A chemist has three different acid solutions. The first solution contains 25% acid, the second contains 35%acid, and the third contains 55% acid.
She created 120 liters of a 40% acid mixture, using all three solutions. The number of liters of 55% solution used is 3 times the number of liters of 35% solution used.

How many liters of each solution was used?

Let x, y, and z be the amounts (in liters, L) of the 25%, 35%, and 55% solutions that the chemist used.

She ended up with 120 L of solution, so

x + y + z = 120 … … … [1]

x L of 25% acid solution contains 0.25x L of acid. Similarly, y L of 35% solution contains 0.35y L of acid, and z L of 55% solution contains 0.55z L of acid. The concentration of the new solution is 40%, so that it contains 0.40 (120 L) = 48 L of acid, which means

0.25x + 0.35y + 0.55z = 48 … … … [2]

Lastly,

z = 3y … … … [3]

since the chemist used 3 times as much of the 55% solution as she did the 35% solution.

Substitute equation [3] into equations [1] and [2] to eliminate z :

x + y + 3y = 120

x + 4y = 120 … … … [4]

0.25x + 0.35y + 0.55 (3y) = 48

0.25x + 2y = 48 … … … [5]

Multiply through equation [5] by -2 and add that to [4] to eliminate y and solve for x :

(x + 4y) - 2 (0.25x + 2y) = 120 - 2 (48)

0.5x = 24

x = 48

Solve for y :

x + 4y = 120

4y = 72

y = 18

Solve for z :

z = 3y

z = 54

.there can be infinite values of m and n that can satisfy the above equation. for each value of m of m or n. there maybe infinite sets of solution for the above

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