,  xy = 21, x(4 + x) = 21, , [1], , 2x(x – 5) + 9(x – 5) = 0, , x2 + 4x – 21 = 0, , (x – 5)(2x + 9) = 0, , (x + 7)(x – 3) = 0, , x = 5 or x = –9/2, , x = –7, 3, , [1], , [1], , So, x = 5 [Because it is given that x is a positive, integer], [1], , [1], , Thus, the required integers are (5 –1), i.e. 4, 5, and 6., [1], , y = –3, 7,  Numbers are –7, –3 or 3, 7, , [1], , 40. 9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0, , 43. Let the smaller number be x and larger number, be y., , Discriminant, D = 81(a + b)2 – 36(2a2 + 5ab + 2b2), , [1], , D = 9[9a2 + 9b2 + 18ab – 8a2 – 8b2 – 20ab], D = 9[a2 + b2 – 2ab], 2, , , , D  9(a  b ), , , , 9(a  b )  9(a  b )2, x, 29, , [1], [1], , y2 – x2 = 88, , ...(i), , y = 2x – 5, , ...(ii), , [1], , In equation (i), (2x – 5)2 – x2 = 88, , [1], , 4x2 – 20x + 25 – x2 = 88, [1], , 3x2 – 20x – 63 = 0, , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , By splitting the middle term,, 3x2, , x, , – 27x + 7x – 63 = 0, , 3x(x – 9) + 7(x – 9) = 0, , [1], , (x – 9)(3x + 7) = 0,  x = 9 and x = –7/3, , [1], , We cannot take negative value because x must, be greater than 5., So, smaller number = 9, And larger number = 2x – 5 = 18 – 5 = 13, 44., , A, , 180 km, , [1], , B, , Distance travelled by train = 180 km, let say, speed = s km/hr, 180, Time taken (t ) , s, , Mathematics, , [1], , 180, s9, , , , t 1, , , , t, , , , 180, 180, 1, s9, s, , 46. Total cost of books = `80, Let the number of books be x., So, the cost of each book = `, , [1], , Cost of each book if he buy 4 more book, , 80, x4, As per given in question :, = `, , [1], , 80, 80, , 1, x x4, , [1], , , , 80 x  320  80 x, 1, x ( x  4), , , , 320, 1, x 2  4x, , [1], ,  x = –20, 16, , [1], , So, the number of books he bought is 16., , s2 + 9s – 1620 = 0, , [1], , Now, the sum of their reciprocals is, , [∵ s cannot negative] [1], , s  36 km/hr, , , , 9xx 1, , x (9  x ) 2, , Taking L.C.M on left side of equality, , , , 9, 1, , 9x  x 2 2, , [1], , 3x – 8 = 2x2 – 3x – 10x + 15, , [1], , 2x2 – 16x + 23 = 0, 16  256  4  2  23, x, 4, , 1, , therefore, 2, , 1, 1, 1, , , x 9x 2, , 1, 1, 3, ,  1, x  , 5., 2x  3 x  5, 2, x  5  2x  3, 1, (2 x  3)( x  5), ,  18 = 9x – x2, , [1], [1], , [1], ,  x2 – 9x + 18 = 0,  (x – 6)(x – 3) = 0,  x = 6, 3, , [1], , [1], , 47. Let the first number be x then the second, number be (9 – x) as the sum of both numbers, is 9., [1], , [1], ,  s2 + 45s – 36s – 1620 = 0, , 16  72, x, 4, , 80, x, , Since, number of books cannot be negative., , 189s + s2 = 180s + 1620, , 45., , [1], ,  (x + 20)(x – 16) = 0, [1], , (189 + s)s = 180s + 1620, , , , [1], ,  x2 + 4x – 320 = 0, , 180, 1, s9, ,  s = –45, 36, , 16  6 2, 4, , , 3 2, x  4 , , 2 , , , It is given if speed had been (s + 9) km/hr, Train would have travelled AB in (t – 1) hrs. [1], , 15, , [1], , If x = 6 then other number is 3., and if x = 3 then other number is 6., , [1], , Hence, numbers are 3 and 6., , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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16, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Chapter - 5 : Arithmetic Progressions, 1., , First term of an AP = p, , 5., , Common difference = q, , 2., , [½], , T10 = p + 9q, , [½], , , , a, , 4,  2a, 5, , 2a , , , , This is an AP with first term 1 and the common, difference 2., [½], Sum of 20 terms = S20, , 4, , a, 2 are in AP, 5, , a, , S20 , [½], , 4, 2, 5, ,  2a , , 3., , The first 20 odd numbers are 1, 3, 5, ..... 39, , T10 = p + (10 – 1)q, , Given, , 6., , 14, 5, , 7, 5, , 1  6q 1 1  6q  1 6q, , , ,  2, 3q, 3q, 3q, 3q, , [½], 7., , Answer (C), , We need to find the value of y., , p, ap 2  bp   2k  ( p  1)d , 2, 2b + 2ap = (2k – d) + pd, , [1], , The first three terms of an AP are 3y - 1, 3y + 5, and 5y + 1, respectively., , We know that if a, b and c are in AP, then :, b–a=c–b, , 2ap + 2b = 2k + (p – 1)d, ,  2b = a + c, , [½], ,  2(3y + 5) = 3y – 1 + 5y + 1, , Comparing terms on both sides,, , [½], ,  6y + 10 = 8y, , 2a  d, ,  10 = 8y – 6y, , 2k – d = 2b, ,  2y = 10, , 2k = 2b + 2a, ,  y=5, , k ab, , Hence the correct option is C., , Common difference = 2a, , 8., , [½], , Answer (C), , If k + 9, 2k – 1 and 2k + 7 are the consecutive, terms of AP, then the common difference will be, the same., , Given common difference of the, ,  (2k – 1) – (k + 9) = (2k + 7) – (2k – 1) [½], , AP = d = 3, ,  k – 10 = 8, , Lets say the first term = a, ,  k = 18, , First term = a + b, 4., , Answer (C), Common difference =, , Lets say first term = k & common difference = d, , , , 20, 2(1)  (20  1)(2)  102  38  400 [½], 2, , Thus, the sum of first 20 odd natural numbers is, 400., , Given an AP which has sum of first p terms, = ap2 + bp, , , , Answer (C), , [½], , 9., , a20 = a + 19d = a + 19 × 3, [½], , a20 – a15 = a + 57 – a – 42, [½], , ...(i), , In an AP a1, a2, a3, a4 ....., an = a1 + (n – 1)d, , = a + 42, , = 15, , Given, a21 – a7 = 84, , = a + 57, a15 = a + 14d = a + 14 × 3, , [½], , d = common difference, , a21 = a1 + 20d, , ...(ii), , a7 = a1 + 6d, , ...(iii), , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Substituting (ii) and (iii) in (i), , Mathematics, , 17, , Last term which is Tn, , a1 + 20d – a1 – 6d = 84, , = a + (n – 1)d, , 14d = 84, , = a + (9)d, , [½], ,  29 = 2 + 9d, , d=6,  Common difference = 6, , [½], , 10. a7 = 4, , d 3, Common difference = 3, , a + 6d = 4 (as an = a + (n – 1)d), , [½], , 14. Two digit numbers divisible by 6 are,, , but d = –4, 12, 18..... 96, a + 6(–4) = 4, , [½], , a + (–24) = 4, , [1], ,  96 = 12 + (n – 1) × 6, [∵ an = a + (n – 1)d], , a = 4 + 24 = 28, Therefore first term a = 28, , [½], ,  n, , 11. Two digit numbers divisible by 3 are, , 96  12,  1  15, 6, , [½], ,  Two digit numbers divisible by 6 are 15. [½], , 12, 15, 18, ....., 99., a = 12, d = 15 – 12 = 3, , [½], ,  Tn = 99, , 15. First three– digit number that is divisible by, 7 = 105, Next number = 105 + 7 = 112, ,  a + (n – 1)d = 99,  12 + (n – 1)3 = 99, , Therefore the series is 105, 112, 119,…, ,  n = 30, , The maximum possible three digit number is 999., ,  Number of two digit numbers divisible by 3, are 30., [½], , When we divide by 7, the remainder will be 5., , 12. Given an AP 3, 15, 27, 39, ....., , Clearly, 999 – 5 = 994 is the maximum possible, three – digit number divisible by 7., , Lets say nth term is 120 more than 21st term, , The series is as follows :, ,  Tn = 120 + T21, , 105, 112, 119, …., 994, , a + (n – 1)d = 120 + (a + 20d), , [1], , Here a = 105, d = 7, , (n – 1)12 = 120 + 20 × 12, , Let 994 be the nth term of this AP., , n – 1 = 30, , an = a + (n – 1)d, ,  31st term is 120 more than 12th term., , [1], ,  994 = 105 + (n – 1)7,  (n – 1)7 = 889, , 13. Given an AP with first term (a) = 2, Last term () = 29, ,  (n – 1) = 127, , Sum of the terms = 155, ,  n = 128, , Common difference (d) = ?, , So, there are 128 terms in the AP., , Sum of the n terms ,  155 , , , n, (a   ), 2, , [½], , n, (2  29), 2, , n  10, , [½], , , , Sum , , , [½], , [½], , n, {first term + last term}, 2, 128, a1  a128 , 2, , 64{105 + 994} = (64)(1099) = 70336, , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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18, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 16. Let a be the first term and d be the common, difference., Given : a = 5, , Substituting value of d in (ii), we have,  2a + 45 = 47, , Sn = 400, ,  2a = 2, , We know :, ,  a=1, , Tn = a + (n – 1)d, , Thus, the given AP is 1, 6, 11, 16,....., ,  45 = 5 + (n – 1)d,  40 = (n – 1)d, And Sn , , ...(i), , [1], , 18. 4th term of an AP = a4 = 0,  a + 3d = 0,  a = –3d, , [½], , On substituting n = 16 in (i), we get :, , = –3d + 24d, , ...[From (i)], , [½], , = 21d, , 40 = (16 – 1)d, , 3 times 11th term of an AP = 3a11, ,  40 = (15)d, , = 3[a + (11 – 1)d], , 40 8, , 15 3, , = 3[a + 10d], , 8, Thus, the common difference is ., 3, , = 3[–3d + 10d], [½], , i.e., the 25th term of the AP is three times its, 11th term., [½], , 5, 2a  4d  7 2a  6d  167, 2, 2,  5a + 10d + 7a + 21d = 167, , 1, 1, 3, 19. Given progression 20, 19 , 18 , 17 , ....., 4, 2, 4, , , , ...(i), , This is an Arithmetic progression because, [½], , Also, S10 = 235, , Common difference, , (d )  19, , 10, , 2a  9d  235, 2,  10a + 45d = 235, , d, ,  2a + 9d = 47, , ...(ii), , Multiplying equation (ii) by 6, we get, 12a + 54d = 282, , [½], ,  a25 = 3a11, , n, 2a  (n  1)d, 2,  S5 + S7 = 167, , Now, Sn , ,  12a + 31d = 167, , = 3 × 7d, = 21d, , 17. S5 + S7 = 167 and S10 = 235, , ...(iii), , [½], , 1, 1, 1,  20  18  19  ......, 4, 2, 4, , 3, 4, , [1], , 3  83  3n, Any nth term an  20  (n  1) , , 4,  4 , Any term an < 0 when 83 < 3n, , Subtracting (i) from (iii), we get, , 23d, , [½], , = a + (25 – 1)d, ,  n = 2 × 8 = 16, , , , ...(i), , 25th term of an AP = a25, , n 400, , 2, 50, , 12a  54d, (  )12a  31d, , [½], ,  a + (4 – 1)d = 0, , n, (a  Tn ), 2, , n,  400  (5  45), 2, ,  d, , [½], , 2a + 9(5) = 47, , Tn = 45, , , ,  d=5, ,  n, , 83, 3, , , 282,  167, ,  n = 28, , , ,  28th term will be the first negative term. [1], , 115, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Mathematics, , 24. First term (a) = 5, , 20. First 8 multiples of 3 are, 3, 6, 9, 12, 15, 18, 21, 24, , Tn = 33, , The above sequence is an AP, , [1], , a = 3, d = 3 and last term l = 24, , n, 8, (a  l )  [3  24]  4(27), 2, 2, Sn = 108, , Sum of first n terms = 123, , , Sn , , [1], , n6, , Let Sn – 1 be sum of (n – 1) terms, tn = Sn – Sn – 1, = (3n2 – 4n) – [3(n – 1)2 – 4(n – 1)], , [½], , = (3n2 – 4n) – [3n2 – 6n + 3 – 4n + 4], , [½], , [1], , Tn = a + (n – 1)d, , [½], , 33 = 8 + (5)d, , d 5, , [1], , 25. Lets say first term of given AP = a, , = 3n2 – 4n – 3n2 + 10n – 7,  tn = 6n – 7, , Common difference = d, , So, required nth term = 6n – 7, 22., , n, a  Tn   123, 2, n, 8  33  123, 2, , [1], , 21. Sn = 3n2 – 4n, , nth, , [½], , Sum of first six terms = 42, , term of 63, 65, 67, ....., , , = 63 + (n – 1)(2), = 63 + 2n – 2, , 6, (2a  5d )  42, 2, 2a + 5d = 14, , = 61 + 2n, , ...(i), , [1], , nth term of 3, 10, 17, ....., = 3 + (n – 1)7, , , , = 7n – 4, , ...(ii), ,  2a = 2d, , 61 + 2n = 7n – 4, , , , n  13, , [1], , 23. Lets assume first term = a, , ad, , [1], , Substituting (ii) in (i),  2a + 5a = 14, a = 2 and d = 2, , Common difference = d, , T13 = a + 12d, , Tm = a + (m – 1)d, , = 2 + 24, , Tn = a + (n – 1)d, Given m.Tm = n.Tn, , ...(ii), , a  9d, 1, , a  29d 3, , Given that nth terms of two AP’s are equal., 65 = 5n, , [1], , 3a + 27d = a + 29d, , [1], , [Using (i) and (ii)], , ...(i), , Also given T10 : T30 = 1 : 3, , = 3 + 7n – 7, , [1], , T13 = 26, , [1], , m(a + (m – 1)d) = n(a + (n – 1)d), , 26. Sum of first ten terms = –150, , ma + m(m – 1)d = na + n(n – 1)d, , Sum of next ten terms = 550, , (m – n)a + d(m2 – m – n2 + n) = 0, , [1], , a(m – n) + d(m – n)(m + n – 1) = 0, , Lets say first term of AP = a, Common difference = d, , (m – n)[a + (m + n – 1)d] = 0, , Sum of first ten terms , , mn,  a + (m + n – 1)d = 0, Tm  n  0, , 19, , 10, [2a  9d ], 2, , –150 = 5[2a + 9d], [1], , 2a  9d  30, , ...(i), , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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20, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , For sum of next ten terms the first term would, be T11 = a + 10d,  550 , , , 10, [2(a  10d )  9d ], 2, , 110  2a  29d, , ...(ii), , 29. Sn = 3n2 + 4n, First term (a1) = S1 = 3(1)2 + 4(1) = 7, S2 = a1 + a2 = 3(2)2 + 4(2) = 20, , [1], , Solving (i) and (ii), , a2 = 20 – a1 = 20 – 7 = 13, So, common difference (d) = a2 – a1 = 13 – 7 = 6, [1], , d = –4, , Now, an = a + (n – 1)d, , a=3,  AP will be 3, –1, –5, –9, –13, ....., , [1], , 27. Given an AP, ,  a25 = 7 + (25 – 1) × 6 = 7 + 24 × 6 = 7, + 144 = 151, [1], 30. Let a be the first term and d be the common, difference of the given AP, , Say first term = a, Common difference = d, , Given :, , Given T4 = 9, , a7 , , 1, 9, , a9 , , 1, 7, , a + 3d = 9, , ...(i), , [1], , Also T6 + T13 = 40, a + 5d + a + 12d = 40, 2a + 17d = 40, , ...(ii), , [1], , a7  a  (7  1)d , , Solving (i) and (ii), a=3, , [1], ,  a  6d , , d=2, ,  AP will be 3, 5, 7, 9, ....., , We know that the nth term of an AP is given by, an = a + (n – 1)d, , a9  a  (9  1)d ,  a  8d , , 2d , , A16 = 1 + 2a8,  a + (16 – 1)d = 1 + 2[a + (8 – 1)d], ...(i), , [1], , Also, it is given that, a12 = 47,  a + (12 – 1)d = 47,  a + 11d = 47, , ...(ii), , [1], , Adding (i) and (ii), we have :, , , , [½], , Hence, a n = a + (n – 1)d = 3 + (n – 1)(4), = 3 + 4n – 4 = 4n – 1, Hence, the nth term of the AP is 4n – 1., , [1], , [½], , 1, in equation (i), we get :, 63, , 1, 63, , a63  a  (63  1)d , , 1, 1  63,  62 , 1, , 63,  63  63, , Thus, the 63rd term of the given AP is 1., , –a + 4 = 1,  a=3, , 1, 7, , 1  1, a   6 , , 63  9, , , 12d = 48, , From (i),, , ...(ii), , 1, 7, , 1, 63, , Putting d , ,  a, ,  d=4, , [1], , 2, 63, ,  d, ,  a + 15d = 1 + 2a + 14d, , ...(i), , Subtracting equation (i) from (ii), we get :, , According to the given information,, ,  –a + d = 1, , 1, 9, , [1], , 28. Let a and d respectively be the first term and the, common difference of the AP., , 1, 9, , [½], , [½], , 31. Here it is given that,, T14 = 2(T8),  a + (14 – 1)d = 2[a + (8 – 1)d],  a + 13d = 2[a + 7d], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), ,  a + 13d = 2a + 14d, , Last term () = 350, ,  –d = a, , ...(i), 6th, , [1], , term is –8., , Common difference (d) = 9, Tn = a + (n – 1)d, , T6 = –8, , = a + (n – 1)d = 350, ,  a + (6 – 1)d = –8, ,  8 + (n – 1)9 = 350, ,  a + 5d = –8,  –d + 5d = –8, ,  Number of terms = 39, , [1], , Sum of the terms, ,  d = –2, Substituting this in eq. (i), we get a = 2, , [1], , , , n, [a  ], 2, , , , 39, [8  350], 2, , Now, the sum of 20 terms,, , n, Sn   2a  (n  1)d , 2, , [1], , = 6981, , 20, S20 , 2a  (20  1)d , 2, = 10[2(2) + 19(–2)], , [1], , 34. Multiples of 4 between 10 and 250 are 12, 16,, ...... 248., [1], We now have an A.P with first term = 12 and, last term = 248, [1], , = 10[4 – 38], = –340, , [1], , 32. Let a 1, a 2 be the first terms and d 1, d 2 the, common differences of the two given AP’s., , n, Thus, we have Sn   2a1  (n  1)d1  and, 2, n, Sn    2a2  (n  1)d 2 , 2, n, 2a1  (n  1)d1 2a1  (n  1)d1, Sn, , [½],  2, , Sn  n  2a  (n  1)d  2a2  (n  1)d2, 2, 2, 2, , , , [1], , n  39, , [∵ Using (i)], ,  4d = –8, , It is given that, , 21, , 33. Given an A.P with first (a) = 8, ,  13d – 14d = 2a – a, Now, it is given that its, , Mathematics, , Sn, 7n  1, , Sn  4n  27, , 2a1  (n  1)d1, 7n  1, , 2a2  (n  1)d 2 4n  27, , Common difference = 4,  248 = 12 + (n – 1)4, [∵ an = a + (n – 1)d], , , [1], , n  60, ,  Multiples of 4 between 10 and 250 are 60. [1], 35. Given : S20 = –240 and a = 7, Consider, S20 = –240, , , 20, (2  7  19d )  240, 2, , [1], , ∵ S  n 2a  (n  1)d , , , n, , 2, , , ...(i), , [½], , To find the ratio of the mth terms of the two, given AP's, replace n by (2m – 1) in equation (i)., , , 2a1  (2m  1  1)d1, 7(2m  1)  1, , 2a2  (2m  1  1)d 2 4(2m  1)  27, , , , 2a1  (2m  2)d1, 14m  7  1, , 2a2  (2m  2)d 2 8m  4  27, , , , a1  (m  1)d1 14m  6, , a2  (m  1)d 2 8m  23, ,  10(14 + 19d) = –240,  14 + 19d = –24, , [1], ,  19d = –38,  d = –2, , [1], , Now, a24 = a + 23d = 7 + 23 × –2 = –39, [∵ an = a + (n – 1)d], [1], , Hence, the ratio of the mth terms of the two AP's, is 14m – 6 : 8m + 23., [1], , Hence, a24 = –39, , [1], , 36. Given AP is –12, –9, –6, ..., 21, First term, a = –12, Common difference, d = 3, , [1], , Let 12 be the nth term of the AP., , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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22, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), ,  t60 = 8 + 59 × 2, , 12 = a + (n – 1)d,  12 = –12 + (n – 1) × 3, , [1], ,  t60 = 8 + 118, ,  24 = (n – 1) × 3, ,  t60 = 126, ,  n=9, , We need to find the sum of the last 10 terms., , Sum of the terms of the AP = S9, , Thus,, , n, [1],  2a  (n  1)d   9  24  8  3   0, 2, 2, If 1 is added to each term of the AP, the sum, of all the terms of the new AP will increase by, n, i.e., 9., , Sum of last 10 terms = Sum of first 60 terms –, Sum of first 50 terms, , , ,  Sum of all the terms of the new AP = 0 + 9, =9, [1], 37. Let a and d be the first term and the common, difference of an AP respectively., , n, [2a  (n  1)d ], 2, , We have :, , ...(i), , [1], , Now, a36 = a + 35d,  Sum of the last 15 terms, , 15, (2a36  (15  1)d ), 2, , [1], , 15, [2(a  35d )  14d ], 2, = 15[a + 35d + 7d], ,  S60 , , 60, 2  8  (60  1)  2, 2, ,  S60 = 30[16 + 59 × 2],  S60 = 4020, , [1], , Similarly,, , 50, 2  8  (50  1)  2, 2, ,  S50 = 25[114],  S50 = 2850, , [1], , Thus the sum of last 10 terms = S60 – S50 =, 4020 – 2850 = 1170, [½], 39. Let there be a value of X such that the sum of the, numbers of the houses preceding the house, numbered X is equal to the sum of the numbers, of the houses following it., That is, 1 + 2 + 3 + ..... + (X – 1) = (X + 1) +, (X + 2) ..... + 49, , , ,  [1 + 2 + 3 + ..... + (X – 1), ,  2565 = 15[a + 42d],  171 = a + 42d, , n, 2a  (n  1)d , 2, ,  S50 = 25[16 + 49 × 2], , 15th term from the last = (50 – 15 + 1)th = 36th, term from the beginning, , , , Sn , ,  S50 , , 10, [2a  9d ], Sum of the first 10 terms , 2,  210 = 5[2a + 9d],  42 = 2a + 9d, , [½], ,  S60 = 30[134], , nth term of an AP, an = a + (n – 1)d, Sum of n terms of an AP, Sn , , [1], , ...(ii), , = [1 + 2 + ..... + X + (X – 1) + ..... + 49], [1], – (1 + 2 + 3 + ..... + X), , [1], , From (i) and (ii), we get,, d=4, , , , X 1, 49, X, [1  X  1] , [1  49]  [1  X ], 2, 2, 2, , a=3, ,  X(X – 1) = 49 × 50 – X(1 + X), , So, the AP formed is 3, 7, 11, 15... and 199. [1], ,  X(X – 1) + X(1 + X) = 49 × 50, , 38. Consider the given AP 8, 10, 12, ..., , [1], ,  X 2 – X + X + X2 = 49 × 50,  2X 2 = 49 × 50, , [1], , Here the first term is 8 and the common, difference is 10 – 8 = 2, ,  X 2 = 49 × 25, , General term of an AP is tn is given by,, ,  X = 7 × 5 = 35, , tn = a + (n – 1)d, , Since X is not a fraction, the value of x satisfying, the given condition exists and is equal to 35. [1], ,  t60 = 8 + (60 – 1) × 2, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 40. Let the numbers be (a –3d), (a – d), (a + d) and, (a + 3d), , Mathematics, , 41. Let the first four terms be a, a + d, a + 2d,, a + 3d, ,  (a – 3d) + (a – d) + (a + d) + (a + 3d) = 32,  4a = 32, , [½], ,  2a + 3d = 20, , [½], , [1], , ...(i), , n, 2a  (n  1) d   280, 2, , (a  3d )(a  3d ) 7, Also,, , (a  d )(a  d ), 15, , 14 , 2a  13d   280, 2,  2a + 13d = 40, ,  8a2 = 128d 2, , [1], , 8a, 888, , 128, 128, , Substituting the value of d in (i), , [1], [½], , a=7, , =4, ,  Sum of n terms , [1], , n, 2a  (n  1)d , 2, , n, 14  (n  1)2, 2, = n2 + 6n, , [½], , , , If d = 2 numbers are : 2, 6, 10, 14, If d = –2 numbers are 14, 10, 6, 2, , ...