A chemist has three acid solutions. The first solution contains 15% acid, the second contains 35% and the third contains 65%. He wants to use all three solutions to obtain a mixture of 228 liters containing 25% acid, using 2 times as much of the 65% solution as the 35% solution. How many liters of each solution should be used?

171 liters of 15% acid

19 liters of 35% acid

38 liters of 65% acid

Step-by-step explanation:

If x is the liters of 15% acid, y is the liters of 35% acid, and z is the liters of 65% acid, then:

x + y + z = 228

0.15x + 0.35y + 0.65z = 0.25 (228)

z = 2y

Solve the system of equations using elimination or substitution. Using substitution:

x + y + 2y = 228

x + 3y = 228

x = 228 - 3y

0.15 (228 - 3y) + 0.35y + 0.65 (2y) = 0.25 (228)

34.2 - 0.45y + 0.35y + 1.3y = 57

1.2y = 22.8

y = 19

x = 228 - 3y = 171

z = 2y = 38

The chemist needs 171 liters of 15% acid, 19 liters of 35% acid, and 38 liters of 65% acid.