CBSE 10th SA 2 2011 Mathematics Solved Exam Paper
MATHEMATICS SOLVED PAPER
CBSE SA – 2
SUBJECT :- MATHEMATICS
CODE – 3
SECTION – A
- – (A) – r
- - (C) – 10/13
- - (A) – 13
- - (B) – 3
- - (C) – 25
- - (A) – 8A
- - (C) – 140o
- - (D) – 35O
- - (C) – 15
- - (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
14 = AD + 9
AD = 14-9
AD = 5 cm
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
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