Machine A produces pencils at a constant rate of 9000

AAPL wrote:GMAT Paper Tests

Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines together must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?

A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4

OA A

To MINIMIZE machine B's operating time, we must MAXIMIZE the time machine A's operating time.
So, let machine A operate for the full 8 hours.
In 8 hours, machine A produces 72,000 pencils [8 hours x 9,000 pencils/hour =72,000 pencils]

So, the number of pencils machine B must make = 100,000 - 72,000 = 28,000

Time = output/rate
So, operating time = 28,000/7000 = 4 hours

Answer: A

Cheers,
Brent

AAPL wrote:GMAT Paper Tests

Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines together must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?

A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4

OA A

If machine B’s operational time is to be minimized, we must maximize the time for machine A to operate. Since each machine can operate for at most 8 hours, we can let machine A operate for the full 8 hours. Since the rate of machine A is 9000 pencils per hour, machine A produces 8 x 9000 = 72,000 pencils, and hence 28,000 pencils are left to be produced.

Thus, it will take machine B 28,000/7,000 = 4 hours to produce the remaining pencils.

Answer: A