By mixing 60% and 10% acid solutions and adding 20 kg of pure water, a 16% acid solution was obtained

By mixing 60% and 10% acid solutions and adding 20 kg of pure water, a 16% acid solution was obtained. If, instead of 20 kg of water, 20 kg of a 20% solution of the same acid were added, then a 24% acid solution would be obtained. How many kilograms of 60% solution were used to obtain the mixture?

Let’s make a short note:

x kg (60%) + y kg (10%) + 20 kg (0%) = x + y + 20 (16%).

x kg (60%) + y kg (10%) + 20 kg (20%) = x + y + 20 (24%).

We convert the percentages to a decimal fraction (divide by 100), we get a system of equations:

0.6x + 0.1y + 20 * 0 = (x + y + 20) * 0.16;

0.6x + 0.1y + 0.2 * 20 = (x + y + 20) * 0.24.

Let’s simplify the equations:

0.6x + 0.1y = 0.16x + 0.16y + 3.2; 0.6x – 0.16x + 0.1y – 0.16y = 3.2; 0.44x – 0.06y = 3.2.

0.6x + 0.1y + 4 = 0.24x + 0.24y + 4.8; 0.6x – 0.24x + 0.1y – 0.24y = 4.8 – 4; 0.36x – 0.14y = 0.8.

Multiply by 100:

44x – 6y = 320;

36x – 14y = 80.

Multiply the first equation by 7, and the second by (-3):

308x – 42y = 2240.

-108x + 42 = 240.

Let’s solve the system using the addition method:

200x = 2000.

x = 10 (kg) – the amount of a 60% solution.