By mixing 40% and 60% acid solutions and adding 20 kg of water, 45% acid solution was obtained.

By mixing 40% and 60% acid solutions and adding 20 kg of water, 45% acid solution was obtained. If, instead of 20 kg of water, 20 kg of a 90% solution of the same acid were added, then a 65% acid solution would be obtained, how many kg of a 40% solution were used.

Let’s introduce variables:

m is the mass of a 40% solution;

n is the mass of a 60% solution.

Mixed m kg of a 40% solution, n kg of a 60% solution and 20 kg of water. The mass of the resulting solution is m + n + 20 kg. The mass of acid in this solution is 0.4m + 0.6n kg.

This solution contains 45% acid. Let’s make an equation.

(0.4m + 0.6n) / (m + n + 20) = 45/100;

(0.4m + 0.6n) / (m + n + 20) = 9/20;

(0.4m + 0.6n) * 20 = 9 * (m + n + 20);

8m + 12n = 9m + 9n + 180;

12n – 9n = (9m – 8m) + 180;

3n = m + 180.

Let’s say you would mix m kg of a 40% solution, n kg of a 60% solution and 20 kg of a 90% solution. The mass of the resulting solution would be m + n + 20 kg. Let’s find out what the mass of acid in this solution would be.

0.4m + 0.6n + 0.9 * 20 = 0.4m + 0.6n + 18 (kg)

This solution contains 65% acid. Let’s make an equation.

(0.4m + 0.6n + 18) / (m + n + 20) = 65/100;

(0.4m + 0.6n + 18) / (m + n + 20) = 13/20;

(0.4m + 0.6n + 18) * 20 = 13 * (m + n + 20);

8m + 12n + 360 = 13m + 13n + 260;

360 – 260 = (13m – 8m) + (13n – 12n);

100 = 5m + n;

n = 100 – 5m;

3n = 300 – 15m.

When we transformed the first equation, we found that 3n = m + 180. Substitute the expression m + 180 for 3n in the second equation.

m + 180 = 300 – 15m;

m + 15m = 300 – 180;

16m = 120;

m = 120/16;

m = 7.5.

So, the mass of a 40% solution is 7.5 kg.

Answer: 7.5 kg.