SIMPLIFICATION

(CLASSROOM SHEET)

1. The value of 2 × 3 ÷ 2 of 3 × 2 ÷ 5. The value of 5 ÷ 5 of 5 × 2 + 2 ÷ 2

(4 + 4 × 4 ÷ 4 of 4 – 4 ÷ 4 × 4) is of 2 × 5 – (5 – 2) ÷ 6 × 2 is :
SSC CGL 4 June 2019 (Morning) SSC CGL 4 June 2019 (Afternoon)
9 19
(a) 8 (b) 1 (a) (b)
5 10
(c) 4 (d) 2
23
7  5 2 1 (c) 19 (d)
2
2. The value of 2 ÷  3 ÷ of 2 
8 6 7 3
9 2 2 3
6. The value of of   of  ÷
 6 1 2  5 15 3 3 2
×   2 of 4 ÷    is :
 7 5 3 9 3 3 3 4 5 5 2 4
SSC CGL 4 June 2019 (Evening)   ÷ of  of   × of 
4 4 4 3 4 2 3 5
1 is :
(a) (b) 4 SSC CGL 6 June 2019 (Afternoon)
4
1 20 4
(c) (d) 5 (a) (b)
23 9 25
3. The value of (5 + 3 ÷ 5 × 5)/(3 ÷ 3 of 6) 18 40
(c) (d)
of (4 × 4 ÷ 4 of 4 + 4 ÷ 4 × 4) is : 125 9
7. The value of 16 ÷ 4 of 4 × [3 ÷ 4 of
SSC CGL 4 June 2019 (Morning)
{4 × 3 ÷ (3 + 3)}] ÷ (2 ÷ 4 of 8) is :
1 1 SSC CGL 6 June 2019 (Morning)
(a) 8 (b) 7
5 3 (a) 6 (b) 9
2 2 (c) 48 (d) 19
(c) 9 (d) 6  1 1 
5 3 8
8. The value of of  5 ÷ 2 of 4 
4. The simplified value of 9 4 3
 2 4  2 4
 1.0025 + 6.25 × 10 –6  ÷  8 ÷ of  of  8 × ÷  is :
3 5 3 5
 0.0025 + 0.95  is :
SSC CGL 7 June 2019 (Morning)
SSC CGL 9 June 2019 (Evening) 1 4
(a) 1 (b)
(a) 1.0025 (b) 1.0525 8 15
(c) 1.0005 (d) 1.0505 1 1
(c) (d)
200 100
9. The value of 4.5 – (3.2 ÷ 0.8 × 5) 15. The value of 6 – 6 ÷ 6 × 6 + (6 ÷ 6
+ 3 × 4 ÷ 6 is  2 11 2
of 6) × 6 –  3  of  ÷ 5 is :
SSC CGL 10 June 2019 (Afternoon) 3 30 3
(a) – 13.5 (b) 4.2 SSC CGL 7 June 2019 (Evening)
(c) – 8.5 (d) 5.7 (a) 0 (b) 2
10. What is the simplified value of 5 (c) – 1 (d) – 2
÷ 10 of 10 × 4 + 4 ÷ 4 of 4 × 10 – 3 2 5 1
(10 – 4) ÷ 16 × 4? 16. The value of 2 ÷ of 1 +
4 3 9 5
SSC CHSL 3 July 2019 (Evening) 2 5 2 1
3 ÷ of 2 is :
(a) 1.2 (b) 2.5 23 6 7 3
(c) 2.1 (d) 58.5 SSC CGL 10 June 2019 (Morning)
11. The simplified value of 5 2
(a) 1 (b) 1
 1 3 8 6 3
 3 –   1 5
5 5 5
(c) 3 (d) 4
1  6  1 1   is : 2 6
1 ÷  –  ÷  17. The value of 3.8 – (4.2 ÷ 0.7 × 3) +
7  7  7 5 
5 × 2 ÷ 0.5 is :
SSC CHSL 5 July 2019 (Afternoon)
SSC CGL 10 June 2019 (Evening)
13 13
(a) (b) (a) 5.8 (b) 18.4
64 16
(c) 21.8 (d) 15.6
13 13
(c) (d) 18. The value of 2.8 + (5.2 ÷ 1.3 × 2) –
8 7
6×3÷8+2
5.75 × 5.75 × 5.75 + 3.25 × 3.25 × 3.25
12. SSC CGL 4 June 2019 (Morning)
57.5 × 57.5 + 32.5 × 32.5 – 57.5 × 32.5
is equal to : (a) 6.45 (b) 4.55
SSC CPO 12 March 2019 (Evening)
(c) 8.45 (d) 10.55
(a) 0.009 (b) 0.0009
19. The value of 7.2 + (8.4 ÷ 0.12 ×
(c) 0.9 (d) 0.09 0.2) – 5 × 3 ÷ 0.05 + 3
13. Find the value of 4 + 144 SSC CGL 11 June 2019 (Morning)

SSC CPO 14 March 2019 (Evening) (a) – 75.8 (b) – 275.8

(a) 14 (b) 12.17 (c) 21.2 (d) – 175.8
(c) 4 (d) 3.74 20. The value of 5.8 + (7.4 ÷ 3.7 × 5) –
1  1 1 1 6 × 2 ÷ 2.5
14. The value of 7 ×  3  4 of 5 
2 5 2 3 SSC CGL 11 June 2019 (Evening)
 5 1  3 (a) 12 (b) 11
+ 11 –   3 – 1  
8 4  ÷ 5 –5÷5×5
 4 (c) 10 (d) 9
of 5 ÷ 25 is :
SSC CGL 7 June 2019 (Afternoon) 21. The value of 3.8 + (8.2 ÷ 4.1 × 2) –
4 × 3 ÷ 1.2
1 1
(a) (b) SSC CGL 12 June 2019 (Morning)
2 10
3 1 (a) 2.2 (b) – 1.2
(c) (d) 1
10 2 (c) 1.2 (d) – 2.2
22. The value of 7.5 + (5.4 ÷ 4.5 × 2) – 3÷ 5 – 5 ÷ (6 – 7) × 8 + 9
8 × 4 ÷ 3.2 28. The value of 4 + 4 × 4 ÷ 4 of 4
SSC CGL 12 June 2019 (Afternoon)
is :
SSC CHSL 2 July 2019 (Morning)
(a) 0.1 (b) – 0.1
1 1
(c) – 0.2 (d) 0.2 (a) (b)
45 18
23. The value of 108 ÷ 36 × 4 + 2.5 × 1 1
4 ÷ 0.5 – 10 (c) (d)
90 3
SSC CGL 12 June 2019 (Evening) 29. The value of 3 × 2 ÷ 3 of 12 – 3 ÷
(a) 18 (b) 16 2 × (2 – 3) × 2 + 3 ÷ 2 of 3 is :
(c) 22 (d) 20 SSC CHSL 2 July 2019 (Morning)

