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They finished 1/4 or 25% work in 20 days.
Therefore , 40 workers will take 20*4 = 80 days to finish the job from scratch.
20 days has already passed. To finish the job in 10 days , workers needed = 8 * 40 = 320.
Additional workers required = 320-40 = 280
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I've solved this differently.

40 workers, 20 days, 1/4
40 workers, 10 days, 1/2 * 1/4 = 1/8

Amount of work incomplete - 3/4

Therefore, if 40 workers complete 1/8th of the work in 10 days, we should have 240 (40*6) workers to complete 6*(1/8) work in 10 days.

Additional workers = 240 - the 40 who are already working = 200 I.e. (C)

Hope this helps.

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Hey Nikhilaery, can you explain how did you arrive at
Quote:
Therefore, if 40 workers complete 1/8th of the work in 10 days, we should have 240 (40*6) workers to complete 6*(1/8) work in 10 days.
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Hi Nikhilaery

Thanks this helps

Quote:
Hi Harsh,

Firstly, the key is "all the workers work at a constant rate".

I hope you understand how I arrived at 40 workers complete 1/8th piece of work in 10 days.

So knowing that 1/4th of work is already completed, we can infer that 3/4th is still pending. If I could arrive at an equation which could tell me how many workers and days will be required to finish 3/4th of pending work, that would suffice my requirement.

40 workers in 10 days complete 1/8th piece of work.

So, let's say, if in the same period I have to double the work being done in 10 days or same time, I'll simply double the number of workers working at a constant rate.

I'll thus say
40 workers x 2 in 10 days will do double the work they do I.e. 2 x 1/8th piece of work

Now, we know that what's pending is 3/4th. And 6 x (1/8) will give us (3/4)

So, if you want to finish the pending work in 10 days, you'll have to increase the workers.

Hence, 40 workers x 6 in 10 days will complete 3/4th of pending work.

Hope this helps.
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Given:
    • Initial number of workers = 40
    • The work must be completed in 30 days
    • After, 20 days, only \(\frac{1}{4}^{th}\) of the work was completed
    • All the workers work at a constant rate

To find:
    • The number of new workers to be recruited, so that the work gets completed on time

Approach and Working:
    • If 40 workers can complete \(\frac{1}{4}^{th}\) of the work in 20 days, they can finish the work in 20 * 4 = 80 days.
      o Implies, total work = 40 * 80 ………… (1)

    • Now, let us assume that the number of new workers = x
    • Thus, 40 + x workers must finish the rest \(\frac{3}{4}^{th}\) of the work in remaining 10 days
      o Implies, total work = \((40 + x) * 10 * \frac{4}{3}\) …………. (2)

    Equating (1) and (2), we get,
    • \(40 * 80 = (\frac{40}{3}) * (40 + x)\)
    • Thus, x = 240 – 40 = 200

Hence the correct answer is Option C.

Answer: C
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e-GMAT Question of the Week #21

A group of 40 workers working together have to complete a piece of work in 30 days. If all the workers work at a constant rate and after 20 days, it was found that only \(\frac{1}{4}^{th}\) of the work was completed, then how many more workers should be recruited so that the work gets completed on time?

    A. 32
    B. 100
    C. 200
    D. 240
    E. 280


Given:
1. A group of 40 workers working together have to complete a piece of work in 30 days.
2. All the workers work at a constant rate and after 20 days, it was found that only \(\frac{1}{4}^{th}\) of the work was completed

Asked: How many more workers should be recruited so that the work gets completed on time?

40 * 20 man-days could complete = 1/4 work
800 man-days are required to complete 1/4 work
3200 man-days will be required to complete the work.

800 man-days were already consumed.
2400 man-days were required in 10 days.
240 men will be needed for 10 days
200 men will be required to be recruited.

IMO C
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let total work = 120
so work done by 40 workers in 20 days is ; 1/4*120 ; 30
for 30 units work done the rate per worker per day was ; 30* 1/20*1/40 ; 3/80
with the same rate per day per worker we now have to complete 3/4 of pending work ; viz 90 units which is to be done in 10 days
so we have 90=3/80 * 10*x
x=240 workers
we already have 40 workers so extra workers required ; 240-40 ; 200
IMO C




EgmatQuantExpert
e-GMAT Question of the Week #21

A group of 40 workers working together have to complete a piece of work in 30 days. If all the workers work at a constant rate and after 20 days, it was found that only \(\frac{1}{4}^{th}\) of the work was completed, then how many more workers should be recruited so that the work gets completed on time?

    A. 32
    B. 100
    C. 200
    D. 240
    E. 280

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Days | work | worker
20 | 1/4 | 40
1 | 1/4 | 40*20
1 | 1 | 40*20*4
10 | 1 | 3200÷10
10 | 3/4 | (320*3)/4
=240
Additional worker needed = 240-40
= 200(ans)

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[A group of 40 workers working together have to complete a piece of work in 30 days. If all the workers work at a constant rate and after 20 days, it was found that only \(\frac{1}{4}^{th}\) of the work was completed, then how many more workers should be recruited so that the work gets completed on time?

    A. 32
    B. 100
    C. 200
    D. 240
    E. 280

Let the daily rate per worker = 1 widget per day, implying that the daily rate for 40 workers = 40 widgets per day.
In 20 days, the number of widgets produced by 40 workers = (daily rate for 40 workers)(number of days) = 40*20 = 800 widgets.

Since these 800 widgets constitute 1/4 of the total job, we get:
\(800 = \frac{1}{4}j\)
\(j = 3200\)

Remaining work = (total job) - (widgets produced in the first 20 days) = 3200 - 800 = 2400.
For 2400 widgets to be produced in the remaining 10 days, the required rate \(= \frac{work}{time} = \frac{2400}{10} = 240\) widgets per day.
Since the daily rate must increase from 40 widgets per day to 240 widgets per day, the number of additional workers that must be recruited = 200.

