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
Spoiler Alert :VeritasPrep
Mock Test Question.
Though I could get this question right, I took more than 3 minutes. Is there any simpler way to solve this one?
Responding to a pm:
Actually, the best method for this question is the one given by Bunuel. I used the same method to give the explanation in the mock too. But you can use weighted average if you wish.
Raisins have 20% water. Water is 100% water. You mix these two to make grapes which has 92% water. (This is another way of saying you remove water from grapes to make raisins)
In what ratio will you mix them?
Wr/Ww = (Cw - Cavg)/(Cavg - Cr) = (100 - 92)/(92 - 20) = 8/72 = 1/9
So every 1 unit of raisin will be mixed with 9 units of water to give 10 units of grapes.
So 10 pounds of raisins will mix with 90 pounds of water to give 100 pounds of grapes.
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