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DIRECTORATE OF MINORITIES, , MINORITIES WELFARE DEPARTMENT, , MATHEMATICS, S.S.L.C Super Notes: - 2020 – 21, -: MENTOR :-, , MAHIBOOB SAB KARATAGI, DIRECTOR, DIRECTORATE OF MINORITIES, GOVT OF KARNATAKA, , -: CO-ORDINATOR :-, , KANTHARAJU K M.A, M.ED, M.PHIL, NODAL OFFICER,, STATE LEVEL QUESTION PAPERS DESIGN COMMITTEE, MINORITY WELFARE DEPARTMENT, PRINCIPAL, MMDRS, SIRA, DIRECTORATE OF MINORITIES, , 1, , Use E-Papers, Save Trees, Above line hide when print out
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CHITTI CREATIONS, , Index, Sl, No., 1, , Marks to, be, expected, 1 mark, , 2, 3, 2 marks, , 4, 5, 6, 7, 8, 9, 10, 11, 12, , 3 marks, , 4 marks, 13, 14, , 4 or 5, marks, , Name of the chapter/content, , Page, no., , Surface area and volumes- Learn All, formulae, Arithmetic progression – finding terms, Arithmetic progression –, sum of nth terms, Coordinate geometry- Problems on, distance formula, Quadratic equations- solving quadratic, equation by formula method, Pair of linear equations in two, variables- solve x & y, Constructions – dividing the line, segment, Constructions – Tangent construction, Statistics – Mean, Median & Mode, Statistics – Ogive graph, Circle- theorem, Pair of linear equations in two, variables- graphical solution(x & y), Constructions – construction of similar, triangles, Triangles – theorem, , 2-3, , Marks, to be, allotted, 1, , 4-5, 6-7, , 2, 2, , 8-9, , 2, , 10-11, , 2, , 12-13, , 2, , 14-15, , 2, , 16-17, 18-21, 22-25, 26, 27-30, , 3, 3, 3, 3, 4, , 31-33, , 4, , 34-37, , 5, , Total, 38+2, **** practice all formulae from every chapter, you will get 2 more, marks at least., , T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 1
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CHITTI CREATIONS, , 2. Arithmetic Progression: nth terms of A.P, 20th, , an= a+(n-1)d, , 1. Find the, term from the last term of the AP: 3, 8, 13, …, 253., Solution: We have, last term = 1 = 253, And, common difference d = 2nd term – 1st term = 8 – 3 = 5, Therefore, 20th term from end = 1 -(20 – 1) × d = 253 – 19 × 5 = 253 – 95, = 158., 2. Find the number of natural numbers between 101 and 999 which are, divisible by both 2 and 5., Solution:, Natural numbers between 101 and 999 divisible by both 2 and 5 are, 110, 120, … 990., so, a1 = 110, d = 10, an = 990, We know, an = a1 + (n – 1)d, 990 = 110 + (n – 1) 10, (n – 1) = 990−11010, ⇒ n = 88 + 1 = 89., 3. Find how many integers between 200 and 500 are divisible by 8., Solution:, AP formed is 208, 216, 224, …, 496, Here, an = 496, a = 208, d = 8, an = a + (n – 1) d, ⇒ 208 + (n – 1) x 8 = 496, ⇒ 8 (n – 1) = 288, ⇒ n – 1 = 36, ⇒ n = 37., 4. How many terms of the AP 18, 16, 14, …. be taken so that their sum is zero?, Solution:, Here, a = 18, d = -2, sn = 0, Therefore, n2 [36 + (n – 1) (- 2)] = 0, ⇒ n(36 – 2n + 2) = 0, ⇒ n(38 – 2n) = 0, ⇒ n = 19., 5. Which term of the AP: 3, 8, 13, 18, … , is 78?, Solution:, Let an be the required term and we have given AP, 3, 8, 13, 18, ….., Here, a = 3, d = 8 – 3 = 5 and an = 78, T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 4
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CHITTI CREATIONS, , Now, an = a + (n – 1)d, ⇒ 78 = 3 + (n – 1) 5, ⇒ 78 – 3 = (n – 1) × 5, ⇒ 75 = (n – 1) × 5, ⇒ 755 = n – 1, ⇒ 15 = n – 1, ⇒ n = 15 + 1 = 16, Hence, 16th term of given AP is 78., Practice:, 6. Find the 9th term from the end (towards the first term) of the A.P. 5,, 9,13,185.(ans: 153)., 7. How many two-digit numbers are divisible by 3?. (ans: 30), 8. Find the middle term of the A.P. 6,13,20,...,216. (Ans;111), 9. Find the 25th term of an arithmetic progression 2, 6, 10, 14, . . . . . . .(ans:, 98), 10. Find the 10th term of arithmetic progression 2, 7, 12 ....... using the, formula.(ans: 47)., , T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 5
