1. (B) volume of Hemisphere = 4 volume of sphere
SSC CGL TIER – 2
MATHEMATICS
Ans is ½ of radius of the Hemisphere
2. (A) Sec
Sec
25 days B work = 25 5 = 125
Remaining work = 180 – 125 = 55
A + B work = 55/11 = 5
= 5 days
9. (D)
12600
10. (B) Total age of 30 student = 430 years
Total age of 35 students = 481 years 3 months
5 New students Age sum =51 years 3 months
 51 year 3 months – 9 years 11 month
= 41 years 4 months
Therefore 4 new students average is 10 years 4
months
Sec
11. (C)
401 : 544
12. (A)
3
13. (A)
3. (D)
Area of ADE =
2
= 80%
DB : AB =
4. (C)
20 – 20 5. (A)
= 4% decreased
14. (A)
1200 : 1025
48 : 41
15.
(D)
= 16.7
16. (C) 54.5
17. (A)
Area Triangle = ½ × Base × height =
6x = 15y
15 : 6
6. (B)
2 mintues Total work 7 × 2 = 14
Remaining work = 24 – 14 = 10
Time of B =
7. (C)
Total work = 90 × 12 × 16
8. (A)
18. (A)
= 15 years
19. (C)
TanB =
Cot B =
and
Sin A = m SinB  SinB =
Cosec B =
=1
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m2 – n2 cos2A = Sin2A = 1 – Cos2A
m2 -1 = n2 cos2A – cos2A
cos2A
20. (D)
0
21. (D)
No – 1 : 4
HCF = 21
No  21 = 84
Large No is 84
22. (C)
1
23. (B)
Ratio =
A: B:C
6 : 9 : 21
A +B : B : C : C + A
+ 5 : 30 : 27
5 : 10 : 9
24. (B)
= 6 km
25. (D)
10 – 10 –
= 1%
26. (B)
= 20 km/hr
27. (D)
28. (D)
hrl =
29.
30.
31.
32.
33.
34.
35.
According to Q. =
(B)90 degree
(C)
440
(A)
1/8
(C)
SinA = 1 – Sin2A
SinA = = cos2A
Squaring on both side
Sin2A = cos4A
Therefore , SinA + Sin2A = cos2A + cos4A = 1
1
(C)
1
(B)14
(B)
(AB + C) = BC = AD
16.2 = 8.5 + AD
7.7 = AD
7.7 cm
36. (B)
A+B=8
B + C = 12
A×C=x
A+B+C =6
Therefore : A + C = 24 days
37. (D)
9
38. (B)
AD2 = BD × CD
AD2 = 3 × 4
AD = 2
39. (D)
Prism Volume = Area of Base × h
= 126 × 9 = 1134 cm3
40. (A)
4
41. (C)
0
42. (C)
a = b=c
43. (C)
Pyramid total surface Area = CSS + Box Area
C.S.A = ½ × 40 × 13
260 + 100 = 360 cm2
44. (B)
8x = 27y
27 : 8
= 25
45. (C)
31% profit
46. (A)
300
= 60 gm
47. (B)
30 5/8 %
48. (A)
10 sides
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49. (B)
Increase by 1
50. (D)
(a-3)3
2
(a-3)3 -
= 8 + 2 × 3 = 14 Ans.
51. (A)
Simple Interest =
16 = x2
x=4
r=8
52. (A)
61.
62.
= 21 cm2
53. (C)
63.
64.
54.
55.
56.
57.
½ × L1 ×L2
L2 = 18
½ × 24 × 18 = 216 cm2
(D)
= 86.1
(D)
42
(B)
1 cm
(B)
Speed =
km/h
=
58. (B)
= 52. 8 km/hr
A:B:C
3:4:5
81.5
59. (A) s
A
A: B: C = 552 : 528 : 506
A/s share =
= 2760 Rs.
60. (B)
P+Q
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Q + R  60/7
(Q + R) 1/3 work = 7 ×6 = 42
Remaining = 18/3 = 6 to P.
Q’s = 4
R’s = 3
Days difference for R and P = 60/3 – 60/6 = 10 days
(C)
52+ d2 = 22 + (d +3)2
21 – 9 = bd
D = 2 diamter = 2
(C)
100 – 140 – 112
12 = 48
=
= 400
(D)
LCM (12, 16, 18 and 21)
 1008
(2000 + n) divisible by 1008.
So, n = 16 = 1 + 6 = 7
(A)
=1
65. (B)
C to from oppdion
Only 18 divisible by 9.
66. (C)
PD = 8 cm
67. (A)
(0.67 – 0.33) = 0.34
68. (D)
PQ =
69. (A)
n – y = 5/1 , n + y = 7/1
n = 6, y = 1
= 12
70. (B)
If n = a sin
y = acos
then, x2 + y2 = a2 + b2 Property
71. (B)
60A = 30B
100B = 40C
A:B=1:2
B:C=2:5
A:B:C=1:2:5
So, A is 500% 0 to C
72. (D) Check from option
7Sin2
then tan
If tan
, = 30
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73. (B) Amount = 17640
= 17640
74. (C)
300
86. (A)
x=
32800
y=
Profit 405 P% =
= 8%
87.
75. (A)
76. (A)
–
=
330
77. (A)
Unit digit = 7 1
=7
78. (A)
bucket
1=
88.
buckets.
Volume of each bucket =
= 18 liters
79. (A)
59
= 72
80. (C)
89.
90.
91.
92.
93.
x + y = 24
3 x2 + 3y2 = 57 × 3
1722 – 5 × 1 = 1717
(B)Total student = 1554
Boys
Girl
4
:
3
888
:
666
- ×
+30
812
696
7
:
6 × = 76
(D)Income
AB
8x – 56 = 12000
5x – 3y = 10000
X = 14000
Difference = 3 × 14000 = 42000
(A)
(D)
(A) 6(1/2)km/hr
(C) 24
(D)
95
81. (C)
35 : 40
2x : 3x
70x + 12 x = 190x
46x + 4x × 55
46x + 220x = 266x
= 40%
82. (D)
83. (B) 1
84. (D)
85. (C)
A = 16 -----6
B = 32 -----3
96
C = 48 -----2
A 6 Days work is = 36
A+C=8
Remaining work =
4 + 8 = 12 days
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2
= 400
94. (D)
n
64 = r2
r = 8 cm
95. (A)
= 10%
97. (C)
P = 10
P = 10
P + H = 10 + 20 = 30
30m
98. (D)864 cm2
99. (B)
x = 20
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Total distance = 4
100.
(B)
S1 + S2 =
x + 8 + x = 27
2x = 27
x = 9 ½ km/hr
20 = 80km
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