Question

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?

The chemist should use

liters of 65 % solution.

liters of 15 % solution,

liters of 35 % solution

Answer 1

Let x, y, and z be the amounts (in liters) of the 15%, 35%, and 65% solutions, respectively.

The chemist wants to end up with 72 L of solution, so

x + y + z = 72

Now,

• x L of 15% solution contributes 0.15x L of acid

• y L of 35% solution contains 0.35y L of acid

• z L of 65% solution contains 0.65z L of acid

so that the total amount of acid in the resulting mixture is 45% of 72 L, or 32.4 L, which means

0.15x + 0.35y + 0.65z = 32.4

It's also stipulated that the chemist uses twice as much of the 65% solution as the 35% solution, which translates to

z = 2y

Substitute this into the other two equations:

x + 3y = 72

0.15x + 1.65y = 32.4

Solve for x in terms of y :

x = 72 - 3y

Solve for y :

0.15 (72 - 3y) + 1.65y = 32.4

10.8 - 0.45y + 1.65y = 32.4

1.2y = 21.6

y = 18

Solve for x and z :

x = 72 - 3 (18) = 18

z = 2 (18) = 36

So, the chemist should use

• 36 L of 65% solution (z)

• 18 L of 15% solution (x)

• 18 L of 35% solution (y)