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John can complete a given task in 20 days. Jane will take [#permalink]

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20 Jul 2008, 05:23

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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?

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20 Jul 2008, 05:36

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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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20 Jul 2008, 07:04

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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.

Re: John can complete a given task in 20 days. Jane will take [#permalink]

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07 Sep 2013, 13:22

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

Re: John can complete a given task in 20 days. Jane will take [#permalink]

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25 Sep 2013, 20:55

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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]

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Re: John can complete a given task in 20 days. Jane will take [#permalink]

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24 Jul 2015, 02:50

My approach is slightly different: John+Jane can complete the work in 7.5 days (20*12)/(20+12). Jane left 4 days before the work got completed...that means they together worked for 3.5 days. Now with rate of 1/20 and 1/12 per day..combined...they can do...2/15 work in 1 day. That means in 3.5 days, then can do -->7/15 of work. So left over work is 8/15 and needs to be be done by John alone. which would take him...(1/20)/(8/15)--->32/3 days... This is taking me nowhere..Can anyone suggest what is wrong in this logic??

Re: John can complete a given task in 20 days. Jane will take [#permalink]

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12 Oct 2015, 16:00

narmit wrote:

My approach is slightly different: John+Jane can complete the work in 7.5 days (20*12)/(20+12). Jane left 4 days before the work got completed...that means they together worked for 3.5 days. Now with rate of 1/20 and 1/12 per day..combined...they can do...2/15 work in 1 day. That means in 3.5 days, then can do -->7/15 of work. So left over work is 8/15 and needs to be be done by John alone. which would take him...(1/20)/(8/15)--->32/3 days... This is taking me nowhere..Can anyone suggest what is wrong in this logic??

i did exactly the same and got stucked too... (just note the following in your last calc. -> (1/20)/(8/15) = 36/3 = 12 days not sure what is wrong, i learned this approach from manhattan and usually works fine

Re: John can complete a given task in 20 days. Jane will take [#permalink]

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09 Nov 2015, 13:45

elch4ngo wrote:

narmit wrote:

My approach is slightly different: John+Jane can complete the work in 7.5 days (20*12)/(20+12). Jane left 4 days before the work got completed...that means they together worked for 3.5 days. Now with rate of 1/20 and 1/12 per day..combined...they can do...2/15 work in 1 day. That means in 3.5 days, then can do -->7/15 of work. So left over work is 8/15 and needs to be be done by John alone. which would take him...(1/20)/(8/15)--->32/3 days... This is taking me nowhere..Can anyone suggest what is wrong in this logic??

i did exactly the same and got stucked too... (just note the following in your last calc. -> (1/20)/(8/15) = 36/3 = 12 days not sure what is wrong, i learned this approach from manhattan and usually works fine

Having the same issue as both of these people with this method.

I think this method, because it seems straight forward in my brain, but this is the second time ive been hung up on a question. And got the answer by choosing the closest answer. But something is off logic wise.

My approach is slightly different: John+Jane can complete the work in 7.5 days (20*12)/(20+12). Jane left 4 days before the work got completed...that means they together worked for 3.5 days. Now with rate of 1/20 and 1/12 per day..combined...they can do...2/15 work in 1 day. That means in 3.5 days, then can do -->7/15 of work. So left over work is 8/15 and needs to be be done by John alone. which would take him...(1/20)/(8/15)--->32/3 days... This is taking me nowhere..Can anyone suggest what is wrong in this logic??

i did exactly the same and got stucked too... (just note the following in your last calc. -> (1/20)/(8/15) = 36/3 = 12 days not sure what is wrong, i learned this approach from manhattan and usually works fine

Having the same issue as both of these people with this method.

I think this method, because it seems straight forward in my brain, but this is the second time ive been hung up on a question. And got the answer by choosing the closest answer. But something is off logic wise.

Can Bunuel or Engr2012 help if possible ?

There is a problem in this logic.

"Jane was indisposed 4 days before the work got over." - implies that Jane left 4 days before the work actually got over. So perhaps only 1 day work was left but Jane left and John alone completed the work in 4 days. After Jane left, the work continued for 4 days. In your logic, John and Jane would have taken total 7.5 days and Jane left after 3.5 days. But then, the leftover work would have been completed in more than 4 days because John was working alone.

In such questions, you need to start from the end. Last 4 days John works alone and completes 4 * (1/20) = 1/5 of the work. So 4/5 of the work should have been completed by the two of them together before Jane left. Their combined rate of work is 1/20 + 1/12 = 8/60 Time taken to complete 4/5 of the work = (4/5)/(8/60) = 6 days.

So total number of days taken to complete the work = 6 + 4 = 10 days.
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Re: John can complete a given task in 20 days. Jane will take [#permalink]

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09 Nov 2015, 23:37

VeritasPrepKarishma wrote:

There is a problem in this logic.

"Jane was indisposed 4 days before the work got over." - implies that Jane left 4 days before the work actually got over. So perhaps only 1 day work was left but Jane left and John alone completed the work in 4 days. After Jane left, the work continued for 4 days. In your logic, John and Jane would have taken total 7.5 days and Jane left after 3.5 days. But then, the leftover work would have been completed in more than 4 days because John was working alone.

In such questions, you need to start from the end. Last 4 days John works alone and completes 4 * (1/20) = 1/5 of the work. So 4/5 of the work should have been completed by the two of them together before Jane left. Their combined rate of work is 1/20 + 1/12 = 8/60 Time taken to complete 4/5 of the work = (4/5)/(8/60) = 6 days.

So total number of days taken to complete the work = 6 + 4 = 10 days.

Re: John can complete a given task in 20 days. Jane will take [#permalink]

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25 Nov 2015, 11:35

Hi guys, I used the same logic but using smart numbers (let's say entire work=120) John 6*20=120 Jane 10*12=120 In 4 days working alone John did: 4 Days * Rate = 4*6=24, so working together they did 120-24=96 --> Rate(A+B)*t=96 -> 16*t=96, t=6. hence the answer is B (John worked 4 days + 6 days they worked together = 10 days)
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Re: John can complete a given task in 20 days. Jane will take [#permalink]

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25 Nov 2015, 11:54

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arjtryarjtry wrote:

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

Let the total task be 60 units

John can complete 3 units of work per day Jane can complete 5 units of work per day

Togather Jane and John can complete 8 units of work per day

Let John and Jane togather worked for " d " days , so work completed by them is 8d units

Then John completes the remaining work in 4 more days , so work done by John in 4 days is ( 4*3) 12 units

Now , we have Total Work = Work completed by John and Jane ( in "d " days ) and work done by John in 4 days

60 = 8d + 12 48 = 8d

or, d =6

So, John and John worked for 6 days and then John worked for 4 days , so they took 10 days to complete the work. _________________

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Re: John can complete a given task in 20 days. Jane will take [#permalink]

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30 May 2016, 03:59

R(john)=1/20 R(Jane)=1/12

Since,jane left the work 4 days before,it means the remaining 4 days john had to work alone. So using work time equation: Work=Rate *time=R(john)*4=1/20*4=1/5 th of work was done by john in the last 4 days. So,4/5th work has to be done when jane and john were working together. 4/5=(1/20+1/12)*T T=10 days.

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Re: John can complete a given task in 20 days. Jane will take
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30 May 2016, 03:59

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