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
