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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, 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)

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

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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)

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

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)

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.

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.

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.
_________________

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

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19 Sep 2012, 16:21

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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

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

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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.

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).
_________________

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

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.

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

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

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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