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

GMATGuruNY wrote: 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.

manik11 wrote:Mitch,
I tried to solve for the weight of raisins by equating the weight of water in both raisins and grapes. So, if raisins are 2 pound water, this 2 pound of water must constitute for 92% of grapes.
My equation looked like this:
2 = (92/100)x
x = 200/92

What am I doing wrong here?

Thanks!
Manik

GMATGuruNY wrote:Raisins are dried grapes: grapes that are dried in the sun so that their water content is greatly REDUCED.
Grapes contain MUCH more water than raisins do.

manik11 wrote:Mitch,
I tried to solve for the weight of raisins by equating the weight of water in both raisins and grapes. So, if raisins are 2 pound water, this 2 pound of water must constitute for 92% of grapes.
My equation looked like this:
2 = (92/100)x
x = 200/92

What am I doing wrong here?

Thanks!
Manik

The portion in red and the equation in blue misrepresent the problem.
As stated above, grapes contain much more water than raisins do.
Thus, the 2 pounds of water in the raisins must be far less than the 92% water content in the grapes.
Thus, the equation in blue is invalid.