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constant rates! tricky

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by Ozlemg » Sat Aug 20, 2011 4:49 am
Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?

A. 4
B. 6
C. 8
D. 10
E. 12

OA : E
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Source: — Data Sufficiency |

by Anurag@Gurome » Sat Aug 20, 2011 5:59 am
Ozlemg wrote:Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?
Say, X takes x days to produce w widgets.
So, Y will take (x - 2) days to produce w widgets.

Also X will take 2x days to produce 2w widgets. So we have to determine the value of 2x.

Now, in 3 days both of them together produce 3*(w/x + w/(x - 2)) widgets.

Therefore, 3*(w/x + w/(x - 2)) = 5w/4 ----> [1/x + 1/(x - 2)] = 5/12 = 10/24 = (4 + 6)/(4*6) = 1/4 + 1/6 = 1/6 + 1/(6 - 2)

Hence, x = 6

Okay! That blue part was completely my intuition. You don't need to follow that.

We have, [1/x + 1/(x - 2)] = 5/12 which is an quadratic equation in x. We can go on solving that but that will take a bit more time. To save that we can utilize the fact that the options are giving the value of 2x. Hence, plug in half the value of each options in that equation and check.

Only Option E works.

The correct answer is E.
Anurag Mairal, Ph.D., MBA
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