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) About Me  (Show Source):
You can put this solution on YOUR website!
TYPE__________PERCENT-CONC_______VARIABLE
Low____________25_________________x
Medium_________40_________________y
High___________80_________________z
-
DESCRIPTION: z%2Fy=3
WANT: 66 liters of 55%

Accounting for volumes
x%2By%2Bz=66
x%2By%2B3y=66
x%2B4y=66

Account for concentrations keeping all as percents
%2825x%2B40y%2B80z%29%2F%2866%29=55

z can be eliminated from this, just as done in volume account.
%2825x%2B40y%2B80%2A3y%29%2F66=55
Divide members by 5.
%285x%2B9y%2B16%2A3y%29%2F66=11
highlight_green%28%285x%2B57y%29%2F66=11%29
5x%2B57y=726

A system of two equations in just x and y is possible.
system%28highlight%28x%2B4y=66%29%2Chighlight%285x%2B57y=726%29%29

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

Equivalent:
system%285x%2B20y=330%2C5x%2B57y=726%29
subtract first equation from second equation to find y.
highlight%28y=10.702%2Aliters%29-------the 40% material

Before even finding x, finding z is possible right now.
z=3%2Ay
z=3%2A10.702
highlight%28z=32.108%2Aliters%29------the 80% material

The original volume sum equation most easily will give x.



Answer by MathTherapy(10552) About Me  (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: highlight_green%2822%29 L

40% acid to mix: highlight_green%2811%29 L

80% acid to mix: highlight_green%2833%29 L