SOLUTION: A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 35% ACID, and the third contains 65%. He wants to use all three solut

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Question 1130639: A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 35% ACID, and the third contains
65%. He wants to use all three solutions to obtain a mixture of 102 liters containing 40% acid, using 2 times as much of the 65% solution as the 35% solution. How many liters of each solution should be used?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
x of 25%
y of 35%
z of 65%
-
z%2Fy=2
z=2y


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He wants to use all three solutions to obtain a mixture of 102 liters containing 40% acid,
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x%2By%2B2y=102
x%2B3y=102

Accounting for the desired percentages and amount of pure acid
25x%2B35y%2B65%282y%29=40%2A102
Simplify this pure-acid equation.
25x%2B%2835%2B2%2A65%29y=40%2A102
25x%2B165y=40%2A102
divide both sides by 5,
5x%2B33y=8%2A102
5x%2B33y=816

Simpler system in variables x and y:
system%28x%2B3y=102%2C5x%2B33y=816%29

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You should get, if continue with Elimination Method,
highlight_green%28y=17%29
From that you should find that highlight_green%28z=34%29.......
and highlight_green%28x=51%29.