## 24 May 2017

### Practice Quantitative Aptitude Questions For NICL AO Prelims 2017

[Dated: 24th May] Practice Quantitative Aptitude Questions For NICL AO Prelims 2017 (Simplification):
Dear Readers, Important Practice Aptitude Questions for  Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the Bank Examination and other Competitive Examination can use this.

Directions (Q. 1-10): What approximate value should come in place of question mark (?) in the following questions?
1). 90.05 + 281 ÷ 4 – 151.06 = 3√?
a)   27
b)   343
c)   216
d)   729
e)   243

2). 17.982 ÷ (4.05)2 × 90.11 ÷ 4.98 = ?
a)   396
b)   336
c)   242
d)   325
e)   365

3). 80.04% of 150.16 + 60.02% of 50.07 = ?
a)   150
b)   125
c)   210
d)   175
e)   140

4). √628 × 17.996 ÷ 15.04 = ?
a)   30
b)   10
c)   5
d)   20
e)   15

5). (1/8) × 121 + (1/5) × 76 - ? = 25
a)   5
b)   45
c)   15
d)   10
e)   25

6). (28.07 × 4.97 + 15 × 6.05) / [(7.03)2 + √256.10 + 13.0001] = ?
a)   2
b)   -2
c)   3
d)   6
e)   4

7). 849 of (11/16.13) of (441.26 / 20.98) ÷ (17.13 / 319.85) = ?
a)   238000
b)   201300
c)   234500
d)   231000
e)   213000

8). √[√(14640) + √? ] = 13
a)   2400
b)   2916
c)   2305
d)   2210
e)   2350

9). 17.156 × (864.63 – 356.34) = ? – 6909.8003
a)   15782
b)   15802
c)   15402
d)   15852
e)   15560

10). 4567.8 – (221 × 9.7) = 5059 - ?
a)   2590
b)   2409
c)   2380
d)   2700
e)   2485

1). D)
3√? = 90.05 + 281 ÷ 4 - 151.06 90
= 90 + 280 ÷4  - 151 = 90 + 70 - 151 = 9
? = 9 x 9 x 9 = 729

2). E)
? = (18)2  ÷ (4)2 × 90 ÷ 5
= (4.5)2 × 18 = 20.25 x 18 = 364.5 = 365

3). A)
? = 80.04% of 150.16 + 60.02% of 50.07 = 80% of 150 + 60% of 50
= 80 x 1.5 + 60 x 0.5 = 120 + 30 = 150

4). A)
? = √628 x 17.996 ÷ 15.04
= 625 × 18 ÷ 15 = 25 × 18 ÷15 = 30

5). A)
(1/8) x 121 + (1/5) x 76 - ? =25
or, ? = (1/8) x 120 + (1/5) x 75 - 25 = 15 + 15 – 25 = 5

6). C)
? = (28.07x4.97 + 15x6.05) /[ (7.03)2 + √256.10 +13.0001]
? =  (28 x 5 + 15 x 6) / [(7)2 + √256 + 13] =(140+90) / (49+16+13)
=  230 / 78

7). D)
? = 849 × 11/16.13 × 441.26/20.98 ÷ 17.13/319.85
= 850 × 11/16 × 441/21 × 320/17 = 50 × 11 × 21 × 20 = 231000

8). C)
13 = √[√(14640) + √? ]
Squaring both sides
or, 169 = {√[√(14640) + √? ]} 2 = √(14640) + √?
or, 169 = √(14640) + √? = 121 + √?
or, √? = 169 – 121 = 48
? = (48)2 = 2304 = 2305

9). E)
? = 17 x (865 — 356) + 6910
= 8653 + 6910 = 15563 = 15560

10). D)
? = 5059 - 4567.8 + (221 x 9.7)
= 5060 - 4570 + (221 x 10) = 5060 — 4570 + 2210 = 7270 - 4570 = 2700