# Homework Help: Algebra Problem, solving for the waterweight of grapes?

1. Apr 4, 2012

### PotentialE

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
anything you can create!

3. 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?

2. Apr 4, 2012

### 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.

3. Apr 4, 2012

### PotentialE

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