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.