24. 21.6 ÷ 3.6 × 2 + 0.25 × 16 ÷ 4 – 6 1 1

(a) 2 (b) – 2
is equal to : 3 3
2 2
SSC CGL 13 June 2019 (Morning) (c) – 3 (d) 3
3 3
(a) 6 (b) 5 30. The simplified value of 3 × 2 ÷ 3
(c) 8 (d) 7 of 2 × 3 ÷ (5 + 5 × 5 ÷ 5 of 5 – 5 ÷
25. 15.2 + 5.8 ÷ 2.9 × 2 – 3.5 × 2 ÷ 0.5 1
10 of ) is :
× 4 ÷ 0.5 – 10 is equal to 5
SSC CHSL 3 July 2019 (Afternoon)
SSC CGL 13 June 2019 (Afternoon)
6 17
(a) 4.8 (b) 3.2 (a) (b)
7 5
(c) 5.2 (d) 5.4 2 30
(c) (d)
3  1  1  1 3   3 59
26. 9 ÷ 2  4 –  2 +   31. The simplified value of
4  6  3  2 4  
is equal to : 7 7 3 4  7 1 1
  of  ÷ –   10 × 7 
5 10 4 9  16 2 5
SSC CGL 13 June 2019 (Evening)
5
15 × is :
(a) (b) 3 12
4 SSC CHSL 3 July 2019 (Afternoon)
39 47 39
(c) (d) 4 (a) (b)
8 8 4
3 3 3 4 49 41
27. The value of ÷ of × + (c) (d)
4 4 4 3 8 4
5 2 5 2 2 5 32. The simplified value of 3 × 6 ÷ 4
 of –  + of  is :
of 6 – 6 ÷ 2 × (4 – 6) + 4 – 2 × 3 ÷ 6
2 5 4 3 3 6
1
SSC CHSL 1 July 2019 (Evening) of is :
3
14 41 SSC CHSL 4 July 2019 (Morning)
(a) (b)
3 9 3 3
(a) 1 (b) 7
4 4
22 50 3 1
(c) (d) (c) 13 (d) 8
3 9 4 3
33. The simplified value of 15 of 8 – 38. The simplified value of
6 + [(27 – 3)÷ 6 – 4] is : 3
46 – of 32 – 6
SSC CHSL 4 July 2019 (Afternoon) 4 is :
3
(a) 114 (b) 120 37 – of (34 – 6)
4
(c) 124 (d) 116 SSC CHSL 8 July 2019 (Evening)
34. The simplified value of 15 of 8 + 19
6 + [(27 – 3) ÷ 6 + 4] is : (a) 2 (b)
16
SSC CHSL 5 July 2019 (Morning) 19
(c) (d) 1
(a) 128 (b) 134 64
39. The simplified value of
(c) 130 (d) 136 3
46 – of 32 – 6
35. The simplified value of 4
3 is :
 1 3 8 11  of (34 – 6)
4
 3   
5 5 5 SSC CHSL 9 July 2019 (Morning)
is :
1  6  1 1  1
1   –    (a) (b) 1
7  7  7 5  7
1 1
SSC CHSL 5 July 2019 (Evening) (c) (d)
4 2
19 19 40. The simplified value of
(a) (b)
7 8 3
46  of 32 – 6
19 19 4
(c) (d) 3 is :
16 64 11  of (34 – 6)
4
 1 3 8
 3    SSC CHSL 9 July 2019 (Afternoon)
5 5 5
36. The value of is 1
1  5  1 1  (a) 1 (b)
1       4
7  7  7 3  1
(c) 2 (d)
SSC CHSL 8 July 2019 (Morning) 2
19 19 1
(a) (b) 41. The simplified value of 2 of
8 16 3
3 2  2 19 1 
19 19    –  4    is :
(c) (d) 5 9 5 20 2 
64 7
SSC CHSL 9 July 2019 (Evening)
 1 3 8 1
 3    (a) 0 (b)
5 5 5 4
37. The value of is
1  5  1 1  1
1       (c) (d) 1
8  8  8 3  2
42. The simplified value of
SSC CHSL 8 July 2019 (Afternoon)
 1  3  4 13  5  6 
19 19 1 of  – 1     ÷   5 
(a) (b)  5  7  15 15 7  7
16 7 is :
19 19 SSC CHSL 10 July 2019 (Morning)
(c) (d)
9 64
 2 1 48. The simplified value of 20 – [2.8 ×
(a) (b) 5 ÷ 0.7 – 3 ÷ 0.9 × 1.5 + 2] is equal
15 3
4 to :
(c) 1 (d) SSC CHSL 11 July 2019 (Evening)
15
1 8 (a) 3 (b) 3.4
43. The simplified value of of ÷ (c) 3.8 (d) 3.6
2 5
 1  5 3 7 2  49. (24 ÷ 6 – 2) + (3 + 2 + 4) is equal
2 –    1    is : to :
 5 16 5 8 3 
SSC CHSL 10 July 2019 (Afternoon) SSC CPO 16 March 2019 (Morning)
(a) 24 (b) 16
2
(a) 1 (b) (c) 20 (d) 12
5
1 14 – 6 × 2
(c) (d) 4 50. is equal to :
5 15 ÷ 3 + 3
44. The simplified value of SSC CPO 16 March 2019 (Morning)

0.01404 2 1
is : (a) 6 (b)
24 2  62 – 144 5 4
SSC CHSL 10 July 2019 (Evening) 4
(c) (d) 2
(a) 3 × 10 –5
(b) 6 × 10 –5 5
–4
(c) 2.4 × 10 (d) 3 × 10–4 5  2  3 4 1 
45. The simplified value of 51. 5  2 – 3  3  9    is
6  3  4 5 2 
 1  1 2 5 1 1 equal to :
1 of  2  1  – 1  + 2
 4 3 5 12  2 3 SSC CPO 12 March 2019 (Evening)
2 1
+  is : 44
7 6 (a) (b) 7
7
SSC CHSL 11 July 2019 (Morning)
43 22
7 3 (c) (d)
(a) (b) 6 3
3 2 52. The value of
7
(c) (d) 1 1  1 5  2 3 4  
6 3 – 2 –  –   –   is :
18.43 ×18.43 – 6.57 × 6.57 5  2  6  5 10 15  
46. The value of 11.86
is :
SSC CPO 12 March 2019 (Evening)
SSC CHSL 11 July 2019 (Morning)
6 9
(a) 23.62 (b) 25 (a) (b)
5 10
(c) 26 (d) 24.12
11 13
47. The simplified value of : (c) (d)
10 5
2 3 14 2  1 1 
÷  7 of 5  1 3 –  3 2 – 2 6   is : 63.5 × 63.5 × 63.5  36.5 × 36.5 × 36.5
3  53.
6.35 × 6.35 + 3.65 × 3.65 – 6.35 × 3.65
SSC CHSL 11 July 2019 (Afternoon) is :
1 SSC CPO 13 March 2019 (Evening)
(a) (b) 1
3 (a) 10,000 (b) 1,00,000
2 (c) 100 (d) 1,000
(c) 2 (d)
3
54. The value of  1 5  3 1 4  
1
60. 5 – 3 –  –     
1  1 5  3 3 7   5  2 6 5 10 15   is
6 –  4 –  –   –   is
5  2  6  5 10 15   equal to :
SSC CPO 13 March 2019 (Evening) SSC CPO 13 March 2019 (Morning)
(a) 2.5 (b) 1.8 21 7
(a) (b)
(c) 2.1 (d) 2.8 10 5
7 8
17  1  5  4 1   (c) (d)
55.   3 –  –  3  9   is 3 3
30  5  6  5 2   61. The value of
equal to :  2 15  4 1  
5
SSC CPO 13 March 2019 (Evening) 3 + 3 3 –  4  5 5  14 2   is :
6   
3 1
(a) (b) SSC CPO 13 March 2019 (Morning)
5 5
37 35
11 10 (a) (b)
(c) (d) 6 6
3 3
19
675 × 675 × 675 + 325 × 325 × 325 (c) 6 (d)
56. is 3
67.5 × 67.5 + 32.5 × 32.5 – 67.5 × 32.5 62. The value of 3 × 3 – [6 – {12+ 15 ÷
equal to : (7 – 2)] is equal to :
SSC CPO 12 March 2019 (Morning) SSC CPO 14 March 2019 (Morning)
(a) 100 (b) 10,000 (a) – 15 (b) 18
(c) 1,000 (d) 1,00,000 (c) 0 (d) 15
3  1  1  1 3   1 5 –5
57. 9 ÷ 2 6  4 3 –  2 2  4   is 63. The value of   is equal to:
4    3 6 8
equal to : SSC CPO 14 March 2019 (Morning)

