Question 719353: 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 68 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?
Answer by josgarithmetic(39618)   (Show Source):
You can put this solution on YOUR website! Let x amount of 20% acid
let y amount of 30% acid
let z amount of 75% acid
"using 2 times as much of the 75% solution as the 30% solution. ",
That means z/y=2/1 or  .
Two essential starting equations:
 and  .
The less advanced way to solve this is to begin with the ratio between z and y and substitute into the "two essential" equations:
The percentage equation:
The solution summation equation:
Our simpler system to solve is now in variables x and y:
Subtract the second from the first, and from resulting 6y=102, obtain  .
Use that result again in the second equation, x=68-3y, x=68-3*17,  .
Now, go back to the given relationship between z and y, z=2y, z=2*17,  .
ANSWER SUMMARY: x=17 liters of 20%, y=17 liters of 30%, z=34 liters of 75% acid.
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