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%soln math Q

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

  1. tRNA

    tRNA Member
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    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
    .2(3)+x=.5(30+x)
    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
    .4(4)+x=.16(4+x)
    then solve for x
     
  2. Ex_EE

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    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.
    (.6)(4)+x=.84(4+x)
    x=6 oz cola
     
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  3. Nasem

    2+ Year Member

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

    Given:
    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%.

    Check:
    (18+6) / (30+18) = 0.50 pulp
    and
    24 / (30+18) = 0.50 water
     
  4. OP
    OP
    tRNA

    tRNA Member
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    thanks EX_EE
     

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