SSC CPO 12 March 2019 (Morning) 1

(a) 1 (b)
15 4
(a) 3 (b) 1
4 (c) – (d) 0
17 4
(c) 4 (d) 64. 3 × 7 + 4 – 6 ÷ 3 – 7 + 45 ÷ 5 × 4 +
4 49 is equal to :
4 3 4 3 1
58. 4 ÷ of 7 + × – is SSC CPO 16 March 2019 (Evening)
5 7 5 10 5 (a) 99 (b) 101
equal to :
(c) 103 (d) 67
SSC CPO 12 March 2019 (Morning)
65. (8 + 4 – 2) × (17 – 12) × 10 – 89 is
7 8
(a) (b) equal to :
5 5 SSC CPO 16 March 2019 (Evening)
14 41 (a) 413 (b) 411
(c) (d)
25 25 (c) 4120 (d) 4150
6.75 × 6.75 × 6.75 + 4.25 × 4.25 × 4.25
59. 0.72 × 0.72 × 0.72 – 0.39 × 0.39 × 0.39
67.5 × 67.5 + 42.5 × 42.5 + 67.5 × 42.5 66.
is equal to : 0.72 × 0.72 + 0.72 × 0.39 + 0.39 × 0.39
SSC CPO 13 March 2019 (Morning) is equal to :
SSC CPO 16 March 2019 (Evening)
(a) 2.5 (b) 0.25
(a) 0.39 (b) 0.36
(c) 0.0025 (d) 0.025 (c) 0.33 (d) 0.45
67. (–4) × (–8) ÷ (–2) +3 × 5 is equal to: 74. 13 ÷ {4 of 2 – 3 + 4 × (6 – 4)} is
SSC CPO 15 March 2019 (Morning) equal to :
(a) – 1 (b) 1 SSC CGL 4 June 2019 (Morning)
(c) 31 (d) – 31 1
(a) – 2 (b) 0
13
3 5  1  8 4 
68.   4   5 – 3   is equal to : (c) 1.3 (d) 1
4 2   75. Find the value of 32 ÷ 4 of 2 × 3
SSC CPO 15 March 2019 (Morning) 5
+ [5 of 6 – {7 of 8(10 + 6 of ÷5
11 3 6
(a) (b) – 1) ÷ 80}] – 7 × 3 ÷ 2.
24 4
SSC MTS 2 August 2019 (Morning)
1 11
(c) (d) (a) 7.5 (b) 17.5
4 12 (c) 12.5 (d) 24.5
69. 15 – {5 + 24 ÷ (3 × 9 – 15)} is equal 76. What is the value of
to : 72 ÷ 9 + 3 – 6 – (2 × 3) + 5of3 – (1 + 5 × 2 – 2)
SSC CPO 16 March 2019 (Afternoon) 8 ÷ 4 + 2 – (6 × 8 ÷ 2) + (7 × 4 – 2 × 2)
1 SSC MTS 2 August 2019 (Morning)
(a) – 2 (b) 11 11 5
3 (a) (b)
1 4 4
(c) 6 (d) 8 15
4 (c) 0 (d)
70. (–4) × (1020 ÷ 85 × 3 – 22) is equal 4
to : 77. What is the value of 7 ÷ 2 – [3 of 7
÷ 4 ÷ {(2 ÷ 5) × (25 ÷ 8) ÷ (5 ÷ 2)}]
SSC CPO 16 March 2019 (Afternoon)
SSC MTS 2 August 2019 (Afternoon)
(a) – 402 (b) – 56 (a) – 8 (b) – 7
(c) 912 (d) 72 (c) – 1 (d) – 9
71. Find the value of 3 1 1 
 78. Find the value of of    +
3 2  4 3 2
× 4 7 –  5  13 + 2 is :
5     2 3 2
 2 –  × 
SSC CPO 14 March 2019 (Evening) 5 2 3
SSC MTS 2 August 2019 (Afternoon)
1
(a) 1 (b) 107 101
3 (a) (b)
2 30 6
(c) 2 (d) 0 109 103
5 (c) (d)
72. Find the value of 7 – {4 × 3 – (–10) 17 25
79. What is the value of
× 8 ÷ (–4)}.
2 9 1 5
SSC CPO 14 March 2019 (Evening) of  
3 4 2 4
(a) –1 (b) 0 1 1  1
(c) 53 (d) 15 1 –   1  
3 4  3
73. 10 – {17 – 12 ÷ (5 + 9 × 2 – 17)} is SSC MTS 2 August 2019 (Evening)
equal to : 7 9
SSC CPO 15 March 2019 (Evening) (a) (b)
10 19
(a) – 5 (b) 5 19 7
(c) (d)
(c) 7 (d) – 7 10 11
 3 9 4 86. Find the value of
80. Find the value of ÷ 2 +
7 21 3 3 9 4 2 27
   of
1 12 25 5 4 32 3 3 16
of   .
2 5 18 9 1 8  4 1 1 .
   2    + 
SSC MTS 2 August 2019 (Evening) 2 3  9 3 6
17 SSC MTS 5 August 2019 (Evening)
(a) 2 (b) 13 10
3
(a) (b)
14 2 3
(c) (d) 4 25 31
3 (c) (d)
81. What is the value of 90 × 3 ÷ 9 + 2 2
4 ÷ 2 × 3 of 4× 8 ÷ (18 × 2 – 4)? 87. Find the value of
SSC MTS 5 August 2019 (Morning)
39  26  22  11 × 2 + 4 × 3
(a) 48 (b) 40 2 of 5 – 3(7 +10 ÷ 2 – 3 × 3)
(c) 36 (d) 42 SSC MTS 6 August 2019 (Morning)

82. If A = 40 ÷ 8 + 5 × 2 – 4 × 5 of 3 61 49
(a) (b)
and B = 24 ÷ 4(4 + 2) + 19 of 2, 2 2
then what is the value of A – B? 39 35
(c) (d)
SSC MTS 5 August 2019 (Morning) 2 2
88. What is the value of (24 + 16 × 5 –
(a) –11 (b) 11
8 of 4) ÷ 84 × 48 ÷ 24 × 6 + 4 + 3?
(c) 13 (d) – 13
SSC MTS 6 August 2019 (Morning)
83. Find the value of 36 ÷ 8 × 4 + 2 ÷
139 156
4 – 1 + 5 of 3 ÷ (4 × 2 – 3) – 3. (a) (b)
3 5
SSC MTS 5 August 2019 (Afternoon)
121 56
(a) 18 (b) 16 (c) (d)
7 3
35 31 89. What is the value of (3 × 4 of 12 ÷
(c) (d)
2 2 2) ÷ 9 × 4 + 4 ÷ 8 + 3 × 2?
84. If A = 7 × 3 ÷ (2 + 4) + 4 – 2, B = 3 SSC MTS 6 August 2019 (Afternoon)
÷ 6 × 4 + 2 – 2 of 3 and C = 6 ÷ 2 + 89 17
4 × 3 – 2 then the value of (A + B (a) (b)
3 2
– C) is :
SSC MTS 5 August 2019 (Afternoon) 94 27
(c) (d)
–16 19 3 2
(a) (b) 90. If A = 8 ÷ 4 × (3 – 1) + 6 × 3 ÷ 2 of
3 2
–19 16 3 and B = 4 ÷ 8 ×2 + 7 × 3, then
(c) (d) the value of A + B is :
2 3
85. What is the value of SSC MTS 6 August 2019 (Afternoon)
(a) 29 (b) 31
3 of 24 ÷ 8 × 3 + 4 ÷ 2 – 4 × 5
. (c) 33 (d) 35
36 ÷12 × 4 ÷ 2 + 5 × (6 – 4) 91. What is the value of (6 of 4 ÷ 16 ×
SSC MTS 5 August 2019 (Evening) 48) ÷ 8 × 4 + 2 × 3 ÷ 6 + 5(6 – 2)?
8 9 SSC MTS 6 August 2019 (Evening)
(a) (b)
15 16 (a) 63 (b) 79
3 3 (c) 67 (d) 57
(c) (d)
10 4
92. What is the value of 97. What is the value of
3 1 1  2 4  1 11  1 2 (49 – 13) ×18 ÷ 9 + 4 ×12 ÷ 6 + 5
 +  + of ÷  – + ×
4  2 16  3 9 3 81  4 3 98 ÷ 14 + 7 × 4 of 6 ÷ 8 + 4
SSC MTS 6 August 2019 (Evening) SSC MTS 7 August 2019 (Evening)
(a) 3 (b) 1 45 37
(c) 2 (d) 4 (a) (b)
23 18
93. Find the value of 85 53
(c) (d)
 (0.7)2  0.14 + (0.6)2 + 0.18 + (0.5)2 ÷ 0.05  32 17
 
