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john can do a piece

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TomB
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john can do a piece [#permalink] New post   03 May 2010, 05:26
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John can do a piece of work in 15 days. If he is joined by jill who is 50% more efficient, in what time will john and jill together finish the work?

a) 10 days b) 6 days c) 18 days d) data insufficient e) 8 days

My approach to the problem is

1/15+1/7.5=1/t
5+10/75=1/t
t=75/15
t=5
Show Spoiler
b

what am i missing?



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lagomez
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Re: john can do a piece [#permalink] New post   03 May 2010, 07:43
TomB wrote:
John can do a piece of work in 15 days. If he is joined by jill who is 50% more efficient, in what time will john and jill together finish the work?

a) 10 days b) 6 days c) 18 days d) data insufficient e) 8 days

My approach to the problem is

1/15+1/7.5=1/t
5+10/75=1/t
t=75/15
t=5
Show Spoiler
b

what am i missing?


I am getting 6 days by doing the following:

1/15 + 1/10 = 5/30 = 6 days

The reason I used 10 is because 7.5x = 15 and x = 2 which is 100% more efficient
10x = 15 x = 50% more efficient
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gurpreetsingh
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Re: john can do a piece [#permalink] New post   03 May 2010, 08:34
TomB wrote:
John can do a piece of work in 15 days. If he is joined by jill who is 50% more efficient, in what time will john and jill together finish the work?

a) 10 days b) 6 days c) 18 days d) data insufficient e) 8 days

My approach to the problem is

1/15+1/7.5=1/t
5+10/75=1/t
t=75/15
t=5
Show Spoiler
b

what am i missing?


Jill is 50% more efficient => 1.5 times John ( its not 50% of john, 50% more => (1+50%) john = 150% of john)

If john can do a piece of work in 15 days => Jill can complete the work in 15/1.5 = 10 days

Now 1/15 + 1/10 = 1/t => t = 6 days.
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Silvers
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Re: john can do a piece [#permalink] New post   03 May 2010, 10:58
1
Kudos
TomB wrote:
John can do a piece of work in 15 days. If he is joined by jill who is 50% more efficient, in what time will john and jill together finish the work?

a) 10 days b) 6 days c) 18 days d) data insufficient e) 8 days

My approach to the problem is

1/15+1/7.5=1/t
5+10/75=1/t
t=75/15
t=5
Show Spoiler
b

what am i missing?


Do it by ratios.

When Jill works the unit work is 1

When Joh joins in the collective unit work is 1 + 1.5 = 5/2

Since total effort is 5/2 the time will be 2/5 of the original time.

2/5*15 = 6 days.
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mbafall2011
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Re: john can do a piece [#permalink] New post   03 May 2010, 11:05
TomB wrote:
John can do a piece of work in 15 days. If he is joined by jill who is 50% more efficient, in what time will john and jill together finish the work?

a) 10 days b) 6 days c) 18 days d) data insufficient e) 8 days

My approach to the problem is

1/15+1/7.5=1/t
5+10/75=1/t
t=75/15
t=5
Show Spoiler
b

what am i missing?



I think you already know what is wrong with your approach. Here is another try:

15 Days john finishes 100%
In the same 15 days JIll will finish 150%

Together in 15 days they will finish 250% of the job.

Therefore how many days will it take to do 100 % which is 6 days.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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gmatclubot
Re: john can do a piece [#permalink]
03 May 2010, 11:05
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