Question

A chemistry teacher needs to mix a 20% salt solution with an 80% salt solution to make 20 qt of a 40% salt solution. How many quarts of each solution should the teacher mix to get the desired result?

Answer 1

• 6 2/3 qt 80%
• 13 1/3 qt 20%

Explanation:

It is often convenient to solve a mixture problem by letting a variable represent the quantity of the higher-concentration contributor to the mix.

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We can let x represent the number of quarts of 80% solution needed. Then (20-x) is the number of quarts of 20% solution needed. The amount of salt in the final mix is ...

0.80x +0.20(20-x) = 0.40(20)

0.60x = 0.20(20) . . . . . . . . subtract 0.20(20) and simplify

x = 20/3 = 6 2/3 . . . . . . . . . divide by 0.60; quarts of 80% solution

(20 -x) = 13 1/3 . . . . . . . . . . amount of 20% solution needed

The teacher should mix 6 2/3 quarts of 80% solution with 13 1/3 quarts of 20% solution.