# CBSE Class 10 Mathematics Solved Question Paper (1999)

## Mathematics

## Question Paper

## 1999

**Question 1)**Evaluate without using trigonometric table: cos 75

^{o}/sin 15

^{o}+ sin 12

^{o}/cos 78

^{o}– cos 18

^{o}/sin 72

^{o}(Marks 2)

**Question 2)**Evaluate (sec2A – 1)(1 – cosec2A ) (Marks 2)

**Question 3)**If tan A = b/a where a and b are real numbers, find the value of sin2 A (Marks 2)

**Question 4)**A cricketer has mean score of 58 runs in nine innings. Find out how many runs are to be scored in the tenth innings to raise the mean score to 61. (Marks 2)

**Question 5)**Find the mean of the following data: 46, 64, 87, 41, 58, 77, 35, 90, 55, 33, 92 If in the data, the observation 92 is replaced by 19, determine the new median. (Marks 2) click for answer

**Question 6)**Harshad purchased a motorcycle for Rs. 42,952 which includes the amount of sales tax. If the tax charged is 12% of the list price, find the list price of the motorcycle. (Marks 2)

**Question 7)**The circumference of the edge of a hemispherical bowl is 132 cm. Find the capacity of the bowl ( Π= 22/7) (Marks 2)

**Question 8)**Find the value of c for which the quadratic equation 4×2 – 2(c + 1)x + (c + 4) = 0 has equal roots. (Marks 2)

**Question 9)**The ages of two girls are in the ratio 5 : 7. Eight years ago their ages were in the ratio 7 : 13. Find their present ages. (Marks 2)

**Question 10)**Find the value of k for which the system of equation 8x + 5y = 9, kx + 10y = 15 has no solution. (Marks 2)

**Question 11)**Find the G.C.D. of 24(4x

^{2}– 9) and 18(2x

^{2}+ 5x – 12) (Marks 2)

**Question 12)**Flow chart. Deleted from the syllabus. (Marks 2)

**Question 13)**In figure ABCD is a cyclic quadrilateral. AE is drawn parallel to CD and BA is produced. If ABC = 92o, FAE = 20o, find BCD. (Marks 2)

*No answer*

**Question 14)**Determine the length of an altitude of an equilateral triangle of side 2a cm. (Marks 2)

**Question 15) I**n the figure, a circle touches all the four sides of a quadrilateral ABCD where sides AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD. (Marks 2) sorry, no answer for this as I do not have diagram for this question>

### ANSWERS

**Answer1)**cos 75

^{o}/sin 15

^{o}+ sin 12

^{o}/cos 78

^{o}– cos 18

^{o}/sin 72

^{o}= cos (90-15)

^{o}/sin 15

^{o}+ sin (90-78)

^{o}/cos 78

^{o}– cos (90-72)

^{o}/sin 72

^{o}= sin 15

^{o}/sin 15

^{o}+ cos 78

^{o}/cos 78

^{o}– sin 72

^{o}/sin 72

^{o}= 1 + 1 – 1 = 1 (ANS)

**Answer2)**(sec

^{2}A – 1)(1 – cosec

^{2}A ) = ((1+tan

^{2}A) – 1)(1 – ( 1 + cot

^{2}A )) = ( tan

^{2}A )( – cot

^{2}A ) = ( tan

^{2}A )( – 1/tan

^{2}A ) = – 1

**Answer2)**tan A = b/a ==> tan

^{2}A = b

^{2}/a

^{2}==> sin

^{2}A/cos

^{2}A = b

^{2}/a

^{2}==> sin

^{2}A/1 – sin

^{2}A = b

^{2}/a

^{2}==> a

^{2}sin

^{2}A = b

^{2}– b

^{2}sin

^{2}A ==> a

^{2}sin

^{2}A + b

^{2}sin

^{2}A = b

^{2}==> sin

^{2}A (a

^{2}+ b

^{2}) = b

^{2}==> sin

^{2}A = b

^{2}/(a

^{2}+ b

^{2})

**Answer 4)**Mean score of the cricketer after nine match = 58 ==> Total score of the cricketer after nine match = 58 x 9 = 522 Suppose the cricketer needs n runs to raise the mean score to 61. ==> Total runs scored by the cricketer after tenth match = 522 + n ==> Average runs scored by the cricketer after tenth match = 522 + n / 10 ==> 522 + n / 10 = 61 ==> 522 + n = 610 ==> n = 88 ==> Answer: Cricketer needs 88 runs to raise the mean score to 61.