(ii), , On subtracting (i) from (ii), we get d = 2, , 2, , d = ±2, , [½], , , ,  15a2 – 135d 2 = 7a2 – 7d 2, , d2, , a + a + d + a + 2d + a + 3d = 40, Sum of first 14 terms = 280, , a=8, , d2 , , 23, , [1], , [½], , Chapter - 6 : Triangles, 1., , Length of the diagonals of a rhombus are 30 cm, and 40 cm., , PQ || BC, , AP 1, , PB 2, , D, , A, , PB 2, , AP 1, , C, , O, , PB, 2, 1 1, AP, 1, , B, , PB  AP 3, , AP, 1, , i.e., BD = 30 cm, AC = 40 cm, , AP 1, , AB 3, ,  OD = OB = 15 cm, OA = OC = 20 cm, , [½], , , In AOD,, OA2 + OD2 = AD2, , A, , [½], , R, , 60°, , AD = 25 cm, Side of rhombus = 25 cm, , 2, , ar( APQ )  AP , 1, ,  , ar( ABC )  AB , 9, N, , 3., , (20)2 + (15)2 = AD2, , 2., , [½], , [½], , L, , 50°, , M, , P, , Q, , Given LMN ~ PQR, In similar triangles, corresponding angles are equal., , P, ,  L = P, , Q, , M = Q, B, , C, , N = R, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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24, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , In LMN,, , A, , 7., , L + M + N = 180°, M = 180° – 50° – 60°, , D, , M = 70°,  Q = 70°, , B, , P, , 4., , 1, , 2, , 1, , E, 2, , C, , DE || BC,  ADE  ABC, T, , S, , , , ar( ABC )  AB , , , ar( ADE )  AD , , Q, R, Given : PT = 2 cm, TR = 4 cm. So, PR = 6 cm, , 3,   ,  1, , As it is given that ST || QR, , , , PST ~ PQR, PS PT ST, , , PQ PR QR, , [½], 2, , Also,, , , 2, , ar( PST )  PS ,  PT    ST , , ,  , , , ar( PQR )  PQ ,  PR ,  QR , 2, , ar( PST )  PT , 2, ,   , ar( PQR )  PR , 6, , Ratio : 1 : 9, , 2, , [By area similarity theorem], , ST || QR, , , , [By AA similarity] [½], , 9, 1, , [½], , E, , 8., , 2, , F, , D, , 2, , 2, , A, , C, B, , In ABE and CFB,, A = C (Opposite angles of a parallelogram), , [½], H, , 5., , [1], AEB = CBF, , A, , C, , (Alternate interior angles as AE || BC), , K, ,  ABE ~ CFB (By AA similarly criterion), [1], , B, Given AHK ~ ABC, , , , AH HK AK, , , AB BC AC, , 9., , D, C, , [½], , Also, we know AK = 10 cm, BC = 3.5 cm and, HK = 7 cm., , 6., , , , AK HK, , AC BC, , , , 10, 7, , AC 3.5, , A, In ABC, , B, , AB2 + AD2 = BD2, , ...(i), , In ABC, , AC  5 cm, , [½], , ar( ABC ) AB 2, , ar( PQR ) PQ 2, , [½], , ...(ii), , In ACD, , (Ratio of area of similar triangle is equal to, square of their proportional sides), 2, , ar( ABC )  1 , 1,   , ar( PQR )  3 , 9, , AC2 + BC2 = AB2, AC2 + CD2 = AD2, , ...(iii), , Subtracting (iii) from (ii), AB2 – AD2 = BC2 – CD2, , ...(iv), , [1], , Adding (i) and (iv), [½], , 2AB2 = BD2 + BC2 – CD2, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , In right PCB,, BP2 = PC2 + CB2, , ...(ii), [1], , On adding equations (i) and (ii), we get, AQ2 + BP2 = AC2 + CQ2 + PC2 + CB2, =, =, , AB2, , +, , CB2), , +, , PQ2, , +, , (CQ2, , +, , To Prove : (Hypotenuse)2 = (Base)2 + (Perpendicular)2, i.e.,, , [½], , B, , [½], [1], , 18. Let the each side of ABC be ‘a’ unit., , A, , a, BD , 3, , ADB = ABC, , Construction : Draw AM  BC :, , and,A = A, , a a a,  , 2 3 6, , [1],  In ABM, , ...(i), , In triangles BDC and ABC, we have, , AD2 = AM 2 + DM 2, , CDB = ABC, , ...(ii), , AM, AB, , and, C = C, , [Common], , So, by AA-similarity criterion, we have, BDC ~ ABC, , 3, a, 2, , , , Now, taking 9(AD)2, 9(AM 2 + DM 2), , [1], , 2, 2, , a 3  a , , 9 ,   ,  2   6  , , , ,  3a, a , 28a, , 9,   9, 4, 36, 36, , , , [Each equal to 90°], , [1], ,  AM = ABsin60°, , 2, , [1], , ...(i), , and in ADM, , 2, , AD AB, , AB AC, ,  AB2 = AD × AC, , AB2 = BM 2 + AM 2, , sin60 , , [Common], , [∵ In similar triangles corresponding sides, are proportional], , C, , D M, , [Each equal to 90°], , ADB ~ ABC, , , In ABM,, , C, , So, by AA-similarity criterion, we have, , A, , B, , D, , Proof : In triangle ADB and ABC, we have, , To prove : 9(AD)2 = 7(AB)2, , DM , , AC2 = AB2 + BC2, , Construction : From B draw BD  AC., , PC2), , [By Pythagoras theorem], , , , 27, , 19. Given : A right-angled triangle ABC in which, B = 90°., , [By Pythagoras theorem], , (AC2, , Mathematics, , DC BC, , BC AC, , [∵ In similar triangles corresponding sides, are proportional],  BC2 = AC × DC, , ...(ii), , Adding equation (i) and (ii), we get, 2, , AB2 + BC2 = AD × AC + AC × DC,  AB2 + BC2 = AC(AD + DC), , 7(AB)2 = 7a2, or, ,  AB2 + BC2 = AC × AC, ,  9(AD2) = 7(AB2), ,  AB2 + BC2 = AC2, , Hence proved., , [1], , [1], , Hence, AC2 = AB2 + BC2, , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , A, , B, , Mathematics, , P, , C, , Q, , 1, R, , , , 2, , BC = QR, 2, , ar( ABC ) AB, BC, AC, , , , ar( PQR ) PQ 2 QR 2 PR 2, , AC = PR, , [1], , Hence, corresponding sides are equal., , Given that ar(ABC) = ar(PQR), , , [1], ,  AB = PQ, , Let ABC be similar to PQR., 2, , AB 2 BC 2 AC 2, , , PQ 2 QR 2 PR 2, , 29, ,  ABC  PQR, , ar( ABC ), 1, ar( PQR ), , (SSS rule), , [1], , Hence proved., , Chapter - 7 : Coordinate Geometry, 1., , A(6, –5), ,  a , 4    6  2 , 5  3 , ,  , , 2 , 2   2, , B(–2, 11), , P(2, p), , a, 8, , 2, 2, , Given P is midpoint of AB, , , 6  2 5  11 , (2, p )  , ,, , 2 ,  2, , [½], , (2, p) = (2, 3), , 2., , a  8, 4., , p3, , [½], ,  ( 6  1)2  (7  5)2, ,  6  1, 6  2    x  4 , y  3 , ,  , , 2   2, 2 ,  2, ,  AB = 13,  2AB = 26, 5., , x4 7,  ,, 2, 2, x=7–4, , 8 y 3, , 2, 2, 8=y+3, , x=3, , y=8–3=5, , Hence coordinates of D = (3, 5), , [½], , Answer (B), It is given that the point P divides AB in the ratio, 2 : 1., , [½], , Using section formula, the coordinates of the, point P are, , On comparing, ,  1 1 2  4 , 1 3  2  6    1 8 , 3  12   (3, 5), ,  , , 2 1   3, 3 ,  2 1, [½], , Hence the coordinates of the point P are (3, 5)., , [½], , [½], , Answer (A), 6., , Given a line segment joining, A(–6, 5) and B(–2, 3), A(–6, 5), , [½], ,  25  144, , B(4, 3), , Let O be the mid-point of diagonals AC and BD, of the parallelogram ABCD and coordinates of D, is (x, y) then, , 3., , Answer (B), , Distance between the points = AB, , O, , A(1, 2), , [½], , Given 2 points are A(–6, 7) and B(–1, –5), , C(6, 6), , D, , [On comparing], , P, , [½], B(–2, 3), , a, Midpoint of A & B is P  , 4 , 2 , , Answer (A), Let the coordinates of the other end of the, diameter be (x, y)., We know that the centre is the mid-point of the, diameter. So, O(–2, 5) is the mid-point of the, diameter AB., , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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30, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , The coordinates of the point A and B are (2, 3), and (x, y) respectively., , 9., , Using mid-point formula, we have,, , 2 , , 2 x,  4  2  x  x  6, 2, , 3y, 5,  10  3  y  y  7, [½], 2, Hence, the coordinates of the other end of the, diameter are (–6, 7)., [½], 7., , Using distance formula, , (OP )  ( x  0)2  ( y  0)2, , [½], , (OP )  x 2  y 2, , [½], , 10. Let the centre be O and coordinates of point A, be (x, y), , Answer (C), , x 1, 2, 2, , [By Mid-point formula], ,  x=3, , [½], , y, , y 4,  –3, 2, , 5, 4, 3, ,  y = –10, , A(1, 3), ,  Coordinates of A = (3, –10), , 2, , x, , 11. Given points (k, 3), (6, –2), (–3, 4) are collinear, , 1, , B, –3 –2 –1, , –1, , [½], , 1, , 2, , 3, , C, , 4, , 5, ,  Area of the triangle formed by these, points = 0, [½], , x, , –2, , 1, k ( 2  4)  6(4  3)  3(3  2), 2, , [½], , From the figure, the coordinates of A, B, and C, are (1, 3), (–1, 0) and (4, 0) respectively., , –6k + 6 – 15 = 0, , [½], , Area of ABC, , k, , y, , , , 1, 1(0  0)  ( 1)(0  3)  4(3  0), 2, , , , 1, 0  3  12, 2, , 1, 15, 2, = 7.5 sq. units, , [½], , 3, 2, , [½], B(x, 5), , 12., , O(2, 3), , A(4, 3), , , , 8., , [½], , Answer (A), It is given that the three points A(x, 2), B(–3, –4), and C(7, –5) are collinear.,  Area of ABC = 0, , 1, x1( y 2  y 3 )  x2 ( y 3  y1)  x3 ( y1  y 2 )  0, , 2, [½], Here, x1 = x, y 1 = 2, x2 = –3, y2 = –4, and, x3 = 7, y3 = –5,  x[–4 – (–5)] – 3(–5 – 2) + 7[2 – (–4)] = 0,  x(–4 + 5) – 3(–5 – 2) + 7(2 + 4) = 0, , OB  (2  x )2  (3  5)2  (2  x )2  4, , [½], ,  2  (2  x )2  4, , [∵ OA = OB (radii)], , 4 = (2 – x)2 + 4, , , x2, , [½], , [½], , AB = 10, , (y +, ,  x = –63, , [½], , 13. Distance between the points A(3, –1) and, B(11, y) is 10 units, , 64 + (y + 1)2 = 100, ,  x + 21 + 42 = 0  x + 63 = 0, , Hence, the correct option is A., , [½], , (3  11)2  ( 1  y )2  10, ,  x – 3 × (–7) + 7 × 6 = 0, , Thus, the value of x is –63., , OA  (2  4)2  (3  3)2  2, , 1)2, , [½], [½], , = 36, , y + 1 = 6 or y + 1 = –6, , [½], , , , [½], , y  7, 5, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 14. It is given that the point A(0, 2) is equidistant, from the points B(3, p) and C(p, 5)., So, AB = AC  AB2 = AC2, , [½], , Using distance formula, we have :,  (0 – 3)2 + (2 – p)2 = (0 – p)2 + (2 – 5)2 [½],  9 + 4 + p2 – 4p = p2 + 9,  4 – 4p = 0, , Mathematics, , A, , P, , AC2, , = (7 –, Now,, 9 + 16 = 25, , 4)2, ,  By section formula,, ,  1( 7)  2(2) 1(4)  2( 2) , ,, Coordinates of P  , , 1 2, 1 2, , , , =, , =p2, , [½], , 3,   , 0   ( 1, 0),  3, , , ...(i) [Pythagoras], ,  , , , 14  2 8  2 , ,, , 3, 3 , , 12 6 ,  , ,   ( 4 , 2),  3 3, , – 8p + 16 + 16, , [1], , 18. Let A(3, 0), B(6, 4) and C(–1, 3) be the given, points of the vertices of triangle., , – 8p + 32, , BC2 = (7 – p)2 + (3 – 3)2 = 49 – 14p + p2 + 0, [1], , From (i), we have, , Now,, , AB  (6  3)2  (4  0)2  (3)2  (4)2,  9  16  25, , 25 = (p2 – 8p + 32) + (p2 – 14p + 49),  25 = 2p2 – 22p + 81, , ...(i), , [½], , BC  ( 1  6)2  (3  4)2  ( 7)2  ( 1)2, ,  2p2 – 22p + 56 = 0, ,  49  1  50, ,  p2 – 11p + 28 = 0, , ...(ii), , [½], , AC  ( 1  3)2  (3  0)2  ( 4)2  (3)2, ,  p2 – 7p – 4p + 28 = 0, ,  16  9  25, ,  p(p – 7) – 4(p – 7) = 0, , ...(iii), , [½], ,  BC2 = AB2 + AC2 and AB = AC, ,  (p – 7)(p – 4) = 0,  p = 7 and p = 4, , [1], ,  2( 7)  1(2) 2(4)  1( 2) , ,, Coordinates of Q  , , 2 1, 2 1, , , , + (3 – 7)2 = (3)2 + (–4)2 =, , =p2 – 14p + 49, , 7  4 4  4 , ,, , 3, 3 , ,  , , , AB2 = (p – 4)2 + (3 – 7)2 = p2 – 8p + 16 + (–4)2, p2, , (–7, 4), , [½], , 15. ABC is right angled at B.,  AC2 = AB2 + BC2, , B, , (2, –2), ,  4p = 4,  p=1, , Q, , 31, , [1], , 16. Given, the points A(x, y), B(–5, 7) and C(–4, 5), are collinear., So, the area formed by these vertices is 0., , , 1, x(7  5)  ( 5)(5  y )  ( 4)( y  7)  0 [½], 2, , , , 1, 2 x  25  5 y  4 y  28  0, 2, , [½], , Hence triangle is isosceles right triangle., , Thus, ABC is a right-angled isosceles triangle., 19. Let the coordinates of points P and Q be P(0, a), and Q(b, 0) respectively., [∵ P on y-axis Q on x-axis], 0b 0a,  , ,, , 2 ,  2, , b a,   , , 2 2, [½], , On comparing with (2, –5), ,  y = –2x – 3, , [½], , b, a,  2 and,  5, 2, 2, b = 4, a = –10, , 17. Since P and Q are the points of trisection of AB,, AP = PQ = QB, , [½], , Coordinates of mid-point of PQ, , 1, 2x  y  3  0, 2,  2x + y + 3 = 0, , , , [½], , Thus, P divides AB internally in the ratio 1 : 2, , Hence coordinates of P = (0, –10), , and Q divides AB internally in the ratio 2 : 1., , Hence coordinates of Q = (4, 0), , [½], , [½], [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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32, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , P, , 20. Given that, PA = PB, , A(1, –3), , By using distance formula, , , , ( x  5)2  ( y  1)2  ( x  1)2  ( y  5)2, , [½], , B(4, 5), , k:1, , 5k  3, 0, k 1, k, , 3, 5, , Squaring on both sides, , , ,  x2 + 25 – 10x + y2 – 2y + 1, , Hence, the required ratio is 3 : 5., , =, , x2, , + 2x + 1 +, , y2, , – 10y + 25, ,  –10x – 2y = 2x – 10y, , Now putting the value of k in (i) and (ii), we get, , [½], [½], , x, ,  8y = 12x,  3x = 2y, , A, , K, , (2, 3), , P, , 1, , A, , 6K  2 3K  3 , ,, Co-ordinates of point P  ,  [½], K 1 ,  K 1, , 3K  3, m, K 1, , ...(ii), ,  6K + 2 = 4K + 4, , [From (i)], ,  7a = 3a + PB,  PB = 7a – 3a = 4a, , [½], ,  K=1, Putting K = 1 in equation (ii), , 3(1)  3, m, 1 1, , x, , , , x, , 68, 12  8, and y , 7, 7, , , , x, , 2, 20, and y , 7, 7, , A, , 24., , Ratio is 1 : 1 and m = 0, [½], , 22. Let P(x, y) divides the line segment joining the, points A(1, –3) and B(4, 5) internally in the ratio, k : 1., Using section formula, we get, , x, , 4k  1, k 1, , …(i), , y, , 5k  3, k 1, , …(ii), , 2(3)  ( 2)(4), ( 4)(3)  (4)( 2), and y , 34, 34, [1], , , , 2 20 ,  The coordinate of P ( x, y )   ,,  [½], 7 ,  7, , [½], , i.e. P is the mid-point of AB, , [1], , Let the point P(x, y) divide the line segment, joining the points A(–2, –2) and B(2, –4) in the, ratio AP : PB = 3 : 4, [½], ,  2K = 2, ,  m=0, , B, ,  AB = AP + PB, , But the co-ordinates of point P are given as, (4, m), ...(i), , P, , AP 3, , AB 7, As, AB = 7a, AP = 3a, , (6, –3), , 6K  2, 4, K 1, , [1], , 23., , B, , (4, m), , 17, and y = 0, 8, , 17, So, coordinates of point P are  , 0 ,  8, , , [½], , 21. Suppose the point P(4, m) divides the line, segment joining the points A(2, 3) and B(6, –3), in the ratio K : 1., , [½], , P(3, 4), B, , R(5, 7), , Q(4, 6), , C, , Consider a ABC with A(x1, y1), B(x2, y2) and, C(x3, y3), P(3, 4), Q(4, 6) and R(5, 7) are the, mid-points of AB, BC and CA. Then,, [½], , 3, , x1  x 2,  x1  x 2  6, 2, , ...(i), , Since, P lies on x-axis. So its ordinate will be, zero., , 4, , y1  y 2,  y1  y 2  8, 2, , ...(ii), , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 4, , x2  x3,  x 2  x3  8 ...(iii), 2, , 5, , y2  y3,  y 2  y 3  12 ...(iv), 2, , Mathematics, , Join AC, Area of Quadrilateral ABCD = ar(ABC), + ar(ADC), Area of triangle ABC , , x  x1, 6 3,  x3  x1  10 ...(v), 2, , 7, , y 3  y1,  y 2  y 1  14 ...(vi), 2, , [½], , 2(x1 + x2 + x3) = 6 + 8 + 10 = 24, ...(vii), , 1 4( 5  2)  ( 3), 2 ( 2  2)  3( 2  5), , , , 1, 4( 3)  3(0)  3(3), 2, , , , 1, 12  0  9, 2, , , , 21, square units, 2, , [½], , From (i) and (vii), we get x3 = 12 – 6 = 6, From (iii) and (vii) we get x1 = 12 – 8 = 4, From (v) and (vii), we get x2 = 12 – 10 = 2 [½], Now, adding (ii), (iv) and (vi), we get, 20(y1 + y2 + y3) = 8 + 12 + 14 = 34,  y1 + y2 + y3 = 17, , ...(viii), , Area of triangle ADC , , [½], , From (ii) and (viii), we get y3 = 17 – 8 = 9, From (iv) and (viii), we get y1 = 17 – 12 = 5, From (vi) and (viii), we get y2 = 17 – 14 = 3 [½], Hence, the vertices of ABC are A(4, 5), B(2, 3),, C(6, 9)., [½], m, , 25., A(–2, 2), , n, P(2, y), , , , 3m  2n 2n  7m , (2, y )  , ,, , mn ,  mn, , 2, , [1], , 3m  2n, mn, , m:n=4:1, y, , 2  7 4, 5, , y, , 30, 5, , A(–4, –2), , 1 4(3  2) , 2 3( 2  2)  3( 2  3), , , , 1, 4(5)  3(0)  3( 5), 2, , , , 1, 20  0  15, 2, , , , 1, 35, 35 , sq. units [1], 2, 2, , [1], , 21 35, , 2, 2, , = 28 sq. units, [½], 1, , 27., A(2, 1), , 2, P, , B(5, –8), , Given :, , y=6, 26., , 1 4  3  (2) , 2 2 2  (2)  3  2  3, ,  Area of quadrilateral (ABCD) , , 2m + 2n = 3m – 2n, , [1], , , , B(3, 7), , Lets say ratio = m : n, , [½], , 1 4  5  (2)  (3), 2  2  (2)  3  2  (5), , , , On adding (i), (iii) and (v) we get,  x1 + x2 + x3 = 12, , 33, , [1], B(–3, –5), , AP 1, , AB 3, , , , AP, 1, , AP  PB 3, , PB = 2AP, D(2, 3), , C(3, –2), ,  AP : PB = 1 : 2, , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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34, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , , , P  , , , C(4, –1), , 30., , By section formula, 22  5 2  8 , ,, , 3, 3 , , D(3, 4), , P = (3, –2), , B(–2, –4), , [1], , Also it is given that P lies on 2x – y + k = 0, A(–3, –1), ,  2(3) – (–2) + k = 0, , Area of quadrilateral ABCD = ar(ABC) + ar(ADC), , k  8, , [1], , We know that,, , 28. Since R(x, y) is a point on the line segment, joining the points, P(a, b) and Q(b, a), , Area of triangle , ,  P(a, b), Q(b, a) and R(x, y) are the collinear., , 1 x2 ( y 2  y 3 )  x2 ( y 3  y1 ), 2  x3 ( y1  y 2 ), [½], , [½],  Area of PQR = 0, , Thus,, , [½], , Area of ABC , , 1, x1( y 2  y 3 )  x2 ( y 3  y1)  x3 ( y1  y 2 )  0, 2, [1], , , 1, a(a  y )  b( y  b )  x(b  a )  0, 2, , , , 1, 9  0  12, 2, , , , 21, sq. units, 2, ,  a2 – ay + by – b2 + x(b – a) = 0,  y(b – a) + x(b – a) = b2 – a2, , Area of ADC , ,  (x + y)(b – a) = (b – a)(b + a),  x+y=a+b, m, , 29., A(–5, 8), , [1], n, B(4, –10), , P(x , 4), , Lets say ratio = m : n, 4m  5n 10m  8n , P ( x, 4)  , ,, , mn ,  mn, , 4, , 10m  8m, mn, , [1], , 1 ( 3)( 4  1)  ( 2), 2 ( 1  1)  4( 1  4), , 1 ( 3)(4  1)  3( 1  1), 2 4( 1  4), , , , 1, 15  0  20, 2, , , , 1, 35, 2, , , , 35, sq. units, 2, , Area of quadrilateral ABCD , , [On equating], , [½], , Hence, area of quadrilateral is 28 square units., C(2, 5), , 31., , m 2, , n 7, , [1], , 4m  5n, mn, , , , 2, m, 4    5 4    5, 7, n, , , x, , 2, m, 1, 1, 7, n, , , , x, , R, , A(2, 1), , Q, , Q, , B(4, 3), , P, Q, R are the mid-points to the sides of the, ABC, , 8  35, 9, ,  x = –3, , 21 35 56, , , 2, 2, 2, , = 28 sq. units, ,  14m = 4n, , We know x , , [1], , Substitute these values in equation (i), we have,, ,  4m + 4n = –10m + 8n, , , , [1], , 4  2 3  1, P  , ,,   (3, 2), 2 ,  2, , [1], , Similarly, Q = (3, 4), R = (2, 3), , [1½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Area of PQR , , , , 1 3(4  3)  3(3  2), 2 2(2  4), , 1, 334, 2, , = 1 sq. unit, 32., , P, , K, A(3, –5), , [½], , [1], , I, , (x, y), , 4K  3, K 1, , y, , 8K  5, K 1, , 4K  3 8K  5, , 0, K 1, K 1, [On putting the values of x and y], ,  K, , [½], , [½], , 1, 2, , [1], , 12k  (6), y, , k 1, ,  24  30 , 2, 12   6 , , 5  6, 5, , , 2, 7, 25, 1, , , 5,  5 , [1], , Thus, the coordinates of the point of division, , 6, are  0,  , [1], 7, , 35. The given points are A(–2, 3) B(8, 3) and C(6, 7)., Using distance formula, we have :, AB2 = (8 + 2)2 + (3 – 3)2, , Hence the value of K , , 1, 2, , [1], ,  AB2 = 102 + 0,  AB2 = 100, , A(1, –3), , 33., , 4 2, , 10 5, , Thus, the y-axis divides the line segment joining, the given points in the ratio 2 : 5, , , ,  4K – 2 = 0, , 2,  K, 4, , 10k  4,  0  10k  4  0, k 1, , [1], , Since P lies on x + y = 0, , ,  10k  4 , 12k  6   0, y, , , k 1 ,  k 1, ,  k, , By using the section formula co-ordinates of, P are., , 35, , 34. Let the y-axis divide the line segment joining the, points (–4, –6) and (10, 12) in the ratio k : 1, and the point of the intersection be (0, y). Using, section formula, we have:, , , , B(–4, 8), , Let the co-ordinates of point P be (x, y), , x, , Mathematics, , BC2, , = (6 –, , 8)2, , [½], + (7 –, , 3)2, ,  BC2 = (–2)2 + 42,  BC2 = 4 + 16, B(4, p), ,  BC2 = 20, , C(–9, 7), , The area of a , whose vertices are (x1, y1),, (x2, y2) and (x3, y3) is, , , , 1, x1( y 2  y 3 )  x 2 ( y 3  y1 )  x3 ( y1  y 2 ) [1], 2, , Substituting the given coordinates, , Area of  , , , , [½], ,  10p + 60 = ±30,  10p = –30 or 10p = –90, ,  CA2 = (–8)2 + (–4)2,  CA2 = 64 + 16,  CA2 = 80, , [½], , BC2 + CA2 = 20 + 80 = 100 = AB2, , [1], , So, by the converse of Pythagoras Theorem,, ABC is a right triangle right angled at C., , [½], , 36. The given points are A(0, 2), B(3, p) and C(p, 5)., It is given that A is equidistant from B and C., , [½], ,  p = –3 or p = –9, Hence the value of p = –3 or –9, , CA2 = (2 – 6)2 + (3 – 7)2, , It can be observed that :, , 1, 1( p  7)  4(7  3)  ( 9)( 3  p ), 2, [½], , 1, ( p  7)  40  27  9 p  15, 2, , [½], ,  AB = AC,  AB2 = AC2, , [½], ,  (3 – 0)2 + (p – 2)2 = (p – 0)2 + (5 – 2)2 [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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36, , Mathematics, ,  9+, , p2, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , + 4 – 4p =, , p2, , +9, ,  4 – 4p = 0,  4p = 4,  p=1, , , , AB  AP 7  3, , AP, 3, , , , BP 4, , AP 3, , , , AP 3, , BP 4, , [1], , Thus, the value of p is 1, , [1], ,  P divides AB in the ratio 3 : 4, , Length of AB  (3  0)2  (1  2)2  32  ( 1)2, [1], , x, , 3  2  4( 2), 3  ( 4)  4( 2), ; y, 34, 34, , 37. The given points are A(–2, 1), B(a, b) and, C(4, –1)., , x, , 68, 12  8, ; y, 7, 7, , Since the given points are collinear, the area of, the triangle ABC is 0., [½], , x, , 2, 20, ; y, 7, 7, ,  9  1  10 units., , , 1, x1( y 2  y 3 )  x 2 ( y 3  y1 )  x3 ( y1  y 2 )  0, 2, , 2 20 ,  The coordinates of P are  ,, , 7 ,  7, , Here, x1 = 2, y1 = 1, x2 = a, y2 = b, x3 = 4,, y3 = –1, , , 1, 2(b  1)  a( 1  1)  4(1  b )  0, 2, ,  AP = BP, , [½], ,  x  (a  b)2   y  (b  a)2, , , , , ,  2a + 6b = 2, ...