Bunuel
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.
Answer: B