Quantitative Aptitude

CWC/FCI Prelims 2019 – Quantitative Aptitude Questions (Day-39)

Dear Readers, Exam Race for the Year 2019 has already started, To enrich your preparation here we are providing new series of Practice Questions on Quantitative Aptitude – Section for CWC/FCI Exam. Aspirants, practice these questions on a regular basis to improve your score in aptitude section. Start your effective preparation from the right beginning to get success in upcoming CWC/FCI Exam.

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Directions (Q. 1 – 5) what approximate value should come in place of (?) in the following questions?

1) (52.13)2 ÷ 13.9 × 7.2 + (44.78)2 = 5 × ? – 35 % of 3097

a) 892

b) 716

c) 954

d) 1028

e) 1150

2) 11 4/5 % of 1499 + 5 2/3 % of 2949 = ?

a) 564

b) 678

c) 252

d) 357

e) 423

3) (4/9) of 1223 = ? – 27.4 × 12.78 ÷ 38.89

a) 420

b) 646

c) 553

d) 395

e) 712

4) 13.9 % of 742 – ? % of 393 = 89.78 – (12.23)2

a) 56

b) 28

c) 62

d) 77

e) 40

5) (1023.77)2 ÷ (255.89)3/2 × 64.12 = (3.92)4 × 4?

a) 3

b) 5

c) 9

d) 7

e) 6

Directions (Q. 6 – 10): Study the following information carefully and answer the given questions:

There are 3 Schools A, B and C having the total strength of 7600, which is in the ratio of 8 : 7 : 4. All the given students were appeared for the examination. Total number of boys appeared for the examination from school A is 62 ½ % of total students in that school. The ratio of boys to that of girls appeared for the examination from school B and C is 3 : 4 and 5 : 3 respectively. The percentage of failed students from School A and C is 24 % and 32 % of appeared students in those schools respectively. Total number of students passed from School B is 2380.

 

6) Find the ratio between the total number of passed girls from school A to that of total number of failed students from school C, if the total number of passed boys from school A is 1592?

a) 52 : 17

b) 38 : 15

c) 105 : 64

d) 117 : 59

e) None of these

7) Find the difference between the total number of girls appeared for the examination from school A to that total number of boys appeared for the examination from school C?

a) 200

b) 350

c) 280

d) 420

e) None of these

8) Total number of failed students from school B is approximately what percentage of total number of passed students from school C?

a) 25 %

b) 55 %

c) 70 %

d) 40 %

e) 85 %

9) Find the difference between the average number of students passed the examination to that of failed the examination from all the given schools together?

a) 1800

b) 1400

c) 2200

d) 2500

e) None of these

10) Total number of boys appeared for the examination from school A is what percentage more/less than the total number of students appeared for the examination from school C?

a) 25 % more

b) 10 % less

c) 10 % more

d) 25 % less

e) 15 % more

Answers :

Direction (1-5) :

1) Answer: a)

522 ÷ 14 × 7 + 452 = 5x – 35 % of 3100

(52*52*7)/14 + 2025 + (35/100)*3100 = 5x

1352 + 2025 + 1085 = 5x

5x = 4462

X = 892.4 = 892

2) Answer: d)

12 % of 1500 + 6 % of 2950 = x

X = (12/100)*1500 + (6/100)*2950

X = 180 + 177 = 357

3) Answer: c)

(4/9)*1224 = x – (27*13)/39

544 + 9 = x

X = 553

4) Answer: e)

14 % of 750 – x % of 400 = 90 – (12)2

(14/100) * 750 – (x/100) * 400 = 90 – 144

105 – 90 + 144 = 4x

159/4 = x

X = 40

5) Answer: a)

(1024)2 ÷ (256)3/2 × 64 = 44 × 4x

(45)2 ÷ (44)3/2 × 43 = 44 × 4x

410 ÷ 46 × 43 = 44 × 4x

410 – 6 + 3 – 4 = 4x

43 = 4x

X = 3

Directions (Q. 6 – 10):

Total students in all the given schools (A, B and C) = 7600

Total students from school A = 7600*(8/19) = 3200

Total students from school B = 7600*(7/19) = 2800

Total students from school C = 7600*(4/19) = 1600

Total number of boys appeared for the examination from school A

= > 62 ½ % of 3200

= > (125/200)*3200 = 2000

Total number of girls appeared for the examination from school A

= > 3200 – 2000 = 1200

Total number of boys appeared for the examination from school B

= > 2800*(3/7) = 1200

Total number of girls appeared for the examination from school B

= > 2800 – 1200 = 1600

Total number of boys appeared for the examination from school C

= > 1600*(5/8) = 1000

Total number of girls appeared for the examination from school C

= > 1600 – 1000 = 600

Total number of students failed the examination from school A

= > 3200*(24/100) = 768

Total number of students passed the examination from school A

= > 3200 – 768 = 2432

Total number of students failed the examination from school C

= > 2800*(32/100) = 896

Total number of students passed the examination from school C

= > 2800 – 896 = 1904

Total number of students passed from School B = 2380

6) Answer: c)

The total number of passed students from school A = 2432

The total number of passed boys from school A = 1592

The total number of passed girls from school A

= > 2432 – 1592 = 840

The total number of failed students from school C = 512

Required ratio = 840 : 512 = 105 : 64

7) Answer: a)

The total number of girls appeared for the examination from school A

= > 37 ½ % of 3200

= > (75/200)*3200 = 1200

The total number of boys appeared for the examination from school C

= > 1600*(5/8) = 1000

Required difference = 1200 – 1000 = 200

8) Answer: d)

Total number of failed students from school B = 420

Total number of passed students from school C = 1088

Required % = (420/1088)*100 = 38.6 % = 40 %

9) Answer: b)

The average number of students passed the examination from all the given schools together

= > (2432 + 2380 + 1088)/3

= > 5900/3

The average number of students failed the examination from all the given schools together

= > (768 + 420 + 512)/3

= > 1700/3

Required difference = (5900/3) – (1700/3) = 4200/3 = 1400

10) Answer: a)

Total number of boys appeared for the examination from school A

= > 2000

Total number of students appeared for the examination from school C

= > 1600

Required % = [(2000 – 1600)/1600]*100 = 25 % more

 

This post was last modified on April 16, 2019 6:18 pm