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

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Joined: 09 Jul 2013
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Ferman can do a job in 6 days and Kelly can do the same job [#permalink]  06 Sep 2013, 10:13
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Question Stats:

72% (02:00) correct 27% (03:05) wrong based on 66 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
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Kudos [?]: 15 [2] , given: 3

Re: Ferman can do a job in 6 days and Kelly can do the same job [#permalink]  06 Sep 2013, 10:44
2
KUDOS
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$Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 616 Followers: 25 Kudos [?]: 402 [0], given: 128 Re: Ferman can do a job in 6 days and Kelly can do the same job [#permalink] 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]  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 Manager Joined: 23 May 2013 Posts: 60 Followers: 0 Kudos [?]: 11 [0], given: 63 Re: Ferman can do a job in 6 days and Kelly can do the same job [#permalink] 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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