SOLUTION: A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 35% and the third contains 75%. He wants to use all three solutions to

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Question 1174998: A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 35% and the third contains 75%. He wants to use all three solutions to obtain a mixture of 64 liters containing 50% acid, using 33 times as much of the 75% solution as the 35% solution. How many liters of each solution should be used?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi
 Question States:
 .25(64 -x -33x) + .35x + .75(33x) = .50(64L)
    .25(64) - .25x - .25(33x) + .35x + .75(33x) = .50(64L)
         .50(33x) + .10x = .25(64L)
               x(.50(33) + .10) = .25(64L)
               x = .25%2864L%29%2F%28.55%2833%29+%2B.10%29%29
 My Point in the Set-up is to leave it intact prior to using Your calculator.
 One can easily check for simplifying errors and less chance for arithmetical errors
Recheck the simplifying and use Your calculator.
Wish You the Best in your Studies.