21 May 2017

Practice Quantitative Aptitude Questions For NICL AO Prelims 2017

Practice Quantitative Aptitude Questions For NICL AO Prelims 2017
[Dated: 21st May] Practice Quantitative Aptitude Questions For NICL AO Prelims 2017 (Data Interpretation & 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.

https://www.youtube.com/channel/UC-2Ga6vFz3l4Q-O0jFVnIIg
Directions (Q. 1-5): Refer to the pie-charts and answer the given questions.
Distribution of number of students (both male and female) in five management institutes in 2013 Total number = 2100

Distribution of number of female students in five management institutes in 2013 Total number : 900

1). 1/3 rd of the number of students (both male and female) in institute C are Science graduates. If the number of male science graduate students in institute C is 32, what percentage of female students in institute C are science graduates?
a)    45
b)    35
c)    30
d)    40
e)    50

2). Number of students (both male and female) in institute A increased by 50% from 2013 to 2014. If the respective ratio of number of male and female students in 2014 in institute A was 4 : 3, what was the number of female students in institute A in 2014?
a)   183
b)   195
c)   171
d)   189
e)   187

3). What is the central angle corresponding to the number of students (both male and female) in institute D?
a)   63.40°
b)   65.20°
c)   71.30°
d)   61.20°
e)   67.60°

4). What is the average number of male students in institutes A, D and E?
a)   242
b)   252
c)   248
d)   246
e)   238

 5). Number of female students in institute B is what percent more the number of male students in institute C?
a)   30
b)   29
c)   32
d)   42
e)   36

6). What is the difference between total number of female students in institutes A and E together and number of students (both male and female) in ,institute B?
a)   215
b)   213
c)   207
d)   203
e)   217

Directions (Q. 7-11): What approximate value should come in place of question mark (?) in the questions given below:
7). √3598.9 x [(10008.99)2 / 10009.001] x 0.4987 = ?
a)   400168
b)   200368
c)   300270
d)   300570
e)   310670

8). 39.05 x 14.95 - 27.99 x 10.12 = (36 + ?) × 5
a)   25
b)   31
c)   125
d)   8
e)   45

9). 68.25 x 170 + 28 x 16.5 —125 x 16.5 = ?
a)   9600
b)   9800
c)   10000
d)   11500
e)   11000

10). 487.532 +2849.029 —675.48 = 743.095 +?
a)   1620
b)   1920
c)   1820
d)   2020
e)   1720

11). 142% of 3915 +2874 = 12600 —?
a)   4615
b)   4565
c)   4260
d)   4090
e)   4165

Explanation With Answer Key:
1). D)
 Total number of students (both male and female) in institute C =2100 x (20 /100) = 420
Total number of science graduates in institute C = 420 x (1/3) =140
Total number of female science graduates in institute C =140 - 32 =108
Total number of female students in institute C =900 x (30/100) =270
Required percentage = (108/270) x 100 = 40%

2). D)
 Total number of students (both male and female) in institute A in year 2014
= 2100 x (14/100) x (150/100) = 441
Total number of females in year 2014
=  3 /(3+4) x 441 = (3/7) x 441 =189

3). D)
 Required central angle = (360 / 100) x 17 =61.2°

4). C)
 Total number of students (both male and female) in institute A, D and E .
=2100x (14+17+25) /100
=2100 x 56/100 =1176
Total number of female students in institute A, D and E = 900 x [(18 + 15 + 15) / 100] = 900 x (48/100) = 432
 Total number of male students in institute A, D and E
= 1176 — 432 =744
Required average = 744/3 = 248

5). C)
 Number of female students in institute B = 900 x (22/100) =198
The number of male students in institute C = 2100 × (20 / 100) – 900 × (30 / 100)
= 420 —270 =150
Required percentage = (198 - 150) / 150 × 100
=  48/150 x100 = 32

6). C)
 Total number of female students in institute A and E together
= 900 x (18 + 15) /100 
= 900 x 33 / 100
=297
Total number of students (both male and female) in institute B
= 2100 x 24 / 100
=504
Required difference =504-297 =207

7). C)
? = √3598.9 x [(10008.99)2 / (10009.001)] x 0.4987
= √3600 x [(10009)2 / 10009] x 0.4987
= 60 x 10009 x 0.5 = 30 x 10009 = 300270

8). A)
39.05 x 14.95 — 27.99 x 10.12 = (36 + ?)5
or, 39 x 15 — 28 x 10 = 180 + 5 x (?)
or, 5 x ? = 585 — 280 — 180 = 585 — 460 = 125
? = 125/5 = 25

9). C)
? = 68.25 x 170 + 28 x 16.5 — 125 x 16.5 = 11602.5 + 462 — 2062.5
= 12064.5 — 2062.5 = 10002 = 10000

10). B)
? = 487.582 + 2849.029 — 675.48 — 743.095 = 488 + 2849 — 675 — 743 = 1919 1920

11). E)
? = 12600 — (142 / 100) x 3915 - 2874

= 12600 — 5560 — 2874 = 4166 = 4165



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