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CHITTI CREATIONS, , 3. Arithmetic Progression: Sum of nth terms., 𝑛, , 𝑛, , 2, , 2, , Sn= [2a + (n − 1)d] & Sn= [a + l], 1. Find the sum of the A.P: 1, 3, 5, …….. 199., Solution: a=1, d=2 and last term l=199, an=a+(n−1)d, ⇒199=1+(n−1)×2, ⇒2n=200, n=100, 𝒏, ∴sum= [a+l], 𝟐, 𝟏𝟎𝟎, , = [1+199], 𝟐, =10000, 2. Find the sum of the series 51+50+49+----------+21., Solution: a=51, d=-1 and last term l=21, an=a+(n−1)d, ⇒21=51+(n−1)×-1, 21=51+1-n, ⇒n=52-21, n=31, 𝒏, ∴sum= [a+l], 𝟐, 𝟑𝟏, , = [51+21] =, 𝟐, =1116, , 𝟑𝟏, 𝟐, , [72], , 3. How many terms of the AP 18, 16, 14, …. be taken so that their sum is, zero?, Solution:, Here, a = 18, d = -2, sn = 0, 𝑛, Therefore, [36 + (n – 1) (- 2)] = 0, 2, ⇒ n(36 – 2n + 2) = 0, ⇒ n(38 – 2n) = 0, ⇒ n = 19, , T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 6
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CHITTI CREATIONS, , 4. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term, is 149., Solution: Given,, Common difference, d = 7, 22nd term, a22 = 149, To find: Sum of first 22 term, S22, By the formula of nth term, we know;, an = a + (n − 1)d, a22 = a + (22 − 1)d, 149 = a + 21 × 7, 149 = a + 147, a = 2 = First term, Sum of nth term is given by the formula;, Sn = n/2 (a + an), = 22/2 (2 + 149), = 11 × 151, = 1661, 5. Find the sum of first 20 natural numbers which are divisible by 4., Solution: The A.P which are divisible by 4 is 4, 8, 12, ……, Here we have to find an. a=4, d=4, an=a+(n-1d), a20=4+19x4, a20=4+76, a20=80., 𝒏, ∴sum= [a+l], 𝟐, 𝟐𝟎, , = [4+80], 𝟐, =10x84, =840., Practice :, 6. Find the sum of first 50 natural numbers which are divisible by 5., 7. Find the sum of : 1+5+9+------------ up to 25 terms., 8. Find the sum of first 30 terms of the A,P 2, 6, 10, ………, 9. How many terms of the A.P. 27,24,21,… should be taken so that their, sum is zero?, 10. Find the sum of 2+5+8+................. to 20 terms using the formula., , T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 7
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CHITTI CREATIONS, , 4. Coordinate geometry:, , Problems on distance formula., , Distance formula = √(x2 − x1)2 + (y2 − y1)2 ., 1. Find the distance between the two points (2, 5) & (7, 6)., Solution: here x1=2, x2=7, y1=5 & y2=6. Put all the values in the given, formula., d= √(x2 − x1)2 + (y2 − y1)2 ., = √(7 − 2)2 + (6 − 5)2 ., = √(5)2 + (1)2 ., = √25 + 1., =√26 sq.units, 2. Prove that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an, isosceles right triangle., Solution:, Let A (7, 10), B(-2, 5), C(3, -4) be the vertices of a triangle., , … [By converse of Pythagoras theorem, ∆ABC is an isosceles right angled triangle. …(ii) From (i) & (ii), Points, A, B, C are the vertices of an isosceles right triangle., 3. Find that value(s) of x for which the distance between the points, P(x, 4) and Q(9, 10) is 10 units. (2011D), Solution:, PQ = 10 …Given, PQ2 = 102 = 100 … [Squaring both sides, (9 – x)2 + (10 – 4)2 = 100…(using distance formula, (9 – x)2 + 36 = 100, (9 – x)2 = 100 – 36 = 64, (9 – x) = ± 8 …[Taking square-root on both sides, T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 8
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CHITTI CREATIONS, , 9 – x = 8 or 9 – x = -8, 9 – 8 = x or 9+ 8 = x, x = 1 or x = 17, 4. Find the distance of the point (-3, 4) from the x-axis., , Solution:, B(-3, 0), A (-3, 4), Here x1=-3, x2=-3, y1=0 & y2=4. Put all the values in the given formula., d= √(x2 − x1)2 + (y2 − y1)2 ., , 5. Find distance between the points (0, 5) and (-5, 0)., Solution:, Here x1 = 0, y1 = 5, x2 = -5 and y2 = 0), , Practice:, 6. Find the distance between the two points(-4, 0) & (0, 3)., 7. Find the distance between the points(-3, 4) from its origin., 8. The point A(3, y) is equidistant from the points P(6, 5) and Q(0, -3)., Find the value of y., 9. Find the distance between the points A(3, 6) and B(5, 7) using, distance formula., 10., Find the distance between the co-ordinate of the points A(2, 3), and B(10, -3)., , T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 9
