SOLUTION: A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 35% and the third contains 55%. They want to use all three solutions to

Algebra ->  College  -> Linear Algebra -> SOLUTION: A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 35% and the third contains 55%. They want to use all three solutions to      Log On


   

Question 1191884: A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 35% and the third contains 55%. They want to use all three solutions to obtain a mixture of 216 liters containing 30% acid, using 3 times as much of the 55% solution as the 35% solution. How many liters of each solution should be used?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
CONC. %           QTY VOLUME L        PURE
  20               216-4v             0.2(216-4v)
  35                 v                0.35v
  55                3v                0.55*3v

  30               216           0.2(216-4v)+0.35v+0.55*3v

0.2%28216-4v%29%2B0.35v%2B0.55%2A3v=0.30%2A216
Solve this for v.
.
.
.

Answer by ikleyn(52775) About Me  (Show Source):
You can put this solution on YOUR website!
.
A chemist has three different acid solutions. The first acid solution contains 20% acid,
the second contains 35% and the third contains 55%.
They want to use all three solutions to obtain a mixture of 216 liters containing 30% acid,
using 3 times as much of the 55% solution as the 35% solution. How many liters of each solution should be used?
~~~~~~~~~~~~~~~~~ 

35% solution:  x liters;

55% solution:  3x liters;

20% acid:      216 - x - 3x = 216 - 4x liters.


The pure acid in ingredients is the same as the pure acid in the mixture

    0.35x + 0.55*(3x) + 0.2*(216-4x) = 0.3*216.


From this equation

    x = %280.3%2A216+-+0.2%2A216%29%2F%280.35%2B0.55%2A3+-+0.2%2A4%29 = 18.


ANSWER.  18 liters of the 35% solution; 3*18 = 54 liters of the 55% solution and (216-4*18) = 144 liters of the 2-% solution.

Solved (using one single equation in one unknown).