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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]

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New post 06 Sep 2013, 11:13
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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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Re: Ferman can do a job in 6 days and Kelly can do the same job [#permalink]

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New post 06 Sep 2013, 11:44
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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]

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New post 06 Sep 2013, 11:47
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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 \(3*[\frac{1}{6}+\frac{1}{8}+r_{Mary}] = 1\)unit 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]

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New post 25 Sep 2013, 06:21
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 \(3*[\frac{1}{6}+\frac{1}{8}+r_{Mary}] = 1\)unit 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]

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New post 08 Nov 2013, 06:07
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]

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New post 06 Dec 2014, 13:38
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Re: Ferman can do a job in 6 days and Kelly can do the same job [#permalink]

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New post 23 Dec 2015, 00:14
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Ferman can do a job in 6 days and Kelly can do the same job   [#permalink] 23 Dec 2015, 00:14
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Ferman can do a job in 6 days and Kelly can do the same job

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