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Math Expert V
Joined: 02 Sep 2009
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Machine Mean produce x units in 3/4 of the time it takes machine N to  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 58% (02:12) correct 42% (02:52) wrong based on 112 sessions

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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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Senior PS Moderator V
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Machine Mean produce x units in 3/4 of the time it takes machine N to  [#permalink]

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Bunuel 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

If Machine N takes 120 minutes to produce x units, machine M will take $$\frac{3}{4}*120 = 90$$ minutes.
Similarly, as Machine N will take $$\frac{2}{3}$$rd of the time to produce the units machine O produces.
$$120 = \frac{2}{3}*x$$ -> $$x = 120*\frac{3}{2} = 180$$ minutes

If 360 units are being produced,
machine N will produce 3 units in a minute,
machine M will produce 4 units in a minute,
and machine O will produce 2 units in a minute.

Therefore, machine N will produce $$\frac{3}{3+4+2} = \frac{3}{9} = \frac{1}{3}$$ of the total units(Option B)
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Intern  B
Joined: 15 Jan 2018
Posts: 7
Re: Machine Mean produce x units in 3/4 of the time it takes machine N to  [#permalink]

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Bunuel 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

Let "t" be the time taken by machine O to produce x units ==> machine O produces 'x/t' units per hour (time unit not mentioned so considering it as per hour).
Since, it is given that machine N takes 2/3 of time taken by machine O ==> time taken by machine N is (2/3)t ==> machine N produces 'x/(2/3)t' units per hour.
Similarly ==> machine M produces 'x/(1/2)t' units per hour.

Now, for the question ==> units produced by machine N per hour/ units produced by all machines per hour simultaneously
units produced by machine N per hour ==> 'x/(2/3)t'
units produced by all machines per hour simultaneously ==> adding 'x/t', 'x/(2/3)t' and 'x/(1/2)t' which is equal to x/(2/9)t
After division the result is '1/3'.
Intern  B
Joined: 09 Nov 2015
Posts: 34
Re: Machine Mean produce x units in 3/4 of the time it takes machine N to  [#permalink]

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Let T mins. be the time taken by Machine N to produce x widgets. Then it follows that:
Machine M will produce (4x/3) widgets in T mins.
Machine O will produce (2x/3) widgets in T mins.
So, total number of widgets produced in T mins. when all three machines are working together is (x+4x/3+2x/3)=9x/3=3x.
Since Machine N's contribution to the total output is x widgets, it produces 1/3 of the total output. Ans: B.
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Re: Machine Mean produce x units in 3/4 of the time it takes machine N to  [#permalink]

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Bunuel 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
Attachment: MACHINES M, N AND O.PNG [ 2.28 KiB | Viewed 628 times ]

Let the total work be 24 units so, efficiency of the machines are
Attachment: MNO Efficiency.PNG [ 2.91 KiB | Viewed 630 times ]

Since all the 3 machines are working simultaneously work done by N is $$\frac{3}{(2+3+4)}$$ = $$\frac{1}{3}$$ , Answer must be (B)
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Intern  B
Joined: 08 Feb 2017
Posts: 5
Re: Machine Mean produce x units in 3/4 of the time it takes machine N to  [#permalink]

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t=time that O needs to produce x units then
N: x/(2/3T)=3x/2t and M:x/(3/4)*(2/3)*t=(2x)/t
so sum=O+N+M=x+3x/2t+2x/t=9x/2t
N/Sum=(3x/2t)/(9x/2t)=1/3

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Re: Machine Mean produce x units in 3/4 of the time it takes machine N to  [#permalink]

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Bunuel 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

let O's time=t
N's time=2t/3
N's rate=3/2t
total rate=2/t+3/2t+1/t=9/2t
N/total rate=(3/2t)/(9/2t)=1/3
B
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Re: Machine Mean produce x units in 3/4 of the time it takes machine N to  [#permalink]

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Bunuel 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

Let t = the time it takes machine O to produce x units; then the time it takes machine N to produce x units = (2/3)t, and the time it takes machine M to produce x units = (3/4)(2/3)t = (1/2)t.

Therefore, the rate of M is x/[(1/2)t] = 2x/t, the rate of N is x/[(2/3)t] = 3x/2t, and the rate of O is x/t. Their combined rate is 2x/t + 3x/2t + x/t = 4x/2t + 3x/2t + 2x/2t = 9x/2t.

If all three machines are working simultaneously, the fraction of the total output that is produced by machine N is:

(3x/2t)/(9x/2t) = 3x/9x = 1/3

Alternate Solution:

Let’s denote the number of units produced by machine O in 12 hours as y.

Since machine N can produce the same number of units in 2/3 of the time, machine N can produce y units in 8 hours.

Since machine M can produce the same number of units in 3/4 of the time necessary for machine N, machine M can produce y units in 6 hours.

Now, let’s suppose all the machines ran for 24 hours. In 24 hours, machine M will produce 4y units; machine N will produce 3y units, and machine O will produce 2y units. The total number of units produced is 4y + 3y + 2y = 9y. The number of units produced by machine N is 3y, which is (3y)/(9y) = 1/3 of all the units produced.

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