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EgmatQuantExpert
e-GMAT Question of the Week #21

A group of 40 workers working together have to complete a piece of work in 30 days. If all the workers work at a constant rate and after 20 days, it was found that only \(\frac{1}{4}^{th}\) of the work was completed, then how many more workers should be recruited so that the work gets completed on time?

    A. 32
    B. 100
    C. 200
    D. 240
    E. 280


Can use ratios

3x as much work needs to be done than has been done.

That would suggest 3x as many workers if they were able to work for the same 20 days.

Since they have only half as much time left, 10 days, that would suggest 2*3 = 6x as many workers are necessary.

Since the original number is 40 and 6*40 = 240, 200 more workers are required

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EgmatQuantExpert
e-GMAT Question of the Week #21

A group of 40 workers working together have to complete a piece of work in 30 days. If all the workers work at a constant rate and after 20 days, it was found that only \(\frac{1}{4}^{th}\) of the work was completed, then how many more workers should be recruited so that the work gets completed on time?

    A. 32
    B. 100
    C. 200
    D. 240
    E. 280


This is a Hidden Plug In: there's no variable listed, but there is some piece of information that we can make up to make our lives easier. Let's make something up to define how much each worker does each day.

Each worker solves one GMAT question per day. In 20 days, 40 workers solve 800 GMAT questions. That's 1/4 of the number of GMAT questions that need to be solved, so the total job is 3200 GMAT questions. 800 are solved already, so we still need to solve 2400. We have 10 days, so we need 240 per day. We already have enough workers to solve 40 per day, so we need to hire 200 additional workers.

Answer choice C.


ThatDudeKnowsHiddenPlugIn
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EgmatQuantExpert
Question of the Week #21

A group of 40 workers working together have to complete a piece of work in 30 days. If all the workers work at a constant rate and after 20 days, it was found that only \(\frac{1}{4}^{th}\) of the work was completed, then how many more workers should be recruited so that the work gets completed on time?

    A. 32
    B. 100
    C. 200
    D. 240
    E. 280



Here is my approach:

Target : 40 workers * 30 days = x (let's assume, required work to be done = x)
Status : 40 workers * 20 days = x/4 --------(1)
Requirement : (p+40) (let's assume total p workers are required to meet new deadline) workers * 10 days = (x-x/4) = 3x/4

work done by 40 workers in next 10 days = x/8
work to be done by p workers in next 10 days = 3x/4 - x/8 = 5x/8

p workers * 10 days = 5x/8

Multiplying each side by 2/5 -

p workers * 10 days * 2/5 = 5x/8 * 2/5
p workers * 10 day * 2/5 = x/4 = 40 workers * 20 days ------(from 1)

p workers = (40*20*5)/(10*2) = 200

Therefore, answer is 200 workers (C)
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I solved it this way, please advise if it works...

20 days worked is 2/3 of the 30 days in which the project is due.

1/4 of the work was done in that time

So if 1/4 of work done in 2/3 of time then x is the amount that would be done in all (3/3 or 1) of the time at the rate the 40 people work.

1/4 ÷ 2/3 = ×/1
Cross multiply and you get 2/3x=1/4*1

Solving you have x=1/4*3/2 = 3/8 of the work being done in 30 days at the rate these people work.

To get all the work done at this rate, you need 30 ÷ 3/8, or 30*8/3 people, to get it done on schedule. (30*8)/3 = 80

So if it take 80 people at this rate over a 30 day period, then during the remaining 10 days (or 1/3 of the time) you would need to have 3x as many, which is 240.

However, since you already have 40 people, you'd subtract that from the 240 needed for a total of 200.

Not sure if I just got lucky so please let me know if my logic is correct.

Thank you...

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EgmatQuantExpert
e-GMAT Question of the Week #21

A group of 40 workers working together have to complete a piece of work in 30 days. If all the workers work at a constant rate and after 20 days, it was found that only \(\frac{1}{4}^{th}\) of the work was completed, then how many more workers should be recruited so that the work gets completed on time?

    A. 32
    B. 100
    C. 200
    D. 240
    E. 280



So I solved it like this,
40 workers, worked --> 20 days, completed -->\(\frac{1}{4}\)th of the total work
So, 1 worker would take --> 20 x 40 = 800 days to complete \(\frac{1}{4}\)th of the total work
Or,
1 worker would, in a day, complete --> \(\frac{1}{4*800 }\)th of the total work
and, we know that, 1- \(\frac{1}{4}\) = \(\frac{3}{4}\)th work is left for the workers to do in 30-20 = 10 days,
using the basic formula, Work = Rate*Time ,
and taking \(x\) as the total number of workers it would take to complete the work in 10 days,
we can write, \(x(\frac{1}{4*800})10 = \frac{3}{4}\),
which gives us \(x\) = 240
so 240 is the total number of workers required to complete the rest of the work in 10 days,
so, additional workers required = 240 - 40 = 200
So C
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There's another easier approach:

Let's take the total work as 1200 units

efficiency of the workers for the first 20 days 900/(20*40) = 0.375
work to be done for the next 10 days = 900/10 = 90 units per day
efficiency of the workers for the next 10 days = 90/0.375 = 240
therefore 200 new workers
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Hi everyone:

40 workers - 20 days - 1/4 of the work
remain 10 days
Since we have twice less time -> so we need to multiply the number of workers by 2 to do the same job.
However, we have 3 times more work
So we need to multiply by 3 the number of workers

-> 40x2x3 = 240 - 4- = 200 new workers
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