 4(2.5 of 4 – 13 × 0.25 × 3)  98. What is the value of
SSC MTS 7 August 2019 (Morning) 2 ÷ 3 × (1 + 3) + 5 – 6
?
25 19 2 of 3 ÷ 5 × 4 + 3 – 2
(a) (b) SSC MTS 7 August 2019 (Evening)
2 2
23 21 36 31
(c) (d) (a) (b)
2 2 89 73
94. Find the value of 25 27
(c) (d)
 1  1 1 2 87 92
1 –    of  
 4   2 2 5 99. Find the value of 2 of 16 ÷ 48 ×
2 1 3 8 12 + 4 ÷ 8 × 16 + (7 – 2) × 25 +
  2 –  15?
5 4 2  5
SSC MTS 8 August 2019 (Morning)
SSC MTS 7 August 2019 (Morning)
23 59
2 5 (a) (b)
(a) (b) 3 3
3 8 49 56
4 5 (c) (d)
(c) (d) 3 3
5 11 100. Find the value
95. Find the value of 1 1 1 1 1 1 1 1
    –     of
 3 3 1 128 2 7 121  15 2  2 2 2 2 2 2 2 2
1 +   of 5      – 
4 21 3 49 3 11 49  14 7 
1 1 
SSC MTS 7 August 2019 (Afternoon)    ?
2 2
69 62 SSC MTS 8 August 2019 (Morning)
(a) (b) 3 1
25 29
(a) (b)
57 59 2 2
(c) (d) 5
41 32 (c) 1 (d)
96. If A = 2 ÷ 3 × 4, B = 3 of 4 + (7 – 2) 2
101. Find the value
and C = 4 + 5 – 6, then what is the
value of A + B + C? 12 of 3 ÷ 6 + 12 × 2 – (2 × 4 – 5)
12 ÷ 3 × 4 + (2 × 4 – 5)
SSC MTS 7 August 2019 (Afternoon)
SSC MTS 8 August 2019 (Afternoon)
85 79 27 23
(a) (b) (a) (b)
3 3 22 17
59 68 27 21
(c) (d) (c) (d)
3 3 19 9
102. find the value 5 of 5 of 5 ÷ 5 + 5 – 110. The value of x in the given
6 ÷ 3 × 4 + 2 + (3 ÷ 6 × 2) equation :
SSC MTS 8 August 2019 (Afternoon) 232 + x = 625
(a) 21 (b) 25
SSC MTS 13 August 2019 (Morning)
(c) 28 (d) 19
(a) 9576 (b) 9124
103. Find the value (9 ÷ 30)2 × 2.4 + 0.3
(c) 9216 (d) 9028
of 12 × (1 – 0.3)2 + 9 × (0.3)2
111. What is the value of
SSC MTS 8 August 2019 (Evening)
 1 13 1 
(a) 3.43 (b) 3.69  2  1 –  × 16 ÷ 4
6 18 6 
(c) 2.79 (d) 2.17
SSC MTS 13 August 2019 (Morning)
104. Find the value 2 of 3 ÷ 3 × 2 + {4
× 3 – (5 × 2 + 3)}. 1
(a) 42 (b) 41
SSC MTS 8 August 2019 (Evening) 72
(a) 3 (b) – 24 134
(c) (d) 63
(c) 6 (d) – 21 9
112. The value of [12 × 5 – {200 – (50 +
95
105. The value of 99 × 99 – 95 is : 247 – 386)}] ÷ 2 is :
99
SSC MTS 13 August 2019 (Afternoon)
SSC MTS 9 August 2019 (Morning)
(a) 162 (b) 161
(a) 9897 (b) 9993
(c) 111 (d) 82
(c) 9999 (d) 9801
113. The value of
106. What is the value of
 1 1 2 3 6 
 5  2  9  4  2  is :
 0.56 × 0.36 + 0.42 × 0.32  3 7 5 8 47 
 
0.8 × 0.21 SSC MTS 13 August 2019 (Afternoon)
(a) 1 (b) 1000
SSC MTS 9 August 2019 (Morning)
(c) 100 (d) 10
3
(a) 1 (b) 25.60 0.10
2 114.  =?
(c) 32 (d) 72.90 8.10
SSC MTS 13 August 2019 (Evening)
107. Find the value of (3576 + 4286 +
27 27
6593) ÷ (201 + 105 + 107). (a) (b)
30 20
SSC MTS 9 August 2019 (Afternoon)
19 27
(a) 35 (b) 31 (c) (d)
27 28
(c) 22 (d) 18
115. Compute (15 + 3 × 1.1) ÷ 0.0003
108. The value of x, 45 × x = 25% of
SSC MTS 13 August 2019 (Evening)
900 is :
(a) 61000 (b) 122000
SSC MTS 9 August 2019 (Afternoon)
(c) 16100 (d) 30500
(a) 16.2 (b) 4
(c) 500 (d) 5 l16. If 7 × 7 of 3 ÷ 3 – 14 × x = 7, then
5 4 3 2 x is equal to :
109. (x ÷ x ) ÷ x = ?
SSC MTS 14 August 2019 (Morning)
SSC MTS 9 August 2019 (Evening)
(a) 1 (b) 2
(a) x2 (b) x3
(c) x –1 (d) x (c) 3 (d) 4
117. What is the value of (1 × 2 + 2 × 3 125. Find the value of
– 3 × 4 + 4 × 5 – 5× 6 + 6 × 7)? (2 of 14 ÷ 7 × 3) + (44 ÷ 11 × 8) –
SSC MTS 14 August 2019 (Afternoon) (12 × 9 ÷ 3).
(a) 28 (b) 25 SSC MTS 16 August 2019 (Evening)
(c) 20 (d) 24 (a) 9 (b) 6
118. The vale of 8 of 3 ÷ 6 + (10 + 2) × (c) 8 (d) 7
3 – 96 ÷ 3 is : 126. Find the value
2 – 2 ÷ 2 × 2 + 2(2 of 2 – 2 – 2 ÷ 2)
SSC MTS 14 August 2019 (Afternoohn)
SSC MTS 19 August 2019 (Morning)
(a) 7 (b) 9
(a) 4 (b) 0
(c) 10 (d) 8 (c) 2 (d) 1
2
119. 1 2 + 16 of 3 – 20 ÷ 4 = ? 127. Find the value of
SSC MTS 14 August 2019 (Evening) 3 61 9 1 4 1 1 4
(a) 240 (b) 180 3 –   of 1 +    .
4 122 2 2 3 3 2 3
(c) 156 (d) 187 SSC MTS 19 August 2019 (Morning)
120. What is the value of the square 155
root of (a) (b) 3
12
[{(100 of 0.9 × 0.8 – 7 × 1.2 ÷ 0.2
+ 5 × 4 – 3 × 2)} ÷ 10 + 1.85] ? 163
(c) (d) 9
SSC MTS 16 August 2019 (Morning) 11
128. Find the value of
(a) 12.25 (b) 2.5
(c) 6.25 (d) 3.5  3 1 6
1 –   of
4 2 10
121. Find the value of (1 × 2 + 3 × 4 +
2 4  1 25
5 × 6 + 7 × 8 – 9 × 10) ÷ 2 of 5?   1 –  of
3 10  5  16
SSC CGL 4 June 2019 (Morning)
SSC MTS 19 August 2019 (Afternoon)
(a) 1 (b) 3
33 49
(c) 2 (d) 4 (a) (b)
175 115
122. If (28 ÷ 4 × 7) + (44 ÷ 4 × 7) – (12 × 29 47
x) = 18, then the value of x is : (c) (d)
175 115
SSC MTS 16 August 2019 (Afternoon) 129. What is the value of
(a) 3 (b) 12 3 ÷ 3 of 3 + 2 ÷ 4 + (4 × 2 – 2) ÷ 12 + 4.
(c) 9 (d) 6 SSC MTS 19 August 2019 (Afternoon)
123. Find the value of 2 × 2 + 4 × 4 + 2 12 16
of 3 × 6 – 7 × (5 + 4 ÷ 2) (a) (b)
5 3
SSC MTS 16 August 2019 (Afternoon) 14 17
(a) 5 (b) 7 (c) (d)
3 6
(c) 4 (d) 6 130. The value of
124. Find the value of (0.4 of 50 × 6 ÷  2 4  2 4  2 4
 8 ÷ of    8    of  8    is :
8) ÷ (12 × 10 ÷ 16) + 5 × 0.2 – 0.01 3 5 3 5 3 5
× 102. SSC MTS 19 August 2019 (Evening)

SSC MTS 16 August 2019 (Evening)

