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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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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.
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Re: Working at their respective constant rates, printing machine X, Y, and [#permalink]
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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]
banksy wrote:
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

Let,Total amount of work= LCM of 9,12,18=36
Amount of work By X machine in 1 hr= 36/9=4
Amount of work By Y machine in 1 hr= 36/12=3
Amount of work By Z machine in 1 hr = 36/18=2

if three machines work together for 1 hr, the amount of work they will complete= 4+3+2=9
we require 36 unit of work to be done
it would take 4 hr to complete 36 unit of work by all three machine
hence,
the amount of work contributed by Z machine =2*4=8

Now, the fraction of the work finished by the machine Z = 8/36 = 2/9

IMO is D
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Re: Working at their respective constant rates, printing machine X, Y, and [#permalink]
Let the total work be LCM of (9, 12, 18), i.e. 36 units.
Thus, total units are done by X in 1 hour = 4 units
Total units are done by Y in an hour = 3 units
Total units are done by z in an hour = 2 units.

If three machines work together,
Proportion of work done by z = 2/(4+3+2) = 2/9

Thus, the correct option is D.
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Re: Working at their respective constant rates, printing machine X, Y, and [#permalink]
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Re: Working at their respective constant rates, printing machine X, Y, and [#permalink]
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