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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