(i), ,  x  (a  b)2   y  (a  b)2, , [1], ,  [x – (a + b)]2 + [y – (b – a)]2, , [1], , = [x – (a – b)]2 + [y – (a + b)]2, , Given :, ,  x2 – 2x(a + b) + (a + b)2, , a–b=1, , ...(ii), , + y2 – 2y(b – a) + (b – a)2, , Subtracting equation (i) from (ii) we get :, , = x2 – 2x(a – b) + (a – b)2, , 4b = 0, , + y2 – 2y(a + b) + (a + b)2, ,  b=0, , [1], ,  –2x(a + b) – 2y(b – a), , Subtracting b = 0 in (ii), we get :, , = –2x(a – b) – 2y(a + b), , a=1, ,  ax + bx + by – ay = ax – bx + ay + by, , Thus, the values of a and b are 1 and 0,, respectively., [1], ,  2bx = 2ay, , 38. Here, P(x, y) divides line segment AB, such that, , AP , , [1], , 39. P(x, y) is equidistant from the points A(a + b,, b – a) and B(a – b, a + b)., ,  –2b – 2 – 2a + 4 – 4b = 0, ,  a + 3b = 1, , [½], ,  bx = ay, m, , 40., P(2, –2), , 3, AB, 7, , ....(proved), , [1], , n, , R  24 , y ,  11 , , Q(3, 7), , Lets say ratio is m + n, , , , AP 3, , AB 7, , Then, , , , AB 7, , AP 3, ,  24 , y    3m  2n , 7m  2n , ,  , , mn ,  11   m  n, , , , AB, 7, 1 1, AP, 3, , [½], , [1], , 24 3m  2n, 7m  2n, , , y, 11, mn, mn, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Mathematics, ,  24(m + n) = 11(3m + 2n), , Ar(ABC ) , , 24m + 24n = 33m + 22n, 2n = 9n, , , m 2, ,  Ratio  2 : 9, n 9, , [1], , y, , 72  29, 11, , y, , 4, 11, , 1, 5  52  12, 2, , , , 1, 35, 2, , , , 35, Sq. units, 2, , Ar( ADC ) , , [1], , , , 41. M is mid-point of diagonals AC and BD, Using mid-point formula,, (–2, 1), A, D, (1, 2), , , , (a, 0), B, , M, C, , (4, b), , 1 b 2,   1 b  2  b  1, 2, 2, , [½], , Ar( ADC ) , , , , Ar(ABCD) , , [1], , 35 109 144, , ,  72 sq. units, 2, 2, 3, , So, AP2 = BP2, , [½], , i.e., (5 – 0)2 + (–2 – y)2 = (–3 – 0)2 + (2 – y)2 [1], 2, ,  9  1  10, ,  25 + y2 + 4 + 4y = 9 + 4 + y2 – 4y,  8y = – 16,  y=–2, Hence, the required point is (0, –2), , [1], , 1 x1( y 2  y 3 )  x 2 ( y 3  y1 ), 2  x3 ( y1  y 2 ), , If A = (x1, y1), B = (x2, y2), C = (x3, y3) are, vertices of ABC., A(–5, 7), , 109, sq. units, 2, , , , We are given that AP = BP, , Side DC  AB  (1  4)2  (2  1)2, , Ar( ABC ) , , 2, , [½], ,  9  1  10, , 42., , 109, , 43. Let the point on y-axis be P(0, y) which is, equidistant from the points A(5, –2) and B(–3, 2)., [½], , Side AD  BC  ( 2  1)  (1  2), , 1, 55  52  2, 2, , [½], , 2 a 1, ,  a 1 2  a  1, 2, 2, , 2, , 1, 5( 5  6)  4( 6  7)  1(7  5), 2, , [1], ,  2 , 1  b    a  1, 2 , ,  , , 2   2, 2, 2, , and, , [1], ,  Area cannot be negative., ,  2  4 , 1  b    a  1, 2  0 , ,  , , 2   2, 2 ,  2, , , , 1 5( 5  6)  4( 6  7), 2 1(7  5), , , , m = 2, n = 9, , 37, , B(–4, –5), , 1, , 44., A(2, 1), , :, P, , :, , 1, , [1], , 1, , Q, , B(5, –8), , Here, AP : PB = 1 : 2, , [½], , , , 1 5  2  2 1 8  2  1 , Coordinates of P  , ,, , 1 2, 1 2, , ,  Coordinates of P = (3, –2), Since, P lies on the line 2x – y + k = 0, , [1], [½], ,  2(3) – (–2) + k = 0, D(4, 5), , C(–1, –6), , Ar( ABCD) = Ar(ABC) + Ar(ADC) ....(i) [½], ,  6+2+k=0,  k = –8, , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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38, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 45. The given vertices are A(x, y), B(1, 2) and, C(2, 1)., It is know that the area of a triangle whose, vertices are (x1, y1), (x2,y2) and (x3, y3) is given, by,  Area of ABC , , 1 x (2  1)  1 (1  y ), 2 2( y  2), , [½], , , , 1, x  1  y  2y  4, 2, , [½], , , , 1, x  y 3, 2, , [½], , x, , 72, 8, , ratio, , 3, : 1, i.e. 3 : 5., 5, , [½], , It is given that the area of the triangle is, 24 sq. unit, Area of the triangle having vertices ( x1, y 1 ),, ( x 2 , y 2 ) and ( x3 , y 3 ) is given by, , , , [1], , 1, x1( y 2  y 3 )  x 2 ( y 3  y1 )  x3 ( y1  y 2 ) [1], 2, ,  24 , , 1 1(2k  ( 5))  ( 4)(( 5), 2 ( 1))  ( k )(( 1)  2k ), ,  2k2 + 3k – 27 = 0, , Sol. Let the Point P(x, 2) divide the line segment, joining the points A(12, 5) and B(4, –3) in the, ratio k : 1, ,  k, ,  (2k + 9)(k – 3) = 0, , The values of k are , , Now, the coordinates of P are (x, 2), , , 4k  12, 3k  5,  x and, 2, k 1, k 1, , [1], , 3k  5, 2, k 1, , 48., ,  k, , 3, 5, , Substituting k , 3, 4   12, 5, x, 3, 1, 5, , [1], , 3, 4k  12, in,  x, we get, 5, k 1, , 9, and 3., 2, , [1], , AD AE 1, , , AB AC 3, , , AB AC, , 3, AD AE, , , , AD  DB AE  EC, , 3, AD, AE, , , , 1, , , , DB EC, , 2, AD AE, , , , AD AE 1, , , DB EC 2, ,  –3k + 5 = 2k + 2,  5k = 3, , [1], , 9, or k  3, 2, , Then, the coordinates of P are, [½], , [1], ,  48 = |(2k + 5) + 16 + (k + 2k2)|, , [1], , 46. Find the ratio in which the point P(x, 2) divides, the line segment joining the points A(12, 5) and, B(4, –3). Also find the value of x. [2014] ...[4], ,  4k  12 , 3k  5 , , , k 1 ,  k 1, , [½], , 47. Take ( x1, y1 )  (1,  1), ( 4, 2k ) and ( k ,  5), ,  x + y – 3 = 12,  x + y = 15, Proved, , , , Also, the point P divides the line segment, joining the points A(12, 5) and (4, –3) in the, , Since the area of ABC is given as 6 sq. units., , 1, x  y 3  6, 2, , 12  60, 35, , Thus, the value of x is 9, , (x + y – 3) will be positive, , , , x, ,  x=9, , 1 x2(y2  y3 )  x2(y3  y1), [½], 2 x3(y1  y2 ), , , , , , DB, EC,  1, 3, AD, AE, ,  AD : DB = AE : EC = 1 : 2, , [½], , So, D and E divide AB and AC respectively in, the ratio 1 : 2., [½], , By using section formula, The coordinates of D is, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Mathematics, ,  1  8 , 5  12    3, 17  and, ,  , , 3 ,  1 2 1 2  , Coordinates of E is,  7  8 , 2  12    5, 14 , ,  , , 3 ,  1 2 1 2  , 4, , 3, , 5, , 4, , 6, , 17, 3, , 14, 3, , 6, , , , 1, (64)  (49), 2, , , , 1, (15), 2, , , , 15, sq. units, 2, , [1], , 39, , ...(ii), , [1], , From (i) and (ii), , ,  4  17  3  14  5  6 , , , 3, 3, , 1, Area of ADE , 2 , 17, 14 , 3  6  5 ,  4 , 3, 3 , , , If three points are collinear then the area of the, triangle will be zero. For any 3 points (x1, y1),, (x2, y2), (x3, y3) Area will be, ,  68  42  90 , , , 3, , 1, , 2, 54  85  56 ,  , , 3, , , , Area , , , , 0, , 1 x1( y 2  y 3 )  x 2 ( y 3  y 1 ), 0, 2  x3 ( y1  y 2 ), , [½], , 1 (k  1)(2k  3  5k )  3k (5k  2k ), [½],  (5k  1)(2k  2k  3), 2, ,  |–3k2 + 3 + 9k2 + 3 – 15k| = 0,  |6k2 – 15k + 6| = 0, , 1 5,  , 2 3, , [1], ,  6k2 – 15k + 6 = 0, , 5, sq. units ...(i), 6, , 4, , 1, , 7, , 4, , 6, , 5, , 2, , 6, , , , , , 0 = |(k + 1)(3 – 3k) + 3k(3k) – 15k + 3|, , 1 200   195 ,  , , , 2 3   3 , , Area of ABC , , [½], , 49. Given A(k + 1, 2k), B(3k, 2k + 3), C(5k – 1, 5k), are collinear., ,  68  14  30 , , , , 1 3, , 2, 85, 56 ,   18 , , , 3, 3 , , , , , 5, Ar(ADE ), 5 2, 1,  6  , , Ar(ABC ) 15 6 15 9, 2, ,  2k2 – 5k + 2 = 0, , [1], , [½], ,  2k2 – 4k – k + 2 = 0,  2k(k – 2) – 1(k – 2) = 0,  (k – 2)(2k – 1) = 0, , [½], , 1, 2, , [½], , 1 (4  5  1 2  7  6), 2  (1 6  7  5  4  2), ,  k  2,, , 1, (20  2  42)  (6  35  8), 2, , Hence the value of k are 2 and, , 1, 2, , [½], , Chapter - 8 : Introduction to Trigonometry, 1., , tan A , , 5, 12, , 2., , sin A cos A, (sin A  cos A)sec A , , cos A cos A, = tanA + 1, , 5, 1, 12, 17, , 12, , sec2(1 + sin)(1 – sin) = k,  sec2(1 – sin2) = k, , [½], ,  sec2 · cos2 = k, , , , , [½], , [½], , cos2 , k, cos2 , ,  k = 1, , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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42, , Mathematics, , , , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , , , 1,  tan2 A  cot A, 1  tan A, , , , Now,, , , , 1,   tan2 A  1 , , , 1  tan A , tan A , , , , 1  tan3 A, tan A(1  tan A), , , , (1  tan A)(1  tan2 A  tan A), tan A(1  tan A), , [1], ,  4sin   cos   1 ,  4sin   cos   1   cos  cos  cos  , ,  , ,  4sin   cos   1  4sin   cos   1  [½],  cos  cos  cos  , , = cotA + tanA + 1 = R.H.S., , [1], , [½], , 5, 4, , 5, 4, 4, , [½], , 2, , Hence proved., 17. L.H.S. = (cosecA – sinA)(secA – cosA), , 1, 1,  ,  sin A  ,  cos A ,  sin A,   cos A, , , (8  5), 4, , (16  5), 4, , 1  sin A 1  cos A , 2, , , , 5, 4, , 5, 3  1, 4, 3  1, , [∵ a3 – b3 = (a – b)(a2 + b2 + ab)], , , , 4 tan   1  sec , 4 tan   1  sec , , , , 2, , sin A cos A, , cos2 A sin2 A, sin A cos A, , , , = sinA·cosA, , ...(i), , [1], , 13, 11, , [½], , 19. Given that,, tan 2A = cot(A – 18°), , 1, R.H.S. , tan A  cot A, , , , , ,  cot(90° – 2A) = cot(A – 18°), [∵ tan = cot(90° – )], , 1, ,  90° – 2A = A – 18°, , sin A cos A, , cos A sin A, , , , [∵, , +, , cos2A, , ...(ii), , L.H.S. = R.H.S.; Hence Proved, , = 1], , 20. L.H.S : (sin + cosec)2 + (cos + sec)2, = sin2 + cosec2 + 2 + cos2 + sec2 + 2, , [1], , 1, 1 , ∵ sin  , and cos  , , cosec, sec  , ,  tan2  , , 9, 16, , [1], , [1], = 1 + 1 + 1 + 4 + tan2+ cot2, , tan2, , sec 2   1 , ,  sec   5, 4, , [∵ cos2 + sin2 = 1], , [½], , We know that,, , , , [1], , = (sin2 + cos2) + (1 + cot2) + (1 + tan2) + 4, , 3, 4, , =1+, , 108, 3, , [1], , 18. Given that,, , sec2, , A, ,  A = 36°, sin2A, , From (i) and (ii), , tan  , , [1], ,  3A = 108°, , 1, sin2 A  cos 2 A, sin A  cos A, , sin A  cos A, , 1, = sinA·cosA, , [1], , [∵ cosec2 + 1 + cot2 and sec2 = 1 + tan2], , 9, 25, , 16 16, , [½], = 7 + tan2 + cot2 = R.H.S., [½], , Hence Proved, , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 21., , cos A, 1 , sin A, 1 , L.H.S :  1 , , ,  1 , , sin A sin A   cos A cos A , ,  , , , , sin A  cos A  1  cos A  sin A  1 , , , sin A, cos A, , , 2, , (sin A  cos A)  (1), sin A · cos A, , Mathematics, , 23. LHS , , sin A  cos A  1, sin A  cos A  1, , , , tan A  1  sec A, tan A  1  sec A, , [½], , 2, , [½], , sin2 A  cos2 A  2sin A · cos A  1, , sin A · cos A, , 1  2sin A · cos A  1, , sin A · cos A, , [½], , [½], = 2 = R.H.S., , 22., , L.H.S. , , , [½], , (Dividing numerator & denominator by cos A) [½], , , ,  tan A  sec A   1,  tan A  sec A   1, , [½], , , ,  tan A  sec A   1  tan A  sec A ,  tan A  sec A   1  tan A  sec A , , [½], , [∵ sin2A + cos2A = 1], , Hence Proved, , 43, , [1], , sin A  2sin3 A, , , , 3, , 2cos A  cos A, , sin A(1  2sin2 A), , [1], , cos A(2cos2 A  1), , 2, 2, 2, sin A  sin A  cos A  2sin A , , [1], , , 2, 2, cos A  2cos A  sin A  cos2 A , [∵ sin2A + cos2A = 1], ,  cos2 A  sin2 A ,  tan A , , 2, 2 ,  cos A  sin A , , Hence proved., , [1], , 2, , , , A  sec 2 A   tan A  sec A , , tan A  sec A  1  tan A  sec A , , [½], , , , 1  tan A  sec A,  tan A  sec A  1  tan A  sec A , , [½], , , , 1(tan A  sec A  1), (tan A  sec A  1)(tan A  sec A), , [½], , , , 1,  R.H.S., secA  tan A, , [½], , [1], , = tanA = R.H.S., ,  tan, , Hence Proved., , Chapter - 9 : Some Applications of Trigonometry, 1., , Answer (C), , 2., , C, , 45°, , A, B, Given AB = 25 m, And angle of elevation of the top of the tower, (BC) from A = 45°, ∵ BAC = 45°, , BC, In ABC, tan 45 , AB, , Answer (B), Let AB be the tower and BC be its shadow. Let,  be the angle of elevation of the sun., According to the given information,, BC  3 AB …(1), , In ABC,, , C, , AB, AB, 1, tan  , , , BC, 3 AB, 3, We know that tan 30 , , , , A, , B, , [Using (1)], , 1, 3, ,  BC = 25 m, ,   = 30°, ,  Height of the tower = 25 m, , Hence, the angle of elevation of the sun is 30°., , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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44, 3., , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Answer (C), , D, , Let AB be the tower and BC be its shadow., , B, , AB  20, BC  20 3, In ABC,, , 75 m, 30°, A, C, Let AB be the tower of height 75 m and C be the, position of the car, , tan  , , tan  , , In ABC,, , AC, AB,  AC = ABcot30°, , tan  , , cot 30 , , , , AC  75 m  3, , , , AC  75 3 m, , 20, , [½], , 20 3, 1, 3, , But, tan  , , 1, 3, ,   = 30°, , Thus, the distance of the car from the base of, the tower is 75 3 m ., 4., , AB, BC, , The Sun is at an altitude of 30°., , A, , 6., , Answer (D), , Ladder, , A, , [½], , Wall, , C, B 60°, 2.5 m, , M, , Let AB be the ladder and CA be the wall., , B, , 60°, 2m, , The ladder makes an angle of 60° with the, horizontal., , N, ,  ABC is a 30° – 60° – 90°, right triangle. [½], , In the figure, MN is the length of the ladder,, which is placed against the wall AB and makes, an angle of 60° with the ground., , Given: BC = 2.5 m, ABC = 60°, , The foot of the ladder is at N, which is 2 m away, from the wall., , Hence, length of the ladder is AB = 5 m., , In right-angled triangle MNB:, , G, , S, , Height of tower TG = 30 m, , Therefore, the length of the ladder is 4 m., Hence, the correct option is D, A, , Length of shadow GS  10 3 m, , tan  , , 30, 10 3, , tan   3, , , B, , [½], , TGS is a right angled triangle, , , C, , , 10 3 m, , Angle of elevation of sun = GST = , , , , 5., , 30 m, , BN, 2, , MN MN, , 1, 2, , 2 MN,  MN = 4 m, , [½], , T, , 7., ,  BN = 2 m, , cos 60 , ,  AB = 5 m, , [½], ,  = 60°, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , A, , 8., , Mathematics, , 10., , 45, , C, , 30°, , 24 - h, , 60°, , D 45°, , B, C, Given CD = 100 m, AB = ?, In ABC, tan 60 , , BC , , [1], , 3, [∵ tan45° = 1], , AL = BD = 15 m and AB = LD = h, So, CL = CD – LD = 24 – h, ,  100, , [1], , 3 1, , , , AB = 237 m, , [1], , Given: Position of kite is B., , 3, ,  24  h , , [1], , O, , 45°, , OB, AB, , A, , 45, AB, , [1], , 3 45, , AB, 2, AB , , 3, , 5 3, , 3  1.732 ], , [1], , O, , 60°, , 200 m, ,  sin 60 , , , , 15, , [1], , 11. Let d be the distance between the two ships., Suppose the distance of one of the ships from, the light house is x meters, then the distance, of the other ship from the light house is, (d – x) meter., , In right angled triangle AOB,, , , , 24  h, 15, , Thus, height of the first pole is 15.34 m., , Kite, B, , sin A , , , ,  h = 15.34, , Required length of string = AB, , 45 m, , 1, ,  h = 24 – 5 × 1.732 [Taking, , Angle of inclination = 60°, , 60°, , 24  h, 15, ,  h  24  5 3, , Height of kite above ground = 45 m, , A, , CL, AL, ,  tan30 , , 100 3, , [1], , In ACL,, , tan30 , , AB = 236.98, , 9., , D, 15 m, Let AB and CD be the two poles, where CD, (the second pole) = 24 m., Let the height of pole AB be h m., ,  3  1, AB ,   100, , 3 , , AB , , h, , BD = 15 m, , BD – BC = CD, , 3, , 24 m, , B, , AB, , AB, , L, , h, , AB, BC, , BD = AB, , AB , , 30°, , A, , 45  2, 3, , 45°, x, , d, In right-angled ADO, we have., , tan 45 , , , , 90, 3, ,  30 3 m, , Hence, the length of the string is 30 3 m . [1], , 60°, B, D d–x, , OD 200, , AD, x, , 200, x,  x = 200, ,  1, , …(i), , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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46, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , In right-angled BDO, we have, tan 60 , 3, , , , OD, 200, , BD d  x, 200, dx, ,  dx , , 200, , 3, Putting x = 200. We have:, , [1], , Speed , , distance, time, , Speed , , 3000, 15, , Speed 200m/s, Converting it to km/hr  200 , , 200, , d  200 , d, , We know that, the aeroplane moves from point, B to D in 15 seconds and the distance covered, is 3000 metres., , 3, , D, , 13., , 200, ,  200, 3,  d = 200 × 1.58,  d = 316 m, , h, , (approx.), , [1], , Thus, the distance between two ships is, approximately 316 m., 12. Let BC be the height at watch the aeroplane is, observed from point A., Then, BC  1500 3, In 15 seconds, the aeroplane moves from point, B to D., B and D are the points where the angles of, elevation 60° and 30° are formed respectively. [1], Let AC = x metres and CE = y metres, , °, 60, 30°, , A, , x, , In CBA,, tan 60 , , y, , [1], , DE, EA, , h, x, , h  3x, , BC, AC, , …(i), , In ABC,, tan30 , , [1], …(i), , 1, 3, , , , AB, BC, , 10, x, , x  10 3, , DE, tan30 , AE, 3, , The angle of depression of the base C of the hill, CD observed from A is 30° and the angle of, elevation of the top D of the hill CD observed, from A is 60°., , 3, , E, , In ADE,, , , , 10 m, 30°, B, C, x, Let CD be the hill and suppose the man is, standing on the deck of a ship at point A., , tan60 , C, , E, , In AED,, , D, , 1500 3, 3, x,  x = 1500 m, , 1, , 60°, x, , A, ,  EAD = 60° and BCA = 30°, , AE = x + y, B, , 18,  720 km/hr [1], 5, , …(ii), , [1], , Substituting x  10 3 in equation (i), we get, , h  3  10 3  10  3  30, , 1500 3, xy, , DE = 30 m, ,  x + y = 1500 × (3) = 4500, , CD = CE + ED = 10 + 30 = 40 m, ,  1500 + y = 4500, , Thus, the distance of the hill from the ship is, 10 3 m and the height of the hill is 40 m. [1], ,  y = 3000 m, , …(ii), , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , T, , 14., , Mathematics, , , , 1 80  x, , x, 3, ,  x = 240 – 3x, , , ,  4x = 240, , D C, F, Given CF = 4 m, , From (i),, , TCF + TDF = 90°, , 1 h, , 3 x, , Let say TCF = , , [1], , TDF = 90° – , ,  h, , In a right angled triangle TCF, TF TF, , CF, 4, , TF = 4tan, , ...(i), , C, , TF, 16, , [1], , TF = 16cot, , A, , ...(ii), , Multiply (i) and (ii), we get, (TF)2 = 64  TF = 8 m,  Height of tower = 8 m, , C, , [1], , D, , 60°, , P, Given AB = 80 m, , 60,  3, BD, ,  BD , , 60, , [2], , 3, ,  BD  20 3, AC, AP, , 1, , h, , 3 x, , [1], …(i), , In right ACE,, , CE,  tan30, AE, , , In BPD,, , BD, PB, , h, 80  x, , h, x, h, 80  x, , CE, 1, , AE, 3, ,  CE , …(ii), , Dividing (ii) by (ii), we get, , 1, 3 , 3, , 60°, B, D, In right ABD,, , , , B, , In APC,, , 3, , E, , AB,  tan60, BD, , Let AP = x m, therefore, PB = (80 – x) m, , tan60 , , 30°, , 60°, , 60 m, , 15. Let AC and BD be the two poles of the same, height h m., , tan30 , ,  20 3 m, , 3, , 16. Let AB be the building and CD be the tower., , tan(90  ) , , 30°, , 60, , Thus, the height of both the poles is 20 3 m, and the distances of the point from the poles, are 60 m and 20 m., [1], , In TDF, , A, , [1], ,  x = 60 m, , DF = 16 m, , tan  , , 47, , [1], , 20 3, 3, , ( AE = BD),  20, , Height of the tower = CE + ED = CE + AB =, 20 m + 60 m = 80 m, Difference between the heights of the tower and, the building = 80 m – 60 m = 20 m, Distance between the tower and the building, [2],  BD  20 3 m, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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48, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , C, , 17., , h, , 30°, 60°, , A, 20 m, , Q, , 18., , M, 20 m, B, , P, , 45°, , h M, , 40 m, X, , 60°, , P, MP = YX = 40 m, , h + 20, , Y, ,  QM = h – 40, C, Let PB be the surface of the lake and A be the, point of observation such that, , In right angled QMY,, , AP = 20 metres. Let C be the position of the, cloud and C be its reflection in the lake., ,  PX = h – 40, , tan 45 , , tan60 , , Then mCAM = 30° and mCAM = 60°, , PX , , Let CM = h. Then, CB = h + 20 and CB = h + 20., , , , CM, AM, , 1, , h, , 3 AM, …(i), , [1], , , , CB  BM, 3, AM, , , , 3, , 19., , h  20  20, AM, , h  20  20, 3, , 3, , 3h  40 3  h, , , , 3h  h  40 3, , X, , P 30°, …(ii), , [1], , h  20  20, , [1], , A, , B, , Y, , 45°, , 45° Q, , Given aeroplane is at height of 300 m,  AB = 300 m and XY || PQ, Angles of depression of the two points P and Q, are 30° and 45° respectively., [1], , 3, ,  3h = h + 40, , XAP = 30° and YAQ = 45°, ,  2h = 40, , XAP = APB = 30°, [Alternate interior angles], ,  h = 20 m, In CMA, sin30 , , [1], , h, , , , 30°, , From equation (i) and (ii), we get, , 3h , , ...