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# Machine M can produce x units in 3/4 of the time it takes

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Manager
Joined: 11 Nov 2006
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Machine M can produce x units in 3/4 of the time it takes [#permalink]

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02 May 2007, 09:54
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Please solve it and explain it... thanks!

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Senior Manager
Joined: 11 Jun 2006
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02 May 2007, 13:44
Use machine O as a starting point since it is the slowest machine, all other machines will be a fraction of O.

Time of M = 3/4 * 2/3 = 6/12
Time of N = 2/3 * 4/4 = 8/12
Time of O = 12/12

The recipricol of the times stated above = the rates; since rates are additive, add...

M + N + O = 9/2

Putting N over the total:

3/2 / 9/2 = 1/3

Answer (B)

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Senior Manager
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Re: Machines M, N and O [#permalink]

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02 May 2007, 18:14
M N O
----------------------------
t/2 2t/3 t (How long it takes each to produce x)

In t: 2x 3x/2 x (How much they produce in t (hs, secs, etc))

What fraction of total output is produced by N? (3x/2)/(2x+3x/2+x) = 1/3 => B.

Last edited by Andr359 on 11 May 2007, 14:25, edited 1 time in total.

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Manager
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02 May 2007, 19:14
Is the answer C?

let o=12 hence n=8 and m = 6.

Total hours =26
n=8

Therefore fraction =8/26 =4/13

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Manager
Joined: 02 May 2007
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Re: Machines M, N and O [#permalink]

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02 May 2007, 21:27
querio 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

Please solve it and explain it... thanks!

(B) it is

M = time for M to produce the total output T:
=> T/M = output M produce in 01 unit of time
M = 3/4N
N = time for N to produce the total output T:
=> T/N = output N produce in 01 unit of time
N = 3/4O
O = time for O to produce the total output T

Output the 03 machines produce in 01 unit of time = T/M + T/N + T/O = T/(3/4N) + T/N + T(3N/2) = 3T/N

Fraction of the total output is produced by machine N = (T/N)/3T/N = 1/3

Last edited by kirakira on 02 May 2007, 22:03, edited 1 time in total.

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Manager
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02 May 2007, 21:57
B - 1/3

Let say it takes T time for O to produce 1 unit -> rate of work 1/T
N's rate of work 1/(2/3 * T) = 3/2T
M's rate of work - 1/(2/3 * 3/4 T) = 2/T

so total rate of work - 1/T + 3/2T + 2/T = 9/2T

N's part is (3/2T) / (9/2T) = 1/3

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Manager
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03 May 2007, 01:20
- time it takes machine O to do the job = t
- time it takes machine N to do the job = 2t/3
- time it takes machine M to do the job = 3/4 * 2/3 * t = t/2

In one hour O+N+M do 1/t + 3/2t + 2/t = 9/2t of the job.

In one hour N do (3/2t) / (9/2t) = 1/3 of the job.

Answer B.

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Manager
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03 May 2007, 04:27
[b]OA[/b] is B

thanks!

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VP
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03 May 2007, 12:59
baer wrote:
Use machine O as a starting point since it is the slowest machine, all other machines will be a fraction of O.

Time of M = 3/4 * 2/3 = 6/12
Time of N = 2/3 * 4/4 = 8/12
Time of O = 12/12

The recipricol of the times stated above = the rates; since rates are additive, add...

M + N + O = 9/2

Putting N over the total:

3/2 / 9/2 = 1/3

Answer (B)

can you please explain why 4/4 ?

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Senior Manager
Joined: 11 Jun 2006
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03 May 2007, 13:56
KillerSquirrel wrote:
baer wrote:
Use machine O as a starting point since it is the slowest machine, all other machines will be a fraction of O.

Time of M = 3/4 * 2/3 = 6/12
Time of N = 2/3 * 4/4 = 8/12
Time of O = 12/12

The recipricol of the times stated above = the rates; since rates are additive, add...

M + N + O = 9/2

Putting N over the total:

3/2 / 9/2 = 1/3

Answer (B)

can you please explain why 4/4 ?

Just to get a common denominator of 12, so that you can add up the rates in the next step. I was just showing the logic of the problem.

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03 May 2007, 13:56
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# Machine M can produce x units in 3/4 of the time it takes

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