Question 923670: A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 30% and the third contains 70%. He wants to use all three solutions to obtain a mixture of 144 liters containing 40% acid, using 3 times as much of the 70% solution as the 30% solution. How many liters of each solution should be used?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A chemist has three different acid solutions.
The first acid solution contains 20% acid, the second contains 30% and the third contains 70%.
He wants to use all three solutions to obtain a mixture of 144 liters containing 40% acid, using 3 times as much of the 70% solution as the 30% solution.
How many liters of each solution should be used?
:
Let a = amt of 20% solution
Let b = amt of 30% solution
then "using 3 times as much of the 70% solution as the 30%."; therefore:
3b = amt of 70% solution
:
A typical mixture equation, using the decimal equiv of %
.2a + .3b + .7(3b) = .4(144)
.2a + .3b + 2.1b = 57.6
.2a + 2.4b = 57.6
We know that a = (144-4b); replace a
.2(144-4b) + 2.4b = 57.6
28.8 - .8b + 2.4b = 57.6
1.6b = 57.6 - 28.8
1.6b = 28.8
b = 28.8/1.6
b = 18 liters of of the 30% solution
then
3(18) = 54 liters of the 70% solution
a = 144 - 4(18)
a = 72 liters of the 20% solution
;
:
See if this checks out
.2(72) + .3(18) + .7(54) = .4(144)
14.4 + 5.4 + 37.8 = 57.6
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