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

__Answer key__2.a)

3.a)

4.e)

5.c)

6.a)

7.c)

8.a)

9.d)

10.b)

__Solution__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

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