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CHITTI CREATIONS, , 5. Quadratic equations:, , Formula method., −𝑏±√𝑏2 −4𝑎𝑐, , Quadratic formula is x=, , 2𝑎, , 1. Solve by using quadratic formula: x2 -3x+1=0., Solution: a=1, b=-3, c=1, −𝑏±√𝑏2 −4𝑎𝑐, , Quadratic formula is x=, , 2𝑎, −(−3)±√(−3)2 −4𝑋1𝑋1, , x=, , 2𝑋1, , 3±√9−4, x=, 2𝑋1, 3±√5, , x=, , 2, 3+√5, x=, 2, , 3−√5, , or x=, 2, 2. Solve the quadratic equation by using the formula: x2-6x-4=0, Solution: a=1, b=-6, c=-4, −𝑏±√𝑏2 −4𝑎𝑐, , Quadratic formula is x=, , 2𝑎, −(−6)±√(−6)2 −4𝑋1𝑋−4, , x=, , 2𝑋1, , 6±√36+16, x=, 2, 6±√52, , x=, x=, , 2, 6+√52, 2, , =, , 6−√52, 2, , or x=, , 3. By using the quadratic formula, find the solutions: 6x2-7x-5=0., Solution: a=6, b=-7, c=-5., −𝑏±√𝑏2 −4𝑎𝑐, , Quadratic formula is x=, , 2𝑎, −(−7)±√(−7)2 −4𝑋6𝑋−5, , x=, , 2𝑋6, 7±√49+120, x=, 12, 7±√169, 7±13, , x=, x=, x=, , 12, 7+13, 12, 20, , 12, 5, , x=, , 3, , T.SHIVAKUMAR, , =, , 12, 7−13, , or x=, , 12, −6, , or x=, or, , x= -, , 12, −1, 2, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 10
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CHITTI CREATIONS, , 4. Solve the quadratic equation by formula: 2x2+11x+5=0., Solution: a=2, b=11, c=5., −𝑏±√𝑏2 −4𝑎𝑐, , Quadratic formula is x=, , 2𝑎, −11±√(11)2 −4𝑋2𝑋5, , x=, , 2𝑋5, −11±√121−40, x=, 10, −11±√81, −11±9, , x=, x=, x=, , 10, −11+9, 10, −2, , 10, −11−9, , or x=, , 10, −20, , or x=, , 10, 1, , x= -, , =, , or, , 5, , 10, , x= - 2, , 5. Solve the quadratic equation using formula: x2-8x+15=0., Solution: a=2, b=11, c=5., −𝑏±√𝑏2 −4𝑎𝑐, , Quadratic formula is x=, , 2𝑎, −11±√(11)2 −4𝑋2𝑋5, , x=, , 2𝑋5, −11±√121−40, x=, 10, −11±√81, −11±9, , x=, x=, x=, , 10, −11+9, 10, −2, 10, 1, , x= -, , 5, , =, , 10, −11−9, , or x=, , 10, −20, , or x=, or, , 10, , x= - 2, , Practice:, Solve the quadratic equation by using formula method, 6. 2x2+x-5=0., 7. x2+2x+1=0., 8. 5x2+31x+6=0., 9. x2-x-30=0., 10. 4x2-11x-3=0., 11. x2+2x-5=0., , T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 11
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CHITTI CREATIONS, , 6. Pair of linear equations in two variables:, , solve x & y., , 1. Solve the equations by elimination method: x+y= -2 & 2x-y= 8., Solution: let the given equations be x+y= -2 & 2x-y= 8., x+y= -2 ---------(1), 2x-y= 8 ----------(2), By eliminating add the above two equations., We get x+y= -2, 2x-y= 8, 3x=6, x=2, put above x value in any one equation we get y value, equation (1) becomes 2+y=-2, y=-2-2, y=-4, 2. Solve: x-y= 1& 2x-3y= 5., Solution: The given two equations are x-y= 1& 2x-3y= 5., x-y= 1 -----------(1), 2x-3y= 5 --------(2), For eliminating, multiple 2 to the equation (1) we get, 2x-2y=2, 2x-3y=5, subtract this two, y=3, put y value in equation (1) we get, , x-(-3)=1, x=-2, , 3. Solve: x-2y=2 & 2x-y=-8., Solution: The given two equations are x-2y=2 & 2x-y=-8., X-2y=2 -----------(1), 2x-y=-8--------(2), For eliminating, multiple 2 to the equation (1) we get, 2x-4y=4, 2x-y=-8, subtract this two, y=-3, put y value, in equation (1) we get, x-2(-3)=1, x=-5, T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 12
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CHITTI CREATIONS, , 7. Constructions :, , Dividing the line segment, , 1. Draw a line segment of length 9cm and divide it in the ratio 2:3., , 2. Draw a line segment of length 7.6cm and divide it in the ratio 5:8., , 3. Draw a line segment of length 8.3cm and divide it in the ratio 2:5., , T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 14
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CHITTI CREATIONS, , 4. Draw a line segment PQ = 8.4 cm. Divide PQ into four equal parts using, ruler and compass., , 5. Draw a line segment of length 7.6cm divide it in the ratio 3:5., , Practice:, 6. Draw a line of length 7cm, divide it in the ratio 2:4., 7. Draw a line segment then divide internally in the ratio of 3:7., 8. Draw a line segment AB=10cm & divide it in the ratio 5:8., 9. Draw a line of length 7.3cm and then divide it in the ratio 4:6., 10., Draw a line segment of AB=8cm and divide it in the ratio 3:2 by, geometrical construction., 11., Construct a tangent to a circle of radius 4cm at any point P on its, circumference., , T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 15
