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

Question

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

A. 20
B. 18
C. 16
D. 12
E. 8

A. 20
B. 18
C. 16
D. 12
E. 8

smyarga · Accepted Answer

Solution 1  
Suppose a job is to make 5000 details. So 1 men in 1 day can do  5000/100*50=1  detail.

After 14 days they make 14*100=1400 details.

After  x  more days they do  1400+x\cdot (100-n)  details. So, we have  5000-1400-x(100-n)=3600-x(100-n)  details to do and  50-14-x=36-x  days.

We have an equation:  3600-x(100-n)=(36-x)*(100-n+2n)

3600-100x+xn=3600+36n-100x-xn

2xn=36n

x=18

So, these  2n  men work for  36-18=18  days.

Solution 2  
After 14 days we have 36 days to finish the work. And after  x  more days we didn't do  x\cdot n  details, because  n  people leave the work.

To make everything on time we hire  2n  people or just  n  people more to finish the undone work in  36-x  days. So,  x\cdot n=(36-x)\cdot n  or  x=18 .

The correct answer is 18.

iamevitaerc · Answer

Step1: Find Each day what % of work is finished ; it requires 50 days to complete which means each day 1/50= 2% work will be completed.
Step 2 : After 14 Days = 14* 2= 28% of work is finished. remaining is 72% ( which means if same 100 men were about to complete the job they would have required 36 more days)
Step3: Let us assume all people have left job. therefore contractor hires 2n people. initially it was taking 36 days to compete for n people. at present 2n people are present to finish job. hence 18 more days.

hak15 · Answer

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

dina98 · 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.

dixitraghav · Answer

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, n men leave. After some more days, the contractor hires 2n more men to complete the job on time. For how many days did these 2n men work?

A. 20
B. 18
C. 16
D. 12
E. 8

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.

Mohammed22Abrar · Answer

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.

keats · Answer

A good question by Veritas. Thanks team.

ShashankDave · Answer

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)

testcracker · Answer

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 .....

TimeTraveller · Answer

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.

ScottTargetTestPrep · Answer

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

manishtank1988 · Answer

Here is my solutions -

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

NaeemHasan · Answer

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.

GmatPoint · Answer

﻿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
﻿

BadukMagician · Answer

Evil question I got stuck at the half-way point.

Matrix:
# -*- r -*- t == w
100 -*- 1/5000 -*- 50 = 1
100 -*- 1/5000 -*- 14 = 1400/5000
 This means that there is 3600/5000 left of work to complete  
So what do we know? 
We know that it we had (100-n) men work for (36-x) amount of days at a rate of (1/5000 per man)
We also know that we had (100-n+2n) men work for (x) amount of days at a rate of (1/5000 per man)

so logically if we should set the sum of those two amounts to the amount of remaining work:

(100-n)(36-x)(1/5000) + (100-n+2n)(x)(1/5000) = 3600/5000

I don't like fractions and I can see that we can multiply the entire thing by 5000, to get rid of the denominator

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

Combining like terms leaves you with

-36n+2nx = 0
Factor out an -n

-n(36 - 2x) = 0

One of the multiples have to equal 0, in this instance we are looking for the amount of days worked so what amount for x would cause (36 - 2x) to be 0?

Our answer is  B. 18

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