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

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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, 04:14
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]

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New post 16 Apr 2014, 01:26
(1/x) of the work is completed by Machine A in one hour.
(1/y) of the work is completed by Machine B in one hour.
Combined rate(one hour) is 1/x + 1/y = (x+y/xy).
fraction of the job that B will not have to complete because of A's help?
(1/x) of the work will be completed by A.
As we are asked about fraction , (1/x) / (x+y/xy) = (y/x+y).
Hence 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 13 Jan 2015, 01:35
1
.......................... Rate ..................... Time ................. Work Done

Machine A ............... \(\frac{1}{x}\) ...................... x ........................ 1 (Assumption)

Machine B ................ \(\frac{1}{y}\) ....................... y ....................... 1

Combined rate ............ \(\frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy}\) ............. \(\frac{xy}{x+y}\) ................... 1

Work done due to A's help = Rate of A * Time required for the combined

\(= \frac{1}{x} * \frac{xy}{x+y} = \frac{y}{x+y}\)

This will be the benefit for B

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 17 Sep 2015, 00:49
I used the following approach but was not able to proceed, could somebody tell me how/whether we can solve further to get the answer
A = 2 hrs (x), B = 4 hrs(y).
A+B = 4/3 hrs.

If B alone is working, it takes 4 hrs. But both working, B has to work only for 4/3 hours.
So B does not work for 4 hrs - 4/3 hrs = 8/3 hours.

I then thought that B did not have to work for 8/3 hours since A pitched in.
So 8/3 hours multiplied by A's rate (1/2)...gives 4/3..

But none of the options give this value. Any reason this approach does not work?...could somebody pls help.
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Re: Machine A can complete a certain job in x hours. Machine B  [#permalink]

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New post 17 Sep 2015, 21:26
Success2015 wrote:
I used the following approach but was not able to proceed, could somebody tell me how/whether we can solve further to get the answer
A = 2 hrs (x), B = 4 hrs(y).
A+B = 4/3 hrs.

If B alone is working, it takes 4 hrs. But both working, B has to work only for 4/3 hours.
So B does not work for 4 hrs - 4/3 hrs = 8/3 hours.

I then thought that B did not have to work for 8/3 hours since A pitched in.
So 8/3 hours multiplied by A's rate (1/2)...gives 4/3..

But none of the options give this value. Any reason this approach does not work?...could somebody pls help.


B did not have to work for 8/3 hrs because A pitched in - great.
So what was the work that B would have done in this 8/3 hrs? Work = Rate*Time = (1/4)*(8/3) = 2/3 (it has to be B's rate)
Option (E) gives you 2/3.
So B did not have to do 2/3rd of the work.
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Re: Machine A can complete a certain job in x hours. Machine B  [#permalink]

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New post 18 Sep 2015, 09:15
VeritasPrepKarishma wrote:
Success2015 wrote:
I used the following approach but was not able to proceed, could somebody tell me how/whether we can solve further to get the answer
A = 2 hrs (x), B = 4 hrs(y).
A+B = 4/3 hrs.

If B alone is working, it takes 4 hrs. But both working, B has to work only for 4/3 hours.
So B does not work for 4 hrs - 4/3 hrs = 8/3 hours.

I then thought that B did not have to work for 8/3 hours since A pitched in.
So 8/3 hours multiplied by A's rate (1/2)...gives 4/3..

But none of the options give this value. Any reason this approach does not work?...could somebody pls help.


B did not have to work for 8/3 hrs because A pitched in - great.
So what was the work that B would have done in this 8/3 hrs? Work = Rate*Time = (1/4)*(8/3) = 2/3 (it has to be B's rate)
Option (E) gives you 2/3.
So B did not have to do 2/3rd of the work.


