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Working at their respective constant rates, printing machine X, Y, and

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Working at their respective constant rates, printing machine X, Y, and  [#permalink]

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New post Updated on: 24 Apr 2015, 08:31
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Working at their respective constant rates, printing machine X, Y, and Z can finish a certain work in 9, 12, and 18 hours. If three machines work together to finish the work, what fraction of the work will be finished by the machine Z?

(A) 4/9
(B) 1/3
(C) 1/4
(D) 2/9
(E) 1/9

Originally posted by banksy on 17 Feb 2011, 15:11.
Last edited by Bunuel on 24 Apr 2015, 08:31, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Working at their respective constant rates, printing machine X, Y, and  [#permalink]

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New post 17 Feb 2011, 19:47
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banksy wrote:
192. Working at their respective constant rates, printing machine X, Y, and Z can finish a certain work in 9, 12, and 18 hours. If three machines work together to finish the work, what fraction of the work will be finished by the machine Z?
(A)
(B)
(C)
(D)
(E)


Since time taken by X, Y and Z is in the ratio 9:12:18 i.e. 3:4:6,
their speeds are in the ratio \(\frac{1}{3}:\frac{1}{4}:\frac{1}{6} = 4:3:2\)
So Z will do \(\frac{2}{(4+3+2)} = 2/9\) of the work
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Re: Working at their respective constant rates, printing machine X, Y, and  [#permalink]

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New post 18 Feb 2011, 01:48
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X; 9 hour-> 1 work
=> 1 hour -> 1/9 work

Y; 12 hour-> 1 work
=> 1 hour-> 1/12 work

Z; 18 hour-> 1 work
=> 1 hour-> 1/18 work

Total work done by all machines in 1 hour;
1/9+1/12+1/18 = (4+3+2)/36=9/36=1/4 work

Jointly;
1/4 work-> 1 hour
1 work-> 1/(1/4) = 4 hours(Jointly X,Y,Z all spent for 4 hours towards completing the job)

We know from before;
Z; 18 hour-> 1 work
=> 1 hour-> 1/18 work

If Z can perform 1/18 work in 1 hour;
In 4 hours, he would have performed
(1/18)*4 work
2/9 work

Ans: 2/9
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Re: Working at their respective constant rates, printing machine X, Y, and  [#permalink]

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New post 17 Feb 2011, 16:02
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banksy wrote:
192. Working at their respective constant rates, printing machine X, Y, and Z can finish a certain work in 9, 12, and 18 hours. If three machines work together to finish the work, what fraction of the work will be finished by the machine Z?
(A)
(B)
(C)
(D)
(E)


In 1 hour X can finish 1/9=4/36 of the work, Y can finish 1/12=3/36 of the work and Z can finish 1/18=2/36 of the work. So in 1 hour out of 4+3+2=9 parts of the work Z finished 2 parts or 2/9 of the work.

Or if Z can do 1 part in some time interval then in the same time interval X can do 2 parts (as its rate is twice the rate of Z), and Y can do 1.5 parts (again as its rate is 1.5 times the rate of Z) so share of Z is 1 out of 1+2+1.5=4.5, or 1/4.5=2/9.

Please provide the answer choices for PS questions.
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Re: Working at their respective constant rates, printing machine X, Y, and  [#permalink]

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New post 03 May 2016, 13:59
The fraction of the work performed by Z is :

Rz / (Rx + Ry + Rz) = (1/18) / (1/9 + 1/12 + 1/18) = (1/18)/(1/4) = 2/9
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Re: Working at their respective constant rates, printing machine X, Y, and  [#permalink]

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New post 29 Nov 2018, 09:22
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Working at their respective constant rates, printing machine X, Y, and   [#permalink] 29 Nov 2018, 09:22
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