math
posted by emmet .
Mary has a solution that is 60% alcohol and another that is 20% alcohol. How much of each should she use to make 100 milliliters of a solution that is 52% alcohol?

let the volume of 60% solution used be x
then the volume of the 20% has to be 100x
solve:
.6x +.2(100x) = .52(100) 
where did you get 100x???

If the both of us together have 100 cents, and I have x, then you have 100x
Mary wants to end up with 100 ml
if x ml are of the 60% type, then
100x ml must be the 40% kind.
notice (x) + (100x) = 100 
and that doesnt answer the question, the question asks how much of each solution she should use

its 20% not 40% percent

if the 60% one is x, then 100x = 10060 which equals 40, not 20

Emmet, look how I defined x.
Once you find x you have the amount of 60% solution used, plug x into 100x and you have the 20% solution.
Do you know how to solve the equation?
(the 40% is a typo, look at my equation, I used 20%)
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