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Machine A can complete a certain job in x hours. Machine B [#permalink]
 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/(x-y) E. y/(x+y)
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Re: Work Prob - Part of the Job done [#permalink]
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/(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. Answer: E.
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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]
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.
Regards.
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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]
24 Jun 2012, 22:11
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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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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]
19 Sep 2012, 17:21
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]
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]
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]
15 Nov 2012, 03:03
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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]
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]
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]
07 Jan 2013, 04:25
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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]
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]
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]
07 Jan 2013, 05:10
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Re: Machine A can complete a certain job in x hours. Machine B [#permalink]
07 Jan 2013, 05:14
Thanks, thanks, thanks!!!! Iam just not that good at re-rewriting mathematical stuff with variables. But with the numerical example I understand it completely. Thank your very much!
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Re: Machine A can complete a certain job in x hours. Machine B
[#permalink]
07 Jan 2013, 05:14
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