(ii), , Thus, PQ is 94.79 m and PX = 94.79 ÷ 1.73, = 54.79 m, [1], , CM, tan60 , AM, , AM , , h, ,  1.73h – h = 40(1.73)  h = 94.79 m, , In AMC we have,, , , , QP, QP,  3, PX, PX, , h – 40 =, , AM  3h, , [1], , ...(i), , 3, From (i) and (ii), we get, , In CMA we have,, , , , …(MY = PX), , In right angled QPX,, , Then CB = CB. Let AM be perpendicular from, A on CB., [1], , tan30 , , QM, h  40,  1, MY, PX, , h, CA, , YAQ = AQB = 45°,  CA = 40 m, , Hence, the distance of the cloud from the, point A is 40 metres., [1], , [1], , In PAB,, , tan30 , , AB, PB, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , PB  300 3 m, , Mathematics, , [1], , , , In BAQ,, , 100 3, m, 3, , QR , , tan30 , , BQ = 300 m,  Width of the river = PB + BQ, , 1, , , , = 300(1+ 3) m, , [1], , 20. Let ships are at distance x from each other., , P, , 3, , 100, QS, , , , , , QS  100 3 m, , , , RS = QS – QR =, , 100 m, Speed, , y, , 45°, , 30°, x, , A, , 100, 1, y, , tan 45 , ,  y = 100 m, , OP, 100, 1, , , OB x  y, 3, , …(ii), , [1], , E, , [1], , 21. Let the light house be PQ and the boat changes, its position from R to S., Here, PQ = 100 m, PRQ = 60° and PSR = 30°., , Q, , In PQR,, , tan60 , , PQ 100, , QR QR, , D, , Height of aeroplane (CD) = 3600 3 m = BE, , = 73.2 m, , 60°, , C, 3600 3 m, , tan60 , , = 100 × 0.732, , R, , [1], , In ABE, ,  x = 100(1.732 – 1), , 30°, , [1], , BAD = 60° and CAD = 30°, , x  100 3  y  100 3  100  100( 3  1), , 100 m, , 200 3 100 3, , 32, 3, , B, , A, , P, , =, , [1], , xy, 100, ,  Ships are 73.2 meters apart., , Distance, Time, , = 57.73 m/min, 22., , x  y  100 3, , 100 3 200 3, , 3, 3, , = 57.73 (approx.) (Using 3  1.732), …(i) [1], , In POB, , S, , [1], , =, , B, , In APO, , 3, , PQ, QS, , 100 3 , , O, , [1], , In PQS,, , AB, tan 45 , BQ, , tan30 , , ...(i), , 49, , AE , , BE, AE, , [1], , BE, tan 60, , AE = 3600 m, , [∵ BE  3600 3 m], , [1], , In ACD, , tan30 , , AD , , CD, AD, , 3600 3, 1, 3, , AD = 10800 m, , [1], ,  BC = AD – AE = 10800 – 3600, , [1], , BC = 7200 m, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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50, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Speed of aeroplane , , distance, time, , [1], , D, , 24., , h, , 7200, ,  240 m/s, 30, , Speed (in km/hr) = 864 km/hour, , [1], , 7m, , C, , 23., , D, , So, AB = AE = 7 m, , 60° 30°, , A, , B, , Given that: AD = 3125 m and, ABC = 60°, , [1], , ABD = 30°, , AD, tan30 , AB, , ...(i), , [1], , AC, AB, [1], , Height of tower  h  7  7(1  3) m, , [1], , A, (h – 10), , 30°, , E, 10, , C, , 60°, , B, , In ABC, tan 60 , , 3125  h, 3, , h, BC, , [1], , [1], , 3, , In ADE,, ...(ii), , [1], , tan30 , , h  10, ED, , [1], , ED  (h  10) 3, ,  3125 3, , h = 3125 × 3 – 3125, , [1], , h, , BC , , Equating equation (i) and (ii), we have, , 3, , [1], , BE = CD = 10 m, , 3125  h, AB, , 3125  h, ,  h7 3m, , Given CD =10 m and BC = ED, , AD  DC, 3, AB, , AB , , [1], , Height of the tower (AB) = h, , ABC, , , , So, DE  AE 3  7 3, , 10, , AB  3125 3, , 3, , [1], , D, , 3125, AB, , tan60 , , tan 60  3  DE /AE, , 25., , In ABD,, , , , [1], , Again in triangle AED,, , Let the distance between the two planes be h m., , 3, , 7m, , Now, in triangle ABC, tan 45° = 1 = AB/BC, , 3125 m, , , , E, , C, B, Let AB be the building and CD be the tower such, that EAD = 60° and EAC = ACB = 45° [1], , h, , 1, , 60°, 45°, , A, , , [1], , h, 3, ,  (h  10) 3, , [1], , 2, h, 3, , h = 6250, , 10 , , Hence, distance between the two planes is, 6250 m., [1], , h  15 m, , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Mathematics, , 51, , In right triangle ABC,, , 26. D, , 30°, , 50, 50, 1,  tan30 , ,  AB  50 3, AB, AB, 3, , C, , h, , [2], , In right triangle BAD,, , 30°, , A, , 60°, , 50 m, , h, h,  tan 60 ,  3  h  3 AB, AB, AB, , B, ,  h  3(50 3)  150 m, Hence, the height of hill is 150 m., , Let the height of hill be h., , [2], , [2], , Chapter - 10 : Circles,  It is an isosceles triangle, , A, , 1., ,  OPQ = OQP, In POQ,, , Q, , R, , POQ + OPQ + OQP = 180°, POQ + 2OPQ = 180°, , B, , C, P, Given BR = 3 cm, AR = 4 cm & AC = 11 cm, , We know that OP  PT, , BP = BR, ,  OPT = 90°, , OPQ = 55°, , AR = AQ, , OPT = TPQ + OPQ, , CP = CQ, , 90° = TPQ + 55°, , (Lengths of tangents to circle from external point, will be equal), , TPQ = 35°, ,  AQ = 4 cm and BP = 3 cm, , 3., , QC + AQ = 11 cm, , B, A, ,  QC = 7 cm, , 40°, ,  PC = 7 cm, , AB and AC are the tangents drawn from external, point A to the circle., ,  BC = 3 + 7, [½], , Answer (D), , O, C, , We know BC = BP + PC, , BC = 10 cm, , [½], , Answer (C), , [½], , As AC = 11 cm, , 2., , [½], ,  OB  AB  OBA = 90°, OC  AC  OCA = 90°, ABCD is a quadrilateral in which sum of opposite, angles is 180°, , P, T, 70°, , O, , Q, , Given POQ = 70°, In POQ, OP = OQ (radii), , i.e.; OBA + OCA = 180°, , [½], ,  ABCD is a cyclic quadrilateral,  BAC + BOC = 180°, BOC = 180° – 40°, , BOC  140, , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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52, 4., , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 6., , Answer (A), , AP  PB (Given), , It is known that the tangents from an external, point to the circle are equal., , CA  AP, CB  BP (Since radius is perpendicular, to tangent), ,  EK = EM, DK = DH and FM = FH ...(i) [½], Perimeter of EDF = ED + DF + FE, , AC = CB = radius of the circle, , = (EK – DK) + (DH + HF) + (EM – FM), [Using (i)], = EK + EM, , Therefore, length of each tangent is 4 cm. [½], 7., , = 2 EK = 2 (9 cm) = 18 cm, Hence, the perimeter of EDF is 18 cm., , [½], , Therefore, APBC is a square having side equal, to 4 cm., , = (EK – DH) + (DH + HF) + (EM – FH), , 5., , Answer (B), , Answer (B), Q, , [½], , P, , R, , Answer (A), Given: AB, BC, CD and AD are tangents to the, circle with centre O at Q, P, S and R, respectively. AB = 29 cm,, , T, , AD = 23, DS = 5 cm and B = 90°, Construction: Join PQ., , It is known that the length of the tangents drawn, from an external point to a circle is equal., , A, , R, D, S, , O, , r, , Q, ,  QP = PT = 3.8 cm, , ...(i), , PR = PT = 3.8 cm, , ...(ii), , From equations (i) and (ii), we get :, , r, , QP = PR = 3.8 cm, , C, , B, P, We know that, the lengths of the tangents, drawn from an external point to a circle are, equal., , Now, QR = QP + PR, = 3.8 cm + 3.8 cm, = 7.6 cm, , DS = DR = 5 cm,  AR = AD – DR = 23 cm – 5 cm = 18 cm, AQ = AR = 18 cm, , [½], , Hence, the correct option is B., 8., , [½], , Answer (B), ,  QB = AB – AQ = 29 cm – 18 cm = 11 cm, , Q, , QB = BP = 11 cm, In PQB,, PQ 2 = QB 2 + BP 2 = (11 cm) 2 + (11 cm) 2 =, 2 × (11 cm)2, , PQ  11 2 cm, , ...(i), , [½], , 46º, , O, , P, , R, , In OPQ,, , Given: •QPR = 46°, , PQ2 = OQ2 + OP2 = r2 + r2 = 2r2, , PQ and PR are tangents., , (11 2)2  2r 2, , Therefore, the radius drawn to these tangents, will be perpendicular to the tangents., , 121 = r2, , So, we have OQ  PQ and OR  RP., , r = 11, Thus, the radius of the circle is 11 cm., , [½], ,  OQP = ORP = 90°, , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Mathematics, , B, , 11., , So, in quadrilateral PQOR, we have, OQP + QPR + PRO + ROQ = 360°, , a, ,  90° + 46° + 90° + ROQ = 360°, , 30°, 30°, , P, ,  ROQ = 360° – 226° = 134°, Hence, the correct option is B., , O, , R, , [radii], , PA = PB, , [length of tangents are equal], , OP = OP, , [Common], ,  PBO and PAO are congruent., , T, , , , So, OPQ = OPT – QPT, , = 30°, , [½], , [½], , 12., , Q, S, , P, , = 120°, , A, , D, , 1,  240, 2, ,  PRQ = 120°, , [½], , P, , Given a parallelogram PQRS in which a circle is, inscribed, We know PQ = RS, , C, 30, °, , QR = PS, B, , O, , (∵ OBBP), , B, , 1, reflexPOQ, 2, , a, 1, , OP 2, , OP = 2a units, , C, , Reflex POQ = 360° – 120° = 240°, , 60,  30, 2, , R, , POQ = 180° – 2QPO = 180° – 60° = 120°, , 30°, , BPO  OPA , , In PBO, sin30 , , = 90° – 60°, , A, , [½], , [By SSS criterion of congruency], , (radius is perpendicular to the tangent), , 10., , A, , OB = OA = a, , OPT = 90°, , , , a, , Given that BPA = 60°, , P, , PRQ , , O, , [½], , Q, , 9., , 53, , Q, , [½], , DP = PA, , ...(i), (tangents to the circle from external, point have equal length), , In ACO,, , Similarly,, QA = BQ, , ...(ii), ,  ACO is an isosceles triangle., , BR = RC, , ...(iii), , CAB = 30°, , DS = CS, , ...(iv), , OA = OC, , [Radii of the same circle], , [Given], ,  CAO = ACO = 30°, , [½], , [angles opposite to equal sides, of an isosceles triangle are equal], , Adding above four equations,, , DP + BQ + BR + DS = PA + QA + RC + CS, (DP + DS) + (BQ + BR) = (PA + QA) + (RC + CS), , PCO = 90°, , [½], , [radius drawn at the point of contact, is perpendicular to the tangent], , 2QR = 2(PQ),  PQ = QR,  PQ = QR = RS = QS, , Now PCA = PCO – ACO,  PCA = 90° – 30° = 60°, , [½], , [½], ,  PQRS is a rhombus, , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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54, 13., , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , R, , D, , In right angled triangle OBP,, , C, , OP2 = OB2 + BP2, , Q, , S, ,  BP2 = OP2 – OB2, = 172 – 52 = 289 – 25 = 264, , [½], ,  BP 2 = 264  BP  2 66 cm, , [½], , B, , A, , P, AB = 6 cm, , 15. Given : ABC is an isosceles triangle, where, AB = AC, circumscribing a circle., , BC = 9 cm, CD = 8 cm, , To prove : The point of contact P bisects the, base BC., , AB, BC, CD, AD, are tangents to the circle, And AP = AS,, BP = BQ, , RD = DS,, , i.e. BP = PC, , and, , Proof : It can be observed that, , CQ = CR, , [½], , Also AB = AP + BP, , ...(i), , BC = BQ + QC, , ...(ii), , CD = RC + DR, , ...(iii), , AD = AS + DS, , ...(iv), , BP and BR; CP and CQ; AR and AQ are pairs, of tangents drawn to the circle from the external, points B, C and A respectively., So, applying the theorem that the tangents, drawn from an external point to a circle are, equal, we get, , [½], , Adding (i), (ii), (iii), (iv), we have, 6 + 9 + 8 + AD = AP + AS + BP + BQ + CQ, + RC + RD + DS, [½], 23 + AD = 2(AP) + 2(BP) + 2(RC) + 2(RD), , BP = BR, , …(i), , CP = CQ, , …(ii), , AR = AQ, , …(iii), , [½], , Given that AB = AC, , 23 + AD = 2(AB) + 2(CD), ,  AR + BR = AQ + CQ, , AD  5 cm, , [½], ,  BR = CQ [from (iii)], , 14. Given : Tangents PA and PB are drawn from an, external point P to two concentric circles with, centre O and radii OA = 8 cm, OB = 5 cm, respectively. Also, AP = 15 cm, To find : Length of BP, , [½], ,  BP = CP [from (i) and (ii)], , [½], ,  P bisects BC., Hence proved., , [½], , 16., , Construction : We join the points O and P., , O, , A, , A, , C, , B, , P, , O, , Given : AB is chord to larger circle and tangent, to smaller circle at C concentric to it., , B, , To prove : AC = BC, , Solution : OA  AP; OB  BP, , Construction : Join OC, , [Using the property that radius is perpendicular, to the tangent at the point of contact of a circle], , Proof : OC  AB, , OP2 = OA2 + AP2 [Using Pythagoras Theorem],  OP = 17 cm, , [½], (∵ Radius is perpendicular to, tangent at point of contact), , In right angled triangle OAP,, , = (8)2 + (15)2 = 64 + 225 = 289, , [1], , [½], [½], ,  AC = BC, , [½], (∵ Perpendicular from, centre bisects the chord), , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 17. Given : AB = 12 cm, BC = 8 cm and AC = 10 cm., Let, AD = AF = x cm, BD = BE = y cm and, CE = CF = z cm, , Mathematics, , 55, , 19. Let us draw the circle with extent point P and, two tangents PQ and PR., , (Tangents drawn from an external point, to the circle are equal in length), , 60°, P, ,  2(x + y + z) = AB + BC + AC = AD + DB, + BE + EC + AF + FC = 30 cm, [½], , Q, O, R, ,  x + y + z = 15 cm, AB = AD + DB = x + y = 12 cm, , We know that the radius is perpendicular to the, tangent at the point of contact., , [½], ,  z = CF = 15 - 12 = 3 cm, ,  OQP = 90°, , AC = AF + FC = x + z = 10 cm, , We also know that the tangents drawn to a, circle from an external point are equally inclined, to the line joining the centre to that point., ,  y = BE = 15 – 10 = 5 cm, , [½], ,  x = AD = x + y + z – z – y = 15 – 3 – 5, = 7 cm, , [½], ,  QPO = 60°, , cos 60 , , Construction : Join OB and OC., , , Draw OA || XY, , B, , [½], , Now, in QPO,, , 18. Let XBY and PCQ be two parallel tangents to a, circle with centre O., , X, , [½], , Y, , PQ, PO, , [½], , 1 PQ, , 2 PO, ,  2PQ = PO, , [½], , 20., , A, , O, P, , C, , P, O, , Q, , Now, XB || AO, , R, Q, ,  XBO + •AOB = 180°, , [½], , (Sum of adjacent, interior angles is 180°), Now, XBO = 90°, (A tangent to a circle is perpendicular, to the radius through the point of contact),  90° + AOB = 180°,  AOB = 180° – 90° = 90°, , We know that the line joining the centre and the, external point is the angle bisector of angle, between the tangents., Thus,, PRO  QRO , , [½], , Similarly , AOC = 90°, AOB + AOC = 90° + 90° = 180°, , Given that PRQ = 120°, , [½], , 120,  60, 2, , [½], , Also we know that lengths of tangents from an, external point are equal., Thus, PR = RQ., , Hence, BOC is a straight line passing through O., , Join OP and OQ., , Thus, the line segment joining the points of, contact of two parallel tangents of a circle, passes through its centre., [½], , Since OP and OQ are the radii from the, centre O,, OP  PR and OQ  RQ., , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Thus, OPR and OQR are right angled, congruent triangles., , x2 + 15x – 54 = 0, , [½], , x2 + 18x – 3x – 54 = 0, , Hence, POR = 90° – PRO = 90° – 60° = 30°, , x(x + 18) – 3(x + 18), , QOR = 90° – QRO = 90° – 60° = 30°, , (x + 18)(x – 3) = 0, , sin QRO  sin30 , , [½], , 1, 2, , As distance cannot be negative, x = 3 cm, AC = 3 + 9 = 12 cm, , PR 1, , OR 2, , AB = AF + FB = 6 + x = 6 + 3 = 9 cm, ,  OR = PR + PR,  OR = PR + QR, , [½], , A, , 21., , x, , x, , 6 cm, , AP = AS, , ...(i), , BP = BQ, , ...(ii), , CR = CQ, , ...(iii), , DR = DS, , ...(iv), , [½], , Adding equations (i), (ii), (iii) and (iv), we get, , E, , F, , [½], , 22. Since tangents drawn from an exterior point to, a circle are equal in length,, , Thus,  OR = 2PR, , O, , AP + BP + CR + DR = AS + BQ + CQ + DS, , [½], ,  (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ), , 9 cm, , [½], , , B, , C, 9 cm, 6 cm D, Let the given circle touch the sides AB and AC, of the triangle at points F and E respectively and, let the length of line segment AF be x., , AB + CD = AD + BC, ,  AB + CD = BC + DA, 23., , Now, it can be observed that:, O, , BF = BD = 6 cm (tangents from point B), , [Proved], , [½], , T, , 60°, , 56, , Q, , 30°, , 2r, , P, , CE = CD = 9 cm (tangents from point C), S, , AE = AF = x (tangents from point A), AB = AF + FB = x + 6, , In the given figure,, , BC = BD + DC = 6 + 9 = 15, , OP = 2r, , CA = CE + EA = 9 + x, , [½], , [Given], , OTP = 90°, [radius drawn at the point of contact, is perpendicular to the tangent], , 2s = AB + BC + CA = x + 6 + 15 + 9 + x =, 30 + 2x, s = 15 + x, , In OTP,, , s – a = 15 + x – 15 = x, , sin OPT , , s – b = 15 + x – (x + 9) = 6, , OPT = 30°, , s – c = 15 + x – (6 + x) = 9, Area of ABC  s(s  a )(a  b )(s  c ), , OT, r, 1, ,   sin30, OP 2r 2, , TOP = 60°, [½], , [½], ,  OTP is a 30° – 60° – 90°, right triangle., , 54  (15  x )( x )(6)(9), , In OTS,, , 54  3 6(15 x  x 2 ), , OT = OS, , 18  6(15 x  x 2 ), ,  OTS = OST, , 324 = 6(15x +, 54 = 15x +, , x2, , x2), , [Radii of the same circle], ,  OTS is an isosceles triangle., [½], , [Angles opposite to equal sides, of an isosceles triangle are equal], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Mathematics, , In OTQ and OSQ, , For the circle AS & AP are tangents, , OS = OT, ,  AS = AP, , [Radii of the same circle], , OQ = OQ, , 57, , ...(i), , Similarly,, [side common to both triangles], , OTQ = OSQ, [angles opposite to equal sides of, an isosceles triangle are equal],  OTQ = OSQ, , [By S.A.S], , [½], , ...(ii), , CQ = CR, , ...(iii), , RD = DS, , ...(iv), , [½], , and BC + AD = BQ + QC + DS + AS ...(vi) [½], ,  TOS = 120°, , BC + AD = BP + CR + RD + AP using (i), (i),, (iii), (iv), , [TOS = TOQ + SOQ, = 60° + 60° = 120°], ,  AB + CD = BC + AD, ,  OTS + OST = 180° – 120° = 60°, , [Using (v)], , Hence proved, , [½], , [½], , P, , 26., , 24., , O, , [½], , Now, AB + CD = AP + PB + CR + RD ...(v), ,  TOQ = SOQ = 60° [C.A.C.T], ,  OTS = OST = 60° ÷ 2 = 30°, , BP = BQ, , x, , B, , A, , V, , 6c, m, , x, O, , 12 cm, , U, 6, , cm, , 9 cm, , 6 cm, , P, , Q, , 12 cm, , AB is the chord, , T, , 9 cm, , R, , We know that OA = OB [radii], , ar(PQR) = ar(POQ) + ar(QOR) + ar(POR), , OBP = OAP = 90°, , 1, 1, 1,  189   OV  PQ   OT  QR   OU  PR, 2, 2, 2, [½], , Join OP and OP = OP, , [Common], , [½], , By RHS congruency, OBP  OAP, , [½], ,  By CPCT, BP = AP, , [½], , In ABP BP = AP, ,  189 = 3(x + 12 + 12 + 9 + 9 + x), , Angles opposite to equal sides are equal,  BAP = ABP, Hence proved., 25., , C, R, , Q, B, A, , [½], , [∵ PV = PU = x, QT = 12 cm and RT = RU, = 9 cm as tangents from external point to a, circle are equal], [½],  63 = 24 + 18 + 2x, , D, S, , 1, 189   6(PQ  QR  PR)  3(PQ  QR  PR) [½], 2, (∵ OT = OV = OU = 6 cm), , P, ,  2x = 21, , 21,  PV  PU, 2, , , , x, , , , PQ  PV  QV  12 , , ABCD is the Quadrilateral, Circle touches the sides at P, Q, R, S, , and PR  PU  UR  9 , , [½], , 21 45, , cm, 2, 2, , 21 39, , cm, 2, 2, , [½], [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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58, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 27. A circle with centre O touches the sides AB, BC,, CD, and DA of a quadrilateral ABCD at the, points P, Q, R, , , , (5)2 = (4)2 + OM 2, , , , OM = 3 cm, , [½], , In OPT and OPM,, , and S respectively., To Prove : AOB + COD = 180°, , MOP  TOP, , [Common angles], , and AOD + BOC = 180°, , OMP   OPT, , [Each 90°], , , , POT ~ MOP, , [By AA similarity] [½], , , , TP, OP, , MP OM, , [½], , , , TP , , 45, 3, , [½], , D, , S, , R, C, , 7, 8O 6, 5, 1, 2 34, , A, , Q, B, , P, CONSTRUCTION, , [∵ OP = 5 cm, PM = 4 cm, MO = 3 cm], , Join OP, OQ, OR and OS., Proof : Since the two tangents drawn from an, external point to a circle subtend equal angles, at the centre., , , , [Sum of all the angles, subtended at a point is 360°],  2(2 + 3 + 6 + 7) = 360° and, 2(1 + 8 + 4 + 5) = 360°, , O, , Given : A circle with centre O and a tangent XY, to the circle at a point P, [½], , [½], , P, , Therefore, OQ is longer than the radius OP of, the circle, That is, OQ > OP., [1], This happens for every point on the line XY, except the point P., [½], So OP is the shortest of all the distances of the, point O to the points on XY., [½], And hence OP is perpendicular to XY., , 5 cm, , M, , [½], , Hence, proved., , O, , 30. Given : l and m are two parallel tangents to the, circle with centre O touching the circle at A and, B respectively. DE is a tangent at the point C,, which intersects l at D and m at E., , 4 cm, , Q, , In OPM,, OP2 = PM2 + OM2, , Q, , Proof : Here the point Q must lie outside the, circle as if it lies inside the tangent XY will, become secant to the circle., [½], , [½], , 4 cm, , P, , [1], , 28. Join OT which bisects PQ at M and, perpendicular to PQ, , T, , X, , Construction : Take a point Q on XY other than, P and join OQ., [½], ,  AOB + COD = 180° and AOD + BOC, = 180°, Hence, proved, , Y, , [½], , and 2 + 3 = AOB, 6 + 7 = COD, 1 + 8 = AOD and 4 + 5 = BOC, , [½], , To Prove : OP is perpendicular to XY., ,  (2 + 3) + (6 + 7) = 180° and, (1 + 8) + (4 + 5) = 180°, , 20, 2,  6 cm, 3, 3, , 29., ,  1 = 2, 3 = 4, 5 = 6 and 7 =  8, Now, 1 + 2 + 3 + 4 + 5 + 6 + 7 +, 8 = 360°, [½], , TP , , [By Pythagoras Theorem], , To prove: DOE = 90°, , [½], , Construction: Join OC., , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Mathematics, , OP = OP, , Proof:, A, , D, , 59, , [common], ,  AOP  BOP, , l, , [1], , [by R.H.S. congruence criterion], O, ,  AP = BP, , C, B, , E, , [corresponding parts of, congruent triangles], , m, , Hence, the length of the tangents drawn from an, external point to a circle are equal., [½], , In ODA and ODC,, OA = OC, , [Radii of the same circle], , 32. In the figure, C is the midpoint of the minor arc, PQ, O is the centre of the circle and, , AD = DC, (Length of tangents drawn from an, external point to a circle are equal], DO = OD, , [Common side], , ODA  ODC, , [SSS congruence criterion], [1], ,  DOA = COD, , ...