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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]

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Updated on: 24 Feb 2014, 13:28
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Question Stats:

82% (02:24) correct 18% (02:44) wrong based on 247 sessions

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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

Originally posted by prasannar on 11 Mar 2008, 04:47.
Last edited by Bunuel on 24 Feb 2014, 13:28, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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30 Nov 2011, 21:49
12
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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11 Mar 2008, 07:06
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1
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.

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18 Dec 2011, 10:24
2
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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Re: Micheal and Adam can do together a piece of work in 20 days.  [#permalink]

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14 Mar 2014, 00:33
5
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

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Re: Micheal and Adam can do together a piece of work in 20 days.  [#permalink]

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29 Mar 2014, 10:41
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
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Re: Micheal and Adam can do together a piece of work in 20 days.  [#permalink]

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10 Oct 2016, 06:48
prasannar wrote:
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

Let the total work be 60 ( LCM of 20,12 & 10 )

Efficiency of Micheal and Adam is $$\frac{60}{20} = 3$$

Quote:
After they have worked together for 12 days.....

Unit of work completed is 12*3 => 36 units
Units of work left is 60 - 36 => 24 units

Quote:
Adam completes the remaining work in 10 days

So, Adam completes 24 units in 10 days..
Efficiency of Adam = 24/10 => $$\frac{12}{5}$$ => $$2.4$$

Efficiency of Micheal and Adam is = 3
Efficiency of Adam is = 2.40
So, Efficiency of Micheal is = 0.60

Thus the time required for Micheal to complete the work is = $$\frac{60}{0.60}$$ => 100

Hence correct answer is (B) 100

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Re: Micheal and Adam can do together a piece of work in 20 days.  [#permalink]

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10 Mar 2019, 08:56
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Re: Micheal and Adam can do together a piece of work in 20 days.   [#permalink] 10 Mar 2019, 08:56
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