# %soln math Q

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

1. ### 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

3. ### 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

4. ### Nasem

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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

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