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# Machine A can complete a certain job in x hours. Machine B

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05 Nov 2011, 12:23
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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)
[Reveal] Spoiler: OA

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Re: Machine A can complete a certain job in x hours. Machine B can complet [#permalink]

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05 Nov 2011, 13:31
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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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05 Nov 2011, 13:34
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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: Machine A can complete a certain job in x hours. Machine B can complet [#permalink]

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05 Nov 2011, 18:02
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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: Machine A can complete a certain job in x hours. Machine B can complet [#permalink]

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05 Nov 2011, 18:09
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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: Machine A can complete a certain job in x hours. Machine B can complet [#permalink]

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26 Dec 2011, 04:37
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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)
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Machine A can complete a certain job in x hours. Machine B [#permalink]

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27 Jan 2012, 15:46
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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)
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Re: Work Prob - Part of the Job done [#permalink]

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27 Jan 2012, 16:20
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docabuzar wrote:
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)

Working together A and B complete the job in $$\frac{xy}{x+y}$$ hours (as time is a reciprocal of rate then take the reciprocal of combined rate, which is $$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$$);

Now, in $$\frac{xy}{x+y}$$ hours A will complete $$\frac{1}{x}*\frac{xy}{x+y}=\frac{y}{x+y}$$ part of the job (rate*time=job) and this will be the part which B will not have to complete.

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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]

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24 Jun 2012, 18:55
Hi Bunuel,

Can you help me understand that when we want to calculate the work just done by A and not B of the total work which in this case we assume it as 1. Then why do we multiply the rate of only A with the total time?

I can't get my head around this concept. Is it one of those things that I just accept and move on.

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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]

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24 Jun 2012, 21:11
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Take an example to understand it.
A takes 2 hrs (= x) to complete a work alone and B takes 2 hrs (= y) to complete a work alone. Together, they will complete the work in 1 hr = [xy/(x+y) = 2*2/(2+2)]
So A does 1/2 of the work in 1 hr (A's rate of work) and B does 1/2 of the work in 1 hr (B's rate of work).

So when they were working together, each needed to work for only 1 hr.
How much work was done by A in that one hr? 1*(1/2) = (1/2)
So because of A's help, B doesn't need to do half of the work i.e. work done by A.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 21 Jul 2010 Posts: 2 Followers: 0 Kudos [?]: 7 [2] , given: 1 Re: Machine A can complete a certain job in x hours. Machine B [#permalink] ### Show Tags 19 Sep 2012, 16:21 2 This post received KUDOS My solution might be very simple: If we pick numbers like x = 3hrs and y = 4hrs then it takes 7hrs to complete two units for both of them. (We can use two units because we are looking for fractional work, not a total quantity #). If it takes 7hrs between the two of them then Ma does 3/7 or the work and Mb does 4/7 of the work. This means that Mb is not doing 3/7 of the work and you can plug in your answer choices to find out which one gives you x=3, y=4 -> 3/7 Intern Joined: 03 Oct 2012 Posts: 10 GMAT 1: 620 Q39 V38 GMAT 2: 700 Q49 V38 Followers: 0 Kudos [?]: 6 [0], given: 23 Re: Machine A can complete a certain job in x hours. Machine B [#permalink] ### Show Tags 15 Nov 2012, 00:45 VeritasPrepKarishma wrote: Take an example to understand it. A takes 2 hrs (= x) to complete a work alone and B takes 2 hrs (= y) to complete a work alone. Together, they will complete the work in 1 hr = [xy/(x+y) = 2*2/(2+2)] So A does 1/2 of the work in 1 hr (A's rate of work) and B does 1/2 of the work in 1 hr (B's rate of work). So when they were working together, each needed to work for only 1 hr. How much work was done by A in that one hr? 1*(1/2) = (1/2) So because of A's help, B doesn't need to do half of the work i.e. work done by A. This explanation is vague since you picked 1/2. We cannot see how it works because it does not matter what you do with x and y it's around 1/2. I see that formula is absolutely correct but don't understand why if the value of 1/x is 2/6 and 1/y is 3/6, the portion which y should not do comes out as 3/5 not 2/5. Senior Manager Joined: 13 Aug 2012 Posts: 464 Concentration: Marketing, Finance GMAT 1: Q V0 GPA: 3.23 Followers: 25 Kudos [?]: 433 [3] , given: 11 Re: Machine A can complete a certain job in x hours. Machine B [#permalink] ### Show Tags 15 Nov 2012, 01:25 3 This post received KUDOS 1 This post was BOOKMARKED $$\frac{1}{x}+\frac{1}{y}=W$$ Combining their efforts they could finish: $$W=\frac{x+y}{xy}$$ This means it will take them t to complete the job. $$\frac{x+y}{xy}(t)=1==>t=\frac{xy}{x+y}$$ The job done by x will be the job that y doesn't have to worry about. $$W=\frac{1}{x}(xy/x+y)==>W=\frac{y}{x+y}$$ _________________ Impossible is nothing to God. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7119 Location: Pune, India Followers: 2129 Kudos [?]: 13625 [2] , given: 222 Re: Machine A can complete a certain job in x hours. Machine B [#permalink] ### Show Tags 15 Nov 2012, 02:03 2 This post received KUDOS Expert's post felixjkz wrote: This explanation is vague since you picked 1/2. We cannot see how it works because it does not matter what you do with x and y it's around 1/2. I see that formula is absolutely correct but don't understand why if the value of 1/x is 2/6 and 1/y is 3/6, the portion which y should not do comes out as 3/5 not 2/5. The explanation was meant for saswani who had trouble understanding "Then why do we multiply the rate of only A with the total time?" It doesn't matter what the numbers are - as long as they are easy to work with - you can make out what's going on. As for your question "why if the value of 1/x is 2/6 and 1/y is 3/6, the portion which y should not do comes out as 3/5 not 2/5." If x = 3 and y = 2, A does 1/3rd work in 1 hr and B does 1/2 work in 1 hr. When they work together, they complete the work in 1/(1/2 + 1/3) = 6/5 hrs In 6/5 hrs, A does (1/3)*(6/5) = 2/5 work and B does (1/2)*(6/5) = 3/5 work So B does NOT need to do 2/5 of the work. Answer still (E). _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]