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CHITTI CREATIONS, , 8. Constructions :, , Tangent construction, , 1. Construct tangents to a circle of radius 5cm such that the angle between, the tangents is 600., , 2. Construct a circle of radius 4.5cm, such that the angle between the two, radii is 1350., , 3. Draw a circle with the help of a bangle. Take a point outside the circle., Construct the pair of tangents from this point to the circle., , 4. Construct a tangent to a circle of radius 4 cm from a point on the, concentric circle of radius 6 cm and measure its length., , T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 16
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CHITTI CREATIONS, , Justification:, In ∆BPO, we have, ∠BPO = 90°, OB = 6 cm and OP = 4 cm, ∴ OB2 = BP2 + OP2 [Using Pythagoras theorem], , 5., , Similarly, BQ = 4.47 cm, , Draw a line segment AB of length 8 cm. Taking A as centre, draw a, circle of radius 4 cm and taking B as centre, draw another circle of, radius 3 cm. Construct tangents to each circle from the centre of the, other circle., , Justification:, On joining BP, we have ∠BPA = 90°, as ∠BPA is the angle in the semicircle., ∴ AP ⊥ PB, Since BP is the radius of given circle, so AP has to be a tangent to the circle. Similarly, AQ, BR, and BS are the tangents., , 6. Construct a pair of tangents to a circle of radius 6.2cm from an external point 3.8 cm, away from the circle., 7. Construct a pair of tangents to a circle of radius 4cm from an external point 4 cm, away from the circle., 8. Construct a tangent to a circle of radius 3.5cm from a point on the concentric circle, of radius 7cm and measure its length., 9. Construct a pair of tangents to a circle of radius 5.5cm at the end point of radii. The, angle between the two radii is 900., T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 17
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CHITTI CREATIONS, , 9. Statistics :, , Mean, Median & Mode., ∑ 𝑓𝑥, , Mean for grouped data, x=, , 𝑛, , ( direct method), 𝑛, −𝑓𝑐, 2, , Median for grouped data, median= LRL+{, , }xh, , Mode for grouped data, Mode=LRL+{, , }x h., , 𝑓𝑚, 𝑓1−𝑓0, , 2𝑓1−𝑓0−𝑓2, , 1. Find the mean, median and mode for the gollowing data., C.I, 10-20 20-30 30-40, f, 5, 2, 3, To find the mean,, C.I, f, x, fx, 10-20 5, 15, 75, 20-30 2, 25, 50, 30-40 3, 35, 105, 40-50 6, 45, 270, 50-60 4, 55, 220, ∑ 𝑓𝑥 = 720, n=20, , 40-50, 6, , 50-60, 4, , ∑ 𝑓𝑥, , x=, , 𝑛, 720, , x=, , 20, , Mean=36, 𝑛, 20, To find the median, first we should find , = = 10, 2, 2, C.I, f, fc, 10-20 5, 5, 20-30 2, 7, 30-40 3, 10, 40-50 6, 16, 50-60 4, 20, n=20, 𝑛, −𝑓𝑐, 2, , Median= LRL+{, =30+{, , 𝑓𝑚, , }xh, , LRL=30, fm=3, fc=7 & h=1, , 10−7, 3, , } x 10 = 30+1x10, , Median = 30+10= 40, T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 18
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CHITTI CREATIONS, , To find the mode, note that f1, f0 & f2., C.I, f, 10-20 5, 20-30 2, 30-40 3 f0, 40-50 6 f1, 50-60 4 f2, Mode=LRL+{, , 𝑓1−𝑓0, , 2𝑓1−𝑓0−𝑓2, 6−3, , }x h,, , LRL=40, f1=6, f0=3 & f2=4., 3, , =40+{, }X10 ⇒40+(5)X10, 12−3−4, =40+6., Mode=46, 2. Find the mean, median and mode for the gollowing data., C.I, 2-6, 7-11, 12-16 17-21 22-26, f, 7, 13, 8, 7, 5, To find the mean,, C.I, f, x, fx, 2-6, 7, 4, 28, 7-11, 13, 9, 117, 12-16 8, 14, 112, 17-21 7, 19, 133, 22-26 5, 24, 120, ∑ 𝑓𝑥 =510, n=40, ∑ 𝑓𝑥, , x=, , 𝑛, 510, , x=, , 40, , Mean=12.75, 𝑛, , To find the median, first we should find , =, 2, C.I, f, fc, 2-6, 7, 7, 7-11, 13, 20, 12-16 8, 28, 17-21 7, 35, 22-26 5, 40, n=40, T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 40, 2, , = 20, , 9916142961, , PAGE 19
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CHITTI CREATIONS, 𝑛, −𝑓𝑐, 2, , median= LRL+{, =7+{, , 𝑓𝑚, 20−13, 7, , }xh, , LRL=7, fm=13, fc=7 & h=5, , } x 5 = 7+5, , Median = 12, To find the mode, note that f1, f0 & f2., C.I, f, 2-6, 7 f0, 7-11, 13 f1, 12-16 8 f2, 17-21 7, 22-26 5, Mode=LRL+{, , 𝑓1−𝑓0, , 2𝑓1−𝑓0−𝑓2, 13−7, , }x h,, , LRL=7, f1=13, f0=7 & f2=8., 6, , =7+{, }X5 ⇒7+(11)X10, 26−7−8, =7+5.4., =12.4., 3. Find the mean, median and mode for the gollowing data., C.I, 1-5, 6-10, 11-15 16-20 21-25, f, 6, 7, 4, 8, 5, To find the mean,, C.I, f, x, fx, 1-5, 6, 4, 24, 6-10, 7, 9, 63, 11-15 4, 14, 56, 16-20 8, 19, 152, 21-25 5, 24, 120, ∑ 𝑓𝑥 =415, n=30, ∑ 𝑓𝑥, , x=, , 𝑛, 415, , x=, , 30, , Mean=13.83, 𝑛, , To find the median, first we should find , =, 2, C.I, f, fc, 1-5, 6, 6, T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 30, 2, , = 15, , 9916142961, , PAGE 20