3 13
(a) 21 (b) 7
(a) 1 (b) 4 5 16
15
(c) 2 (d) 3 (c) 1 (d)
64
131. The value of 165 – [135 – {84 ÷ 4 137. In the given equation, find the
of 3 – (16 – 18 ÷ 3)}] is : value of x?
SSC MTS 19 August 2019 (Evening) (5 )2 + (6)2 + (30)2 = (x)2
1 SSC MTS 21 August 2019 (Morning)
(a) 81 (b) 36 (a) 53 (b) 37
3
(c) 27 (d) 83 (c) 41 (d) 31
132. Find the value of 27 × 6 ÷ 24 + 6 138. Find the value of 56 + (4)3 – 3 ×
÷ 2 of 3 + 30 ÷ 24 × 18 – 9 ÷ 54 of (3)2
3 × 216 SSC MTS 21 August 2019 (Morning)
SSC MTS 20 August 2019 (Morning) (a) 93 (b) 79
(a) – 77.75 (b) 18.25 (c) 76 (d) 83
(c) – 69.75 (d) 26.25 139. Find the value of
133. Find the value of
1 1 3
1 2  1  5 3 7 1 3  3  –
of 1  5 –    1  1 of  16 2 4
7 5  2  32 5 8 3 16  
SSC MTS 21 August 2019 (Afternoon)
SSC MTS 20 August 2019 (Morning)
3
17 27 (a) 1 (b) 1
(a) (b) 4
135 32 1 1
27 6 (c) 1 (d) 1
(c) (d) 2 4
160 27 140. Approximate value of
1 3 1 2 (4488 ÷ 11.01 – 7.98) ÷ 15.99 is :
134. Find the value of   1 of –
4 4 4 5 SSC MTS 21 August 2019 (Afternoon)
 1  3 14 2  1 1   (a) 2.5 (b) 26
   of  1 –  3 – 2  
6 7 5 3  2 6   (c) 25 (d) 2.6
SSC MTS 20 August 2019 (Afternoon) 141. Find the value of
3 1 1
(a) (b) 7 ÷ 14 of 2 – 7 × 7 ÷ 49 + of (14
8 8 3
÷ 7 + 7) + 7 – 14 ÷ 2
2 1 SSC MTS 21 August 2019 (Evening)
(c) (d)
3 3
1
135. Find the value of 4 × 2 ÷ 4 of (4 + (a) 3 (b) 2
4 ÷ 4 of 4) – (4 ÷ 4 of 2 × 4). 4
1 1
SSC MTS 20 August 2019 (Afternoon) (c) – 1 (d) 4
4 4
1 142. Simplify
(a) –1 (b) 6
2
–26 1 1  1 3 1 2 4 2
6   5  of  – 8 ×  of 1
(c) (d) 4 8  4 7 2  3 5 3
17 4
136. Find the value of 26 – [(2 of 6 ÷ 3) SSC MTS 21 August 2019 (Evening)
– 93 – {17 – (14 – 2)}]. 15
SSC MTS 20 August 2019 (Evening) (a) (b) 3
4
(a) 100 (b) 120 15
(c) 110 (d) 90 (c) – 3 (d) –
4
143. The value of 3 ÷ 21 of 3 × 7 + 24 9 1
× 6 ÷ 18 – 3 ÷ 2 + 3 – 2 × 3 ÷ 6 is : (a) (b)
8 18
SSC MTS 22 August 2019 (Morning) 8
(c) 18 (d)
1 1 9
(a) 6 (b) 9 1 1 1 1
3 2 149. If x =   .......  ,
12.13 13.14 14.15 23.24
5 1 1 1 1 1
(c) 8 (d) 12 y =   ....... 
6 2 36.37 37.38 38.39 71.72
1 1 1 1 x
144. The value of 3  4 of 5 – 2 then is equal to :
5 2 3 3 y
 3  4 13  1 SSC CHSL 10 July 2019 (Evening)
of  – 1 –   1 is :
7 15 30 5
1 1
SSC MTS 22 August 2019 (Morning) (a) (b)
3 24
1 1
(a) 1 (b) 1 1
3 6
(c) (d) 3
2 7 72
(c) 1 (d) 1
15 15 150. If (1.25)(1 – 6.4 × 10–5) = 1.2496 +
145. The value of a, then a is equal to :
2 3 14 2  1 1 1 3 1 SSC CGL 4 June 2019 (Morning)
1   of  1 –  3 – 2    of 
3 7 5 3  2 6 2 2 2
(a) 0.0016 (b) 0.00016
SSC MTS 22 August 2019 (Afternoon)
(c) 0.0032 (d) 0.00032
1 1
(a) 3 (b) 2 151. The value of
6 2
7  8  8  8 of 8  8  8  4 of 4
2
(c) 1 (d) 1 4 ÷ 4 of 4  4  4  4 – 4  4 of 2
3
SSC CGL Tier-II (11 September 2019)
146. the value of 72 ÷ 6 of 12 + 4 × (5 –
3) of 2 ÷ 4 – 2 is : (a) 7.8 (b) 4.6
SSC MTS 22 August 2019 (Afternoon) (c) 8.7 (d) 6.4
(a) 5 (b) 4 152. The value of 22.4  11.67 – 33.59 is :
(c) 0 (d) 3
SSC CGL Tier-II (11 September 2019)
5  2 of 3  3 of 2  3
147. The value of 9  72 ÷ 3 – 2  (3 – 2) – 3 (a) 0.32 (b) 0.412
a (c) 0.31 (d) 0.412
is , where a and b are prime
b
 6 1 2 1
numbers. The value of (b – a) is : 153. The value of  2 of 4   × 1
7 5 3 9
SSC MTS 22 August 2019 (Evening)
(a) 7 (b) 3 3 2 1 1
(c) 5 (d) 4 ÷   2 of   is :
4 3 2 4
 1 3 1
148. The value  5  of  
of SSC CGL Tier-II (11 September 2019)
4 7 2
(a) 5 (b) 8
 1 3 1  1 3 1
 5      5    is : 1 1
4 7 2 4 7 2
(c) (d)
SSC MTS 22 August 2019 (Evening) 8 5
154. The value of (c) 3 – 2  5 (d) 2– 3– 5
3 3
(253)  (247) 159. The value of 0.56 – 0.723  0.39  0.7
25.3  25.3 – 624.91  24.7  24.7 is :
is 50 × 10k, where the value of k SSC CGL Tier-II (12 September 2019)
is : (a) 0.154 (b) 0.154
SSC CGL Tier-II (11 September 2019) (c) 0.158 (d) 0.158
(a) 3 (b) 4 160. The value of 9 × 6 ÷ 24 + 8 ÷ 2 of
(c) 2 (d) – 3 5 – 30 ÷ 4 of 4 + 27 × 5 ÷ 9 is :
155. If  2 5– 3  × k = – 12, then SSC CGL Tier-II (12 September 2019)
647 243
what will be the value of k? (a) (b)
40 8
SSC CGL Tier-II (11 September 2019)
493 259
(c) (d)
(a)  2 5 3  8 8

(b)  2 5 3 2 – 10  161. The value of 28  10 3 –

(c)  2 5– 3 2  5  7 – 4 3 is closest to :

(d)  2 5 3 2 – 5  SSC CGL Tier-II (13 September 2019)

