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Weighted averages - Grapes/Raisins

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by gmattesttaker2 » Sun May 04, 2014 9:01 pm
Hello,

Can you please assist with this:

If grapes are 92% water and raisins are 20% water by weight, then how much did a quantity of raisins, which currently weighs 10 pounds, weigh when all the raisins were grapes? (Assume that the only difference between their raisin-weight and their grape-weight is water that evaporated during their transformation.)


OA: 100 pounds


Thanks a lot for your help - Sri
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by Tushar14 » Sun May 04, 2014 9:28 pm
Hi Sri,
As we know only water evaporates, then the weight of non-water substance in grapes and raisins is the same. Thus 0.08*{weight of grapes}=0.8*{weight of raisins} --> 0.08x = 0.8*10 --> x = 100.

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by GMATGuruNY » Mon May 05, 2014 2:53 am
If grapes are 92% water and raisins are 20% water, then how much did a quantity of raisins, which currently weighs 10 pounds, weigh when all the raisins were grapes? (Assume that the only difference between their raisin-weight and their grape-weight is water that evaporated during their transformation.)

A) 25 pounds

B) 46 pounds

C) 92 pounds

D) 100 pounds

E) 146 pounds
Since raisins are 20% water, the 10 pounds of raisins are composed of 2 pounds of water and 8 POUNDS OF OTHER STUFF.

Since grapes are 92% water, the 8 POUNDS OF OTHER STUFF in the original grapes must have constituted 8% of the grapes:
8 = (8/100)x
x = 800/8 = 100.

The correct answer is D.
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