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CHITTI CREATIONS, , 6-10, 11-15, 16-20, 21-25, , 7, 4, 8, 5, n=30, , 13, 17, 25, 30, 𝑛, −𝑓𝑐, 2, , median= LRL+{, =11+{, , 𝑓𝑚, 15−13, , }xh, , LRL=11, fm=4, fc=13 & h=5, , } x 5 = 11+2.5, , 4, , Median = 13.5, To find the mode, note that f1, f0 & f2., C.I, f, 1-5, 6, 6-10, 7, 11-15 4 f0, 16-20 8 f1, 21-25 5 f2, Mode=LRL+{, , 𝑓1−𝑓0, , 2𝑓1−𝑓0−𝑓2, 8−4, , }x h,, , LRL=16, f1=8, f0=4 & f2=5., 4, , =16+{, }X5 ⇒16+(7)X10, 16−4−5, =16+5.71., =21.71., Practice:, Find the mean, Median and Mode for the following data., C.I, 0-20, 20-40, 40-60, 60-80, f, 3, 4, 2, 7, C.I, f, , 3-13, 12, , 13-23, 9, , C.I, f, , 2-6, 5, , 7-11, 7, , 12-16, 4, , 17-21, 8, , 22-26, 6, , C.I, f, , 1-5, 1, , 6-10, 2, , 11-15, 4, , 16-20, 1, , 21-25, 2, , T.SHIVAKUMAR, , 23-33 33-43, 8, 13, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 43-53, 5, , 9916142961, , 80-100, 4, , 53-63, 3, , PAGE 21
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CHITTI CREATIONS, , 10. Statistics : Ogive graph., 1. Convert the following as less than type then draw its ogive., Marks, No.of, students, , 0-10, 10, , 10-20, 10, , 20-30, 6, , 30-40, 4, , 40-50, 4, , 50-60, 4, , 60-70, 6, , 70-80, 6, , 80-90, 10, , 2., , T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 22
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CHITTI CREATIONS, , 3. Draw a ‘less than type’ ogive for the following frequency distribution., Solution:, , T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 23
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CHITTI CREATIONS, , 4., , Solution:, , T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 24
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CHITTI CREATIONS, , 5., , Practice:, , 6., , 7., , 8., 9., T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 25
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CHITTI CREATIONS, , 11., , Circle: Theorems., 1. Prove that “the tangent at any point of a circle is perpendicular to the radius, through the point of contact”., Solution:, , Given: a circle C(0, r) and a tangent l at point A., To prove: OA ⊥ l, Construction: Take a point B, other than A, on the tangent l. join OB. Suppose OB, meets the circle in C., Proof: We know that, among all line segment joining the point O to a point on l,, the perpendicular is shortest to l., OA=OC (Radius of the same circle), Now, OB=OC+BC., ∴ OB>OC, ⇒OB>OA, ⇒OA<OB, B is an arbitrary point on the tangent l. Thus, OA is shorter than any other line, segment joining O to any point on l., Here OA ⊥ l., 2. Prove that “the lengths of the tangent drawn from an external point to the, circle are equal”., Solution:, Given: A circle with center O. PA & PB are two tangents drawn from an external, point P., , To prove: PA=PB, Construction: Join OA, OB & OP., Proof: It is known that a tangent is at any point of a circle is perpendicular to the, radius through the point of contact., OA⊥PA & OB⊥PB, In ∆OPA & OPB, ⨽OPA=⨽OPB, OA=OB, (radii), OP=OP, (common), Hence ∆OPA is congruent to ∆OPB. Therefore AP=PB., T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 26
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CHITTI CREATIONS, , 12. Pair of linear equations in two variables:, , Graphical solution., , 1. Solve by graphically: x-y=4 & x+y=10., Solution: x-y=4-----------(i) & x+y=10---------(ii), From equation (i), we have the following table:, , From equation (ii), we have the following table:, , Plotting this, we have, , Here, the two lines intersect at point (7,3) i.e., x = 7, y = 3., T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 27