(a) 7.2 (b) 6.1
 1 6 3 (c) 6.5 (d) 5.8
156. The vale of 1  2 of 5  ÷
3 7 5 162. The value of
 2 1 1 3 2 5 1 0.47  0.503 – 0.39  0.8 is
 6  4 of 5  ×   2  of 1 
5 2 3 4 3 9 5 SSC CGL Tier-II (13 September 2019)
= 1 + k, where k lies between : (a) 0.615 (b) 0.615
SSC CGL Tier-II (12 September 2019)
(c) 0.625 (d) 0.625
(a) – 0.07 and – 0.06
(b) – 0.08 and – 0.07 163. The value of
(c) – 0.06 and – 0.05 2 10 52 3
–  is :
(d) – 0.05 and – 0.04 5 2– 7 52 7 2
157. The value of SSC CGL Tier-II (13 September 2019)
(0.545)(0.081)(0.51)(5.2) (a) 2  2 (b) 2 5
is :
(0.324)3  (0.221)3 – (0.545)3
(c) 2 (d) 7
SSC CGL Tier-II (12 September 2019)
(a) – 1 (b) 1 164. Find the value of 24 × 2 ÷ 12 + 12
(c) 3 (d) – 3 ÷ 6 of 2 ÷ (15 ÷ 8 × 4) of (28 ÷ 7 of
5) is :
158. The expression
SSC CGL Tier-II (13 September 2019)
10  2  6  15  10  1 8
(a) 4 (b) 4
SSC CGL Tier-II (12 September 2019) 6 75
2 32
(a) 3  2  5 (b) 3– 2 5 (c) 4 (d) 4
3 75
165. A student was asked to find the 1 9
(a) – (b) –
4 1 1  1 4 3 12 8
value of 9  11 of  1 1  
6  3
×
9 3 5 5 1 19
1 2 4 2 (c) – (d) –
2 of  of . His answer was 3 8
6 3 3 3 170. The value of
1 36  42of 6  7  24  6  18  3 
19 . What is the difference
4 (2 – 6) – (4  3  2)  8
between his answer and the 21  3 of 7
correct answer?
SSC CGL 3 March 2020 (Evening)
SSC CGL Tier-II (13 September 2019)
1
3 2 (a) 7 (b)
(a) 7 (b) 6 7
4 3 1 1
1 1 (c) 7 (d) 8
(c) 7 (d) 6 2 2
2 3 171. The value of
166. The value of
7   4  3(2 – 2  2  5) – 8  5
4.64  5.4 4  (24.84)2 2  2 of (4  4  4 of 4)
is :
is :
4.6 2  5.4 2  24.84 SSC CGL 3 March 2020 (Morning)
SSC CGL Tier-II (13 September 2019)
1
(a) 24.42 (b) 24.24 (a) 26 (b) 25
2
(c) 25.42 (d) 25.48
167. If '+' means '–', '–' means '+', '×' 1
(c) 8 (d) 24
means '÷' and '÷' means '×', then the 2
42  12  3  8  2  15 172. The value of
value of is :
82– 493 1 2 1  1 7 9  9
5 ÷ 3 of +  5 – 7 ÷ 9 ×
SSC CGL 3 March 2020 (Morning) 2 3 4  9 8 20  11
5 15 1
5 ÷ 5 of – 10 ×10 ÷ 20
(a) – (b) – 10
3 19
15 5 SSC CGL 4 March 2020 (Afternoon)
(c) (d)
19 3 4 9
 1 (a) 1 (b) 1
168. The value of 18  2 of  × 5 10
4
4 1
2 3 5 2 3 3 (c) 3 (d) 9
  ×     of  is : 5 2
3 4 8 3 4 4
173. The value of
SSC CGL 3 March 2020 (Afternoon)
8 ÷ [(8 – 3) ÷ {(4 ÷ 4 of 8) + 4 – 4 × 4 ÷ 8} - 2]
2 5 is
(a) 10 (b) 8 8 × 8 ÷ 4 - 8 ÷ 8 of 2 - 7
3 8
:
7 7
(c) 16 (d) 2 SSC CGL 4 March 2020 (Evening)
8 64
5 3 1 17 8
169. The value of –   6  is (a) (b)
2 2 2 8 3
equal to : 16 2
(c) (d)
SSC CGL 3 March 2020 (Evening) 170 17
174. The value of 179. Solve the following 113 × 87 = ?
2 11 2 1 1 3 4 SSC CGL 6 March 2020 (Afternoon)
3 ÷ of - of 2 ÷ × 4
3 30 3 4 2 5 5 (a) 10000 (b) 10169
2 1 3 3 1 1 is (c) 10026 (d) 9831
of 7 ÷ - ×1 ÷ 2
5 2 4 4 2 4
[54  (5  2)  8]  13
SSC CGL 4 March 2020 (Morning) 180. The value of
48  4  3  8  2
6 4
(a) 2 (b) 3 is :
7 7
SSC CGL 6 March 2020 (Evening)
2 10
(c) 2 (d) 141 141
9 21 (a) (b)
175. The value of 106 127
3 7 1 3  1 1 1 89 89
 1  1 of   3  4 of 5  (c) (d)
5 8 3 16  5 2 3 106 127
1 1 1 1 181. The value of 151 – 1492 is:
2
× 2    is :
2 2 8 4 SSC CGL 7 March 2020 (Morning)
SSC CGL 5 March 2020 (Afternoon) (a) 300 (b) 400
2
1 5 (c) 2 (d) 600
(a) 4 (b) 5 182. The value of 3 – (9 – 3 × 8 ÷ 2) is
3 6
SSC CGL 7 March 2020 (Morning)
1 1
(c) 5 (d) 4 21
6 8 (a) – 21 (b)
2
1 1
176. The value of – 1 +   2  5 is : (c) 0 (d) 6
4 2 183. The value of
SSC CGL 5 March 2020 (Evening)
1  1 3 1 2 1 4 2 11
(a) 2 (b) 5
1 ÷  4 ÷ of 8  - ×1 ÷ of1 +
8  4 5 2 5 3 5 3 20
is :

17 7 SSC CGL 7 March 2020 (Afternoon)

(c) (d) –
4 2 1 1
177. If '+' means '–', '–' means '+', '×' (a) 3 (b) 1
2 2
means '÷' and '÷' means '×', then
1 1
[(30 × 5) + (84 × 6)] ÷ 5 (c) 1 (d) 3
2  4 8
the value of  ÷ 18  - (4 ÷ 2) is
3 184. Solve the following expression
SSC CGL 6 March 2020 (Morning) 5.6 – {2 + 0.6 of (2.1 – 2.6 × 1.12)
(a) 1 (b) – 2 SSC CGL 7 March 2020 (Evening)
(c) 2 (d) – 1 (a) 4.0871 (b) 4.0872
178. Solve the following (c) 7.7112 (d) 7.7113
4 1 185. The value of 1800 ÷ 20 × {(12 – 6)
÷ × 2 -1 = ? + (24 – 12)} is :
3 6
SSC CGL 9 March 2020 (Morning)
SSC CGL 6 March 2020 (Afternoon)
(a) 2720 (b) 1720
(a) 8 (b) – 2
(c) 840 (d) 1620
(c) 3 (d) 15
186. Solve the following expression (a) 40 (b) 45
11 + 11 × 11 – 11 ÷ 11 (c) 50 (d) 42
SSC CGL 9 March 2020 (Afternoon) 194. The value of (72 + 34) ÷ 2 + [{75 ÷
(a) 121 (b) 22 15) + 6} × 2] is :
(c) 11 (d) 131 CHSL 13/10/2020 (Evening)
187. The value of 515 × 485 is :
(a) 74 (b) 75
SSC CGL 9 March 2020 (Evening)
(c) 86 (d) 78
(a) 249775 (b) 250225
(c) 20825 (d) 200825 195. If '+' means '÷', '×' means '+', '÷'
188. The value of (26 – 13 × 2) ÷ 2 + 1 means '–' and '–' means '×', then
is : what will be the value of the
following expreswsion?
SSC CGL 9 March 2020 (Evening)
(a) 0 (b) 1 18 + 3 – 5 × 6 ÷ 4
CHSL 14/10/2020 (Morning)
26
(c) (d) 14 (a) 30 (b) 32
3
189. What is the value of the following (c) 15.5 (d) 13.5
– 15 + 90 ÷ [89 – {9 × 8 + (33 – 3 × 196. The value of 4 + [3 {35 + (42 + 10
7)}] ÷ 2 × 3 – 40) + 7] is :
CHSL 12/10/2020 (Morning) CHSL 14/10/2020 (Morning)
(a) 3 (b) 2 (a) 157 (b) 167
(c) 4 (d) 5 (c) 185 (d) 163
190. Find the value of 2.1 + 2.55 ÷ [63 197. The value of
– {7.5 × 8 + (13 – 2.5 × 5)}].
23
CHSL 12/10/2020 (Afternoon) 5 3 +7 2 - 6 - is :
(a) 2.8 (b) 2.0 2+ 3+ 6
(c) 3.0 (d) 3.1 CHSL 14/10/2020 (Afternoon)
191. The value of [0.9 – {2.3 – 3.2 – (7.1 (a) 0 (b) 16
– 5.4 – 3.5)}] is : (c) 12 (d) 10
CHSL 12/10/2020 (Evening)
 4  11 13  2 
(a) 0 (b) 1.8 198. The value of 5 9   4  6   ÷
(c) 2.6 (d) 0.18  
192. If the following interchanges are  3 4 5 4
2

