Question 1162778
A chemist has three different acid solutions.
 The first acid solution contains 15% acid,
 the second contains 30%
 and the third contains 70%.
He wants to use all three solutions to obtain a mixture of 81 liters containing 55% acid, using 3 times as much of the 70% solution as the 30% solution.
 How many liters of each solution should be used?
:
let a = amt of 15% solution
let b = 30% solution
then, since they want 3 times as much of 70% solution:
let 3b = amt of 70% solution
and then since we know the total is 81, we know that:
a = (81-4b) the 15% solution
:
.15a + .30b + .70(3b) = .55(81)
.15a + .30b + 2.1b = 44.55
.15a + 2.5b = 44.55
replace a with (81-4b)
.15(81-4b) + 2.4b = 44.55
12.15 - .60b + 2.4b = 44.55
-.6b +  2.4b = 44.55 - 12.15
1.8b = 32.4
b = 32.4/1.8
b = 18 liters of 30% solution
then
3(18) = 54 liters of 70% solution
and
81 - 18 - 54 = 9 liters of 15% solution
:
:
:
Check see if this works:
.15(9) + .30(18) + .70(54) = .55(81)
1.35 + 5.4 + 37.8 = 44.55