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

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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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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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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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New post 05 Nov 2011, 19:02
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: Machine A can complete a certain job in x hours. Machine B can complet [#permalink]

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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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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 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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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]

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New post 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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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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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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New post 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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\(\frac{1}{x}+\frac{1}{y}=W\)

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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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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New post 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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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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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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New post 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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New post 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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New post 07 Jan 2013, 05:10
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, 05:10

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