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# PS-machine

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PS-machine [#permalink]  11 Jan 2009, 17:31
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Machine M can produce x units in 3/4 of the time it takes machine N to produce x units. Machine N can produce x units in 2/3 the time it takes machine O to produce x units. If all three machines are working simultaneously, what fraction of the total output is produced by machine N?

(A) 1/2
(B) 1/3
(C) 4/13
(D) 8/29
(E) 6/33
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Manager
Joined: 04 Jan 2009
Posts: 243
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Re: PS-machine [#permalink]  11 Jan 2009, 17:50
Let the following be true:
M makes x in time t
Then the following follows:
N makes x in 4t/3
O makes x in 3/2(4t/3) = 2t
M:N:O = 1:4/3:2=3:4:6
So N=4/(3+4+6)=4/13.
vscid wrote:
Machine M can produce x units in 3/4 of the time it takes machine N to produce x units. Machine N can produce x units in 2/3 the time it takes machine O to produce x units. If all three machines are working simultaneously, what fraction of the total output is produced by machine N?

(A) 1/2
(B) 1/3
(C) 4/13
(D) 8/29
(E) 6/33

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tusharvk

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Location: Toronto
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Re: PS-machine [#permalink]  13 Jan 2009, 08:48
Expert's post
vscid wrote:
Machine M can produce x units in 3/4 of the time it takes machine N to produce x units. Machine N can produce x units in 2/3 the time it takes machine O to produce x units. If all three machines are working simultaneously, what fraction of the total output is produced by machine N?

(A) 1/2
(B) 1/3
(C) 4/13
(D) 8/29
(E) 6/33

For combined rates problems, if you can work out what each machine does in the same amount of time, the problem will almost always become easy. Since x is unimportant here, rather than talk about 'producing x units', I'll simply talk about 'completing one job'. Say it takes M t hours to complete one job. Then we know, from the ratios provided:

M completes one job in t hours
N completes one job in 4t/3 hours
O completes one job in 2t hours

We can easily work out what each would do in 4t hours:

M completes 4 jobs in 4t hours
N completes 3 jobs in 4t hours
O completes 2 jobs in 4t hours

So if they work together for the same amount of time, M does 4/9 of the work, N does 3/9 = 1/3 of the work, and O does 2/9 of the work. The answer is 1/3.
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Re: PS-machine [#permalink]  13 Jan 2009, 11:32
Exactly ! For combined work problems we cannot simply compare ratios as done in the first explanation.
Thanks Ian.

Now I did the following way: (reverse method)
Let O take t hours
Let N take (2/3)*t hours
Let M take 2/3t(3/4) = (1/2)*t hours

Now using the formula: 1/M + 1/N + 1/O = 1/T ( T = combined hours )

we get, (9/2t) = 1/T.
t = 9T/2 => N = (2/3)*t = (2/3)*(9T/2)

Hence, N = 3 T .
Now, the question is to find fraction of total output T that is produced by N,
so do we have to find T/N ?? or N/T ?
Also please let me know if the above approach is wrong.
Re: PS-machine   [#permalink] 13 Jan 2009, 11:32
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