made in signs and numbers, which 7 11 of 8 5  1 7  3  is :

equation sould be correct?
CHSL 14/10/2020 (Afternoon)
Interchange
Signs : + and – 1 1
(a) (b)
Numbers : 15 and 5 81 61
CHSL 13/10/2020 (Morning) 1 1
(a) 4 × 30 – 15 ÷ 5 + 12 = 21 (c) (d)
71 91
(b) 12 × 30 – 15 ÷ 5 + 4 = 37 199. Find the value of x in
(c) 4 × 30 – 15 ÷ 5 + 12 = 27 3
15625  x  4
(d) 4 × 30 – 5 ÷ 15 + 14 = 21
CHSL 14/10/2020 (Evening)
193. Evaluate
45 – 5 of (6.3 ÷ 9) + 7 × 0.5 (a) 625 (b) 343
CHSL 13/10/2020 (Afternoon) (c) 441 (d) 81
200. The value of 27 + [3(50 – 20)+168 (a) 1 (b) 2
÷ 4 + 2 – 11 × 2] (c) 3 (d) 0
CHSL 15/10/2020 (Morning) 208. Evaluate [7 + 7 × (7 + 7 ÷ 7)] + 7÷ 7
(a) 245 (b) 139 CHSL 20/10/2020 (Morning)
(a) 10 (b) 5
(c) 149 (d) 239
(c) 63 (d) 64
201. Find the value of 225 – [42 – {25 –
209. The value of
(18 – 18  13) }]. 1 1

CHSL 15/10/2020 (Evening) 4 1 8 8  51  5
 3 of 5
(a) 222 (b) 231 5 4 13 1  1 5 6 is :
(c) 223 (d) 244 5 8
202. The value of 72 – 3(2 + 24 ÷ 4 × 3 CHSL 20/10/2020 (Afternoon)
– 2 × 2) + 8 is : 2 7
CHSL 16/10/2020 (Morning) (a) (b)
15 30
(a) 72 (b) 32 1 1
(c) 36 (d) 24 (c) (d)
15 30
203. The value of 10 – [121 ÷ (11 × 11) 210. What is the value of – 77 + 800 ÷
– (–4) – {3 – (8 – 1)}] is : [83 – { 8 × 9 + (18 – 3 × 5)}]?
CHSL 16/10/2020 (Afternoon) CHSL 20/10/2020 (Evening)
(a) – 1 (b) 1 (a) 24 (b) 23
(c) 0 (d) 19 (c) 26 (d) 25
204. What is the value of 211. The value of
(9 + 3 – 16 ÷ 4 + 10) + {(3 + 5 × 2 ÷ 1 1 1
10)} × (18 – 4 of 5)?  
5 5 5  4 1  105 is :
CHSL 16/10/2020 (Evening) 1 1 1 5
(a) 15 (b) 10  of
5 5 5
(c) 5 (d) 8 CHSL 21/10/2020 (Morning)
205. Simplify the following : (a) 0 (b) 2
4  1 1  1  1 1   (c) 10 (d) 5
4  2  1 –     212. The value of
5  5 2  4  4 5  
CHSL 19/10/2020 (Morning) 3  1  1 
75  15  3 of 5  7   78  3 
(a) 1 (b) 3 5  14  3 
(c) 2 (d) 4 CHSL 26/10/2020 (Evening)
206. Find the value of 3 5
(a) (b)
 3  3  7 12
309 ÷    of (25  35) 12  . 3 5
 2 4  (c) (d)
CHSL 19/10/2020 (Afternoon) 8 8
(a) 8 (b) 16 213. The value of
(c) 12 (d) 4 1 5  1 1 2 1
2  of  5  2    3 is:
207. The value of 36 9  10 5 5 5
CHSL 26/10/2020 (Evening)
3  1   3 11  
1 - 3 ÷ 6 -  2 -   3 5
4  8   4 12   is : (a) (b)
7 12
CHSL 19/10/2020 (Evening) 3 5
(c) (d)
8 8
214. The value of [(3 + 5 – 4) + (17 – 3 (a) 0 (b) 2
× 4)] + [4 ÷ 2 – 16 ÷ 4 + 3] is : (c) 4 (d) 1
CHSL 26/10/2020 (Afternoon) 220. The value of
(a) 16 (b) 10 8 – [8 – (5 + 8) – {8 – (8 – 5 + 8 )} +
10] is :
(c) 12 (d) 4
CHSL 19/03/2020 (Morning)
215. The value of
(a) 5 (b) 20
1  9  5 1  5 1   (c) 0 (d) 10
3 –        
3  4  4 13  2 3   is : 221. Simplify the expression
CHSL 26/10/2020 (Evening) 25 – [16 – {14 – (18  8  3) }]
CHSL 19/03/2020 (Afternoon)
(a) 10 (b) 0
(a) 14 (b) 15
(c) 5 (d) 1
(c) 17 (d) 16
216. What will come at place of 222. The value of 7 + [44 ÷ 4 + {9 × 2 –
132  12  x  3  3 14 ÷ 7} + 5 × 2] is :
x, (x < 10) for
52  6  4  x 2
= 1?
CHSL 19/03/2020 (Evening)
CHSL 17/03/2020 (Morning) (a) 55 (b) 33
(a) 2 (b) 3 (c) 67 (d) 44
(c) 4 (d) 1 
223. The value of 2.4  0.6  3  0.16 
217. If x is the square of the number
2 1 3 2

× 0.27  0.83  0.16  is : 
when  of 6   of 1 is CGL Mains Tier-II (15/10/2020)
5 4 7 7
(a) 0.814 (b) 0.11
1
divided by 11 , then the value of (c) 1.1 (d) 1.36
4
81x is :
1 32 2 32 2
CHSL 18/03/2020 (Morning) 224. If   = a +b 2 ,
4 8 32 2 32 2
(a) 36 (b) 16
then what is the value of
(c) 9 (d) 4 (3a + 4b)?
6
3  CGL Mains Tier-II (15/10/2020)
(2x  3)  4  2 (3x  7)  5
 2 3    1
    2  3   (a) 99 (b) 97
218. If      =   3    , 2
 3   
    1
(c) 98 (d) 98
then the value of 3
2  42x is :
4 4 2  1 1 
CHSL 18/03/2020 (Morning) 225. The value of  7 of 2 5 ×1 3 -  3 2 - 2 6  
 
(a) 5 (b) 6
 1 1 1
(c) 3 (d) 4 ÷  3 ÷ 4 of 5  is ?:
5 2 3
4
219. If = ab 2x 3d 6, CGL Mains Tier-II (15/10/2020)
1 2  3
1
where a, b, c, d are natural (a) 10 (b) 7
2
number, then the value of a + b +
c + d is : 1
(c) 15 (d) 1
CHSL 18/03/2020 (Morning) 3
  1875 1200  175 CGL 2019 Tier-II (16/11/2020)
226. Let x =  3888  768  × ,
1792 (a) 5 (b) 7
then x is equal to : (c) 9 (d) 3
CGL Mains Tier-II (15/10/2020) 1 1 1 1
232. The value of 3  4 of 5  ÷
7 4 5 2 3 8
(a) (b)
12 9 1 1 1 1 1 1 
of      is :
5 5 2 4 4 2 8 4
(c) (d)
12 9 CGL 2019 Tier-II (16/11/2020)
227. The expression
53 13
15  10  5  is equal
(a)
60
(b)
15
10  20  40  5  80 7 3
to : (c) (d)
8 4
CGL Mains Tier-II (15/10/2020) 233. The value of 4 ÷ 12 of [3 ÷ 4 of {(4
(a) 5  2 2 (b) 5  2 2 – 2)× 6 ÷ 2}] – 2 × 6 ÷ 8 + 3 is :
CGL 2019 Tier-II (16/11/2020)

