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Mixture problem

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Mixture problem

by topspin360 » Thu Jan 30, 2014 8:11 pm
Can someone please show how to do this using algebraic approach? Just when I thought I had mixtures down..

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

Thanks!
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by GMATGuruNY » Thu Jan 30, 2014 9:39 pm
topspin360 wrote:Can someone please show how to do this using algebraic approach? Just when I thought I had mixtures down..

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 must constitute 8% of the grapes:
8 = (8/100)x
x = 800/8 = 100.

The correct answer is D.
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