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tricky gmat prob

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tricky gmat prob

by quantskillsgmat » Tue Dec 27, 2011 11:37 pm
Q three pieces of weight 9/2 lbs,27/4 lbs and 36/5 lbs respectively are to be divided into parts of equal weight.further each part must be as heavy as possible. If one such part is served for each guest ,then what is maximum number of guests that can be entertained.
a)54 b)72 c)20 d)none
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by shankar.ashwin » Wed Dec 28, 2011 2:47 am
Convert all fractions to a common base.

90/20 - 135/20 - 144/20

To divide them to equal parts (heaviest) we need to take GCF of the numerator.

GCF of 90,135 and 144 = 9

So each part should be divided into 9/20.

Total such parts that can be divided = (90+135+144)/9 = 41 D IMO
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