(c) 5 3  2 2  
(d) 10 3  2 5  1 1
228. The value of 3 ÷ 18 of 3 × 6 + 21 (a) 4 (b) 7
6 6
× 6 ÷ 18 – 3 ÷ 2 + 3 – 3 ÷ 9 of 3 × 1 1
9 is : (c) 2 (d) 3
3 3
CGL Mains Tier-II (15/10/2020)
73 5 73 5
47 41 234. The value of 
(a) (b) 3 5 3 5
6 9
lies between :
15 29 CGL 2019 Tier-II (16/11/2020)
(c) (d)
9 6
(a) 2 and 2.5 (b) 3 and 3.5
0.0203  2.92 (c) 1.5 and 2 (d) 2.5 and 3
229. The value of ÷
0.7  0.0365  2.9
82 3
2 2 235. If  a 3  b , then the
(12.12)  (8.12) 3 35
is :
(0.25)2  (0.25)(19.99) value of a + b is equal to :
CGL Mains Tier-II (15/10/2020) CGL 2019 Tier-II (18/11/2020)
(a) 0.05 (b) 0.5 (a) 18 (b) 15
(c) 0.1 (d) 0.01 (c) 16 (d) 24
230. The value of 0.57  0.432  0.35 is :
236. If x =  3  38 74 3
CGL 2019 Tier-II (16/11/2020)
where x > 0, then the value of x is
(a) 0.494 (b) 0.498 equal to :
(c) 0.498 (d) 0.494 CGL 2019 Tier-II (18/11/2020)
(a) 2 (b) 3
231. If 11  3 8  a  b 2 , then what
(c) 4 (d) 1
is the value of (2a + 3b)?
 8  2 15 1 242. The value of
237. The value of 5 – 
4 8  2 15 1 1 3 3 1 3
is equal to : 3  2 of 1     1  is :
3 2 5 8 7 4
CGL 2019 Tier-II (18/11/2020) CPO 23/11/2020 (Evening)
2 55 35
(a) (b) 1 (a) (b)
3 24 24
1 1 5 25
(c) (d) (c) (d)
2 4 24 24
238. Evaluate the following : 243. The value of
5 – [96 ÷ 4 of 3 – (16 – 55 ÷ 5)]
1 1 3 3 1 3
CGL 2019 Tier-II (18/11/2020) 2  2 of 1     1  is :
3 2 5 8 7 4
(a) 0 (b) 3 CPO 24/11/2020 (Morning)
(c) 2 (d) 4
35 25
239. The value of 8 – 3 ÷ 6 of 2 + (4 ÷ 4 (a) (b)
24 24
1 1 1
of ) ÷ 8 + (4 × 8 ÷ )× . 5 29
4 4 8 (c) (d)
24 24
CPO 23/11/2020 (Morning)
 1  1 3 1
244. The value of  5  2 of    5  of 
7 7 2 4 7 2
(a) – (b)
4 4  1 7 9  11
÷  5 9  7 8  9 20   21 is :
97 97
(c) – (d)
4 4 CPO 24/11/2020 (Morning)
3 (a) – 2 (b) 8
40 - of 32
4 35 15
240. The value of 3 is : (c) (d)
37 - of (34 - 6) 24 28
4
2 1 4 4 1
CPO 23/11/2020 (Morning) 245. The value of  of   1
3 10 9 5 9
(a) 1 (b) 0
8 3 3 1
1 1     is :
(c) – (d) 15 4 4 2
2 2
CPO 24/11/2020 (Evening)
241. The value of
49 25
 1 3 1  1 7 9  (a) (b)
 5  of    5  7  9  × 12 6
4 7 2 9 8 20 
17 14
11  1 (c) (d)
 9 3
 5  2 of  is :
21 2 246. The value of – 7 ÷ [5 + 1 ÷ 2 – {4 +
CPO 23/11/2020 (Evening) (4 of 2 ÷ 4) + (4 ÷ 4 of 2)}] is :
35 CPO 24/11/2020 (Evening)
(a) (b) 0 7
24
(a) 7 (b)
15 2
(c) (d) – 2 7
28
(c) – (d) – 7
2
 3 249. The value of
40  of 32
4 2 1 4 4 1 8 3 1
247. The value of is :  of   1    is
3 3 10 9 5 9 15 4 2
37  of (34  6)
4 CPO 23/11/2020 (Morning)
CPO 25/11/2020 (Morning)
49 17
3 1 (a) (b)
(a) 1 (b) 2 12 9
29 29 29 14
3 9 (c) (d)
(c) –1 (d) 1 6 3
29 29 250. The value of 7 ÷ [5 + 1 ÷ 2 – {4 +
248. The value of 1 – 3 ÷ 6 of 2 + (4 ÷ 4 (4 of 2 ÷ 4) + (5 ÷ 5 of 2)}] is :
1  1 1 CPO 23/11/2020 (Morning)
of ) ÷ 8 +  4  8    is :
4 4 8
7
CPO 23/11/2020 (Morning) (a) 7 (b)
2
69 69 7
(a) – (b) (c) – (d) – 7
4 4 2
7 7
(c) (d) –
4 4
 Answer Key
1.(d) 2.(d) 3.(c) 4.(b) 5.(b) 6.(d) 7.(a) 8.(c) 9.(a) 10.(a)
11.(a) 12.(d) 13.(c) 14.(d) 15.(d) 16.(c) 17.(a) 18.(d) 19.(b) 20.(b)
21.(d) 22.(b) 23.(c) 24.(d) 25.(c) 26.(c) 27.(d) 28.(c) 29.(d) 30.(a)
31.(a) 32.(b) 33.(a) 34.(b) 35.(d) 36.(a) 37.(c) 38.(d) 39.(d) 40.(c)
41.(a) 42.(c) 43.(d) 44.(a) 45.(c) 46.(b) 47.(b) 48.(a) 49.(d) 50.(b)
51.(b) 52.(c) 53.(a) 54.(c) 55.(d) 56.(d) 57.(a) 58.(d) 59.(d) 60.(a)
61.(c) 62.(b) 63.(c) 64.(b) 65.(b) 66.(c) 67.(a) 68.(d) 69.(d) 70.(b)
71.(c) 72.(d) 73.(a) 74.(d) 75.(d) 76.(b) 77.(b) 78.(a) 79.(c) 80.(c)
81.(c) 82.(d) 83.(c) 84.(c) 85.(b) 86.(b) 87.(d) 88.(c) 89.(d) 90.(a)
91.(d) 92.(a) 93.(d) 94.(b) 95.(d) 96.(d) 97.(c) 98.(c) 99.(a) 100.(c)
101.(c) 102.(b) 103.(c) 104.(a) 105.(d) 106.(d) 107.(a) 108.(d) 109.(d) 110.(c)
111.(c) 112.(c) 113.(b) 114.(c) 115.(a) 116.(c) 117.(a) 118.(d) 119.(d) 120.(b)
121.(a) 122.(c) 123.(b) 124.(c) 125.(c) 126.(c) 127.(a) 128.(a) 129.(b) 130.(d)
131.(c) 132.(b) 133.(a) 134.(b) 135.(c) 136.(b) 137.(d) 138.(a) 139.(c) 140.(c)
141.(b) 142.(d) 143.(c) 144.(d) 145.(a) 146.(c) 147.(d) 148.(c) 149.(d) 150.(d)
151.(d) 152.(d) 153.(a) 154.(a) 155.(b) 156.(a) 157.(a) 158.(a) 159.(a) 160.(a)
161.(c) 162.(d) 163.(c) 164.(a) 165.(a) 166.(d) 167.(b) 168.(c) 169.(d) 170.(a)
171.(b) 172.(b) 173.(b) 174.(a) 175.(c) 176.(b) 177.(b) 178.(d) 179.(d) 180.(a)
181.(d) 182.(d) 183.(b) 184.(b) 185.(d) 186.(d) 187.(a) 188.(b) 189.(a) 190.(c)
191.(a) 192.(a) 193.(b) 194.(b) 195.(b) 196.(b) 197.(c) 198.(a) 199.(c) 200.(b)
201.(b) 202.(b) 203.(b) 204.(b) 205.(b) 206.(d) 207.(a) 208.(d) 209.(d) 210.(b)
211.(a) 212.(a) 213.(d) 214.(b) 215.(b) 216.(d) 217.(a) 218.(a) 219.(c) 220.(c)
221.(d) 222.(d) 223.(c) 224.(c) 225.(a) 226.(c) 227.(c) 228.(a) 229.(a) 230.(c)
231.(d) 232.(a) 233.(a) 234.(a) 235.(a) 236.(a) 237.(b) 238.(c) 239.(d) 240.(a)
241.(d) 242.(b) 243.(d) 244.(b) 245.(a) 246.(a) 247.(a) 248.(b) 249.( c) 250.(d)