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Machine A can produce balls at a constant rate of 2 balls pe

by Max@Math Revolution » Tue Mar 20, 2018 12:41 am
[GMAT math practice question]

Machine A can produce balls at a constant rate of 2 balls per hour, and machine B can produce balls at a constant rate of 3 balls per hour. If at least one of machine A and machine B produces balls at any time, what is the smallest possible number of hours that machine A and machine B must work together at their constant rates to produce 70 balls in 20 hours?

A. 5hrs
B. 6hrs
C. 7hrs
D. 8hrs
E. 9hrs
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by GMATGuruNY » Tue Mar 20, 2018 4:53 am
I believe that the following reflects the intent of the problem.
Machine A produces balls at a constant rate of 2 balls per hour, and machine B produces balls at a constant rate of 3 balls per hour. 70 balls must be produced over the next 20 hours. If at least one of the two machines works at all times, what is the least number of hours that the two machines must work together?

A. 5hrs
B. 6hrs
C. 7hrs
D. 8hrs
E. 9hrs
Since A's rate = 2 balls per hour and B's rate = 3 balls per hour, the combined rate for A+B working together = 5 balls per hour.
We can PLUG IN THE ANSWERS, which represent the number of hours A and B work together.
Since the question stem asks for the LEAST number of hours, start with the smallest answer choice.
To MINIMIZE the number of hours that the two machines work together, we must MAXIMIZE the rate when only ONE machine works.
Since B is faster than A, the machine working at all times must be B.
When the correct answer is plugged in, 70 balls will be produced in 20 hours.

A: 5 hours for A+B together, implying 15 hours for B alone, for a total of 20 work-hours
Since A+B produce 5 balls per hour, the number of balls produced by A+B in 5 hours = rt = 5*5 = 25 balls.
Since B produces 3 balls per hour, the number of balls produced by B in 15 hours = rt = 3*15 = 45.
Total balls produced = 25+45 = 70.
Success!

The correct answer is A.
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by Max@Math Revolution » Wed Mar 21, 2018 11:45 pm
=>

Since machine B has the faster working rate, it must work for the entire period to minimize the number of hours that machine A has to work.
When machine B works for 20 hours, it produces 60 balls. This means that machine A must produce 70 - 60 = 10 balls, and it must work for 5 hours, since the work rate of machine A is 2 balls per hour.

Therefore, the answer is A.

Answer: A
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by Jeff@TargetTestPrep » Thu Mar 22, 2018 2:52 pm
Max@Math Revolution wrote:[GMAT math practice question]

Machine A can produce balls at a constant rate of 2 balls per hour, and machine B can produce balls at a constant rate of 3 balls per hour. If at least one of machine A and machine B produces balls at any time, what is the smallest possible number of hours that machine A and machine B must work together at their constant rates to produce 70 balls in 20 hours?

A. 5hrs
B. 6hrs
C. 7hrs
D. 8hrs
E. 9hrs
To minimize the time both the machines are working together, we must maximize the time that the faster of the two machines, machine B, works alone. Let's assume that machine B works for the entire 20 hours, producing 20 x 3 = 60 balls. The remaining 70 - 60 = 10 balls must be produced by machine A, and it will take 10 / 2 = 5 hours for machine A to produce them. Thus, the two machines must work for a minimum of 5 hours together.

Answer: A

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