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CHITTI CREATIONS, , 2. Show graphically the given system of equations, 2x + 4y = 10 and 3x + 6y = 12 has no solution., Solution: 2x+4y=10-----------(i) & 3x+6y=12---------(ii), From equation (i), we have the following table:, , From equation (ii), we have the following table:, , Plot the points D (2, 1), E (0, 2) and F (4,0) on the same graph paper., Join D, E and F and extend it on both sides to obtain the graph of the, equation 3x + 6y = 12., , We find that the lines represented by equations 2x + 4y = 10 and 3x +, y = 12 are parallel. So, the two lines have no common point. Hence,, the given system of equations has no solution., T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 28
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CHITTI CREATIONS, , 3. Draw the graph of 2x + y = 6 and 2x – y + 2 = 0., Solution:, We have, 2x + y = 6-----------(i), 2x-y=-2 -------------(ii), From equation (i), we have the following table:, , From equation (ii), we have the following table:, , 4. Draw the graph of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0., Solution: we have x-y=-1 ----------(i), 3x+2y=12-------(ii), From equation (i), we have the following table:, , T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 29
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CHITTI CREATIONS, , 13. Constructions : Constructions of similar triangles., This construction is depends on two type of fractions, one is proper and, another is improper fraction. Let’s see both in different examples., 1. Construct a triangle with sides 4cm, 5cm & 6cm and then another, 𝟐, triangle whose sides are of the corresponding sides of the first, 𝟑, triangle., , 2. Construct a triangle with sides 5cm, 6cm & 7cm and then another, 𝟕, triangle whose sides are of the corresponding sides of the first, 𝟓, triangle., Solution:, , 3., , Construct a right angled triangle with sides 3cm & 4cm and then, 𝟓, another triangle whose sides are of the corresponding sides of the, 𝟑, first triangle., T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 31
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CHITTI CREATIONS, , Solution:, , 4., , Construct a triangle ABC with base AB=5cm, ⨽ABC=600 & BC=7cm, 𝟕, and then another triangle whose sides are of the corresponding, 𝟓, sides of the first triangle., Solution:, , 5. Construct a triangle ABC with AB=5cm, ⨽ABC=600 & BC=6cm and, 𝟑, then another triangle whose sides are of the corresponding sides of, 𝟒, the first triangle., T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 32
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CHITTI CREATIONS, , Solution:, , Practice:, 6. Draw a triangle ABC with side BC=6cm, ∟B=600, ∟A=100 Then construct a, 1, triangle a triangle whose sides are times the corresponding sides of, 3, ∆ABC., 7. Draw a triangle PQR with side QR=5cm, ∟Q=450, ∟P=1050.Then construct, 5, a triangle a triangle whose sides are times the corresponding sides of, 2, ∆PQR., 8. Construct an isosceles triangle whose base is 5cm and altitude 3cm and, 2, then another triangle whose sides are times the corresponding sides of, 5, the isosceles triangle., 9. Construct a triangle with sides 3.5cm, 4cm & 5cm and then another, 3, triangle whose sides are of the corresponding sides of the first triangle., 5, 10. Construct a triangle with sides 3cm, 4cm & 6cm and then another, 7, triangle whose sides are of the corresponding sides of the first triangle., 4, 11. Construct a right angled triangle with sides 5cm & 6cm and then, 1, another triangle whose sides are 2 of the corresponding sides of the first, 2, triangle., , T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 33
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CHITTI CREATIONS, , 4. Area Of Similar Triangle:, Prove that “The ratio of the areas of two similar triangles is equal to the, square of the ratio of their corresponding sides”., , T.SHIVAKUMAR, , MMDRS, HARAPANAHALLI, , Use E-Papers, Save Trees, Above line hide when print out, , 9916142961, , PAGE 37
