By mixing 30% and 60% acid solutions and adding 10 kg of pure water, a 36% acid solution was obtained.

By mixing 30% and 60% acid solutions and adding 10 kg of pure water, a 36% 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 41% acid solution would be obtained. How many kilograms of the 30% solution were used to make the mixture?

To solve the problem, we will immediately introduce variables.

Let m be the mass of the first solution, n – the mass of the second solution.

Based on the conditions of the problem, we will compose and solve a system of two equations:

0.3 * m + 0.6 * n = 0.36 * (m + n + 10);

0.3 * m + 0.6 * n + 5 = 0.41 * (m + n + 10);

0.3 * m + 0.6 * n = 0.36 * m + 0.36 * n + 3.6;

0.3 * m + 0.6 * n + 5 = 0.41 * m + 0.41 * n + 4.1;

0.24 * n = 0.06 * m + 3.6;

0.11 * m = 0.19 * n + 0.9;

0.06 * m = 0.24 * n – 3.6;

m = 4 * n – 60;

0.11 * (4 * n – 60) = 0.19 * n + 0.9;

0.44 * n – 6.6 = 0.19 * n + 0.9;

0.25 * n = 7.5;

n = 30 kg.

m = 4 * 30 – 60 = 60 kg – the mass of the first solution.