(i), , Similarly, OEB  OEC,  EOB = COE, , [½], [½], , ...(ii), , [½], , Now, AOB is a diameter of the circle. Hence, it, is a straight line., , AB is tangent to the circle through point C., We have to show the tangent drawn at the, midpoint of the arc PQ of a circle is parallel to, the chord joining the end points of the arc PQ., We will show PQ || AB., , So, arc PC = arc CQ., , [½], , O, P, , Q, , A, B, C, This shows that PQC is an isosceles triangle., ,  COD + COE = 90°,  DOE = 90°, Hence, proved., , [½], ,  PC = CQ, , From (i) and (ii), we have:, 2COD + 2COE = 180°, , [½], , It is given that C is the midpoint point of the, arc PQ., , DOA + COD + COE + EOB = 180° [½], , [½], , [½], , 31. Let AP and BP be the two tangents to the circle, with centre O., , Thus, the perpendicular bisector of the side PQ, of PQC passes through vertex C., The perpendicular bisector of a chord passes, through the centre of the circle., [½], , A, , P, , [1], , O, , So the perpendicular bisector of PQ passes, through the centre O of the circle., [½], , B, To Prove : AP = BP, , Thus perpendicular bisector of PQ passes, through the points O and C., , Proof :, , [½], , In AOP and BOP,, OA = OB, , [½], , AB is the tangent to the circle through the, point C on the circle., , [radii of the same circle], , OAP = OBP = 90°, ,  PQ  OC, , [1], , [since tangent at any point of, a circle is perpendicular to the, radius through the point of contact], ,  AB  OC, , [½], , The chord PQ and the tangent AB of the circle, are perpendicular to the same line OC.,  PQ || AB., , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), ,  AB – AR = AC – AQ [Using (ii)], , Mathematics, ,  The lengths of the tangents drawn from an, external point to a circle are equal., [½], , [½], , BR = CQ,  BP = CP, , (Using (ii), (iii)], , [½], , D, , 36. PT and TQ are two tangent drawn from an, external pant T to the circle C(O, r), , Q, B, , S, A, , P, , P, , Let AB touches the circle at P. BC touches the, circle at Q. DC touches the circle at R.AD., touches the circle at S., [½], , O, Q, , Then, PB = QB ( Length of the tangents drawn, from the external point are always equal), , To prove : PT = TQ, Construction : Join OT, , [½], , Similarly,QC = RC , , Proof: We know that, a tangent to circle is, perpendicular to the radius through the point of, contact, [½],  OPT = OQT = 90°, , [½], , AP = AS, DS = DR, , [½], , [½], , Now,, , In OPT and OQT,, , AB + CD, , OT = OT, , [Common], , OP = OQ, , [Radius of the circle], , = AP + PB + DR + RC, , [½], , = AS + QB + DS + CQ, , [½], , [½], , OPT = OQT = 90°, , = AS + DS + QB + CQ, ,  OPT OQT [RHS congruence criterion], , = AD + BC, , [½],  PT = TQ, , C, , R, , So BP = CP, shows that BC is bisected at the, point of contact., [½], , T, , 61, , Hence, Proved, , [CPCT], , [½], , Chapter - 11 : Constructions, 1., ,  The desired point is P which divides AB in, 3 : 2., , Given a line segment AB = 7 cm, , A, , B, , P, , 2., , C, , Given, AP 3, AP, 3,  ,   5 AP  3 AP  3PB, AB 5, AP  PB 5, ,  2AP = 3PB, , , A, , [1], , P, A2, , A3, , A, , 5 cm, , B, , Steps :, , AP 3, , PB 2, , A1, , 4 cm, , A4, , B, A5, , 2) Draw an arc with radius 4 cm to cut ray SA, at C. Join BC to form ABC., [1], , X, , 1) Draw a line segment AB = 5 cm, Draw a, ray SA making 90° with it., , 3) Draw a ray AX making an acute angle with, AB, opposite to vertex C., , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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62, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 4) Locate 5 points (as 5 is greater in 5 and 3),, A 1 , A 2 , A 3 , A 4 , A 5 , on line segment AX, such that AA = A1A2 = A2A3 = A3A4 = A4A5, , 6. Join B 4 C and now draw a line from B 3, parallel to B4C so that it cuts BC at C, 7. From C draw a line parallel to AC and cuts, AB at A, , 5) Join A3B. Draw a line through A5 parallel to, A3B intersecting line segment AB at B., 6) Through B, draw a line parallel to BC, intersecting extended line segment AC at, C. AB'C' is the required triangle., [1], , 8. A'BC' is the required triangle, , B, , A, , P, A1, , 5 cm, , A2, , B, , B, , [2], , A3, , A4, , 3) Mark four points B1, B2, B3, B4 on BX such, that BB1 = B1B2 = B2B3 = B3B4, 4) Join B4C and draw a line parallel to B4C, from B3 such that it cuts BC at C., , A5, , 5) Form C draw another line parallel to AC, such that it cuts AC at A., [1], , C, , 6) ABC is the required triangle., , 8 cm, , A, , B, , A, 6 cm, Given ABC which is a right angled triangle, B = 90°, , A, B, , Steps :, , 2. Draw an arc with radius 6 cm from B so that, it cuts BX at A, , 5., , 8 cm, , B2, , B3, , P, , A, , A, If APB = 60°, , B1, , B2, , B3, , B4, , O, , A, , C, , C, , B, , X, , 8 cm, , 30°, , [2], x, Pair of a circle with radius = 3 cm inclined to, each other with angle 60°, , 3. Now join AC to form ABC, , B, , C, , 45°, , B1, , 1. Draw line segment BC = 8 cm, draw a ray, BX making an angle 90° with BC, , 6 cm, , C, , 2) Draw a ray BX making acute angle with BC, on the opposite side of vertex A, , x, 3., , 30°, 8 cm, , 1) Draw a ABC with BC = 8 cm, B = 45° &, C = 30°, , Q, , R, , 45°, , Steps :, , C, S, 4 cm, , A, , 4., , C, , [1], , C, , 60°, , [As AOBP is a cyclic quadrilateral], , [2], , Then AOB = 180 – 60°, , B4, , Y, 4. Draw a ray BY by making an acute angle, with BC, opposite to vertex A, 5. Locate 4 points B1, B2, B3, B4, on BY such, that BB1 = B1B2 = B2B3 = B3B4, , = 120°, , [½], , Tangents can be constructed in the following, manner:, Step 1, Draw a circle of radius 3 cm with center O., , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Mathematics, , 8) From C draw a line parallel to AC such, that it cuts AB at A thus ABC is the, required triangle, [1], , Step 2, Take a point A on the circumference of the circle, and join OA. Draw a perpendicular to OA at, point A., , , , Step 3, Draw a radius OB, making an angle of 120° with, OA., , 7., , Draw a perpendicular to OB at point B. Let both, the perpendicular intersect at point P. PA and PB, are the required tangents at an angle of 60°. [1], , It is given that A = 105°, C = 30°., , The steps of construction are as follows:, 1. Draw a line segment BC = 6 cm., 2. At B, draw a ray BX making an angle of 45°, with BC., , P, , 60, °, , AB AB BC 5, , , , AB AC BC 3, , Using angle sum property of triangle, we get,, B = 45°, , Step 4, , 3. At C, draw a ray CY making an angle of 30°, with BC. Let the two rays meet at point A., , B, , 4. Below BC, make an acute angle CBZ., , 60° 120°, O, , A, , 63, , A, , [1½], , 5. Along BZ mark three points B1, B2, B3 such, that BB1 = B1B2 = B2B3., 6. Join B3C., , 6., , 7. From B2, draw B2C || B3C., , A, , B, , 8. From C draw CA || CA, meeting BA at the, point A., [1], , 5 cm, , 3 cm, , Then ABC is the required triangle., , C, , 4 cm, x, , X, , Y, , A, 3 cm, , A, , A, , A, B, B1, , 4 cm, , B2, , C, , B3 B, 4, B5, , B, , C, , 45°, B1, , y, , 6 cm, , B2, [2], , Steps :, , 30°, C, B3, , C, , [2], , Z, , 8., , 1) Draw BC = 4 cm, 2) Draw a ray BX such that XBY = 90°, 3) Take compass with radius 3 cm and draw an, arc from B cutting BX at A, 4) Join A and C to from ABC, 5) Draw a ray BY opposite side of A such that, CBY is acute angle, 6) Along BY mark 5 equidistant points B1, B2,, B3, B4, B5 such that BB1 = B1B2 = B2B3 =, B3B4 = B4B5, 7) Join B5 to C and draw a line parallel to B5C, from B3 such that it cuts BC at C, , A, P, , 2, , 4, , M, , O, , B, , [2], , Steps of construction :, 1. Draw two concentric circle with centre O and, radii 4 cm and 6 cm. Take a point P on the, outer circle and then join OP., , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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64, , 9., , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 2. Draw the perpendicular bisector of OP. Let, the bisector intersects OP at M., , 10. 1. Construct the ABC as per given, measurements., , 3. With M as the centre and OM as the radius,, draw a circle. Let it intersect the inner circle, at A and B., , 2. In the half plane, of AB which does not, contain C, draw. AX such that BAX is an, acute angle., , 4. Join PA and PB. Therefore, PA and PB are, the required tangents., [1], , 3. Along AX mark 8 equidistant points B1, B2, …, B 8 such that B 1 B 2 = B 3B 4 = B 4 B 5 =, B5B6 = B6B7 = B7B8, , Follow the given steps to construct the figure., 1. Draw a line BC of 8 cm length., , 4. Draw B6 B ., , 2. Draw BX perpendicular to BC., , 5. Through B 8 draw a ray B 8B parallel to, B6 B . to intersect AY at B., , 3. Mark an arc at the distance of 6 cm on BX., Mark it as A., , 6. Through B draw a ray BC parallel to BC, to intersect AZ at C., , 4. Join A and C to get ABC., , Thus, ABC is the required triangle., , 5. With B as the centre, draw an arc on AC., , Z, , 6. Draw the bisector of this arc and join it, with B. Thus, BD is perpendicular to AC., 7. Now, draw the perpendicular bisector of BD, and CD. Take the point of intersection of, both perpendicular bisector as O., , C, C, , A, , 8. With O as the centre and OB as the radius,, draw a circle passing through points B, C, and D., 9. Join A and O and bisect it Let P be the, midpoint of AO., 10. Taking P as the centre and PO as its, radius, draw a circle which will intersect the, circle at point B and G. Join A and G., Here, AB and AG are the required tangents, to the circle from A., [1], , X, , B2, , B3, , B4, , B5, , B6, , B, Y, , [2½], , P, B7, , B8, , X, , 11. Steps :, (i) Take a point O on the plane of the paper, and draw a circle of radius OA = 4 cm., (ii) Produce OA to B such that OA = AB =, 4 cm., (iii) Draw a circle with centre at A and radius AB., , (v) Join BP and BQ to get the required, tangents., [2], , G, , D, , B, , B1, , 30° 6 cm 60° B, 8 cm, , (iv) Suppose it cuts the circle drawn in step (i), at P and Q., , A, , P, , P, , 30, °, , 6 cm, , [1½], , O, , 8 cm, , B, , C, , [2], , 30°, 30°, , 60°, , 60° 60°, O, A 4 cm, , Q, , [2], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Mathematics, , A, , 12., , 45°, , (ii) Draw an arc with radius 5 cm from B so that, it cuts BX at A., , 105°, , 30°, C, 7 cm, In the ABC, A + B +C =180°, B, , (iii) Now join AC to form ABC., (iv) Draw a ray BY making an acute angle with, BC opposite to vertex A., ,  C = 30°, , (v) Locate 4 points B1, B2, B3, B4 on BY such, that BB1 = B1B2 = B2B3 = B3B4., , Steps :, 1. Draw BC  7 cm with help of a ruler, , (vi) Join B 4 C and now draw a line from B 3, parallel to B4C so that it cuts BC at C'., , 2. Take a protractor measure,  angle 45° from, point B and draw a ray BX, 3. From point C, make angle 30° with help of, protractor such that BCY = 30°, , , 4. Now both BX and CY intersect at a, point A, , (vii) From C' draw a line parallel to AC and cuts, AB at A'., (viii) A'BC' is the required triangle., , C, , 8. Draw a line through C parallel to CA which, intersects BA at A, [1½], , 6c, , 45º, A, , 5 cm B, , A1, , B1, , A5, , 13., , 30°, , C, , 7 cm, , B2, , B3, , A, , B4, , C, , [2½], , z, , x, , X, , Steps :, (i) Construct ABC such that AB = 5 cm,, CAB = 45° and CA = 6 cm., , [2½], , (ii) Draw any ray AX making an acute angle, with AB on the side opposite to the vertex C., (iii) Mark points A1, A2, A3, A4, A5 on AX such, that AA1 = A1A2 = A2A3 = A3A4 = A4A5., , 5 cm, B, , [2½], , A4, , A, 45°, , B, , A3, , x, , A, , C, , A2, , ABC is the required triangle., , y, , m, , 6. Along the ray BZ mark 4 points B1, B2, B3,, B4 such that BB1 = B1B2 = B2B3 = B3B4, 7. Now join B4 to C and draw a line parallel to, B4C from B3 intersecting the line BC at C’, , [1½], , 14., , 5. Draw a ray BZ making an acute angle with, BC, , B, , 65, , 60°, B1, B2, , B3, , C, , C, , 6 cm, , (iv) Join A5B., (v) Through A 3, draw a line parallel to A 5B, intersecting with AB at B., , B4, , y, , Steps :, (i) Draw a line segment BC = 6 cm, draw a ray, BX making 60° with BC., , (vi) Through B, draw a line parallel to BC, intersecting with AC at C., Now, ABC is the required triangle whose sides, 3, of the corresponding sides of ABC. [1½], are, 5, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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66, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Chapter - 12 : Areas Related to Circles, 4., , 1., , Answer (B), Let r be the radius of the circle., From the given information, we have, , A, , B, , 2r – r = 37, , Given diameter of semicircular protractor, (AB) = 14 cm, , d , Perimeter of a semicircle      d, 2, ,  r(2) – 1 = 37 cm, , 22 , ,  1  37 cm,  r 2, 7, , , , [½], ,  r, ,  14 ,  Perimeter of protractor      14,  2 , , 2., ,  Circumference of the circle, [½], ,  2r  2 , , Answer (A), D, , 5., , C, , E, , [½], , P, , A, , A, , 22,  7 cm  44 cm, 7, , C, , D, , O, , F, , 37,  37 cm, 7, ,  r = 7 cm, , 22 14, , ,  14, 7, 2, , = 36 cm, , [½], , B, , B, , Given a square ABCD with side = 14 cm, , Given OE = OF = a, , AB = CD = BC = AD = 14 cm, , Side of the square circumscribing the circle = 2a, , Semicircles APB and CPD with diameter = 14 cm, , [½], , Perimeter of shaded region = AD + BC + arc(CPD), + arc(APB), [½], , •, ,  Perimeter of square = 4 × 2a = 8a units. [½], , 3., , Answer (B), , Length of arcCPD are , , Diameters of two circles are given as 10 cm and, 24 cm., , 180, 22 14, 2, ,  22, 360, 7, 2, , [½], , Radius of one circle = r1 = 5 cm, , Length of arcAPB = CPD = 22 cm, , Radius of one circle = r2 = 12 cm, , Perimeter of Shaded region = 14 + 14 + 22 + 22, = 72 cm, , According to the given information,, Area of the larger circle    r1     r2 , 2, , 2, , [½], , = (5)2 + (12)2, = (25 + 144), = 169, = (13)2, , 6., , [½], , [½], , Given, OABC is a square of side 7 cm, i.e. OA = AB = BC = OC = 7 cm,  Area of square OABC = (side)2 = 72 = 49, sq.cm, [½], Given, OAPC is a quadrant of a circle with, centre O., ,  Radius of larger circle = 13 cm, ,  Radius of the sector = OA = OC = 7 cm., , Hence, the diameter of larger circle = 26 cm [½], , Sector angle = 90°, , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), ,  Area of quadrant OAPC , , , , Mathematics, , In OAB :, , 90,  r 2, 360, , OB2, , (20)2, , +, , 67, , (20)2, ,  OB  20 2 cm, , 77, sq.cm, 2, = 38.5 sq. cm [½], , Radius of the circle, r  20 2 cm, , , , [½], , Area of quadrant OPBQ, ,  Area of shaded region = Area of Square, (OABC)-Area of quadrant (OAPC), 7., , +, , AB2 =, , = 2 × (20)2, , 2, 1 22, ,  7, 4 7, , = (49 – 38.5)sq. cm = 10.5 sq. cm, , =, , OA2, , [½], , Dimension of the rectangular card board, = 14 cm × 7 cm., Since, two circular pieces of equal radii and, maximum area touching each other are cut from, the rectangular card board, therefore, the, diameter of each of each circular piece is, , , , 90,  r 2, 360, , , , 90,  3.14  20 2, 360, , , , 1,  3.14  800 cm2, 4, , , , , , 2, , cm2, , = 628 cm2, , [1], , Area of square OABC = (Side)2 = (20)2 cm2, , 14,  7 cm, 2, , = 400 cm2,  Area of the shaded region = Area of, quadrant OPBQ – Area of square OABC, = (628 – 400) cm2, , 7 cm, , = 228 cm2, 14 cm, , 9., , 7, cm, 2,  Sum of area of two circular pieces, , B, , [½], , D, , Radius of each circular piece , , A, , 2, , 22 49, 7,  2     2 , ,  77 cm2, 7, 4, 2, , C, , Given AC = AB = 14 cm, [1], , Area of the remaining card board, = Area of the card board - Area of two circular, pieces, , BC  142  142  14 2 cm, Area of shaded region = Area of semi-circle –, (Area of quadrant ABDC – Area of ABC),  Area of ABC , , = 14 cm × 7 cm – 77 cm2, = 98 cm2 – 77 cm2, = 21 cm2, 8., , [1], , Area of Quadrant ABDC , , Let us join OB., , 1 22, , 14 2  154 cm2, 4 7, , [1], , Q, , Area of segment BDC = ar(Quadrant ABDC), – ar(ABC ), , B, , C, , 1,  14  14  98 cm2, 2, , = 154 – 98, = 56 cm2, , [½], , Area of semicircle with diameter BC, 90°, O, , A, 20 cm, , P, , , , 2, , 1  BC , 1 22 1, ,   14 2  14 2,   , 2  2 , 2 7 4, , = 154 cm2, , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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68, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Area of shaded region = Area of semicircle of, diameter BC –, Area of segment BDC, , 12. Area of minor segment, = Area of sector AOB – Area of AOB, B, , = 154 – 56, [1], 12, 0°, , = 98 cm2, 10. Let a be the side of equilateral triangle, , 3a2,  49 3;, 4, a2 = 49 × 4;, , O, , A, , a = 7 × 2 = 14 cm, , Given, , Radius of circle = 14/2 = 7 cm, , AOB = 120°, , [1], , OA = OB = 14 cm, O, 60, °, 60, °, , 60 °, , 14 cm, , 14 cm, 7 cm, , 30°, A, , Area of the first circle occupied by triangle, = area of sector with angle 60°., , , , 60r 2 22 1, 77, ,  77 , cm3, 360, 7 6, 3, , , , [½], , O = 60° B = 30°, D = 90°, , = area of triangle- area of all the 3 sectors, , OD = 7 cm, ,  49 3  77 = 49(1.732) – 77, , DB  7 3 cm, [1], , 11., , 3.5, , D, 3.5, ,  Area of AOB , , 1,  AB  OD, 2, , , , 1,  14 3  7, 2, ,  49 3, , B, , = 84.77, , Given AB = 14 cm and AC = BD = 3.5 cm,  DC = 7 cm, , Area of minor segment , , [1], , 616,  84.77, 3, , [1], , A, , P, , 13., , 2, 2,    3.5 2 ,   14 , 7, ,        2 ,  2  2  , 2 2 , 22, , , ,  196 49 49 , , ,  86.625 cm2, 4  2, 2, 4 , , [1], cm2, , = 120.56 cm2, , Area of shaded region = Area of semicircle AB, + Area of semicircle CD –2 (Area of semicircle, AC), [1], , , , 1 22, 616, 2, ,  14  , [1], 3 7, 3, , In ODB,, , Area of triangle not included in the circle, , C, , 120,  r 2, 360, , Draw OD  AB, , [½], , = 7.868 cm2, , B, , Area of sector AOB , , Area of all the 3 sectors = 77/3 × 3 = 77 cm2, , A, , 30°, D, , O, [1], , 30°, B, , Q, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Area