A chemist has three different acid solutions. The first acid solution contains 15 % 15% acid, the second contains 35 % 35% and the third contains 80 % 80% . They want to use all three solutions to obtain a mixture of 190 190 liters containing 30 % 30% acid, using 2 2 times as much of the 80 % 80% solution as the 35 % 35% solution. How many liters of each solution should be used

First solution = 133L

Second solution = 19L

Third solution = 38L

Explanation:

As we want to make 190L of a 30%=0.3 solution we can write:

190L*0.3 = a*0.15 + b*0.35 + c*0.80

Where a, b and c are the volume of first, second and third acid solutions, respectively.

The volume of c is twice volume of b:

c = 2b

And the volume of the 3 solutions is equal to 190L:

190L = a+b+c

190L = a+3b

190L*0.3 = (190-3b)*0.15 + b*0.35 + 2b*0.80

57 = 28.5-0.45b+0.35b+1.6b

28.5 = 1.5b

b = 19L

c = 2*19L

c = 38L

a = 190L - 19L - 38L

a = 133L

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