Payment Processing...

Quantitative Aptitude DI - Oct 14 2020

Here we are providing new series of Quantitative Aptitude Questions for upcoming exams, so the aspirants can practice it on a daily basis.

Study the following data carefully and answer the questions:

The given line graph shows number of days taken by the A, B, C, D and E to complete the work.

1) A started the work and left after x days and then F work for y days and then left the work. After F left, D complete the remaining work in 3 days and ratio of the value of x to y is 1:2, find the value of y?

a) 3 days

b) 8 days

c) 5 days

d) 12 days

e) 16 days

2) A, B and C together can starts the work. If C works with 75% of his efficiency and A worked 25% more efficiently. All together gets the total wages Rs.6000. Find the individual wage of A?

a) Rs.1200

b) Rs.1500

c) Rs.1800

d) Rs.2000

e) None of these

3) A, D and F starts works on alternatively, starting from F, then A and then D. If A works with (3/4)th of his efficiency, D works with (1/2)th of his efficiency and F works with 66(2/3)% more than his efficiency, in how many days the work will be completed?

a) 20(7/9) days

b) 21(7/9) days

c) 22(7/9) days

d) 23(7/9) days

e) 19(7/9) days

4) E worked for x days and then left the work, A and C together can complete the remaining work in (x + 2.5) days. If E complete 33(1/3)% of the work in x days, find the value of x?

a) 5 days

b) 8 days

c) 10 days

d) 12 days

e) 16 days

Answers :

1) Answer: D

LCM of 30, 24, and 10 = 360

Hence A= 360/30 = 12, F = 360/24 = 15, D = 360/10 = 36

Work done by A = 12x, F = 15y, D = 36 * 3 = 108

12x + 15y + 108 = 360

x/y = 1/2

y = 2x

12x + 15 * 2x + 108 = 360

42x = 252

x = 6

y = 2 * 6 = 12 days

2) Answer: B

LCM of 30, 12, 18 = 360

Hence A = 360/30 = 12, B = 360/12 = 30 and C = 360/18 = 20

75% of C’s efficiency = 15

A worked 25% more efficiently = 15

B = 30

A, C and B = 15: 15: 30= 1: 1: 2

Total wage = 6000

A’s wages = 1/4 * 6000 = Rs.1500

3) Answer: A

F works with 66% more efficiently = 14.4

A works with 3/4 of his efficiency = 40

D works with 1/2 of his efficiency = 20

LCM of 14.4, 40, 20 = 288

Work done by A and D and F for 3 days = 7.2 + 14.4 + 20 = 41.6

After 18 days = 41.6 * 6 = 249.6

Remaining work = 288 – 249.6 = 38.4

After 2 days F and A complete = 20 + 7.2 = 27.2

Remaining work = 38.4 – 27.2 = 11.2

D complete remaining work = 11.2/14.4 days

Total time = 20(7/9) days

4) Answer: A

LCM of 15, 18, 30 = 360

Hence E = 360/15 = 24, C= 360/18 = 20, A = 360/30 = 12,

Work done by E = 24x

Work done by A and C in (x + 2.5) days = (A + C) = 20 + 12 = 32

A and C together can complete 66(2/3)% of the work.

24x/(32 * (x + 2.5)) = (100/3)/(200/3)

48x = 32x + 32 * 2.5

16x = 32 * 2.5

x = 5 days