Bunuel wrote:

Machine A produces parts twice as fast as machine B does. Machine B produces 100 parts of product X in 40 minutes. If each machine produces parts at a constant rate, how many parts of product Y does machine A produce in 6 minutes, if each part of product Y takes 3/2 times of the time taken to produce each part of product X?

(A) 45

(B) 30

(C) 25

(D) 20

(E) 15

1. Find rates of Machines B and A in hours.

Machine B rate: B makes 100 parts in 40 minutes. 40 minutes is \(\frac{2}{3}\) of one hour.

B's rate = \(\frac{100}{(\frac{2}{3})}\) =

\(\frac{100}{1} * \frac{3}{2}\) =

\(\frac{300}{2}\) or 150 parts in 1 hour. B = \(\frac{150}{1}\)

A's rate for product X: A produces parts twice as fast as B does.

In one hour Machine A will produce \(\frac{150}{1}\) * 2 = 300 parts of product X in 1 hour -->

\(\frac{300}{1}\)2. Find rate of Machine A for product Y.

To produce each part of Y takes \(\frac{3}{2}\) times as long as it does to produce each part of product X.

Machine A produces 300 parts in one hour for product X. \(\frac{3}{2}\) times as long means 1.5 hours for the same amount of parts.

A's rate for product Y is \(\frac{300}{1.5} = \frac{3000}{15}\) = 200 Y parts in one hour =

\(\frac{200}{1}\)3. Find how many parts of Product Y that Machine A can make in 6 minutes.

Time in hours:

\(\frac{6 min}{60 min}\) =

\(\frac{1}{10}\) hourA's rate for producing Y is

\(\frac{200}{1}\) parts per hour

R*T = W --- >\(\frac{200}{1}\)

* \(\frac{1}{10}\) =

20 partsAnswer D

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