vinukk wrote:A contractor has two parallel contracts, contract 1 and contract 2, on which work needs to be finished in a specified number of days. He puts all his group of workers on contract 1. To be able to finish contract2 also on time he withdraws one worker each day, from second day onwards, from work on contract 1.

Work on contract 1 finishes when the last worker is withdrawn. Had no worker been withdrawn at any stage, work on contract 1 would have finished in 55% of the time it actually took to finish the work. What is the total number of workers in the contractors group?

5
10
15
20
25

I've added answer choices, which the GMAT would provide.

Let the rate for each worker = 1 unit per day.

We can plug in the answers, which represent the total number of workers.
Contract 2 is irrelevant, since there is no constraint on the amount of time required to complete Contract 2.
All that matters here is Contract 1.

The number of workers for Contract 1 decreases by 1 each day.
Result:
The number of workers from one day to the next constitutes a list of DESCENDING CONSECUTIVE INTEGERS.
For example, if there are a total of 3 workers, we get:
3 workers the first day.
2 workers the second day.
1 worker the third and LAST day.
Total days = 3.
Total work produced over the 3 days = 3+2+1 = 6 units.

As the example above illustrates:
The total number of days is equal to the total number of WORKERS.
The total amount of work is equal to the SUM of the consecutive integers.

With consecutive integers:
Average = (biggest + smallest)/2.
Sum = (number)(average).

When the correct answer choice is plugged in:
(faster time with no workers withdrawn)/(actual time with workers withdrawn) = 55%,

Answer choice C: 15
Total time = number of workers = 15 days.
Total work = (number)(average) = 15 * (15+1)/2 = 120 units.
If no workers are withdrawn, the time needed for all 15 workers to complete 120 units = 120/15 = 8 days.
Faster time/actual time = 8/15 ≈ 53%.
Eliminate C.

To increase the ratio of faster time to slower time, the faster time needs to INCREASE, implying that -- when all of the workers are present -- the work must be completed more SLOWLY.
To DECREASE the rate, the total number of workers must DECREASE.

Answer choice B: 10 workers
Total time = number of workers = 10 days.
Total work = (number)(average) = 10 * (10+1)/2 = 55 units.
If no workers are withdrawn, the time needed for all 10 workers to complete 55 units = 55/10 = 5.5 days.
Faster time/actual time = 5.5/10 = 55%.
Success!

The correct answer is B.

Nice explaination.
can it be solved without using options?

vinukk wrote:Nice explaination.
can it be solved without using options?

The sum of the first n positive integers = n(n+1)/2.

Let n = the number of workers.

As shown in my post above:
Actual time = number of workers = n.

Total work = sum of the descending consecutive integers from n to 1 = sum of the first n positive integers = n(n+1)/2.
Faster time with all n workers = (total work)/(total workers) = [n(n+1)/2] / n = (n+1)/2.

Since (faster time)/(actual time) = 55/100 = 11/20, we get:
[(n+1)/2] / n = 11/20

(n+1)/2n = 11/20

20n + 20 = 22n

20 = 2n

n = 10.