Question

A chemist has three different acid solutionsThe first acid solution contains 25% acid, the second contains 45\% and the third contains 65%. He wants to use three solutions to obtain a mixture of 224 liters containing 35% acid, using 3 times as much of the 65% solution as the 45% solution. How many liters of each solution should be used?

The chemist should use __ liters of 25% solution, ___ liters of 45% solution, and ____ liters of 65% solution

Answer 1

• 160 L of 25%
• 16 L of 45%
• 48 L of 65%

Explanation:

You want the number of liters of 25%, 45%, and 65% acid solution required to make a mixture that is 224 L of a 35% acid solution, using 3 times as much of the 65% solution as of the 45% solution.

Setup

There are numerous ways this mixing problem can be formulated. The calculator display in the attachment shows one of them: 3 equations in 3 unknowns.

Here, we choose to use one variable. Let x represent the amount of 45% solution. Then 3x is the amount of 65% solution, and (224 -4x) is the amount of 25% solution. The amount of acid in the final mix is ...

0.25(224 -4x) + 0.45x + 0.65(3x) = 0.35·224

Solution

The equation can be simplified to ...

1.4x +56 = 78.4

1.4x = 22.4

x = 16 . . . . . . . . . liters of 45% solution

3x = 48 . . . . . . . . liters of 65% solution

224 -4x = 224 -64 = 160 . . . . liters of 25% solution

The chemist should use ...

• 160 L of 25%
• 16 L of 45%
• 48 L of 65%

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