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Ferman can do a job in 6 days and Kelly can do the same job

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Ferman can do a job in 6 days and Kelly can do the same job [#permalink] New post 06 Sep 2013, 10:13
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Question Stats:

71% (02:55) correct 29% (03:00) wrong based on 94 sessions
Ferman can do a job in 6 days and Kelly can do the same job in 8 days. They both undertake the job for $640. With the help of Mary, they finished it in 3 days. How much was paid to Mary?

A. $75
B. $80
C. $85
D. $100
E. $120
[Reveal] Spoiler: OA
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Joined: 14 Aug 2013
Posts: 35
Location: United States
Concentration: Finance, Strategy
GMAT Date: 10-31-2013
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Re: Ferman can do a job in 6 days and Kelly can do the same job [#permalink] New post 06 Sep 2013, 10:44
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Ferman takes 6 days to complete the work, so he does 1/6th of the work in one day
similarly kelly does 1/8th of the work in one day.
Given they worked for 3 days=> 3(1/6+1/8)=7/8th of the work is done by Ferman and Kelly in 3 days
Remaining 1/8th of the work is done with the help of Mary.
So Mary share is proportional the amount of work done=> 1/8th of 640=80$
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Re: Ferman can do a job in 6 days and Kelly can do the same job [#permalink] New post 06 Sep 2013, 10:47
Expert's post
aparnaharish wrote:
Ferman can do a job in 6 days and Kelly can do the same job in 8 days. They both undertake the job for $640. With the help of Mary, they finished it in 3 days. How much was paid to Mary?

A. $75
B. $80
C. $85
D. $100
E. $120


Rate of doing work for Ferman =\frac{1}{6} and for Kelly =\frac{1}{8}. Also, as Time*Rate = Work,

we have 3*[\frac{1}{6}+\frac{1}{8}+r_{Mary}] = 1unit of work

Thus,r_{Mary} = \frac{1}{3}-(\frac{1}{6}+\frac{1}{8}) \to r_{Mary} = \frac{1}{24}

Thus, work done by Mary in 3 days : 3* \frac{1}{24} = \frac{1}{8} units of work, and as the payment is directly proportional to the work done, the payment for her =\frac{640}{8}= 80$

B.
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Re: Ferman can do a job in 6 days and Kelly can do the same job [#permalink] New post 25 Sep 2013, 05:21
mau5 wrote:
aparnaharish wrote:
Ferman can do a job in 6 days and Kelly can do the same job in 8 days. They both undertake the job for $640. With the help of Mary, they finished it in 3 days. How much was paid to Mary?

A. $75
B. $80
C. $85
D. $100
E. $120


Rate of doing work for Ferman =\frac{1}{6} and for Kelly =\frac{1}{8}. Also, as Time*Rate = Work,

we have 3*[\frac{1}{6}+\frac{1}{8}+r_{Mary}] = 1unit of work

Thus,r_{Mary} = \frac{1}{3}-(\frac{1}{6}+\frac{1}{8}) \to r_{Mary} = \frac{1}{24}

Thus, work done by Mary in 3 days : 3* \frac{1}{24} = \frac{1}{8} units of work, and as the payment is directly proportional to the work done, the payment for her =\frac{640}{8}= 80$

B.


Why isn't Mary's rate 1/x ? as in 1 work per x days
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Re: Ferman can do a job in 6 days and Kelly can do the same job [#permalink] New post 08 Nov 2013, 05:07
Skag55 wrote:
Ferman can do a job in 6 days and Kelly can do the same job in 8 days. They both undertake the job for $640. With the help of Mary, they finished it in 3 days. How much was paid to Mary?


Why isn't Mary's rate 1/x ? as in 1 work per x days


you can take it as 1/x too. then work done in 1 day will be

1/6+1/8+1/x = 1/3
1/x = 1/24
so Mary does 1/24 work in 1 day. Hence in 3 days she does 3/24 = 1/8 of work.
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Re: Ferman can do a job in 6 days and Kelly can do the same job   [#permalink] 08 Nov 2013, 05:07
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Ferman can do a job in 6 days and Kelly can do the same job

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