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Challenging Rate Problem - Help w/ Alternate Solution

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Challenging Rate Problem - Help w/ Alternate Solution

by Ryan Ziemba » Fri Sep 03, 2010 7:47 pm
I found the example below in a Kaplan practice quiz and the only solution offered involves backsolving. Can anyone provide a mathematical solution? By the time backsolving occurred to me, I was already too deep into the math and resorted to poor guesswork.

Machine A and machine B are each used to manufacture 660 sprockets. It takes machine A 10 hours longer to produce 660 sprockets than machine B. Machine B produces 10% more sprockets per hour than machine A. How many sprockets per hour does machine A produce?

a) 6
b) 6.6
c) 60
d) 100
e) 110

Answer: D
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by Rahul@gurome » Fri Sep 03, 2010 8:25 pm
Solution:
Let machine A produce x sprockets in 1 hour and let machine B produce y sprockets in 1 hour.
It takes machine A 660/x hours to produce 660 sprockets and it takes machine B 660/y hours to produce 660 sprockets.
Or 660/x - 660/y = 10.
Or 66*(y-x) = xy.
Also y = (110/100)*x = (11/10)*x
Or 66*(x/10) = (11/10)*(x^2).
Or x = 6 sprockets.

The correct answer is (A).
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by this_time_i_will » Sat Sep 04, 2010 1:09 am
The trick in such questions is to find the relationship b/w two given information pieces. The first sentences gives us information about number of sprockets produced and time taken to produce them. The second statement talks about number of sprockets prodcued per hour.
A short thinking would tell us that it is possible to convert the infomration in first sentence to sprockets produced per hour, thus making both sentence in same unit. i..e.. sprockets produced per hour.

Let B takes x hrs. for producing 660 sprockets.
Sentence I: A per hr = 660/x+10.
B per hr = 660/x.

Now useing info form second sentence:
660/x = 1.1(660/(x+10))...solving this equation should be a breeze, giving x = 100 hrs.

A's per hour = 660/(x+10) = 660/110 = 6
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by Ryan Ziemba » Sat Sep 04, 2010 6:12 am
ah. applying the per hour approach makes sense to me. Thanks!
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