Algebra Problem, solving for the waterweight of grapes?

[PotentialE]

Homework Statement

Fresh grapes contain 80% water by weight, whereas dried grapes contain 15% water. How many pounds of dried grapes can be obtained from 34 pounds of fresh grapes?

Homework Equations

anything you can create!

The Attempt at a Solution

To make a regular grape a dried grape, 65% of its water weight must be deducted
waterweight of 34lbs of grapes = 34*.8 = 27.2
65% of 27.2 = 17.68
34 - 17.68 = 16.32lbs

but it says that this is wrong, where is the fault in my logic?

 

[RoshanBBQ]

I'm not sure how you know 65% of its water must be deducted. I would, instead, think about it this way:
grape = base + water
dried grape = base + water

The base in these 2 equations is the same. You can find base using equation 1 and the fact that 80% is water. You can find water in equation 2 by using base previously found in combination with the fact that 15% of the total will need to be water.

 

[PotentialE]

grape = base + water
dried grape = base + water

b = 34 - w
b = dg - wdg

w = (.8)(34) = 27.2
b = 6.8

6.8 = dg - (.15)(dg)
6.8 = .85dg
dg = 8lbs

thank you! that's correct