SOLUTION: This is actually a real-life situation. I'm trying to mix 90% alcohol with 50% alcohol so that I get a solution of at least 70%. (the 50% is a wintergreen that helps the whole t

Question 487281: This is actually a real-life situation. I'm trying to mix 90% alcohol with 50% alcohol so that I get a solution of at least 70%.
(the 50% is a wintergreen that helps the whole thing smell less clinical.)
I am a band director who did very poorly in math many years ago, and am trying to mix a solution of alcohol in order to disinfect mouthpieces that multiple kids will play on as they try instruments.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
mix 90% alcohol with 50% alcohol so that I get a solution of at least 70%.
:
You must decide how much of the resulting solution you want
As an example, let's say you want 10 quarts of the 70% solution
:
Let x = amt of 90% alcohol required
then
(10-x) = amt of 50% alcohol required
:
A mixture equation
.90x + .50(10-x) = .70(10)
.9x + 5 - .5x = 7
.9x - .5x = 7 - 5
.4x = 2
x =
x = 5 quarts
so this works very easily; 5 qt of each will give a 70%
:
From this you can say, mix equal amts of the two
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