# CBSE 10th SA 2 Mathematics Solved Exam Paper

MATHEMATICS SOLVED PAPER

CBSE SA – 2

SUBJECT :- MATHEMATICS

CODE – 3

SECTION – A

1. – (A) – r
2. – (C) – 10/13
3. –  (A) – 13
4. – (B) – 3
5. – (C) – 25
6. – (A) – 8A
7. – (C) – 140o
8. – (D) – 35O
9. – (C) – 15
10. – (A) – m, m+3

SECTION – B

11. a = 203, d = 7, l = 497,

a + (n-1) d

or 203 + (n-1) 7 = 497

or 203 + 7n – 7 = 497

or 7n + 196 = 497

or 7n = 497 – 196

or 7n = 301

n = 301 / 7

= 43

12. Total no. of ticket = 40

No. of ticket is multiple of 5 = 8

P (getting ticket multiple of 5) = 8/40

= 1/5

13. root (11-3) square + (y+1) square = 10

root over (8) square +(y+1) square = 10

root over (64) + (y square + 1+2y) = 10

root over y square + 2y + 65 = 10

squaring both sides ,

y square + 2y + 65 = 100

y square + 2y – 35 = 0

y square + 7y – 5y – 35 = 0

y (y + 7) – 5 (y + 7) = 0

(y + 7) (y – 5) = 0

y = -7, y = 5

14. volume of cube = 27 cm cube

a cube = 3 cm

surface area of 1 cube =  6 a square

= 6*9

= 54 cm square

surface area of both cubes = 2*54

= 108 cm square

area of square = side square

= 3 square

= 9 cm square

surface area of resulting cuboid = 108-9

= 99 cm square

16. as we know that,

AB + CD = AD + BC

6 + 8 = AD + 9

18. area of triangle = 1/2* base * height

= 1/2 *14*24

= 7*24 = 158 cm square

area of quadrant = 1/4*22/7 * 7*7

= 5.5*7 = 38.5 cm square

area of sector = 22/7 *7*7*45/360

= 19.25 cm square

area of shaded region = 158 – 38.5 + 19.25

= 158 – 57.75

= 101.25 cm square

19. PA = PB

PA square = PB square

root over (6 – 3) square + (5-y) square

CBSE 10th SA 2 Mathematics Solved Exam Paper
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