**Answer 5)**Arrange the data in following way. 33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92 Total count of data n = 11 (odd number) Median = ((n + 1) / 2)th term = 6th term = 58 When 92 is replaced by 19 Then data will be be arranged in following order: 19, 33, 35, 41, 46, 55, 58, 64, 77, 87, 90 n = 11 (odd number) Median = ((11 + 1) / 2)th term = 6th term = 55 So the new median is 55.

**Answer 6)**Assume the list price of the motorcycle is : p ==> So tax charged amount: 12p/100 ==> So purchace price of motorcycle : p + 12p/100 ==> p + 12p/100 = 42952 ==> (100p + 12p)/100 = 42952 ==> 112p/100 = 42952 ==> 112p = 42952 x 100 ==> p = 4295200/112 ==> p = 38350 ==> So list price of the motorcycle is Rs 38350.00

**Answer 7)**Assume the radius of the hemispherical bowl is r. ==> So circumference of the edge of a hemispherical bowl = (22/7)(2r) ==> (22/7)(2r) = 132 ==> 44r = 132 x 7 ==> r = (132 x 7)/44 ==> r = 21 ==> Volume of hemispherical bowl = (1/2) x (4/3)Π(radius)

^{3}==> Volume of hemispherical bowl = (1/2) x (4/3) X (22/7) x (21)

^{3}==> Volume of hemispherical bowl = (2/3) X (22/7) x 21 x 21 x 21 ==> Volume of hemispherical bowl = 2 x 22 x 7 x 3 x 21 ==> Volume of hemispherical bowl = 19404

**Answer 8)**Find the value of c for which the quadratic equation 4x

^{2}– 2(c + 1)x + (c + 4) = 0 has equal roots. Using following formulae, if ax

^{2}+bx+c=0 then x = (-b ±Ö (b

^{2}– 4ac))/2a 4×2 – 2(c + 1)x + (c + 4) = 0 ==> x = (2(c + 1) ±Ö ((2(c + 1))

^{2}– 4 . 4 . (c + 4)))/(2 . 4) ==> Equal roots if ((2(c + 1))

^{2}– 4 . 4 . (c + 4)) = 0 ==> 4c

^{2}+ 8c + 4 – 16c -64 = 0 ==> 4c

^{2}– 8c -60 = 0 ==> c

^{2}– 2c -15 = 0 ==> (c – 5)(c + 3) = 0 ==> c = 5 OR c = -3

**Answer 9)**The ages of two girls are in the ratio 5 : 7. Eight years ago their ages were in the ratio 7 : 13. Find their present ages. (Marks 2) Assume current age of first girl is n1. Assume current age of second girl is n2. ==> n1/n2 = 5/7 ==> 7n1 = 5n2 ==> 7n1 – 5n2 = 0 ……….. (i) Eight years ago age of first girl is n1 – 8. Eight years ago age of second girl is n2 – 8. ==> (n1 – 8)/(n2 – 8) = 7/13 ==> 13n1 – 104 = 7n2 – 56 ==> 13n1 – 7n2 = 48 …………. (ii) ==> 91n1 – 65n2 = 0 ……..( by multiplying (i) by 13 ) ……… (iii) ==> 91n1 – 49n2 = 336 ……..( by multiplying (ii) by 7 ) ……. (iv) ==> Subtract (iv) minus (iii) ==> 16n2 = 336 ==> n2 = 21 ==> Now 7n1 = 5 n2 ==> 7n1 = 5 x 21 ==> n1 = 15 So current age of first girl is 15. So current age of second girl is 21.

**Answer 10)**Find the value of k for which the system of equation 8x + 5y = 9, kx + 10y = 15 has no solution. ==> equation 8x + 5y = 9, kx + 10y = 15 will have no solution if ==> 8/k = 5/10 ==> k = 80/5 ==> k = 16

**Answer 11)**Find the G.C.D. of 24(4x

^{2}– 9) and 18(2x

^{2}+ 5x – 12) (Marks 2) G.C.D meand Greatest Common Divisor First simplify 24(4x

^{2}– 9) ==> = 24(2x – 3)(2x + 3) ………………….(i) Now simplify 18(2x

^{2}+ 5x – 12) ==> = 18(2x – 3)(x + 4) …………………..(ii) ==> By comparing (i) and (ii) ==> G.C.D = 6(2x – 3)

**Answer 14)**Determine the length of an altitude of an equilateral triangle of side 2a cm. (Marks 2) Assume length of the altitude of an equilateral triangle is : h ==> length of a side of the triangle : 2a ==> (2a)

^{2}= h

^{2}= (a)

^{2}==> h

^{2}= 4a

^{2}– a

^{2}==> h

^{2}= 3a

^{2}==> h = aÖ3

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