Question

A chemist needs 12 L of a 40% alcohol solution. She must mix a 20% solutions and a 50% solution. How many liters of each will be required to obtain what she needs?

Answer 1

• 8 L of 50%
• 4 L of 20%

Explanation:

Let x represent the quantity in liters needed of 50% solution. Then 12-x is the quantity of 20% solution, and the total amount of alcohol in the mix is ...

0.50x + 0.20(12 -x) = 0.40(12)

(0.50 -0.20)x = 12(0.40 -0.20) . . . . . subtract 12×0.20

x = 12(0.40 -0.20)/(0.50 -0.20) = 8

12-x = 4

The chemist needs 8 L of 50% solution and 4 L of 20% solution.

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If you pay attention to the numbers in the final expression and recognize where they come from, you find the generic answer to a mixture problem like this:

amount of highest contributor = (total amount)(difference between mix and least contributor) / (difference between contributors)

50% solution needed = (12 L)(40-20)/(50-20) = (12 L)(2/3) = 8 L