VeritasPrepKarishma
Thank you. You helped me understand where I was going wrong.
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Re: Machine A can complete a certain job in x hours. Machine B  [#permalink]

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New post 21 Nov 2015, 13:40
rate of A=1/x
rate of A+B=(x+y)/xy
(1/x)/[(x+y)/xy]=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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New post 21 Nov 2015, 14:07
2
The earlier posters have demonstrated the two methods (Algebraic and Input-Output) for solving a question type I call Variables in the Answer Choices.

If you'd like more information on these approaches, we have some free videos:
- Variables in the Answer Choices - http://www.gmatprepnow.com/module/gmat- ... /video/933
- Tips for the Algebraic Approach - http://www.gmatprepnow.com/module/gmat- ... /video/934
- Tips for the Input-Output Approach - http://www.gmatprepnow.com/module/gmat- ... /video/935

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

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New post 27 Sep 2016, 01:21
substituting values for x and y helped me here

x = 2 and y =3

ratio will be [1/2]/[5/6] = 3/5

substituting x and y values in each option, Only E satisfies the condition. Hence 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 31 Jan 2017, 03:42
1) First we need to find A and B's respective rates. A: \(r1*x=1; r1=\frac{1}{x}\); B: \(r2*y=1; r2=\frac{1}{y}\)
2) The rate of A and B working together is \(\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}\)
3) Now we need to find the time that A and B will be working together: \(\frac{y+x}{yx}*T=1; T=\frac{yx}{y+x}\)
4) Now we can find the fraction of work that A will do: \(\frac{1}{x}*\frac{yx}{y+x}=\frac{y}{y+x}\)

The correct answer is E.
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Machine A can complete a certain job in x hours. Machine B  [#permalink]

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New post 28 May 2017, 23:21
enigma123 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)



Machine A can complete a job in X hours.
Machine B can complete a job in Y hours.
Thus, A and B working together can complete a job in \(\frac{xy}{x+y}\) hours.

what fraction of the job that B will not have to complete because of A's help means that A will complete the whole work that requires \(\frac{xy}{x+y}\) hours
Machine A's rate=1/x
W=R X T
W = \(\frac{1}{x}\) X \(\frac{xy}{x+y}\)
W = \(\frac{y}{x + y}\)

HENCE, 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 03 Jul 2017, 06:28
If we would have substracted the work done by B from the total work then we have a different answer. Can some one tell me why substracting is incorrect ?? If we deduct the work done by B then we have work done by A right ? What is it that im missing here ??
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Re: Machine A can complete a certain job in x hours. Machine B  [#permalink]

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New post 17 Aug 2019, 07:51
and I was confused with the question.
I took time taken by both A and B is 2.
So, in one hour both can do 1/2 work.
Together they can do the full work in 1 hour.
So the amount of work reduced for B is 1/2
Only option E gives us 1/2 when we substitute the value of x and y as 2.
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Re: Machine A can complete a certain job in x hours. Machine B  [#permalink]

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New post 17 Aug 2019, 08:43
enigma123 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)


Given: Machine A can complete a certain job in x hours. Machine B can complete the same job in y hours.

Asked: 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?

Work done by Machine A & Machine B together in 1 hours
1/x + 1/y = x+y/xy
Total hours needed to finish the work = xy/(x+y)

Ratio of work done by Machine A to work done by Machine B = 1/x : 1/y = y:x
the fraction of the job that B will not have to complete because of A's help = Machine A's contribution to the job = y/(x+y) * 1

IMO 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 04 Apr 2020, 08:49
ALTERNATIVE APPROACH.

Work not done by B = work done by A
So we want the fraction of the job done by A.

Machine A does the job in X hours.
Machine B does the job in Y hours.
The fraction of the Job done by one machine has to be the ratio of a partial time (X or Y) to total time (X+Y)... so it could be either X/(X+Y) or Y/(X+Y). Which one is the right one?
To choose which one is the fraction of the job done by A let's think in this way:

More time (greater X=slower) brings lower fraction of job done---> so when X increases we want the fraction to decrease----> thus Y/(X+Y) is the right one.
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Re: Machine A can complete a certain job in x hours. Machine B   [#permalink] 04 Apr 2020, 08:49

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