of the shaded region, = Area of sector POQ – Area of sector AOB, , , , 30 22, ,  72  3.52, 360 7, , [1], , , , [1], , 77, , cm2, 8, , [1], , , l,  2r, 360, , [½], , = 31.4 cm, , [1], , 17. It is given that ABC is an equilateral triangle of, side 12 cm., , ,  r 2, 360, , OR  AB, A, , [½], , = 231, , cm, , [1], , 15. AB and CD are the diameters of a circle with, centre O.,  OA = OB = OC = OD = 7 cm (Radius of, the circle), [½], Area of the shaded region, = Area of the circle with diameter OB + (Area of, the semi-circle ACDA – Area of ACD), [1], , 22 49 1 22, 1, ,  ,  49   14  7, 7, 4 2 7, 2, , 77, ,  77  49, 2, = 66.5 cm2, 16. Radius of Semicircle PSR , , Radius of Semicircle RTQ , , B, , C, 12 cm, , Let the radius of the circle be r cm., Area of AOB + Area of BOC + Area of AOC, = Area of ABC, [½], , 1, 1, 1,  AB  OR   BC  OP   AC  OQ, 2, 2, 2, , , [½], , [1], , 1,  10 cm  5 cm, 2, [½], 1,  3  1.5 cm [½], 2, , 1,  7 cm  3.5 cm [½], 2, Perimeter of the shaded region = Circumference, of semicircle PSR + Circumference of semicircle, [½], RTQ + Circumference of semicircle PAQ, , Radius of semicircle PAQ , , Q, , r O r, r, , , , 2, , 1, 7 1, ,          72   CD  OA , 2, 2 2, , , R, , cm, , 60 22, ,  21 21, 360 7, , cm2, , [½], , 12, , , , Draw., OQ  AC, , [1], , (ii) Area of the sector , , Join OA, OB and OC., OP  BC, , 60, 22, 2,  21, 360, 7, , = 22 cm, , , , = 3.14 × 10 cm, , 12, , , , 1, 1, 1, ,    2  5    2 1.5    2  3.5   cm, 2, 2, 2, , , , Construction:, , 14. The arc subtends an angle of 60° at the centre., (i), , 69, , = (5 + 1.5 + 3.5) cm, , ,  , , , R 2 , r 2 , 360,  360, , , , Mathematics, , 3, 2,   side , 4, , 1, 1, 1, 3, 2,  12  r   12  r   12  r ,  12 , 2, 2, 2, 4, ,  3, , 1, 3,  12  r ,  12  12, 2, 4, ,  r  2 3  2  1.73  3.46, , [1], , Therefore, the radius of the inscribed circle is, 3.46 cm., Now, area of the shaded region = Area of ABC, – Area of the inscribed circle, ,  3, 2, ,  12    2 3, 4, , , , , , , 2, , 2,  cm, , ,  36 3  12  cm2, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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70, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , = [36 × 1.73 – 12 × 3.14], = [62.28 – 37.68], , cm2, , cm2, , = 24.6 cm2, , [1], , Therefore, the area of the shaded region is, 24.6 cm2., , , , 1 2 1, r   BC  AC, 2, 2, , , , 1, 1, 2,  3.14   6.5    5  12, 2, 2, , [½], , = 66.3325 – 30, = 36.3325 cm2, , 18. Radius of the circle = 14 cm, Central Angle,  = 60°,, , Thus, the area of the shaded region is, 36.3325 cm2., [½], , Area of the minor segment, , 20. Area of the region ABDC, , , 3 2, ,  r 2 , r, 360, 4, , = Area of sector AOC – Area of sector BOD [½], , [1], , 60, 3, ,   (14)2 ,  142, 360, 4, , , , 40 22, 40 22, ,  14  14 , , 77, 360 7, 360 7, , 1 22, 2,  ,  14  14  3   7 , 6 7, , , , 1, 1,  22  14  2   22  7  1, 9, 9, , 22  14, ,  49 3, 3, , , , 22,   28  7 , 9, , 22  14 147 3, , , 3, 3, , , , 22,  21, 9, , , , 154, cm2, 3, , , , 308  147 3, cm2, 3, , [1], , Area of circular ring, ,  Area of the major segment, ,   14 ,  616 , , 2, , , ,  308  147 3 , 2, ,  cm, , 3, , , , , ,  1540  147 3 / 3 cm2, , [1], , = 22 × (28 – 7), = 22 × 21, = 462 cm2, , [1], , [½], ,  Area of shaded region, , 19. Diameter, AB = 13 cm, , = Area of circular ring, – Area of region ABDC, , 13,  6.5 cm,  Radius of the circle, r , 2, , ∵ ACB is the angle in the semi-circle.,  ACB = 90°, , 22, 22,  14  14 , 77, 7, 7, , = 22 × 14 × 2 – 22 × 7 × 1, , 1, 308  147 3 , 3, , , , [½], , = 462 –, [½], , Now, in ACB, using Pythagoras theorem, we, have, AB2 = AC2 + BC2, , =, , 154, 3, , 1232, cm2, 3, , [½], , C, , 21., , S1, , (13)2 = (12)2 + (BC)2, (BC)2 = (13)2 – (12)2 = 169 – 144 = 25, , BC  25  5 cm, , [1], , Now, area of shaded region, = Area of semi-circle ABC – Area of (ACB) [½], , A, , S2, , S4, B, , D, S3, , Given that AB = BC = CD = 3 cm, , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Mathematics, , A, , Circle C has diameter = 4.5 cm, Semicircle S1 has diameter = 9 cm, , [½], , 9, 3, 3,  4.5    3 ,        ,     [½], 2  2 , 22, 22,  2  22, , , ,   81   9, , 16, 8, , O, C, , C, ,         , ,   62,  62,  62,  62, ,  12 , , , ,  4, 4, 4, 4, , 2, , [½], , , , , [1], , = 122 –  × 36, , C2, , 60º, , R, , Area of the shaded region = Area of square, ABCD – (Area of sector APEC + Area of sector, PFQB + Area of sector RGQC + Area of sector, RHSD), [1], , C1, , 22., , G, , D, , 2, , , , = 12.375 cm2, , Q, H, , Area of shaded region, 2, , F, , S, , = Area of S1 – Area of (S2 + S4) – Area of C +, Area of S3, [1], , 2, , B, , E, , Area of shaded region, , 2, , P, , 71, , = 144 – 113.04, = 30.96 cm2, , [1], , D, , A, , D, , B, , 6, , 24., , Given OC = OD = 21 cm, , C, , 8, O, , A, , OA = OB = 42 cm, , B, , Area of ACDB region, = Area of sector OAB – Area sector OCD, , , 60, 60, 2, 2,    42  ,    21, 360, 360, , , , 1 22, ,  21 63, 6 7, , In right triangle ADC, D = 90°, , [½], , AC2 = AD2 + DC2 [By Pythagoras theorem] [½], [½], , = 62 + 82 = 100, AC = 10 cm, , [½], , 2(AO) = 10, , = 11 × 63 = 693 cm2, , AO = 5 cm, , [½], ,  Radius (r) = 5 cm, , Area of shaded region, , [½], , Area of the shaded region, = Area of c1 – Area of c2, – Area of ACDB region, , = Area of the circle – Area of rectangle [½], , [½], , = r2 – l × b, , = (42)2 – (21)2 – 693, , = 3.14(5)2 – 6 × 8, , 22, ,  63  21  693, 7, , = 3,465 cm2, , = 78.5 – 48 = 30.5, [1], , [½], cm2, , [½], , A, , 25., , 23. Given that ABCD is a square and P, Q, R and, S are the mid-points of AB, BC, CD and DA, respectively, , O, , and AB = 12 cm,  AP = 6 cm, , [P bisects AB], , B, , C, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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72, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , AC = 24 cm, BC = 10 cm, , [½], , = (441) [0.57], = 251.37 cm2, , AB  242  102, AB = 26 cm, , Area of the flower bed ORQ = Area of the flower, bed OPS, , [1], , Diameter of circle = 26 cm, , = 251.37 cm2, , Area of shaded region, = Area of semicircle – Area of ABC, , , , , 1, 13 2   24  10, 2, 2, , , , 3.14,  169  120, 2, , = 145.33, , [1], , Total area of the two flower beds, , [1], , = Area of the flower bed ORQ + Area of the, flower bed OPS, , [½], , = 251.37 + 251.37, = 502.74 cm2, , cm2, , [1], , 26. PQRS is a square., , [1], , 27. Perimeter of shaded region = AB + PB + arc, length AP, (i) [½], , r,  2r , 360, 180, , So each side is equal and angle between the, adjacent sides is a right angle., , Arc length AP , , Also the diagonals perpendicularly bisect each, other., , In right angled OAB,, , In PQR using pythagoras theorem,, , tan  , , AB,  AB  r tan , r, , PR2 = (42)2 + (42)2, , sec  , , PR 2  2  42 , , OB,  OB  r sec , r, , OB = OP + PB, , PR2, , =PQ2, , OR , , +, , QR2, , 1, 42, PR ,  OQ, 2, 2, ,  r sec = r + PB, , [1], , Area of sector ORQ , , , 2, , , , 1 42 42, , , 2, 2, 2, ,  r tan   r sec   r , , 28., , 21 cm, , A, , 2, ,  42    ,      1,  2  2 , , [1], , B, 7 cm, , 2, , = Area of sector ORQ – Area of the ROQ, 2, , [1], , r, 180, , , , ,  r  tan   sec  ,  1, 180 , , , O, , [1], , 7 cm, , Area of the flower bed ORQ, , 2, , ...(iv), , = AB + PB + arc (AP), , 1,  RO  OQ, 2, , 1  42   42 ,  ,  , 2  2   2 , , [ OB = rsec], , Perimeter of shaded region, , Area of the ROQ , ,  42 ,  ,  2 , , [½], , Substitute (ii), (iii) and (iv) in (i), we get, , 1 2, r, 4, 1  42 , , , 4  2, , [½], , (iii) [½], ,  PB = r sec  – r, , From the figure we can see that the radius of, flower bed ORQ is OR., , (ii), , C, , D, , Area of shaded region = Area of rectangle – Area, of semicircle, [1],  21 14 , , = 217, , cm2, ,  7, , 2, , 2, , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Mathematics, , Perimeter of shaded region, = AB + AD + CD + length of arc BC,  21  14  21 , , Area of sector BPQ , , [1], , 180, 22,  2, 7, 360, 7, , = 78 cm, , [1], , , , 77, cm2, 12, , , , 60 22, 2, ,   3.5 , 360 7, , , , 77, cm2, 12, , P, , 29., C, , B, , R, , [1], , [1], , Similarly areas of other sectors PCR and, 77, cm2, RAQ , [1], 12, Area of shaded region, , Q, A, , = ar(ABC) –3 (area of BPQ), , Given that all circles have radii = 3.5 cm,  AB = BC = AC = 7 cm, ABC is an equilateral triangle area of, , ABC , , 60 22, 2, ,   3.5 , 360 7, , 73, , 3,  49 cm2, 4, , [1], , , , 49 3 3  77 , , 4, 12, , , , 49 3  77 7,  7 3  11, 4, 4, , , , [1], , , , [1], , Chapter - 13 : Surface Areas and Volumes, 1., , Surface area of sphere = 616 cm2, 4r2 = 616, , 4, , 4., [½], , Let the original radius and the height of the, cylinder be r and h respectively., , 22 2,  r  616, 7, , Volume of the original cylinder = r2h, , r  7 cm, , [½], , Radius of the new cylinder , , R, , 2., , Answer (C), , r, 2, , Height of the new cylinder = h, 2, , r 2 h, r , Volume of the new cylinder     h , [½], 4, 2, , I, r, , Required ratio , , Given slant height () = 4 cm, Perimeters of circular ends:, , r 2 h, 1,  42   1: 4, r h 4, , 2r = 6 cm, , 3., , 2R = 18 cm, , [½], , C.S.A = () (r + R) = 4 × 12 = 48 cm2, , [½], , Answer (B), Largest cone that can be cut from a cube has, the, Diameter = side of cube, , [½], , Height = side of cube, , , radius , , 4.2,  2.1 cm, 2, , Volume of the new cylinder, Volume of the original cylinder, , 5., , [½], , Answer (B), Let r and h be the radius and the height of the, cylinder, respectively., Given: Diameter of the cylinder = 4 cm,  Radius of the cylinder, r = 2 cm, Height of the cylinder, h = 45 cm, , [½], , Volume of the solid cylinder = r2h =  × (2)2 ×, [½], 45 cm3 = 180 cm3, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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74, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Suppose the radius of each sphere be R cm., Diameter of the sphere = 6 cm, , 8., , Given volume of a hemisphere  2425, ,  Radius of the sphere, R = 3 cm, , , , Let n be the number of solids formed by melting, the solid metallic cylinder., •, ,  n × volume of the solid spheres, , Volume of a hemisphere , ,  n, , 4, R 3  180, 3, , , , 2 3 4851, r , 3, 2, ,  n, , 4, R 3  180, 3, , , , 2 22 3 4851, , r , 3 7, 2, , 180  3, 5, 4  27, Thus, the number of solid spheres that can be, formed is 5., [½], ,  n, ,  r3 , , , Volume of cube = 27 cm3, , r , , [½], , If two cubes are joined end to end the resulting, figure is cuboid,  3 cm , 3c, , m, , Volume of frustum , , , [½], , , , 22,  21 102  202  10  20, 73, , , , [1], , , [1], , Cost of milk = 15.4 × 30, = `462, , = 2 × (6 × 3 + 3 × 3 + 3 × 6) cm2, 10., , = 2 × (18 + 9 + 18), , , , , h R 2  r 2  rR, 3, , = 15400 cm3 = 15.4, , [½], , 7., , [1], , = 22(700) cm3, , Surface area of resulting cuboid = 2(lb + bh + hl), , = 2 × 45 = 90, , 22 21 21,  ,  693 sq.cm, 7, 2 2, , 10, , height = h = 3 cm, , cm2, , [½], , h = 21 cm, , i.e., length = l = 6 cm, breadth = b = 3 cm, , 3, , 20, , 9., , , , 3 cm, , 4851 3 7  21 ,  , , 2, 2 22  2 , , 21, cm, 2, ,  2, , Side  3 27 cm, , 3 cm, , 2 3, r, 3, , So, curved surface area of the hemisphere = 2r2, ,  Volume of cube = (side)3 = 27 cm3, , Side = 3 cm, , 4851, cm3 [½], 2, , Now, let r be the radius of the hemisphere, , = Volume of the solid cylinder, , 6., , 1, cm3, 2, , C, , [1], 6 cm, , O, , D, , [½], , Cone: height = 20 cm, Base radius = 5 cm, , 7 cm, , Cone is reshaped into a sphere, •, ,  Volume of cone = volume of sphere, , [1], , 1, 4, 2, 3,   5   20     r , 3, 3, , A, , r3 = 5 3,  r = 5 cm, , [1], , O, , B, , Given: Radius of cylinder = radius of cone =, r = 6 cm, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Mathematics, ,  Radius of the hemispherical bowl = Radius, of the cylinder, , Height of the cylinder = height of the cone, = h = 7 cm, [½], Slant height of the cone = l  72  62, ,  85 cm, , r , , [½], , [1], ,  Height of the cylinder, h = 13 cm – 7 cm =, 6 cm, [1], , Curved surface area of the cylinder + area of the, base of the cylinder + curved surface area of, the cone, , Total surface area of the vessel = 2 (curved, surface area of the cylinder + curved surface, area of the hemisphere) (Since, the vessel is, hollow), ,  Total surface area of the remaining solid, = (2rh + r2 + rl), [1], , = 2(2rh + 2r2) = 4r(h + r), , 22, 22 2 22,  2, 67 , 6 ,  6 85, 7, 7, 7, ,  4, , 792 132, , 85, 7, 7, ,  377.1 , , 14, cm  7 cm, 2, , Total height of the vessel = 13 cm, , Total surface area of the remaining solid =, ,  264 , , 75, , 22,  7   6  7  cm2, 7, , = 1144 cm2, 3.5 cm, , 13., , 132, 85 cm2, 7, , [1], , [1], , 11. Volume of the conical heap = volume of the sand, emptied from the bucket., h, , Volume of the conical heap, 1 2, 1, r h  r 2  24 cm3, 3, 3, , ...(i), , (height of the cone is 24), , [1], 3.5 cm, , Volume of the sand in the bucket = r2h, = (18)2 × 32 cm3, , ...(ii), , [1], , Radius of the cylinder = Radius of each, hemisphere = r = 3.5 cm, [½], , Equating (i) and (ii),, 1 2, 2, r  24   18   32, 3, ,  r2 , , 18 2  32  3, 24, , Height of the cylinder, h = 10 cm, , [½], , Volume of wood in the toy = Volume of the, cylinder – 2 × Volume of each hemisphere, , [½], ,  r 2 h  2 , ,  r = 36 cm, 12., , 13 cm, , [1], , 4 , ,  r 2  h  r , 3 , , , , 7 cm, , 2 3, r, 3, , 22, 4, 2, ,   3.5   10   3.5 , 7, 3, , , , = 38.5 × (10 – 4.67), , [1], , = 38.5 × 5.33, = 205.205 cm3, , [½], , 14. For the given tank, Let the radius and height of cylinder be r cm and, h cm respectively., , Diameter = 10 m, , Diameter of the hemispherical bowl = 14 cm, , Depth, H = 2 m, , Radius, R = 5 m, [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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76, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , In triangle COB, , Internal radius of the pipe, 20, 1, cm  10 cm , m, 2, 10, , r , , [½], , Rate of flow of water = v = 4 km/h = 4000 m/h, Let t be the time taken to fill the tank., , [½], , So, the volume of water flows through the pipe, in t hours will equal to the volume of the tank.,  r2 × v × t = R2H, , 1, Hence, the time taken is 1 hours, 4, , [½], , C, , 30°, 20 cm, , x, , 10 cm, , O, ,  R, , R, 20, , 20, 3, , cm, , R, , [½], , , , 1,  R 2H  x 2 h, 3, , , , 2, 2, ,  10 , 1   20 , .20, .10 , , , , , , , 3  3 ,  3, , , , , , 1  8000 1000 , , , 3  3, 3 , , , , 1  7000 , , 3  3 , , , , 1,   7000, 9, , , , 7000, , 9, , Q, , P, , A, , , , 3, ,  V , , 25  2  100, 1, 1, 4000, 4, , S, , 1, , Volume of the frustum, V , , 2, , 15., , , , [1], , 2,  1,     4000  t   5   2,  10 , ,  t, , R, CO, , tan30 , , [½], , The volumes of the frustum and the wire formed, are equal., , B, , Let ACB be the cone whose vertical angle ACB, = 60°. Let R and x be the radii of the lower and, upper end of the frustum., , 7000 ,  1 ,   l ,   Volume of wire  r 2 h , 24, 9,  , , Here, height of the cone, OC = H = 20 cm, ,  l, , Height CP = h = 10 cm, , [½], , 2, , 7000,  24  24, 9, , Let us consider P as the mid-point of OC., ,  l = 448000 cm = 4480 m, , After cutting the cone into two parts through P., , Hence, the length of the wire is 4480 m., , OP , , 16. Diameter of the tent = 4.2 m, , 20,  10 cm, 2, , Also, ACO and OCB , , [½], , 1,  60  30, 2, , Now, in triangle CPQ, , x, , 3 10, 10, 3, , cone, , = 2.8 m, , =, , [½], , Slant height of the conical part, l, ,  2.82  2.12, , 1, , x, , Height of the conical part, h, , cylinder, ,  h 2cone  r 2, , x, tan30 , 10, , , , Radius of the tent, r = 2.1 m, Height of the cylindrical part of tent, h, 4m, , After cutting cone CQS from cone CBA, the, remaining solid obtained is a frustum., , , , [½], , cm, ,  2.82  2.12, [½], , = 3.5 m, , [½], , Curved surface area of the cylinder = 2rh, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), ,  2, , = 22 × 0.3 × 8 = 52.8 m2, , [½], , 2, 9,  183 , 3, 10,  72 , 32  h, , [½], , 2, 9,  18  18  18 , 3, 10,  h, 32  72, , Curved surface area of the conical tent, , 22,  2.1 3.5  23.1 m2, 7, , Total area of cloth required for building one tent, = Curved surface area of the cylinder + Curved, surface area of the conical tent, = 52.8 + 23.1, = 75.9, , 77, , 2, 90,    183 , 100,  72  3,   32  h, , 22,  2.1 4, 7, ,  rl , , Mathematics, ,  h = 5.4 cm, , [½], , Height of the small cylindrical bottle = 10.8 cm, , m2, , [½], , 18. Side of the cubical block, a = 10 cm, , Cost of building one tent = 75.9 × 100 = ` 7590, , Largest diameter of a hemisphere = side of the cube, , Total cost of 100 tents = 7590 × 100, , Since the cube is surmounted by a hemisphere,, , = ` 7,59,000, , Diameter of the hemisphere = 10 cm, , Cost to be borne by the associations, , , , 759000,  3,79,500, 2, , Radius of the hemisphere, r = 5 cm, [½], , It shows the helping nature, unity and, cooperativeness of the associations., , Total surface area of the solid = Total surface, area of the cube – Inner cross-section area of, the hemisphere + Curved surface area of the, hemisphere, = 6a2 – r2 + 2r2, , 17. Internal diameter of the bowl = 36 cm, , [1], , = 6a2 + r2, , Internal radius of the bowl, r = 18 cm, 2, 2, Volume of the liquid, V  r 3     183 [½], 3, 3, , = 6 × (10)2 + 3.14 × 52, = 600 + 78.5 = 678.5 cm2, Total surface area of the solid = 678.5 cm2 [1], , Let the height of the small bottle be ‘h’, Diameter of a small cylindrical bottle = 6 cm, , 19. Number of cones = 504, , Radius of a small bottle, R = 3 cm, , Diameter of a cone = 3.5 cm, , Volume of a single bottle = R2h =  ×32 × h [½], , Radius of the cone, r = 1.75 cm, , Number of small bottles, n = 72, , Height of the cone, h = 3 cm, , Volume wasted in the transfer , , 10 2,     183, 100 3, [½], , Volume of liquid to be transferred in the bottles, , , 2, 10 2,    183 ,     183, 3, 100 3, , 2, 10 , ,     183  1 , , 3,  100 , , , 2, 90,    183 , 3, 100, , [½], , Volume of a cone, , , 1 2, r h, 3, , , , 1,  3.5 ,  ,  3, 3,  2 , , , , 1, 3.5 3.5,  , ,  3 cm3, 3, 2, 2, , 2, , [½], , Volume of 504 cones, [½], , Number of small cylindrical bottles, , , , [1], , Volume of the liquid to be transferred, [½], Volume of single bottle, ,  504 , , 1, 3.5 3.5,  , ,  3 cm3, 3, 2, 2, , [½], , Let the radius of the new sphere be ‘R’., Volume of the sphere , , 4, R 3, 3, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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78, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Volume of 504 cones = Volume of the sphere [½], 504 , , , , = 10.5, , 1, 3.5 3.5, 4,  , ,  3  R 3, 3, 2, 2, 3, ,  10.5 , , 504  1   3.5  3.5  3  3,  R3, 3 2 2 4 , ,  R3 , , 8  27  7, 64, , 21. Let the radius of the conical vessel = r1 = 5 cm, Height of the conical vessel = h1 = 24 cm, , 237, 4, ,  R, , 21,  10.5 cm, 2, , Let the water rise upto the height of h2 cm in the, cylindrical vessel., Now, volume of water in conical vessel = volume, of water in cylindrical vessel, [1], , 1 2, r1 h1  r22 h2, 3, , r12 h1  3r22 h2, , Radius of the new sphere = 10.5 cm, V, , 20., , [½], , Radius of the cylindrical vessel = r2, , 3, ,  R, , [1], , Total cost of the canvas at the rate of ` 500 per, m2 = `(500 × 33) = `16500., [1], , 504  3  49, 64, , 7  8  9  3  72,  R , 64, ,  R3 , , 22, 7, , = 33 m2, , 3, , [1½], , 5 × 5 × 24 = 3 × 10 × 10 × h2, 2.8 m, , A, , = 4.2 + 6.3, , h2 , , B, , 1.5 m 1.5 m, , 5  5  24,  2 cm, 3  10  