# Another mixture one

• Apr 23rd 2008, 11:32 AM
j_dryden
Another mixture one
How man ounces of a 50% alcohol solution must be mixed with 80 ounces of a 20% alcohol solution to make a 40% alcohol solution?
• Apr 23rd 2008, 01:28 PM
elizsimca
Quote:

Originally Posted by j_dryden
How man ounces of a 50% alcohol solution must be mixed with 80 ounces of a 20% alcohol solution to make a 40% alcohol solution?

Don't have time to explain, about to run out door, but perhaps someone else can explain.

Be warned, I'm not sure if this is correct, this is just how I would set it up:

$(.40)(80+x)=(.50x)+80(.20)$
• Apr 23rd 2008, 01:46 PM
icemanfan
Quote:

Originally Posted by elizsimca
Don't have time to explain, about to run out door, but perhaps someone else can explain.

Be warned, I'm not sure if this is correct, this is just how I would set it up:

$(.40)(80+x)=(.50x)+80(.20)$

I agree with this method for solving the problem. This is equating the amount of alcohol present in each of the partial solutions with the amount of alcohol present in the mixed solution. The $(.40)(80+x)$ represents the total number of ounces present in the mix after combining the two solutions, multiplied by the final percentage of 40% to arrive at the total amount of alcohol present.