CBSE Class 10 Mathematics Solved
Question Paper Year 1999
Question 1) Evaluate without using trignometric table: cos 75o/sin
15o + sin 12o/cos 78o - cos 18o/sin
72o (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 4x2 -
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(4x2 - 9) and 18(2x2
+ 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) In 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 75o/sin 15o + sin 12o/cos 78o -
cos 18o/sin 72o
= cos (90-15)o/sin 15o + sin (90-78)o/cos 78o
- cos (90-72)o/sin 72o
= sin 15o/sin 15o + cos 78o/cos 78o
- sin 72o/sin 72o
= 1 + 1 - 1
= 1 (ANS)
Answer2)
(sec2A - 1)(1 - cosec2A )
= ((1+tan2A) - 1)(1 - ( 1 + cot2A ))
= ( tan2A )( - cot2A )
= ( tan2A )( - 1/tan2A )
= - 1
Answer2)
tan A = b/a
==> tan2A = b2/a2
==> sin2A/cos2A = b2/a2
==> sin2A/1 - sin2A = b2/a2
==> a2sin2A = b2 - b2sin2A
==> a2sin2A + b2sin2A = b2
==> sin2A (a2 + b2) = b2
==> sin2A = b2/(a2 + b2)
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)