10, , [1], , Thus, the water will rise upto the height of 2 cm, in the cylindrical vessel., 22. Radius of sphere = r = 6 cm, 2.1 m, , Volume of sphere, , , D, , 1.5 m 1.5 m, 3m, , C, , 4 3 4, 3, r     6   288 cm3, 3, 3, , Let R be the radius of cylindrical vessel., Rise in the water level of cylindrical vessel, , For conical portion, we have, , 5, 32, cm , cm, 9, 9, , r = 1.5 m and l = 2.8 m, , h3, ,  S1 = Curved surface area of conical portion, , Increase in volume of cylindrical vessel, ,  S1 = rl, = 4.2m2, ,  R 2 h  R 2 , , [½], , 32 32, , R 2, 9, 9, , [½], , Now, volume of water displaced by the sphere is, equal to volume of sphere, , For cylindrical portion, we have, r = 1.5 m and h = 2.1 m,  S2 = Curved surface area of cylindrical portion,  S2 = 2rh, = 2 ×  × 1.5 × 2.1, , [½], , [½], , = 6.3m2, Area of canvas used for making the tent = S1 + S2, , , , 32, R 2  288, 9, , , , R2 , , 288  9,  81, 32, , [1], [½], ,  R = 9 cm,  Diameter of the cylindrical vessel = 2 × R =, 2 × 9 = 18 cm, [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 23. Given canal width = 5.4 m, Depth = 1.8 m, , [½], , Water flow speed = 25 km/hr, , Total surface area of the article = curved surface, area of the cylinder + 2 (Curved surface area of, a hemisphere), [1], , = 2r(h + 2r), , 25  40, , 60, , [½], , 50, km, 3, ,  2, , 22,  3.5 10  2  3.5  cm2, 7, [1], , 26. Given, , 50,  5.4  1.8  1000, 3, , = 162 × 103 m3, , , , [1], , 3.5 m, , Area irrigate with 10 cm of water standing, , , [1], , = 22 × 17 cm2 = 374 cm2, , Volume of water flows through pipe, , , 79, , = 2rh + 2 × 2r2, , Distance covered by water in 40 minutes, , , , Mathematics, , 162  103, 10  102, , = 162 ×, , 104, , 12 m, , m2, , [1], , 24. Volume of cuboid = 4.4 × 2.6 × 1, = 11.44 m3, , Base radius = 12 m, [½], , Height = 3.5 m, Volume , , 5, , , , , , 30, , 35, , Base diameter = 24 m, , 1 2, r h, 3, , [½], , 1 22, ,  12  12  3.5, 3 7, , = 22 × 4 × 12 × 0.5, = 264 × 2, = 528 m3, , Length = l, Inner radius = 30 cm, , [½], , 2 = 156.25, , Volume of cuboid = volume of cylindrical pipe, ,   156.25  12.5 m, , , ,   l  352  302, 100  100  100, , , , l = 102.05 km, , [½], , Curved surface area = r, [1], , l = 10.205 × 104 cm, , 25., ,  2 = 122 + 3.52 = 144 + 12.25, , Outer radius = 35 cm, , 11.44 , , [1], , [1], , r, , 22, 150  22,  12  12.5 ,  471.428 m2, 7, 7, 27. Width of the canal = 6 m, Depth of the canal = 1.5 m, Length of the water column formed in, = 5 km or 5000 m, , r, , Let r be the radius of the base of the cylinder, and h be its height. Then,, , [1], , 1, hr, 2, [½], , 1, hr, 2, = Volume of cuboid of length 5000 m, width, 6 m and depth 1.5 m., ,  Volume of water flowing in, , = 5000 × 6 × 1.5 = 45000 m3, , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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80, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , On comparing the volumes,, Volume of water in field = Volume of water, coming out from canal in 30 minutes., [½], Irrigated area × standing water = 45000., , 45000, Irrigated Area , [∵ 1 m = 100 cm] [½], 8, 100, , , 45000  100, = 5,62,500 m2 [½], 8, , Radius (r2) of base of cylindrical tank = 40 cm, = 0.4 m, [½], Let the cylindrical tank be filled up to h m in 30, minutes., Volume of water filled in tank in 30 minutes is, equal to the volume of water flowed out in 30, minutes from the pipe, [1], , , 4.2 cm, , 28., , Volume of water that flows in 30 minutes from, pipe = 30 × 0.0024m3 = 0.072 m3, [½], ,   r22  h  0.072, ,  0.42 × h = 0.072, , [½], ,  0.16h = 0.072, 5 cm, ,  h, , 5 cm, , 0.072, 0.16, ,  h = 0.45 m = 45 cm, , Therefore, the rise in level of water in the tank, in half an hour is 45 cm., , 5 cm, , The total surface area of the cube = 6 × (edge)2, = 6 × 5 × 5 cm2 = 150 cm2, [1], Note that the part of the cube where the, hemisphere is attached is not included in the, surface area., So, the surface area of the block = TSA of cube, – base area of hemisphere + CSA of hemisphere, , = 150 – r2 + 2r2 = (150 + r2) cm2, , = (150 + 13.86) cm2 = 163.86 cm2, , [½], Diameter of lower end of bucket = 10 cm,  Radius (r1) of lower end of bucket = 5 cm, [½], , [1], , , ,  r1  r2 2  h2, , , , 15  5 2  242, , [1], ,  102  242  100  576, ,  676  26 cm, ,    r1  r2  l  r22, [½], ,    r12    0.012  0.0001 m2 [½], Speed of water = 0.4 m/s s = 0.4 × 60, = 24 metre/min, Volume of water that flows in 1 minute from pipe, = 0.0024, , [1], , =(15 + 5)26 + (5)2, = 520 + 25 = 545 cm2, , Area of cross-section, , m3, , [1], , Area of metal sheet used to make the bucket, ,  Radius r1 of circular end of pipe, , = 24 × 0.0001, ,  Radius (r1) of upper end of bucket = 15 cm, , Slant height (l) of frustum, , 29. Diameter of circular end of pipe = 2 cm, , 2, , m  0.01 m, 200, , 30. Diameter of upper end of bucket = 30 cm, , [1], ,  22 4.2 4.2 ,  150 cm2  , , , cm2, 2, 2 ,  7, , m3, , [½], , [½], , Cost of 100 cm2 metal sheet = `10, Cost of 545 cm2 metal sheet, `, , 545  3.14  10,  ` 171.13, 100, , [½], , Therefore, cost of metal sheet used to make the, bucket is ` 171.13., , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 2.5 cm, , 31., , Mathematics, , 81, , Diameter of the cylindrical part = Diameter of the, conical part = 4.2 cm,  Radius  of the cylindrical part = Radius , of the conical part = 2.1 cm, [½], Slant height (l) of the conical part, , 11 cm, , , ,  2.12   2.8 2, , Height (h) of the conical vessel = 11 cm, ,  4.41  7.81 cm, , Radius (r1) of the conical Vessel = 2.5 cm, ,  12.25 cm, , Radius (r 2 ) of the metallic spherical balls, 0.5, ,  0.25 cm, [½], 2, , = 3.5 cm, , Let n be the number of spherical balls = that, were dropped in the vessel., Volume of the water spilled = Volume of the, spherical balls dropped, [½], , 2,  Volume of cone = n × Volume of one, 5, spherical ball, , , 2 1 2, 4,  r1 h  n  r23, 5 3, 3, , cm, , [½], , Total surface area of the remaining solid = Curved, surface area of the cylindrical part +Curved, surface area of the conical part + Area of the, cylindrical base, = 2rh + rl + r2, , [1], , 22, 22, 22, , ,  2,  2.1 2.8 ,  2.1 3.5 ,  2.1 2.1 cm2, 7, 7, 7, , , , [1], , = (36.96 + 23.1 + 13.86) cm2, [1], [½], ,  r12 h  n  10r23, , = 73.92 cm2, , [½], , Thus, the total surface area of the remaining, [½], solid is 73.92 cm2, 33. Height of the cylinder (h) = 10 cm, Radius of the base of the cylinder = 4.2 cm [½], ,  (2.5)2 × 11 = n × 10 × (0.25)3,  68.75 = 0.15625 n, , [½], , Volume of original cylinder = r2h, 22, 2,   4.2   10, 7, ,  n = 440, , , , Hence, the number of spherical balls that were, dropped in the vessel is 440., , = 554.4 cm3, , Sushant made the arrangement so that the, water that flows out, irrigates the flower beds., This shows the judicious usage of water., , [½], , [½], , Volume of hemisphere , , [1], , , , 32. The following figure shows the required cylinder, and the conical cavity, , 2 3, r, 3, , [½], , 2 22, 3, ,   4.2 , 3 7, , = 155.232 cm3, , [½], , Volume of the remaining cylinder after scooping, out hemisphere from each end, , 4.2 cm, , Volume of original cylinder – 2 × Volume of, hemisphere, 2.8 cm, , = 554.4 – 2 × 155.232, , [½], , = 243.936 cm3, The remaining cylinder is melted and converted, to a new cylindrical wire of 1.4 cm thickness., Given Height (b) of the conical Part = Height (h), of the cylindrical part = 2.8 cm, , So they have same volume and radius of new, cylindrical wire is 0.7 cm., , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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82, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), 2, , Volume of the remaining cylinder = Volume of, the new cylindrical wire, 243.936 = r2h, , 22  20  x, 2,      3.5, 100, 2, , [½], , 22, 243.936 ,  0.7 2 h, 7, h = 158.4 cm,  The length of the new cylindrical wire of, 1.4 cm thickness is 158.4 cm, [½], , [1½], , x = 2.5 cm, , [½], ,  Rainfall is of 2.5 cm, , [½], , 36. Let r1 = 5 cm and r2 15 cm are radii of lower, and upper circular faces., D, C , 15 cm , , E, , 24 cm, , 24 cm, , 34. Height of conical upper part = 3.5 m, and radius, = 2.8 m, (Slant height of cone)2 = 2.12 + 2.82, A, , = 4.41 + 7.84, Slant height of cone  12.25  3.5 m, ,    r1  r2    r12, , Curved surface area of cylindrical base + curved, surface area of conical upper part, [½], , Slant height of frustum = l =, , = 2rh + rl, , 22,  2.8  10.5, 7, = 92.4 m2, , (i), ,  r2  r1 2  h2, , [½], , l  102  242, , [½], ,  100  576, , , , l  676, , [½], , l = 26 cm, , [½], , So, the canvas used for one tent is 92.4 m2, , Metal required = (5 + 15) 26 + (5)2, , Thus, the canvas used for 1500 tents, , =  × 20 × 26 +  × 25, , = (92.4 × 1500) m2, , [½], , 15  5 2  242, , l, , = r(2h + l), 22,  2.8  7  3.5 , 7, , B, , Metal sheet required = Area of curved surface +, Area of Base, , [½], , The canvas used for each tent, , , , 5 cm, , [½], , = 5(4 × 26 + 5), , [½], , = 5 (109), , Canvas used to make the tents cost ` 120 per, sq. m, So, canvas used to make 1500 tents will cost, ` 92.4 × 1500 × 120, [½], , 22,  109, 7, = 1712.85 cm2, , The amount shared by each school to set up, the tents, , There is a chance of breakdown due to stress, on ordinary plastic., [1], , , , 92.4  1500  120, ` 332640, 50, , [½], ,  5, , 37., , 8 cm, , 20, , The amount shared by each school to set up, the tents is ` 332640., , [1], , 20, , h, , The value to help others in times of troubles is, generated from the problem., [½], 35. Water from the roof drains into cylindrical tank, 12cm, , Volume of water from roof flows into the tank of, the rainfall is x cm and given the tank is full we, can write,, [½], Volume of water collected on roof = volume of, the tank, [1], , Let the height of the bucket be h cm and slant, height be l cm., Here r1 = 20 cm, r2 = 12 cm, , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , And capacity of bucket = 12308.8 cm, , 3, , Slant height of the frustum  I , , We know that capacity of bucket, , , , h 2, (r1  r22  r1r2 ), 3, ,  3.14 , , Mathematics, , [½], , h,  400  144  240, 3, ,  20  8 2  162, , I, , I  122  162, I  144  256, I  400, , h, So we have  3.14  784  12308.8, 3, , I = 20 cm, [½], , [½], , Slant height is 20 cm, Now,, , 12308.8  3, h, 3.14  784, , Surface area = [r2 + (R + r) × l], , = 15 cm, , [½], , =, , 22/7[82, , , , 22, 64  560, 7, , , , 22,  624, 7, , , , 13728, 7, , Now, the slant height of the frustum,, , l  h 2  (r1  r2 )2,  15  8, ,  R  r 2  h 2, [½], , h,  3.14   784, 3, , 2, , 83, , [½], , 2, ,  289, , [½], , = 17 cm, , [½], , [½], , = 1961.14 cm2, , Area of metal sheet used in making it, = r22 + (r1 + r2), , + (20 + 8) × 20], , [1], , [½], , 39. Apparent capacity of the glass = Volume of, cylinder, [½], , = 3.14 × [144 + (20 + 12) × 17], = 2160.32 cm2, , [½], , 5 cm, , 38. Radius of the bigger end of the frustum (bucket), of cone = R = 20 cm, [½], Radius of the smaller end of the frustum (bucket), of the cone = r = 8 cm, [½], Height = 16 cm, , 10 cm, , [½], , Volume = 1/3rh, , [R2 + r2 + R × r], , = 1/3 × 22/7 × 16, , [202 + 82+ 20 × 8], , = 352/21, , [400 + 64 + 160] [½], , [½], , = (352 × 624)/21, , Actual capacity of the glass = Volume of, cylinder – Volume of hemisphere, [½], , = 219648/21, , Volume of the cylindrical glass = r2 h, , = 10459.43 cu. cm, , [½], , [½], , = 3.14 × (2.5)2 × 10, = 3.14 × 2.5 × 2.5 × 10, , Now,, , = 3.14 × 6.25 × 10, , R, , = 196.25, , [½], , cm3, , Volume of hemisphere , h, , , r, , 2 3, r, 3, , [½], , 3, 2,   2.5 , 3, , = 32.7 cm3, , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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84, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Apparent capacity of the glass = Volume of, cylinder = 196.25 cm3, , Let the radius of lower end of the frustum be, r = 8 cm, [½], , Actual capacity of the glass, , Let the radius of upper end of the frustum be, R = 20 cm, [½], , = Total volume of cylinder – volume of, hemisphere, [1], = 196.25 – 32.7, = 163.54, , cm3, , Hence, apparent capacity = 196.25, , cm3, , Let the height of the frustum be h cm, , [½], , Volume of the frustum, , [½], , , 3 73216, h R 2  r 2  Rr  10459 , [1], 3, 7, 7, Therefore, substituting the value of R and r., , , , [½], , Actual capacity of the glass = 163.54 cm3 [½], 40., , , , , , 22 1, 73216,  h 202  82  20  8 , 7 3, 7, , 7 cm, , h  400  64  160  , , 10.5 cm, , 6 cm, , , , 73216 7, , 3, 7, 22, , h × 624 = 9984, 9984,  16 cm, 624, Total surface area of the container, h, , 10 cm, ,   R  r , , Given, internal diameter of the cylinder = 10 cm, Internal radius of the cylinder = 5 cm, , [½], , and height of the cylinder = 10.5 cm, Similarly, diameter of the cone = 7 cm, , [½], , Radius of the cone = 3.5 cm and Height of the, cone = 6 cm, (i) Volume of water displaced out of cylindrical, vessel = volume of cone, [1], 1,  r 2 h, 3, , [½], , 22,  5  5  10.5  77, 7, = 825 – 77 = 748 cm3, 41., , , , 22,  20  8 , 7, , , , 22, 22,  28 122  162 ,  64, 7, 7, , , , 22, 22,  28 144  256 ,  64, 7, 7, , , , 22, 22, 28  400  64 ,  28  20  64 , 7, 7, , , ,  20  8 2  162 , , 22 2, 8, 7, , [½], , , , 22, 22,  560  64    624, [½], 7, 7, Cost of 1 cm square metal sheet is 1. 40 `, , Cost of required sheet =, 22,  624  1.40  2745.60 `, 7, , [½], , , , [1], , , , 1 22,  ,  3.5  3.5  6  77 cm3, [1], 3 7, (ii) Volume of water left In the cylindrical vessel, = volume of cylinder – volume of cone [1], , = R2H – Volume of cone, ,  R  r 2  h2  r 2, , [1], , [1], , 42., h, , [1], , 20 cm, , 16 cm, , ?, , Radius of base of the cone = r = 21 cm, , [½], , Let the height of the cone be h cm, 8 cm, , Volume of the cone = 2/3 volume of the, hemisphere, , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 1 2, 2 2, r h   r 3, 3, 3 3, , Mathematics, , Here, R = 28 cm and r = 21 cm,, Volume of frustum = 28.49 L, , r r 2  h 2  2r 2, , = 28.49 × 1000 cm3, = 28490 cm3, Now, volume of frustum , , [1], , 22, 22,  21 212  282  2 ,  21 21, 7, 7, , [1], ,  66  441  784  2772, = 66 × 35 + 2772, , = 2310 + 2772 = 5082 cm2, , [1], , , , = 15000 m/hr, , , , 22 h, 282  28  21  212  28490, 73, , , , 22, h  1813  28490, 21, ,  h, , Speed of water = 15 km/hr, , , , h 2, R  Rr  r 2, 3, , , , 43. Let the level of water in the pond rises by 21 cm, in t hours., , Diameter of pipe = 14 cm , , [1], , [½], , 4, 4,  h  r   21  28 cm, [½], 3, 3, Surface area of the toy = lateral surface area of, cone + curved surface area of hemisphere [1], , , , 85, , [½], , , , [1½], , [1], , [½], , 28490  21,  15 cm, 22  1813, , Hence the height of bucket is 15 cm., , 14, m, 100, , [1], , [1], , O, , 45., , 7, [½], m, 100, Volume of water flowing out of the pipe in 1, hour = r2h, [½], ,  Radius of the pipe, r , , 2, , 22  7, , , m  15000 m, 7  100 , = 231 m3, , , A, , 7 cm, , B, , [1], ,  Volume of water flowing out of the pipe in t, [½], hours = 231t m3, , Radius of hemi-sphere = 7 cm, , [½], , Volume of water in the cuboidal pond, , Radius of cone = 7 cm, , [½], , Height of cone = diameter = 14 cm, , [½], ,  50 m  44m , , = 462 m3, , 21, m  Volume of cuboid = lbh , 100, [1], , Volume of water flowing out of the pipe in t hours, = Volume of water in the cuboidal pond, [1],  231t = 462, 462,  2 hrs, 231, Thus, the water in the pond rise by 21 cm in 2, hours., [1], ,  t, , 44., , R, , h, , r, , Volume of solid = Volume of cone + Volume of, hemi-sphere, [1], , , 1 2, 2, r h  r 3, 3, 3, , [1], , , , 1 2, r  h  2r , 3, , [½], , , , 1 22, ,  49 14  14 , 3 7, , , , 1 22, ,  49  28, 3 7, , [1], , , , 22  7  28 4312, , cm3, 3, 3, , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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86, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Chapter - 14 : Statistics, 1., , Class, ,  f1  f0, ,  Mode  I  , h,  2 f1  f0  f 2 , , , , Class marks, 10 + 25, , 10 – 25, , 2, 35 + 55, , 35 – 55, , 2, , = 17.5, , [½], , 16  10, , ,  30  ,   10,  2  16  10  12 , , [½], , = 45, ,  30 , , 4., Class, , 2., , Frequency Cumulative frequency, , 10 – 15, , 49, 133, , 15 – 20, 20 – 25, , 63, 15, , 25 – 30, 30 – 35, , 6, 7, , 35 – 40, , 4, , 40 – 45, , 2, , 277, 279, , 45 – 50, , 1, , 280, , 5 – 10, , [1], , 49, , 3, , –6, , –3, , –9, , 13 – 15, , 14, , 6, , –4, , –2, , –12, , 260, 266, , 15 – 17, , 16, , 9, , –2, , –1, , –9, , 17 – 19, , 18, , 13, , 0, , 0, , 0, , 19 – 21, , 20, , f, , 2, , 1, , f, , 21 – 23, , 22, , 5, , 4, , 2, , 10, , 23 – 25, , 24, , 4, , 6, , 3, , 12, , 273, , N 280, ,  140, 2, 2, , h = 2; A = 18, N = 40 + f,fiui = f – 8, X  18, [½], , [½], , N, is, 2, 182 and the corresponding class is 10 – 15., , 1, ,  Mean  A  h   fi ui , N, , , , Thus, 10 – 15 is the median class such that, l = 10, f = 133, F = 49 and h = 5, , [½], , 2 f  8, 40  f, , Here, 30 – 40 is the modal class, and I = 30,, h = 10, [½], , [½], , 0, , [½], , f–8=0, , [1], [½], , [1], ,  1, 18  18  2 ,  f  8 ,  40  f, , , N, ,  2 F ,  140  49 , Median  l  ,   h  10  , 5, f,  133 , , , , , , 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50, 50 - 60, 60 - 70, , [1], , fi = 40 + f, , We have, , Frequency, 8, 10, 10  f0, 16  f1, 12  f2, 6, 7, , 2, , fi ui, , 12, ,  We have N = 280, , Class, , x i – 18, , 11 – 13, , The cumulative frequency just greater than, , 3., , ui =, , [½], , 182, 245, , fiui = f – 8, , = 13.42, , [½], , 6,  10 = 30 + 6 = 36, 10, , Mid values Frequency di = xi, xi, fi, –18, , Class, , Let N = total frequency, , , , [1], , f=8, 5., , Daily, income, , [½], Frequency Income Cumulative, less than frequency, , 100 – 120, , 12, , 120, , 12, , 120 – 140, , 14, , 140, , 26, , 140 – 160, , 8, , 160, , 34, , 160 – 180, , 6, , 180, , 40, , 180 – 200, , 10, , 200, , 50, , [1], , Using these values we plot the points (120, 12), (140, 26) (160, 34), (180, 40) (200, 50) on the, axes to get less than ogive, [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 7., , y, Cumulative frequency, , Mathematics, , 50, , Marks Number of, students, , (180, 40), (160, 34), , 30, 20, , 2, , Less than 5, , 2, , 5-10, , 5, , Less than 10, , 7, , 10-15, , 6, , Less than 15, , 13, , 15-20, , 8, , Less than 20, , 21, , 20-25, , 10, , Less than 25, , 31, , 25-30, , 25, , Less than 30, , 56, , 30-35, , 20, , Less than 35, , 76, , 35-40, , 18, , Less than 40, , 94, , 40-45, , 4, , Less than 45, , 98, , 45-50, , 2, , Less than 50, , 100, , (140, 26), , [2], , 10, , (120, 12), , O, , 120, , 140 160 180, Number of workers, , 6., , 200, , Class, , Frequency, , Cumulative, Frequency, , 0 – 10, , f1, , f1, , 10 – 20, , 5, , 20 – 30, 30 – 40, , 9, , 5 + f1, 14 + f1, , 40 – 50, 50 – 60, 60 –70, , x, , [1], , 26 + f1 + f 2, , 3, , 29 + f1 + f 2, , 2, , 31 + f1 + f 2, Y, , Total = 40 = n, , (45, 98), , 100, , 80, , f1 + f2 = 40 – 31 = 9, , ...