A contractor hires 100 men to finish a job in 50 days. After 14 days,

Total work by 50 days =50*100=5000units
Work done for 14 days=14*100=1400units
WOrk remaining=3600units
The remaining work is done as follows----- {(100-n)*(36-x)---n being no. of workers left job and x ,no. of days work done by 2n men and 36 no. of days left} + {(100+n-2n)*x }
equating the above to 3600 yields 18 as our answer

First attempt took 6 mins long. I believe it's because I was confused about how many days in total this job took even after making changes.

The rate of each worker is (1/5000).

After 14 days, 3600/5000 of the job is left.

If x = some days that 100 - n workers worked for,

3600 = (100 - n) x + (100 - n + 2n) (36 - x)

This will result in x = 18. 36 - 18 = 18. So, it took 18 days to complete the job with 2n additional workers.

Let us say that "n = 1". Also, let us assume that "some = 1000000".

Therefore, the question should now read as:

A contractor hires 100 men to finish a job in 50 days. After 14 days, 1 man leaves. After 1000000 more days, the contractor hires 2 more men to complete the job on time. For how many days did these 2 men work?

So during those 1000000 days, 99 men were still working, correct? So the work would have been finished before hiring 2 more men. So the 2 men need to work 0 days.

Please help me understand what mistake I have made.

To be able to complete the work in time the contractor needs to hire men before the ones already working complete it (in delayed time). I think this is the unstated and understood part of the question. Extrapolation of given conditions help you arrive at an answer sometimes but not in this case. Liked your line of thought though.

A good question by Veritas. Thanks team.

Logical approach..here it goes..

We will do this with the law of averages...
For the first 14 days some work was done..
left days? 36.

Now for some days \(100-n\) people were working...

after some days \(2n\) people joined..or in other words..

we then had a total of \(100+n\) men with us.
We know that every man does the same amount of work everyday(the efficiency is constant).

Now think about this..
the same 100 people were there when we had \(n\) less men, and when we had \(n\) more men. What comes in your mind? we had to employ \(n\) more men than usual to make up for the work that we lost when we had exactly \(n\) less men. These days have to be the same..why? because if we have these \(n\) extra than usual(100 men) men work for more days than when we had \(n\) less men, then we will have more than needed work done, and vice versa.

This brings us to think that there were less men and more men for the same number of days..which is?

36/2 = 18 days

(B)

hi

the solution 1 is beautifully done...

anyway, would you please speak something more about solution 2....? Is it not such that, since it is assumed that the work was completed on time, then (36-x)*2n = the remaining job..? So how can this expression equals x*2n....?

thanks in advance .....

Total job = \(100m*50d = 5000md\) (\(md\)=man-days)

Job done in 14 days = \(100m*14d = 1400md\). Remaining job = \(5000md - 1400md = 3600md\).

Let the number of days that \((100-n)\) men worked be \(k\), then \((100-n)*k + (100+n)*(50-14-k) = 3600\).

\(100k - nk + 3600 - 100k + 36n - nk = 3600\).

\(36n - 2nk = 0\)

\(2nk = 36n\)

\(k = \frac{36}{2} = 18.\)

So, the \(2n\) men worked for \((36-18) = 18\) days.

We are given that 100 men will finish a job in 50 days; thus, the rate of the 100 men is 1/50 and the rate of each man is (1/50)/100 = 1/5000.

This is what we know:

The 100 men work for the first 14 days. Then n men leave and the (100 - n) men work for some number of days. Then 2n men join in (so now there are 100 - n + 2n = 100 + n men) and they all work for the remaining days and complete the job on time, in 50 days.

If we let x = the number of days the (100 - n) men work (after the first 14 days but before the 2n men join in), then (100 + 2n) men (after the 2n men join in) work for (36 - x) days, since the total number of days for the job is still 50. Thus, we can create the following equation:

100(1/5000)(14) + (100 - n)(1/5000)(x) + (100 + n)(1/5000)(36 - x) = 1

The 1 on the right-hand side of the equation represents the complete job and each of the addends on the left-hand side represents the fraction of the job the 100 men, (100 - n) men, and (100 + n) men contribute at different stages of the job, respectively.

Multiplying the equation by 5000, we have:

100(14) + (100 - n)(x) + (100 + n)(36 - x) = 5000

1400 + 100x - nx + 3600 - 100x + 36n - nx = 5000

5000 + 36n - 2nx = 5000

36n = 2nx

36 = 2x

18 = x

Since the extra 2n men actually work (36 - x) days, they work 36 - 18 = 18 days.

Answer: B

Here is my solutions -






Please let me know if i am doing anything wrong!!!

Let, the 2n people were hired d days after the n people leave.

nd=n(36-d)
d=36-d
d=18

This is so, because half of the 2n people will work just to cover the gap created by the leave of the n people.

Considering the total work to be 5000 units.
(100)*(50) = 5000 units.
Assuming after n people left 100 - n workers worked for x days.
For the remaining 50 - (14+x) days = 36 - x days 100+n workers worked as 2n workers were added.
Hence this can be represented as the total work done on time.
100*(14) + (x)*(100-n) + (36 - x)*(100 + n) = 5000.
3600 = 100x - x*n + 3600 + 36n - 100x - n*x.
36n = 2nx.
x = 18 days


Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.