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APPLICATION OF TRIGONOMETRY, 1. Two poles of equal heights are standing opposite each other on either side of the road,, which is 80m wide. From a point between them on the road, the angles of elevation of, the top of the poles are 60° and 30°, respectively. Find the height of the poles and the, distances of the point from the poles., , 2. From the top of a 7 m high building, the angle of elevation of the top of a cable tower, is 60° and the angle of depression of its foot is 45°. Determine the height of the tower., , 3. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of, depression of two ships are 30° and 45°. If one ship is exactly behind the other on the, same side of the lighthouse, find the distance between the two ships., , Use E-Papers, Save Trees, Above line hide when print out
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4. From a point on the ground 40 m away from the foot of a tower, the angle of elevation, of the top of a tower is 30°,the angle of elevation of the top of water tank on top of the, tower is 45°. Find (i) height of the tower (ii) depth of the tank, , 5. A tree is broken over by the wind forms a right angled triangle with the ground. IF the, broken parts makes an angle of 60°,with the ground and the top of the tree is now 20, m from its base, how tall was the tree., , 6. There is a small island in the middle of a 100 m wide river and a tall tree stands on the island, . P and Q are points directly opposite to each other on two banks and in the line with the tree ., If the angle of elevation of the top of the tree from P and Q are respectively 30o and 45o find, the height of the tree, , Use E-Papers, Save Trees, Above line hide when print out
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PRACTICE PAPER, 1. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of, elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high,, find the height of the building., 2. Two poles of equal heights are standing opposite each other on either side of the road, which is, 80m wide. From a point between them on the road, the angles of elevation of the top of the poles, are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the, poles., 3. From a point 20m away from the foot of a tower, the angle of elevation of top of the tower is, 30o, Find the height of the tower., 4. An electric pole is 10m high. A steel wire tied to the top of the pole is affixed at a point on the, ground to keep the pole upright. If the wire makes an angle of 45o with the horizontal through, the foot of the pool, find the length of the wire., , Use E-Papers, Save Trees, Above line hide when print out
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STATISTICS, 1. Find the Mean, Median & Mode of the following data., CI, f, , 1-10, 2, , CI, 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, N = 30, ƩfX, , Mean =, , 11-20, 5, f, 2, 5, 7, 5, 6, 3, 2, , X, 5, 15, 25, 35, 45, 55, 65, , 31-40, 5, , fX, 10, 75, 175, 175, 270, 165, 130, , f, 2, 5, 7, 5, 6, 3, 2, , 1000, , 41-50, 6, , 51-60, 3, , CI, 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, N = 30, , ƩfX = 1000, , /N =, , CI, 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, , 21-30, 7, , f, 2, 5, 7, 5, 6, 3, 2, , 61-70, 2, fc, 2, 7, 14, 19, 25, 28, 30, , N, , /2 = 30/2 = 15, /2 – fc, ×h, fm, , N, , Median = l +, , = 30 + 15 – 14 × 10, 5, , 1, , = 30 + /5 × 10, = 30 + 2, Median = 32, , /30 = 33.3, Mode = l +, , f0, f1, f2, , f1 - f0, ×h, 2f1 – f2 – f0, , 7-5, , = 20 + 2 . 7 – 5 - 5 × 10, = 20 + 2/4 × 10 = 20 + 20/4 = 20 + 5 = 25, , 2. Find the Mean, Median & Mode of the following data., C I 25-35 35-45 45-55 55-65 65-75, f, 7, 8, 16, 10, 9, CI, 25-35, 35-45, 45-55, 55-65, 65-75, N = 50, Mean =, , ƩfX, , f, 7, 8, 16, 10, 9, , /N =, , X, 30, 40, 50, 60, 70, , fX, 210, 320, 800, 600, 630, , ƩfX = 2560, 2560, , CI, 25-35, 35-45, 45-55, 55-65, 65-75, N = 50, , f, 7, 8, 16, 10, 9, , fc, 7, 15, 31, 41, 50, , /50 = 51.2, , N, , /2 = 50/2 = 25, /2 – fc, ×h, fm, , N, , Median = l +, , = 45 + 25 – 15 × 10, 16, , 10, , = 45 + /16 × 10, = 45 + 100/16, Median = 45 + 6.25 = 51.25, , Use E-Papers, Save Trees, Above line hide when print out