(i), , 70, , Median = 32.5, , [Given], ,  Median Class is 30 – 40,  = 30, h = 10, cf = 14 + f1, f = 12, , [½], , Y-axis 1 cm = 10 units, , (30, 56), , 50, 40, (25, 31), (20, 21), , 20, , (10, 7), , (15, 13), (Median = 28.8), , (5, 2), , [½], , 3 = 6 – f1, [½], , On putting in (i),, f1 + f2 = 9, , =6, , 60, , 0, , 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75, , [∵ f1 = 3], [½], , X, , Marks, , Locate, , f1 = 3, , X-axis 1 cm = 10 units, , (35, 76), , 10, , 10, (20  14  f1 ), 12, , f2 = 9 – 3, , Scale, , 30, , , , h, , ,  40  (14  f ) , 1 , , 32.5 = 30   2,   10, , 12, , 2.5 =, , [1], , Cumulative Freq uency, , f1 + 5 + 9 + 12 + f2 + 3 + 2 = 40, , [1], , (50, 100), , (40, 94), , 90, , n,   cf, Median =    2,  f, , [2], , Let us now plot the points corresponding to the, ordered pairs (5, 2), (10, 7), (15, 13), (20, 21),, (25, 31), (30, 56), (35, 76), (40, 94), (45, 98),, (50, 100). Join all the points by a smooth curve., , 26 + f1, , 12, f2, , Cumulative, frequency, , 0-5, , (200, 50), , 40, , Marks less, than, , 87, , n 100, ,  50 on Y-axis, 2, 2, , From this point draw a line parallel to X-axis, cutting the curve at a point. From this point,, draw a perpendicular to X-axis. The point of, intersection of perpendicular with the X-axis, determines the median of the data., Therefore median = 28.8, , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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8., , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Class, , Frequency, , Class, mark (xi), , xifi, , 0 – 20, , 6, , 10, , 60, , 20 – 40, , 8, , 30, , 240, , 40 – 60, , 10, , 50, , 500, , 60 – 80, , 12, , 70, , 840, , 80 – 100, , 6, , 90, , 540, , 100 – 120, , 5, , 110, , 550, , 120 – 140, , 3, , 130, , fi = 50, , Mean, , [1], , 12  10, , ,  60  ,   20, 2, , 12, , 10, , 6, , , = 65, 9., , Weight, , Cumulative, (More than type), , 390, , More than 38, , 35, , fi xi = 3120, , More than 40, , 32, , More than 42, , 30, , More than 44, , 26, , More than 46, , 21, , More than 48, , 7, , More than 50, , 3, , More than 52, , 0, ,  xi fi, ,  fi, , ,  f1  f0 , Mode  I  , h,  2f1  f0  f2 , , 3120, 50, , = 62.4, , [1], , Class, , f, , Less than, cumulative, frequency, , 0 – 20, , 6, , 6, , 20 – 40, , 8, , 14, , 40 – 60, , 10, , 24, , 60 – 80, , 12, , 36, , 80 – 100, , 6, , 100 – 120, 120 – 140, , [1], [2], , Weight (in kg), Upper class limits, , Number of students, (Cumulative frequency), , Less than 38, , 0, , Less than 40, , 3, , Less than 42, , 5, , 42, , Less than 44, , 9, , 5, , 47, , Less than 46, , 14, , 3, , 50, , Less than 48, , 28, , Less than 50, , 32, , More than 52, , 35, ,  n = fi = 50, , n,  25, 2,  Median class = 60 – 80, , n,   c.f, Median  I   2,  f, , , [1], , [2], Taking upper class limits on x-axis and their, respective cumulative frequency on y-axis its, ogive give can be drawn as follows:, y, , , , h, , , , 35, 30, ,  25  24 , Median  60  ,   20,  12 , Median = 61.66, , [1], , Less than, , 25, 20, 15, 10, 5, , Mode :, Maximum class frequency = 12,  Model class = 60 – 80, , Cumulative, frequency (cf), , 88, , 0, , [1], , More than, x, 38 40 42 44 46, Upper class limits, , 48 50 52, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Mathematics, , Here, n = 35, So,, n,  17.5, 2, There is a intersection point of less than and, more than ogive mark that point A whose, ordinate is 17.5 and its x-coordinate is 46.5., Therefore, median of this data is 46.5., [2], 10., , Class, , fi, , Class, mark(xi), , Fixi, , 0 – 10, , 4, , 5, , 20, , 10 – 20, , 4, , 15, , 60, , 20 – 30, , 7, , 25, , 175, , 30 – 40, , 10, , 35, , 350, , 40 – 50, , 12, , 45, , 540, , 50 – 60, , 8, , 55, , 440, , 60 – 70, , 5, , 65, , 325, , f i = 50, , N,  25, 2, Cumulative frequency just greater than 25 is 37.,  Median class 40–50, , N,   C.f, Median     2, f, , , , [1], , N = 50, Cf = 25, f = 12, h = 10, ,  25  25 , Median  40  ,  10  40  0,  12 , , Median  40, [1], , Class, , Frequency, , Cumulative, frequency, , 0 – 10, , 4, , 4, , 10 – 20, , 4, , 8, , 20 – 30, , 7, , 15, , 30 – 40, , 10, , 25, , 40 – 50, , 12, , 37, , 50 – 60, , 8, , 45, , 60 – 70, , 5, , 50, , , , h, , , , Here  = 40, , f ixi = 1910, , 1910, mean ,  38.2, 50, , 89, , [1], , Mode :, Maximum frequency = 12 so modal class 40 – 50, ,  f1  f0 , mode    , ,  2f1  f0  f2 , Here  = 40, h = 10, f0 = 10 f1 = 12 f2 = 8, , 12  10, , , Mode = 40  ,   10, 2, 12, 10, 8, , , , , , Mode = 40 + 3.33, , N = 50, , = 43.33, , [1], , [2], , Chapter - 15 : Probability, 1., , 2., , Total possible outcomes = 6, Outcomes which are less than 3 = 1, 2, , [½], , 1, 3, , Total possible outcomes = {HH, HT, TH, TT}, Number of total outcomes = 4, , 2, Probability , 6, , , Two coins are tossed simultaneously, , Favourable outcomes for getting exactly, One head = {HT, TH}, [½], , Probability , , 2 1, , 4 2, , [½], [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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90, 3., , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , A card is drawn from well shuffled 52 playing, cards so total no of possible outcomes = 52, , Let event E be defined as ‘getting a prime, number less than 23’., [½], , Number of face cards = 12, , n(E) = 8, , Number of Red face cards = 6, Probability of drawing , , , , 6, 52, , 3, A red face card , 26, , 4., , [½], , , , [½], 8., , Answer (C), , [½], , Answer (D), , Probability of drawing an ace card, , {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),, , Number of ace, 4, , Total cards, 52, ,  1, , (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),, [½], , (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),, , 4 12, , 52 13, , (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),, (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),, , [½], , (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}, , Answer (C), , Favourable outcomes are given below:, {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2),, (6, 4), (6, 6)}, , Favourable outcomes = {(1, 1), (2, 2), (3, 3),, (4, 4), (5, 5), (6, 6)}, [½], , Total number of favourable outcomes = 9, ,  Required probability, Number of favourable outcomes 6 1, , , , Total number of outcomes, 36 6, , [½], , •  Probability of getting an even number on both, dice, , Answer (A), S = {1, 2, 3, 4, 5, 6}, Let event E be defined as ‘getting an even, number’., n(E) = {2, 4, 6}, , , P E  , , , , [½], , Total number of outcomes = 36, , When two dice are thrown together, the total, number of outcomes is 36., , 7., , 8, 4, , 90 45, , Number of aces in deck of cards = 4, , Probability that the card is not an Ace, , 6., , Number of favourable outcomes, Number of possible outcomes, , Possible outcomes on rolling the two dice are, given below :, , , , 5., , P E  , , [½], , 9., , , , Total number of favourable outcomes, Total number of outcomes, , , , 9, 1, , 36 4, , [½], , Answer (C), Total number of possible outcomes = 30, , Number of favourable outcomes 3, , Number of possible outcomes, 6, , Prime numbers from 1 to 30 are 2, 3, 5, 7, 11,, 13, 17, 19, 23 and 29., , 1, 2, , Total number of favourable outcomes = 10, , [½], , Answer (C), S = {1, 2, 3,..90}, n(S) = 90, The prime number less than 23 are 2, 3, 5, 7,, 11, 13, 17, and 19., , [½], ,  Probability of selecting a prime number from, 1 to 30, , , Total number of favourable outcomes, Total number of outcomes, , , , 10 1, , 30 3, , [½], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , Mathematics, , 14. The total number of outcomes is 50., , 10. Two dice are tossed, S = [(1, 1), (1, 2), (1, 3),(1, 4), (1, 5), (1, 6),, , Favourable outcomes = {12, 24, 36, 48}, , (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),, ,  Required probability, Number of, favourable outcomes, 4, 2, , , , Total number, 50 25, of outcomes, , (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),, (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),, (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),, (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)], , [½], , Total number of outcomes when two dice are, tossed = 6 × 6 = 36, , [1], , Total number of possible outcomes = 52, , Favourable events of getting product as 6 are:, , Total number of kings and queens = 4 + 4 = 8, , (1 × 6 = 6), (6 × 1 = 6), (2 × 3 = 6), (3 × 2 = 6), , Therefore, there are 52 – 8 = 44 cards that are, neither king nor queen., [1], Total number of favourable outcomes = 44, , Favourable events of getting product as 6 = 4, 4, 1, ,  P(getting product as 6) , 36 9, ,  Required probability = P(E), , [½], , , , 11. There are 26 red cards including 2 red queens., Two more queens along with 26 red cards will be, 26 + 2 = 28,  P(getting a red card or a queen) , , 28, 52, , Favourable outcomes, 44 11, , , [1], Total number of outcomes 52 13, , 16. Rahim tosses two coins simultaneously. The, sample space of the experiment is {HH, HT, TH,, and TT}., Total number of outcomes = 4, , [½], ,  P(getting neither a red card nor a queen), , Outcomes in favour of getting at least one tail on, tossing the two coins = {HT, TH, TT}, [1], , 28 24, 6, , , 52 52 13, , Number of outcomes in favour of getting at least, one tail = 3, ,  1, , [½], , 12. Probability of selecting rotten apple, , , , , Number of rotten apples, Total number of apple, , 0.18 , , •, ,  Probability of getting at least one tail on, tossing the two coins, , [½], , , Number of rotten apples, 900, , Number of favourable outcomes 3, , Total number of outcomes, 4, , 13. A ticket is drawn at random from 40 tickets, , n(s) = 36, (i) A = getting a doublet, , Total outcomes = 40, , A = {(1, 1), (2, 2) ......., (6, 6)}, , Out of the tickets numbered from 1 to 40 the, number of tickets which is multiple of 5 = 5, 10,, 15, 20, 25, 30, 35, 40, , n(A) = 6, , , P ( A) , , = 8 tickets,  Favorable outcomes = 8,  Probability , , , [1], , n( A) 6, 1, , , n(S ) 36 6, , [1], , (ii) B = getting sum of numbers as 10, B = {(6, 4), (4, 6), (5, 5)}, , 8, 40, 1, 5, , [1], , 17. Sample space = S = {(1, 1) (1, 2)...,(6, 6)}, , Number of rotten apples = 900 × 0.18 = 162 [½], , •, , [1], , 15. Let E be the event that the drawn card is neither, a king nor a queen., , i.e. (1, 6), (6, 1), (2, 3), (3, 2), , •, , 91, , n(B) = 3, [1], , P (B ) , , 1, n(B ) 3, , , n(S ) 36 12, , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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92, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 18. An integer is chosen at random from 1 to 100, P (5 will come up on either side), , Therefore n(S) = 100, (i) Let A be the event that number chosen is, divisible by 8, , P (5 will not come up)  1 , ,  A = {8, 16, 24, 32, 40, 48, 56, 64, 72, 80,, 88, 96},  n(A) = 12, , , , 11, 36, , 11, 36, , 25, 36, , (ii) P (5 will come at least once) , , Now, P (that number is divisible by 8), , 22. Total number of cards , , 12, 6, 3, , , , 100 50 25, , P ( A) , , [1], , 3, 25, ,  P(A’) = 1 – P(A), 3, 25, , 35  1, 1, 2, , = 18, , P(prime number less than 15) , , P ( A) , , 22, 25, , [1], , [1], , Product of outcomes will be 12 for, (2, 6), (6, 2), (3, 4) and (4, 3)., , So, the number of favourable outcomes = 2, , Probability , , 2 1, , 8 4, , [1], , 4, 1, , 36 9, , [1], , 24. A disc drawn from a box containing 80, , [1], , Total possible outcomes = 80, , Hence, probability of losing the game = 1 – P(E), 1 3, , 4 4, , [1], , Number of cases where the disc will be, numbered perfect square = 8, Perfect squares less than 80, , 20. Total outcomes = 1, 2, 3, 4, 5, 6, , 8, 1, , 80 10, 25. Total number of outcomes = 52, , Probability , , Numbers lie between 2 and 6 = 3, 4, 5, 3 1, , 6 2, , (ii) P (Numbers lie between 2 and 6) , , [1], , = 1, 4, 9, 16, 25, 36, 49, 64, , Prime numbers = 2, 3, 5, , (i) P (Prime Numbers) , , 5, 18, , 1, [1], 18, 23. Two dice are rolled once. So, total possible, outcomes = 6 × 6 = 36, [1], , Number of favourable cases = 4, ,  1–, , [1], , (ii) Favourable outcomes = {15}, , The favourable outcomes to the event E 'Same, result in all the tosses' are TTT, HHH., [1], , P (E ) , , [1], , P(a number divisible by 3 and 5) , , 19. Total possible outcomes are (HHH), (HHT),, (HTH), (THH), (TTH), (THT), (HTT), (TTT) i.e., 8., , , , 1, 36, , [1], , (i) Favourable outcomes = {3, 5, 7, 11, 13}, , (ii) Let ‘A’ be the event that number is not, divisible by 8., ,  1, , 11, 36, , (iii) P (5 will come up on both dice) , , n( A ),  P ( A) , n(S ), , [1], , [1], 3 1, [1], , 6 2, , 21. Total outcomes = 6 × 6 = 36, (i) Total outcomes when 5 comes up on either, dice are (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5,, 6) (6, 5) (4, 5) (3, 5) (2, 5) (1, 5), , [1], , (i) Probability of getting a red king, Here the number of favourable outcomes = 2, Probability , , , , Number of, favourable outcomes, 2, , Total number, 52, of outcomes, 1, 26, , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 12, 3, , 52 13, , 93, , 27. When three coins are tossed together, the, possible outcomes are, , (ii) Favourable outcomes = 12, Probability , , Mathematics, , [1], , HHH, HTH, HHT, THH, THT, TTH, HTT, TTT,  Total number of possible outcomes = 8, , (iii) Probability of queen of diamond., Number of queens of diamond = 1, hence, , (i) Favourable outcomes of exactly two heads, are HTH, HHT, THH, , Probability, ,  Total number of favourable outcomes = 3, , Number of favourable outcomes 1, , , Total number of outcomes, 52, , [1], , 26. Here the jar contains red, blue and orange balls., ,  P(exactly two heads) , , 3, 8, , [1], , Let the number of red balls be x., , (ii) Favourable outcomes of at least two heads, are HHH, HTH, HHT, THH, , Let the number of blue balls be y., ,  Total number of favourable outcomes = 4, , Number of orange balls = 10, ,  P(at least two heads) , , Total number of balls = x + y + 10, Now, let P be the probability of drawing a ball, from the jar, P(a red ball) , , , x, x  y  10, ,  Total number of favourable outcomes = 4, , Probability of drawing a black ball, (i), , [1], , P (B ) , , Next,, , x, 15  x, , P (W ) , , 1, y, , 3 x  y  10, , 15, 15  x, , Given that P(B) = 3P(W), ,  3y = x + y + 10, , , (ii), , [1], , Probability of drawing a white ball, , y, P(a blue ball) , x  y  10, , [1], , Multiplying eq. (i) by 2 and adding to eq. (ii), we, get, , 6 x  2y,  x  2y, 5x, , [1], , Let say there be x black balls., ,  3x – y = 10, ,  2y – x = 10, , 4 1, , 8 2, , 28. Bag contains 15 white balls., ,  4x = x + y + 10, , , , [1], , (iii) Favourable outcomes of at least two tails, are THT, TTH, HTT, TTT, ,  P(at least two tails) , , 1, x, , 4 x  y  10, , 4 1, , 8 2, ,  20,  10,  30, , [1], , x, 3  15, , 15  x 15  x, , x = 45, , [1], , Number of black balls = 45, 29. The group consists of 12 persons.,  Total number of possible outcomes = 12, Let A denote event of selecting persons who are, extremely patient., ,  x=6, ,  Number of outcomes favourable to A is 3.[1], , Substitute x = 6 in eq. (i), we get y = 8, , Let B denote event of selecting persons who are, extremely kind or honest. Number of persons, who are extremely honest is 6. Number of, persons who are extremely kind is 12 – (6 + 3), =3, [1], , Total number of balls = x + y + 10 = 6 + 8 +, 10 = 24, Hence, total number of balls in the jar is 24. [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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94, , Mathematics, , Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), ,  Number of outcomes favourable to, B = 6 + 3 = 9., (i), , Number of outcomes favourable to A, Total number of possible outcomes, , P  A , , , , (ii) P  B  , , , , 3, 1, , 12 4, , [1], , Each of the three values, patience, honesty and, kindness is important in one‘s life., 30. Total number of cards = 49, (i) Total number of outcomes = 49, The odd numbers from 1 to 49 are 1, 3, 5, 7, 9,, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33,, 35, 37, 39, 41, 43, 45, 47 and 49., Total number of favourable outcomes = 25, , Total number of, favourable outcomes 25, , Total number, 49, of outcomes, , Total number of favourable outcomes = 1, Total number of, favourable outcomes, 1, , , Total number, 49, of outcomes, , [1], , 31. Let S be the sample space of drawing a card, from a well-shuffled deck., n(S) = 52, (i) There are 13 spade cards and 4 ace's in a, deck. As ace of spade is included in 13, spade cards, so there are 13 spade cards, and 3 ace's., A card of spade or an ace can be drawn in, = 16 ways, Probability of drawing a card of spade or an, 16, 4, , [1], 52 13, (ii) There are 2 black king cards in a deck a, card of black king can be drawn in = 2 ways, , ace , ,  Required probability, , , We know that there is only one even prime, number which is 2.,  Required probability, , [1], , Number of outcomes favourable to B, Total number of possible outcomes, , 9, 3, , 12 4, , (iv) Total number of outcomes = 49, , [1], Probability of drawing a black king , , 2, 1, , 52 26, , (ii) Total number of outcomes = 49, The number 5, 10, 15, 20, 25, 30, 35, 40, and 45 are multiples of 5., The number of favourable outcomes = 9, , [1], , Can be drawn in = 44 ways, Probability of drawing a card which is neither a, 44 11, , [1], 52 13, (iv) There are 4 King and 4 Queen cards in a, deck., , Jack nor a King , , (iii) Total number of outcomes = 49, The number 1, 4, 9, 16, 25, 36 and 49 are, perfect squares., Total number of favourable outcomes = 7, , So there are 4 + 4 = 8 cards which are either, King or Queen., ,  Required probability, , A card which is either a King or a Queen can, be drawn in = 8 ways, , Total number of favourable outcomes, , Total number of outcomes, 7, 1, , , 49 7, , (iii) There are 4 Jack and 4 King cards in a, deck., So there are 52 – 8 = 44 cards which are, neither Jacks nor Kings. A card which is neither, a Jack nor a King., ,  Required probability, Total number of, favourable outcomes, 9, , , Total number, 49, of outcomes, , [1], , So, probability of drawing a card which is either, [1], , a King or a Queen , , 8, 2, , 52 13, , [1], , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

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Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10), , 32. x is selected from 1, 2, 3 and 4, , Both outcomes odd = 9 cases, Total favourable cases = 9 + 9 =18, , y is selected from 1, 4, 9 and 16, Let A = {1, 4, 9, 16, 2, 8, 18, 32, 3, 12, 27, 48,, 36, 64} which consists of elements that are, product of x and y., [2], P(product of x and y is less than 16), , Probability that , , 18, 36, , Sum of the even outcomes is, , 1, ., 2, , [1], , (ii) Product of outcomes is even, [1], , This is possible when,  Both outcomes are even, , 7, , 14, 1, , 2, , 95, , Similarly,, , 1, 2, 3, 4, , Number of outcomes less than 16, , Total number of outcomes, , Mathematics, ,  First outcome even & the other odd,  First outcome odd & the other even, [1], , Number of cases where both outcomes are, even = 9, [1], , 33. Two dice are thrown together total possible, outcomes = 6 × 6 = 36, , Number of cases for first outcome odd and, the other even = 9, , (i) Sum of outcomes is even, , Number of cases for first outcome even and, the other odd = 9, , This can be possible when, , Total favourable cases = 9 + 9 + 9 = 27, ,  Both outcomes are even, , Probability , ,  Both outcomes are odd, For both outcomes to be even number of, cases = 3 × 3 = 9, [1], , , , 27, 36, , 3, 4, , [1], , ‰ ‰ ‰, , Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456