Acid mixture

• Let A be the original amount of 40% solution, B for 10% solution and C for the 30% solution.
• We can find out that: $\frac{40A}{100} + \frac{10B}{100} = \frac{30C}{100} => 4A + B = 3C$
• Also, we have $A + B = C$ ; so we conclude: $4A + B = 3C => 3A + (A + B) = 2C + (C) => 3A = 2C => A = \frac{2C}{3}$
• Moving forward, we have the second part of the question, in which we can afford: $\frac{30C}{100} + \frac{20 \times 300}{100} = \frac{25(C + 300)}{100}$ .
• Solving that, we have: $30C + 6000 = 25C + 7500 => 5C = 1500 => C = 300m³$
• Using C = 300m³ , we have $A = \frac{2C}{3} => A = \frac{2 \times 300}{3} => A = 200m³$ .
• Finally, we can conclude that the original amount of 40% acid solution was 200m³ .

let x=amount of 40% acid solution, y = amount of 10% acid solution... .4x + .1y = .3(x+y) .3(x+y) +.2(300) = .25(x + y + 300) then solve for x = 200.