Question 962703
let a = amt of 10%
let b = amt of 15%
let c = amt of 30%
:
A chemist wants to mix three different solutions to create 100 milliliters of a solution that is 13.5% alcohol.
Solution A is 10% alcohol, 
solution B is 15% alcohol and 
solution C is 30% alcohol. 
The amount of solution A that is used must be twice the amount of solution B that is used.
a = 2b
a + b + c = 100
replace a with 2b
2b + b + c = 100
3b + c = 100
then
c = (-3b+100)
the mixture equation
.10a + .15b + .30c = .135(100)
replace a with 2b, replace c with (-3b+100)
.10(2b) + .15b + .30(-3b+100) = 13.5
.20b + .15b - .90b + 30 = 13.5
-.55b = 13.5 - 30
-.55b = -16.5
b = -16.5/-.55
b = + 30 ml of 15% solution
then
2(30) = 60 ml of 10% solution
and
100-30-60 = 10 ml of 30% solution
:
:
Confirm this, see if:
.10(60) + .15(30) + .30(10) = .135(100)