SOLUTION: A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 40% and the third contains 80%. He wants to use all three solutions to
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Question 1013775: A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 40% and the third contains 80%. He wants to use all three solutions to obtain a mixture of 90 liters containing 60% acid, using 3 times as much of the 80% solution as the 40% solution. How many liters of each solution should be used?
You can put this solution on YOUR website! Call the amount used of the 40% solution, x.
Then the amount of 80% used would be 3x.
Thus the amount of 20% acid would be 90 - 3x - x = 90 - 4x.
The set up is like this
.40x + .80(3x) + .20(90-4x) = .60(90)
.4x + 2.4x + 18 - .8x = 54
2x = 36
x = 18 L at 40%
3x = 54 L at 80%
90-4x = 18 L at 20%
