SOLUTION: A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 40% and the third contains 80%. He wants to use all three solutions to

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Question 982911: A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 40% and the third contains 80%. He wants to use all three solutions to obtain a mixture of 66 liters containing 55% acid, using 3 times as much of the 80% solution as the 40% solution. How many liters of each solution should be used?

Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
TYPE__________PERCENT-CONC_______VARIABLE
Low____________25_________________x
Medium_________40_________________y
High___________80_________________z
-
DESCRIPTION:
WANT: 66 liters of 55%

Accounting for volumes




Account for concentrations keeping all as percents


z can be eliminated from this, just as done in volume account.

Divide members by 5.




A system of two equations in just x and y is possible.


Solve for x and y any way you want or know.

Equivalent:

subtract first equation from second equation to find y.
-------the 40% material

Before even finding x, finding z is possible right now.


------the 80% material

The original volume sum equation most easily will give x.



Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 40% and the third contains 80%. He wants to use all three solutions to obtain a mixture of 66 liters containing 55% acid, using 3 times as much of the 80% solution as the 40% solution. How many liters of each solution should be used?
25% acid to mix:  L
40% acid to mix: L
80% acid to mix: L
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