Help with this question:: Machine A produces pencils at a

[lebronge]

Machine A produces pencils at a constant rate of 9,000 pencils per hour, and machine B produces pencils at a constant rate of 7,000 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 and 2/3
c) 5 and 1/3
d) 6
e) 6 and 1/4

This is from the GMAT Prep. I do not have the official answer. The way I solved it was to see what Machine A can produce in 8 hours at its rate, which is 9,000 x 8 = 72,000. 100-72 = 28,000, and 28,000/7,000 = 4 hours. Thus, I would pick A.

Is this the correct answer? It seems to easy of an solution. Please let me know.

[Blue_Lotus]

The Question asks us to use Machine B the least amount of time. This means we have to use Machine A
the maximum number of times to achieve the goal.
Now it is given that a machine can operate a maximum of 8 hours.
Therefore the number of pencils produced in 8 hours by Machine A would be = 9000*8 = 72000

The remaining needs to be produced by machine B.
Remaining pencils that needs to be produced = 100000-72000 = 28000

The time taken by Machine B would be 28000/7000 = 4 hours

Hope this clarifies your doubt

[RonPurewal]

The way I solved it was to see what Machine A can produce in 8 hours at its rate, which is 9,000 x 8 = 72,000. 100-72 = 28,000, and 28,000/7,000 = 4 hours. Thus, I would pick A.

Is this the correct answer? It seems to easy of an solution. Please let me know.

you've just discovered one of the golden secrets of the gmat, which is that a great many problems can become 'easy' if they are REPHRASED in the right way. here, you have rephrased the problem in a way that makes it much easier to solve: namely, minimizing the # of hours on machine b = maximizing the # of hours on machine a.

speaking of rephrasing: if the problem had instead asked for the lowest TOTAL number of hours possible, then your approach would be exactly the same as your approach here (because, if you want to take the least amount of time, you should maximize the time spent on the most efficient machine).

by the way, it's too easy, not 'to easy'.