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Machine M can produce x units in 3/4 of the time it takes machine N to
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Updated on: 02 Jul 2015, 13:38
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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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Originally posted by vscid on 11 Jan 2009, 17:31.
Last edited by Bunuel on 02 Jul 2015, 13:38, edited 1 time in total.
Renamed the topic, edited the question and added the OA.




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Re: Machine M can produce x units in 3/4 of the time it takes machine N to
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03 Jul 2015, 00:18
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 We can avoid the entire cumbersome calculation of fraction by choosing a RIGHT NUMBER for time and x Let, x = 24 Units Let, Time taken by Machine O to produce x(=24) units = 12 minsi.e. Time taken by Machine O to produce x(=24) units = 12 minsi.e. Time taken by Machine N to produce x(=24) units = (2/3)*12 = 8 minsi.e. Time taken by Machine M to produce x(=24) units = (3/4)*8 = 6 minsMachine O produces 24 units in 12 minsi.e In 1 min time, the No. of Units produced by Machine O = 24/12 = 2 Unitsi.e In 1 min time, the No. of Units produced by Machine N = 24/8 = 3 Unitsi.e In 1 min time, the No. of Units produced by Machine M = 24/6 = 4 Unitsi.e. Total Units Products by all three machine (M, N and O) together) = 2+ 3+ 4 = 9 unitsFraction of Total Work Done by Machine N (in 1 Min) = Units Produced by N / Total Units produced = 3/9 = 1/3 Answer: Option B
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Re: Machine M can produce x units in 3/4 of the time it takes machine N to
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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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Re: Machine M can produce x units in 3/4 of the time it takes machine N to
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13 Jan 2009, 08:48
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: Machine M can produce x units in 3/4 of the time it takes machine N to
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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.



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Re: Machine M can produce x units in 3/4 of the time it takes machine N to
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02 Jul 2015, 14:04
We will take all rates in terms of N machine. Suppose N Machine takes ’n’ Hours Therefore M machine takes 3/4n hours and O machine takes 3/2 n Hours (n=3/2O) All machines together in 1 hour will do= 4x/3n+x/n+2x/3n= 3x/n So all three machines works at the rate of 3x/n and machine N works at the rate of x/n. Output produced by machine N/Total output= x/n/3x/n= 1/3 Thanks
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Machine M can produce x units in 3/4 of the time it takes machine N to
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Updated on: 18 Jul 2016, 02:53
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 Another way to look at the problem..
Time ratio for M and N
M:N = 3:4....Efficiency(or work done in the same time) Ratio will be the inverse of this..m:n = 4:3
Similarly..Eff. Ratio for N and O n:o = 3:2
Overall m:n:o = 4:3:2
Fraction of work done by N?
\(\frac{3}{4+3+2}\)
\(=\frac{1}{3}\)
(B)
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Originally posted by ShashankDave on 17 Jul 2016, 06:40.
Last edited by ShashankDave on 18 Jul 2016, 02:53, edited 1 time in total.



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Machine M can produce x units in 3/4 of the time it takes machine N to
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17 Jul 2016, 18:07
let m,n,o times=3/2,2,3 m,n,o rates=2/3,1/2,1/3 combined rate=3/2 (1/2)/(3/2)=1/3



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Re: Machine M can produce x units in 3/4 of the time it takes machine N to
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03 Feb 2017, 07:02
Machine O: time y; rate o; y*o=x; o=x/y Machine N: time 2/3y; rate n; 2/3y*n=x; n=3x/2y Machine M: time 3/4*2/3*y=y/2; rate m; y/2*m=x; m=2x/y
Total rate: x/y+3x/2y+2x/y=9x/2y Rate of n/Total rate=3x/2y / 9x/2y= 1/3
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Re: Machine M can produce x units in 3/4 of the time it takes machine N to
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03 Feb 2017, 20:32
I simply plugged in numbers for this question. Let's say X= 100 M= 45 mins to produce 100 units N= 60 mins to produce 100 units O= 90 mins to produce 100 units Combined (1/45) + (1/60) + (1/90) = 9/180, which we flip to give us 20 mins that it will take for the 3 machines combined to produce 100 units. From here we just use 20/N total time so 20/60 to get 1/3 which is choice B.



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Re: Machine M can produce x units in 3/4 of the time it takes machine N to
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04 Feb 2017, 08:07
Let, M takes 3 hours So, N takes 4 hours And O takes 6 hours LCM of 3, 4, and 6 is 12 In 12 hours M produces 4x In 12 hours N produces 3x In 12 hours O produces 2x In 12 hours, total production by M, N, and O is 9x out of which N produces 3x, i.e., 1/3.
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Re: Machine M can produce x units in 3/4 of the time it takes machine N to
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05 Feb 2017, 05:20
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 Let the time taken be as under  O = 12 N = 8 M = 6 Assume total Work be 24 units... Thus efficiency of the 3 machines are as under  O = 2 N = 3 M = 4 Total efficiency is 9 ( ie, 2 + 3 + 4 ) units/hr So, work done by N is 3/9 = 1/3 Hence, answer will be (B) 1/3
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Re: Machine M can produce x units in 3/4 of the time it takes machine N to
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22 Apr 2018, 19:38
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