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Question of the Week- 21 (A group of 30 men and 10 women working ....)

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Question of the Week- 21 (A group of 30 men and 10 women working ....)  [#permalink]

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New post 02 Nov 2018, 03:54
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Question Stats:

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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

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Re: Question of the Week- 21 (A group of 30 men and 10 women working ....)  [#permalink]

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New post 02 Nov 2018, 07:51
1
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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Re: Question of the Week- 21 (A group of 30 men and 10 women working ....)  [#permalink]

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New post 02 Nov 2018, 08:18
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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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Re: Question of the Week- 21 (A group of 30 men and 10 women working ....)  [#permalink]

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New post 05 Nov 2018, 12:37
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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Question of the Week- 21 (A group of 30 men and 10 women working ....)  [#permalink]

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New post 05 Nov 2018, 18:17
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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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Re: Question of the Week- 21 (A group of 30 men and 10 women working ....)  [#permalink]

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New post 05 Nov 2018, 21:32
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EgmatQuantExpert wrote:

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


Hi,

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

=> \(\frac{1}{4}\) of the work = 20*40 workerdays = 800 workerdays.

Remaining work = \(\frac{3}{4}\) of the work = 3* 800 = 2400 workerdays.

Remaining work has to be completed in 10 days. => Number of workers required = \(\frac{2400}{10}\) = 240 workers.

40 workers are already working, hence the additional number of required workers = 240 - 40 = 200 workers. Answer: (C).

Thanks.
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Re: Question of the Week- 21 (A group of 30 men and 10 women working ....)  [#permalink]

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New post 06 Nov 2018, 05:28
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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Re: Question of the Week- 21 (A group of 30 men and 10 women working ....)  [#permalink]

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New post 11 Nov 2018, 19:40
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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Re: Question of the Week- 21 (A group of 30 men and 10 women working ....)  [#permalink]

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New post 16 Sep 2019, 00:17
EgmatQuantExpert wrote:
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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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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Re: Question of the Week- 21 (A group of 30 men and 10 women working ....)  [#permalink]

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New post 16 Sep 2019, 04:34
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 wrote:
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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Question of the Week- 21 (A group of 30 men and 10 women working ....)  [#permalink]

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New post 20 Sep 2019, 17:51
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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Question of the Week- 21 (A group of 30 men and 10 women working ....)  [#permalink]

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New post Updated on: 21 Sep 2019, 03:06
EgmatQuantExpert wrote:
[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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Originally posted by GMATGuruNY on 20 Sep 2019, 19:02.
Last edited by GMATGuruNY on 21 Sep 2019, 03:06, edited 1 time in total.
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Re: Question of the Week- 21 (A group of 30 men and 10 women working ....)  [#permalink]

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New post 20 Sep 2019, 22:50
You may do everything correctly in this question and still get the question wrong if you don't read the question carefully. While answering GMAT questions always check whether or not you are answering the right question or not?
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Re: Question of the Week- 21 (A group of 30 men and 10 women working ....)   [#permalink] 20 Sep 2019, 22:50
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