SOLUTION: Please help and show all steps and work.Thanks A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 30% and the third con

Question 1001675: Please help and show all steps and work.Thanks
A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 30% and the third contains 75%. He wants to use all three solutions to obtain a mixture of 64 liters containing 50% acid, using 2 times as much of the 75% solution as the 30% solution. How many liters of each solution should be used?

My answers I came up with is 34,65,29.2
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
Variables x, y, z for low, medium, high conc. materials.

Simplify the system to something convenient or comfortable.

Obvious substitution for z.

Using elimination subtracting the '640' equation from the '64' equation,
for the 30% material.

You can determine x for the 20% material and the z for the 75% material yourself.

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

Please help and show all steps and work.Thanks
A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 30% and the third contains 75%. He wants to use all three solutions to obtain a mixture of 64 liters containing 50% acid, using 2 times as much of the 75% solution as the 30% solution. How many liters of each solution should be used?

My answers I came up with is 34,65,29.2

Sorry, but that's incorrect!
Amount of 20% acid solution to mix: L
Amount of 30% acid solution to mix: L
Amount of 75% acid solution to mix: L
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