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A chemist has three different acid solutions. The first acid 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 72 liters containing 45 % acid, using 2 times as much of the 65 % solution as the 35 % solution. How many liters of each solution should be used?

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MathPhys

Answer:

18 liters 15% acid

18 liters 35% acid

36 liters 65% acid

Step-by-step explanation:

Let's say x is the volume of 15% acid, y is the volume of 35% acid, and z is the volume of 65% acid.

x + y + z = 72

z = 2y

The amounts of acid add up to the final amount.

0.15x + 0.35y + 0.65z = 0.45 (72)

Substitute the expression for z into both equations.

x + y + 2y = 72

0.15x + 0.35y + 0.65 (2y) = 0.45 (72)

Simplify:

x + 3y = 72

0.15x + 1.65y = 32.4

Solve for x in the first equation, then substitute into the second.

x = 72 − 3y

0.15 (72 − 3y) + 1.65y = 32.4

Solve for y.

10.8 − 0.45y + 1.65y = 32.4

1.2y = 21.6

y = 18

Now find x and z.

x = 72 − 3y = 18

z = 2y = 36

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