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# Two separate Machines

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Two separate Machines [#permalink]  05 Nov 2011, 12:23
00:00

Difficulty:

(N/A)

Question Stats:

40% (02:31) correct 60% (01:12) wrong based on 7 sessions
Machine A can complete a certain job in x hours. Machine B can complete the same job in y hours. If A and B work together at their respective rates to complete the job, which of the following represents the fraction of the job that B will not have to complete because of A's help?

A) (x – y)/ (x + y)
B) x / (y – x)
c) (x + y) / xy
D) y / (x – y)
E) y / (x + y)

Apologies as official answer was not provided with the question.
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Re: Two separate Machines [#permalink]  05 Nov 2011, 13:31
1
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Since Machine A can complete the job in 'x' hours, the job completed by Machine A in 1 hour will be 1/x

Similarly, since Machine B completes the job in 'y' hours, Machine B will complete 1/y of the job in 1 hour

Hence the job completed in 1 hour when A and B work together at their respective rates is:
1/x + 1/y = (x+y)/xy

Since A completes 1/x of this job, 1/x of the total job is what Machine B will not have to complete because of Machine A's help.

So, the required quantity is (1/x) of (x+y)/xy i.e. (1/x)/(x+y)/xy = xy/x(x+y) = y/(x+y)

Hence, in my opinion, the answer should be option E.
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Re: Two separate Machines [#permalink]  05 Nov 2011, 13:34
Not very sure but just thought of the following way:-

A's rate is 1/x
B's rate is 1/y.

when they work together, then work is done in time xy/(x+y).
In this time, work done by b is x/(x+y), by a is y/(x+y) and total work is 1. So the work which b did not do that is the work done by a should be represented as the fraction of total work
a's work/total work = [y/x+y] / 1 = y/(x+y)

Option E
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Re: Two separate Machines [#permalink]  05 Nov 2011, 18:02
Plugging in numbers,
Let A rate = 3
B rate = 2
To finish a work of 10,
B would work 5 hours to finish it, but only 4 hours with help of A. So the fraction left is 2/5

Plug in numbers
y/(x+y) = 2/5

Hence E

Hope that helps
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Re: Two separate Machines [#permalink]  05 Nov 2011, 18:09
Another approach is algebraic.

Machine A completes a job in x hours so the rate is 1/x
Machine B complets a job in y hours so the rate is 1/y
Collectively 1/x + 1/y = x + y /(xy) for both working together

To get the fraction of what B doesn't have to complete because of A's help is simply to get the fraction of A's workrate when they are both working together

Hence (1/x)/(x+y)/xy = (1/x * xy)/(x+y) = y/(x+y)
Hope that helps again!
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Re: Two separate Machines [#permalink]  24 Dec 2011, 02:15
YOU CAN NOT WRIGHT ANY PROBLEM AS GMAT PREP. PROBLEM

AS MOST OF THE PROBLEMS YOU HAVE POSTED ARE NOT OF GMAT PREP.
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Re: Two separate Machines [#permalink]  24 Dec 2011, 09:41
Dear Enigma123
I apologize for making a harsh comment
Didn't mean to warn you against anything

Regards
Aseem
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Re: Two separate Machines [#permalink]  26 Dec 2011, 04:37
1
KUDOS
job completed in 1 hour when A and B work together at their respective rates is:
1/x + 1/y = (x+y)/xy

to complete job total time taken = xy/(x+y)

in this time job completed by B= 1/y * xy/(x+y) = x/(x+y)

rest is done by A = 1- X/(x+y)=y/(x+y)
Re: Two separate Machines   [#permalink] 26 Dec 2011, 04:37
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