24 May 2017

Practice Quantitative Aptitude Questions For NICL AO Prelims 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.

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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

Solution With Answer Key:
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




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