How many gallons of each of a 60% acid solution and an 80% acid solution must be mixed to produce 50 gallons of a 74% acid solution?

Short way:
the ratio required for the 60% and 80% to get a 74% solution is
(80-74):(74-60) = 6:14 = 3:7
To make 50L, we need
50L*(3/(3+7))=15L of 60%
50L*(7/(3+7))=35L of 80%

Longer way
Let
x=volume of 60% solution (L), then
50-x=volume of 80% solution.
We need 50L of the mixture
so using CV=C1V1+C2V2,
50(0.74)=0.6x+0.8(50-x)
Solve for x
x=(0.8*50-50*0.74)/(0.8-0.6)
=3/0.2
=15L (of 60% solution)
50-x=50-15=35L (of 80% solution)

Still longer method:
Let
x=volume of 60% solution
y=volume of 80% solution
using total volume of mixture
x+y=50(1)
then using CV=C1V1+C2V2
0.6x+0.8y=50(0.74)
=> 0.6x+0.8y=37  (2)
Solve system of equations (1) & (2)
(1)-1.25(2)
x+y-0.75x-y = 50-37*1.25 = 
0.25x=3.75
x=3.75/0.25=15L
x+y=50 => y=50-x = 50-15=35L
So 15L of 60% and 35L of 80% as before.

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