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07 Jan 2013, 03:09
Can someone please explain the formula xy / x+y to state the work done together. I thought I had a well understanding of this concept, however when I saw this question, I lost it completely. Thanks in advance
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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]

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07 Jan 2013, 03:22
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The amount of work that A can do in 1 hour ( or the rate at which A works ) is 1/x, since A takes x hours to complete the whole work. Similarly the amount of work that B can do in 1 hour is 1/y. Now, total work that they can do, when working together, in one hour is 1/x + 1/y.
You can also write it as 1/W = 1/x + 1/y = (x + y ) / xy. 1/W is the rate at which they are working.
Hence, to complete the total work they need xy / (x + y) hours.
Please let me know if anything is not clear.
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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]

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07 Jan 2013, 03:25
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KevinBrink wrote:
Can someone please explain the formula xy / x+y to state the work done together. I thought I had a well understanding of this concept, however when I saw this question, I lost it completely. Thanks in advance

Check here: two-consultants-can-type-up-a-report-126155.html#p1030079 and here: work-word-problems-made-easy-87357.html

Hope it helps.
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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]

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07 Jan 2013, 03:35
Can someone explain this step a little bit further 1/W = 1/x + 1/y = (x + y ) / xy?
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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]

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07 Jan 2013, 03:57
Iam really confused right now, I understand everything completely except for this essential part. I went through the stuff you told me Bunuel and I understand everything stated in there. But how do you come from 1/x + 1/y --> (x+y) / xy??? Thanks in advance
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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]

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07 Jan 2013, 04:10
KevinBrink wrote:
Iam really confused right now, I understand everything completely except for this essential part. I went through the stuff you told me Bunuel and I understand everything stated in there. But how do you come from 1/x + 1/y --> (x+y) / xy??? Thanks in advance

You mean you don't get how 1/x + 1/y equals to (x+y)/xy?

The same way as $$\frac{1}{2}+\frac{1}{3}=\frac{3+2}{2*3}=\frac{5}{6}$$.
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Re: Machine A can complete a certain job in x hours. Machine B   [#permalink] 07 Jan 2013, 04:10

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