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Machine A can complete a certain job in x hours. Machine B can complet [#permalink]
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05 Nov 2011, 13:23
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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)
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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, 14:31
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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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Re: Machine A can complete a certain job in x hours. Machine B can complet [#permalink]
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05 Nov 2011, 14:34
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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, 19:02
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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, 19:09
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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, 05:37
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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, 16:46
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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/(xy) E. y/(x+y)



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Re: Work Prob  Part of the Job done [#permalink]
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27 Jan 2012, 17:20
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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/(xy) 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. Answer: E.
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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, 19:55
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.
Thanks for your help in advance.
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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, 22:11
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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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]
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19 Sep 2012, 17:21
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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



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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]
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15 Nov 2012, 01:45
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.



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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]
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15 Nov 2012, 02:25
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\(\frac{1}{x}+\frac{1}{y}=W\) 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. \(W=\frac{1}{x}(xy/x+y)==>W=\frac{y}{x+y}\)
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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]
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15 Nov 2012, 03:03
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).
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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, 04:09
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, 04:22
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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, 04:25



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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, 04:35
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, 04:57
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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