# Quantitative Aptitude Practice Test for SBI PO Prelims

QUANTITATIVE APTITUDE PRACTICE TEST
Directions(1-5): Study the following pie chart carefully and answer the given questions.
Percentage of different types of employees in a company in two consecutive years
Total no. of employees = 42,980

Total no. of employees = 48,640
1.In 2005 the total no. of which of the following types of pairs of employees was approximately equal to A type of employees in 2006?
a)B and C
b)A and C
c)D and E
d)C and D
e)C and F

2.From 2005 to 2006 in the case of which of the following types of employees the change was maximum?
a)B
b)D
c)C
d)A
e)None of these

3.What was the approximate difference in the number of B type of employees during 2005 and 2006?
a)2285
b)2325
c)2085
d)2620
e)1825

4.If the no.D type of emplyees in 2006 was 5000, what would have been its approximate percentage in the company?
a)8%
b)12%
c)14%
d)16%
e)10%

5.The no. of A type of employees in 2006 was approximately what percent of the no. A type of employees in 2005?
a)115%
b)140%
c)125%
d)130%
e)95%

Directions(6-10): Study the following information and answer the given questions.
Percentage of students in various courses (A, B, C, D, E, F) and percentage of girls out of these.
Total students = 1200 (800 girls + 400 boys)
Total Girls = 800
Percentage of Girls in courses

6.For Course D, what is the respective ratio of boys and girls?
a)3:4
b)4:5
c)3:5
d)5:6
e)None of these

7.For which pair of courses is the number of boys the same?
a)E and F
b)A and D
c)C and F
d)B and D
e)None of these

8.For Course E, the number of girls is how much percent more than the boys for Course E?
a)250
b)350
c)150
d)80
e)None of these

9.For which course is the number of boys the minimum?
a)E
b)F
c)C
d)A
e)None of these

10.How many girls are in Course C?
a)44
b)16
c)40
d)160
e)None of these

1.d)
2.a)
3.a)
4.e)
5.c)
6.a)
7.c)
8.a)
9.d)
10.b)

Solution
1.No. of A type of employees in 2006 = 0.22 × 48640 = 10700 (approx)
In 2005,
A = 0.20 × 42980 = 8596
B = 0.06 × 42980 = 2579
C = 0.10 × 42980 = 4298
D = 0.15 × 42980 = 6447
E = 0.27 × 42980 = 11605
F = 0.22 × 42980 = 9455
C + D = 4298 + 6447 = 10745

2.In 2006,
A = 10700
B = 0.10 × 48640 = 4864
C = 0.11 × 48640 = 5350
D = 0.09 × 48640 = 4377
E = 0.27 × 48640 = 13133
F = 0.21 × 48640 = 10214
% change during 2005-2006,
A = [(10700 - 8596)/8596] × 100 = 24.5% (approx)
B = [(4864 - 2579)/2579] × 100 = 88.6% (approx)
C = [(5350 - 4298)/4298] × 100 = 24.5% (approx)
D = [(6447 - 4377)/6447] × 100 = 32.1% (approx)
E = [(13133 - 11605)/11605] × 100 = 13.2% (approx)
F = [(10214 - 9455)/9455] × 100 = 8% (approx)
The % change was maximum for B.

3.(4864 - 2579) = 2285

4.(5000/48640) × 100 = 10% (approx)

5.(10700/8596) × 100 = 125% (approx)

6.For Course D,
No. of girls = 30% of 800 = 240
No. of students = (35 × 1/100 × 1200) = 420
Therefore, no. of boys = (420 - 240) = 180
Required ratio = 180 : 240 = 3 : 4

7.Number of boys,
in Course E = (12% of 1200 - 14% of 800) = (144 - 112) = 32
in Course F = (13% of 1200 - 14% of 800) = (156 - 112) = 44
in Course A = (20% of 1200 - 30% of 800) = (240 - 240) = 0
in Course D = (35% of 1200 - 30% of 800) = (420 - 240) = 180
in Course C = (5% of 1200 - 2% of 800) = (60 - 16) = 44
Obviously pair C and F is the answer.

8.For Course E,
No. of girls = (14% of 800) = 112
No. of boys = 32 Required% = (80/32) × 100 = 250

9.In Course A, the number of boys the minimum.

10.No. of girls in Course C = 2% of 800 = 16 