By mixing 24% and 67% acid solutions and adding 10 kg of pure water, a 41% acid solution was obtained.

By mixing 24% and 67% acid solutions and adding 10 kg of pure water, a 41% acid solution was obtained. If, instead of 10 kg of water, 10 kg of a 50% solution of the same acid were added, then a 45% acid solution would be obtained. How many kilograms of the 24% solution were used to make the mixture?

1. The mass of the first solution with a concentration of K1 = 24% is equal to: M1 kg;

2. The mass of the second solution with a concentration of K2 = 67% is equal to: M2 kg;

3. Concentration of the third solution: K3 = 41%;

4. The acid balance of this solution:

K1 * M1 + K2 * M2 = K3 * (M1 + M2 + 10);

0.24 * M1 + 0.67 * M2 = 0.41 * (M1 + M2 + 10);

5. In the fourth solution (the concentration of K4 = 0.45% was obtained), the solution was added
concentration K5 = 50% M5 = 10 kg;

6. Balance of the fourth solution:

K1 * m1 + K2 * M2 + K5 * M5 = K4 * (M1 + M2 + M5);

0.24 * M1 + 0.67 * M2 + 0.5 * 10 = 0.45 * (M1 + M2 + 10);

7. Subtract the first from the second equation:

5 = 0.04 * (M1 + M2 + 10) = 0.04 * (M1 + M2) + 0.4;

M1 + M2 = (5 – 0.4) / 0.04 = 115 kg;

M2 = (115 – M1) kg;

0.24 * M1 + 0.67 * (115 – M1) = 0.41 * (115 + 10);

77.05 – 51.25 = 0.43 * M1;

M1 = (77.05 – 51.25) / 0.43 = 60 kg.

Answer: to obtain a mixture, they took 60 kg of a 24% solution.