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CI, 25-35, 35-45, 45-55, 55-65, 65-75, , f, 7, 8, 16, 10, 9, , 16 - 8, , f1- f0, , f0, f1, f2, , Mode = l + 2f – f – f × h = 45 +2 . 16 – 10 - 8× 10 = 45 + 8/14 × 10, 1, 2, 0, = 45 + 80/14 = 45 + 5.7 = 50.7, , For Practice : Find the Mean, Median & Mode of the following data., CI, 1-5, 6-10 11-15 16-20 21-25, f, 3, 5, 4, 2, 6, CI, f, CI, f, , 16-20, 5, , 21-25, 6, , 100-120 120-140, 8, 9, , 26-30, 8, 140-160, 7, , 31-35, 4, 160-180, 5, , Drawing an O-give Curve, 3. Draw a less than type O-give for the following data., CI, f, Less than fc, 0-10, 2, 2, 10-20, 5, 7, 20-30, 7, 14, 30-40, 5, 19, 40-50, 6, 25, 50-60, 3, 28, 60-70, 2, 30, , 4. Draw a More than type O-give for the following data, CI, f, More than fc, 0-10, 2, 30, 10-20, 5, 28, 20-30, 7, 23, 30-40, 5, 16, 40-50, 6, 11, 50-60, 3, 5, 60-70, 2, 2, , Use E-Papers, Save Trees, Above line hide when print out, , 36-40, 7, 180-200, 6
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5. Draw Less than type O-give for the following data., CI, fc, < 10, 5, < 20, 11, < 30, 19, < 40, 30, < 50, 42, < 60, 48, < 70, 55, , 6. Draw a More than type O-give for the following data., CI, fc, > 10, 50, > 20, 40, > 30, 35, > 40, 26, > 50, 20, > 60, 13, > 70, 8, , 7. If the mean of 4, x, 6, 9 is 6, find the value of x., Mean =, , 6=, , Sum of all terms, Number of terms, , 19 + x, 4, , =, , 4+x+6+9, 4, , 24 = 19 + x, , x = 24-19, , x=6, , 8. Write the Modal Class in the following data. 9. Write the Median in 2, 5, 9, 7, 4, 11, 1, CI, 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, , f, 2, 5, 7, 5, 6, 3, 2, , as 7 is the highest frequency, in the data,, 20-30 is the modal class, , Use E-Papers, Save Trees, Above line hide when print out, , 1, 2, 4, 5, 7, 9, 11, here middle term is, 5., Hence 5 is Median
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Arithmetic Progression, 1. General form of A.P is, a, a + d, a + 2d, a + 3d … … … .., 2. nth term of an A.P is, an = a + (n − 1)d, 3. nth term from last of an A.P is, l − (n − 1)d, 4. Common difference of an A.P is, d = a2 − a1 (or) d = a3 − a2, 5. The sum of first ‘n’ positive integer, Sn =, , n(n+1), 2, , 6. The sum of ‘n’ odd natural numbers, Sn = n2, 7. The sum of ‘n’ even natural numbers, Sn = n(n + 1), 8. Sum of first ‘n’ terms of an A.P is, n, , Sn = [2a + (n − 1)d], 2, , 9. Sum of AP, if first and last terms are given, Sn =, , n, 2, , [a + an ], , (or), , Sn =, , n, 2, , [a + l], , 10. In any progression, Sn − Sn−1 = an, , DIRECTORATE OF MINORITIES, , 1, , Use E-Papers, Save Trees, Above line hide when print out
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1. Find 10th term of the sequence an = 2n − 5, Soln: an = 2n − 5, a10 = 2(10) − 5, a10 = 20 − 5, , ∴ a10 = 15, Drill work, , 1. If the nth term of an AP 𝑎𝑛 = 3𝑛 − 2. Find the 9th term., 2. If the nth term of an AP 𝑎𝑛 = 24 − 3𝑛. Find the 2th term., , 3. If the nth term of an AP 𝑎𝑛 = 5𝑛 + 3. Find the 3th term., 2. Find the 10th term of AP 2, 7, 12, ……… using formula., Soln: a = 2, d = 5 , n = 10, , Alternate Method, , w.k.t an = a + (n − 1)d, a10 = 2 + (10 − 1)5, a10 = 2 + 9 x 5, , a10 = a + 9d, a10 = 2 + 9 X 5, a10 = 2 + 45, ∴ 𝐚𝟏𝟎 = 𝟒𝟕, , a10 = 2 + 45, , ∴ 𝐚𝟏𝟎 = 𝟒𝟕, Drill work, 1. In an AP 21, 18, 15, ………... find 35th term., 2. In an AP 3, 8, 13, ………... find 12th term., 3. In an AP 10, 7, 4, ………... find 18th term., 4., , DIRECTORATE OF MINORITIES, , 2, , Use E-Papers, Save Trees, Above line hide when print out
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3. Find the sum of 2+5+8+……... to 20 terms, Soln: a = 2, d = 3, n = 20 and Sn = ?, w.k.t Sn =, S20 =, , 20, 2, , n, 2, , [2a + (n − 1)d, , [2 x 2 + (20 − 1) x 3], , S20 = 10[4 + 19 x 3], S20 = 10[4 + 57], S20 = 10 x 61, ∴ 𝐒𝟐𝟎 = 𝟔𝟏𝟎, Drill work, 1. Find the sum of first 20 terms of an AP 3, 7, 11, 15, ……., 2. Find the sum of first 25 terms of an AP 5, 10, 15, 20, ……., 3. Find the sum of first 18 terms of an AP 2, 7, 12, ……., 4. Find the sum of: 1+5+9+ ……. up to 25 terms., 5. Find the sum terms of an AP 2, 7, 12, ……., , DIRECTORATE OF MINORITIES, , 3, , Use E-Papers, Save Trees, Above line hide when print out
