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Algebra Problem, solving for the waterweight of grapes?

  1. Apr 4, 2012 #1
    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. jcsd
  3. Apr 4, 2012 #2
    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.
  4. Apr 4, 2012 #3
    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
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