SDN members see fewer ads and full resolution images. Join our non-profit community!

%soln math Q

Discussion in 'DAT Discussions' started by tRNA, Apr 23, 2007.

  1. tRNA

    tRNA Member 10+ Year Member

    Apr 28, 2005
    how many gallons of pure pulp must be added to 30 gallons of a 20% solution of orange pulp to make it a 50% solution?
    this is the formula you use:

    if x=number of gallons we have to add
    then solve for x

    Gary mixed 4oz of rum with cola. The rum is 40% alcohol by volume. If the drink is to be 16% alcohol by volume, how much cola is needed?
    i don't understand why can't we use that same equation above to solve this one, ie why is setting up the equation this way is wrong??:

    if x=number of oz of cola we have to add
    then solve for x
  2. SDN Members don't see this ad. About the ads.
  3. Ex_EE

    Ex_EE 2+ Year Member

    Sep 23, 2006
    It doesn't work the way you set it up because the way you have x you are solving for adding more alcohol. Here is how it should be. Everything needs to be converted to percent cola.
    x=6 oz cola
  4. Nasem

    Nasem 2+ Year Member

    Aug 30, 2006
    Lansing, MI
    This is a cool problem, I stopped going my general chemistry homework just to work it out..... heres how I went about solving it:

    20% pulp ---meaning---> 6 gallons of pulp
    80% remaining ---meaning---> 24 gallons of water
    This makes sense, if you do 6 / 30 = 0.20 (20%)
    and 24 / 30 = 0.80 (80%)

    You want:
    (6 + x) / (30 + x) = 0.50
    solve this and you get x=18 which is the amount of pure pulp you add to the solution to get it to become 50%.

    (18+6) / (30+18) = 0.50 pulp
    24 / (30+18) = 0.50 water
  5. tRNA

    tRNA Member 10+ Year Member

    Apr 28, 2005
    thanks EX_EE

Share This Page