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Micheal and Adam can do together a piece of work in 20 days.

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Micheal and Adam can do together a piece of work in 20 days. [#permalink] New post 11 Mar 2008, 03:47
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Micheal and Adam can do together a piece of work in 20 days. After they have worked together for 12 days Micheal stops and Adam completes the remaining work in 10 days. In how many days Micheal complete the work separately.

A. 80 days
B. 100 days
C. 120 days
D. 110 days
E. 90 days
[Reveal] Spoiler: OA

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Last edited by Bunuel on 24 Feb 2014, 12:28, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: Time & Work-2 [#permalink] New post 11 Mar 2008, 06:06
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Suppose rate of work for per day Mike = M & Rate of work per day for Adam = A
Together in a Day they can finish = A + M units
Total Work Done in 20 Days = 20(A+M)
Total Work Done in 12 Days = 12(A+M)
Work Done by Adam in 10 Days = 10A

Since total work is same we can say that 20(A+M) = 12(A+M) + 10A => 8M = 2A => A = 4M
So total work done by them together in 20 Days = 20(4M+M) = 100M
Since Mike does M unit of work per day, it will take him 100 Days to finish up the work.

Answer B.
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Re: Time & Work-2 [#permalink] New post 30 Nov 2011, 20:49
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Rate of both = 1/20
Together they do = 1/20*12 = 3/5

Left work = 1 - 3/5 = 2/5

Adam completes 2/5 work in 10 day
so he took 10*5/2 = 25 days to complete the left work alone.
Thus the rate of adam is 1/25

Rate of Micheal = 1/20 - 1/25 = 1/100
Thus micheal takes 100 days to complete the whole work.
ans. B.
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Re: Time & Work-2 [#permalink] New post 18 Dec 2011, 09:24
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I got another way of the same problem
1/20*12+10x=1 [Rate of Adam is x]
x = 1/25
so, rate of Micheal =1/20-1/25 = 1/100

So Micheal take 100 days
Ans. B
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100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html
Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html
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Re: Micheal and Adam can do together a piece of work in 20 days. [#permalink] New post 24 Feb 2014, 12:23
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Re: Micheal and Adam can do together a piece of work in 20 days. [#permalink] New post 13 Mar 2014, 23:33
Both have completed \frac{12}{20} work, means \frac{8}{20} work is pending

Adam completed \frac{8}{20} work in 10 days

Rate of Adam = \frac{8}{(20 * 10)} = \frac{8}{200}

So, rate of Michael = \frac{1}{20} - \frac{8}{200}

= \frac{2}{200}

= \frac{1}{100}

So Michael completes the work in 100 Days

Answer = B
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Re: Micheal and Adam can do together a piece of work in 20 days. [#permalink] New post 29 Mar 2014, 09:41
I`ve done in 5 steps:

1 - M+A = X/20 where X is arbitrarily 100, so M+A rate is 5/day of a total of 100.

2 - (M+A) *12 = 60. 100-60 = 40

3 - A completed 40 in 10 days, so 4/day.

4 - M+A = 5, M + 4 = 5, M = 1.

5 - 1 (rate) * 100(total work) = 100 days
Re: Micheal and Adam can do together a piece of work in 20 days.   [#permalink] 29 Mar 2014, 09:41
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