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John can complete a given task in 20 days. Jane will take

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John can complete a given task in 20 days. Jane will take [#permalink] New post 20 Jul 2008, 04:23
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John can complete a given task in 20 days. Jane will take only 12 days to complete the same task. John and Jane set out to complete the task by beginning to work together. However, Jane was indisposed 4 days before the work got over. In how many days did the work get over from the time John and Jane started to work on it together?

A. 6
B. 10
C. 8
D. 7.5
E. 3.5

Last edited by Bunuel on 07 Sep 2013, 12:27, edited 1 time in total.
Added the OA and moved to PS forum.
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Re: time and work [#permalink] New post 20 Jul 2008, 04:36
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10 days. Here is the explanation
for 4 days Jane didn't work, hence john would complete (4/20=1/5 of the work). Therefore Jane and John together will complete 4/5 of the work.
Now John and Jane together would complete the work in (1/20+1/12=15/2 days).
Therefore they would take 6 days to complete 4/5 of the work, after which John worked for another 4 days. Hence in total they took 6+4=10 days to complete.
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Re: time and work [#permalink] New post 20 Jul 2008, 04:53
R * T = 1

John's rate = 1/20
Jane's rate = 1/12

(1/20)T + (1/12)(T-4) = 1
T/20 + (T-4)/12 = 1
(12T + 20T - 80)/240 = 1
32T = 320
T=10
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Re: time and work [#permalink] New post 20 Jul 2008, 05:03
john's rate = 1/20
jane's rate = 1/12
combined rate = 1/20+1/12 = 2/15

let the work finished in x days
both woked for (x-4) days amd John alone worked for 4 days

(x-4)*2/15 + 4*1/20 = 1
2x - 8 + 3 = 15
x= 10 answer .............
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Re: time and work [#permalink] New post 20 Jul 2008, 06:04
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I have a different point of view, solution can be too long, however, it is easy to think and solve with this method.

We know that John completes the task in 20 days, Jane completes the task in 12 days.
We can assume that total of this work is 60 units (which is the least common multiply)

From here; John finishes 3 units of the task in a day (60/20=3)
Jane finishes 5 units of the task in a day (60/12=5)

First 4 days, john complete 4*3=12 units of the task. Remaining part is 48 units (60-12=48)
Both of them complete 3+5=8 units of the task in a day. Accrodingly, they can complete 48 units of tasks in 6 days.

Then, the answer is 4+6=10...
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Re: time and work [#permalink] New post 20 Jul 2008, 19:51
explaination plz...................
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Re: John can complete a given task in 20 days. Jane will take [#permalink] New post 07 Sep 2013, 12:22
john takes 20d to complete the work so number of units each day =3
jane takes 12d to complete the work so number of units each day =5
since jane didnt work for last 4 days it means number of units done in last 4 days are =3*4=12
so john and jane work together for (60-12)/3+5 = 6days
thus total days = 6+4 =10days
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Re: John can complete a given task in 20 days. Jane will take [#permalink] New post 25 Sep 2013, 19:55
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Let it takes 'x' days to finish of the work

John's rate = 1/20 units per day
Jane's rate = 1/12 units per day

Total work done = 1 unit

Since John should work full for 'x' days
Jane work for 'x-4' days

Total work done => X/20 + (x-4)/12 = 1

Solving for X we get X=10
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Re: John can complete a given task in 20 days. Jane will take   [#permalink] 25 Sep 2013, 19:55
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