Question

A manager has two parallel projects, project 1 and project 2 in his contract. To finish both the projects on the stipulated time, he started withdrawing 1 person from project 1 every day from the second working day until the work got finished the last worker was withdrawn. Had no worker been withdrawn then work would have finished in 55% of the time it took to finish the work. What is the total number of workers who participated in project 1?
1. 10
2. 12
3. 14
4. 16
5.

Answer 1

Correct Answer - Option 1 : 10

Calculation:

Had all workers participated on all days then they would have completed the work in 55% of the actual time or

in other words, it would have taken 55 man-days to complete the work.

Now, 1 man is withdrawn every day from the second day, so it took 100 man-days to complete the work.

100 man-days can have 10 men work for 10 days (As one man is withdrawn every day from the second day)

That is, On day 1, there were 10 men working

On day 2, there were 9 men working

On day 10, there was only one man working

Total number of man-days 

⇒ 10 × (10 + 1)/2 = 55 man-days 

This means the contractor had 10 men in his group.

If M₁ men can do W₁ in D₁ by working H₁ hours per day and M₂ men can do W2 in D2 by working H2 hours per day then 

⇒  M₁D₁H₁/W₁ = M2D2H2/W2

If A can do a piece of work in p days and B can do it in q days, Then A and B together can complete the same in 

⇒  pq/(p + q)