Three solutions contain a certain acid. The first contains 10% acid, the second 30%, and the third 50%. A chemist wishes to use all three solutions to obtain a 90-liter mixture containing 26% acid. If the chemist wants to use twice as much of the 50% solution as of the 30% solution, how many liters of each solution should be used?
A)43.8 of 10% , 15.4 of 30% , 30.8 of 50%
B)46.8 of 10% , 14.4 of 30% , 28.8 of 50%
C)49.8 of 10% , 13.4 of 30% , 26.8 of 50%
D)45.8 of 10% , 15.4 of 30% , 28.8 of 50%
E)46.8 of 10% , 28.8 of 30% , 14.4 of 50%

Question

Quizplus · Accepted Answer

Let x be the amount of 30% solution. Then 2x is the amount of 50% solution. The rest is 10% solution, which is 90 - 3x. Setting up the equation for acid content: 0.1(90 - 3x) + 0.3x + 0.5(2x) = 0.26(90). Solving gives x = 14.4, 2x = 28.8, and 90 - 3x = 46.8.

Three Solutions Contain a Certain Acid