Question 1191884
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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?
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<pre>
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 = {{{(0.3*216 - 0.2*216)/(0.35+0.55*3 - 0.2*4)}}} = 18.


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

Solved (using one single equation in one unknown).