Test Level 1: Time & Work - 2 (August 27) - CAT MCQ

Work done by Q

Assume each man could do 'x' work in 1 day.
So, total work = 10 × 40x = 400x
But they completed the work in 34 days (40 - 6).
Assume after 'd' days, they increased their efficiency.
After increasing the efficiency, they could do 1.25x work in 1 day.
So, 400x = 10 × d × x + 10 (34 - d)1.25x.
⇒ 40 = d + (34 - d)1.25
⇒ 0.25d = 2.5  d = 10
Thus, they increased their efficiency 10 days after the start of the work.
Hence, answer option 4 is correct.

Suppose the first pipe takes A minutes, the second pipe takes B minutes, the third pipe takes C minutes and the fourth pipe takes D minutes to fill the pool.

Add equations (ii), (iii) and (iv): 

The fifth pipe can empty the tank in 30 minutes.
So, combined time taken by all the five pipes

He will complete the work in 20 days. Hence, he will complete ten times the work in 200 days.

A’s one day work will be 5%, while B will do 6.66 % of the work in one day. Hence, their total work will be 11.66% in a day.
In 8 days they will complete → 11.66 x 8 = 93.33%
This will leave 6.66% of the work. This will correspond to 4/7 of the ninth day since in 6.66/11.66 both the numerator and the denominator are divisible by 1.66.

N + A = 10% 
N = 8.33% 
Hence A = 1.66% → 60 days.

24 man days + 18 women days = 20 man days + 28 woman days
→ 4 man days = 10 woman days.
→ 1 man day = 2.5 woman days
Total work = 24 man days + 18 woman days = 60 woman days + 18 woman days = 78 woman days.
Hence, 1 man + 1 woman = 3.5 women can do it in 78/3.5 = 156/7 = 22(2/7) days.

A = 10%, B = 5% and Combined work is 20%. Hence, C’s work is 5% and will require 20 days.

A = 20%, B = 10% and A + B + C = 50%. Hence, C = 20%. Thus, in two days, C contributes 40% of the total work and should be paid 40% of the total amount.

Let the time taken by Arun be ‘t’ days. Then, time taken by Vinay = 2t days. 1/t + 1/2t = 1/7 → t = 10.5