By mixing 14% and 50% acid solutions and adding 10 kg of pure water, a 22% acid solution was obtained

By mixing 14% and 50% acid solutions and adding 10 kg of pure water, a 22% acid solution was obtained. If instead of 10 kg of water 10 kg of a 50% solution of the same acid were added, you would get a 32% acid solution. How many kilograms 14 -percentage solution used to obtain the mixture?

Suppose that x kg was taken from a 14% solution, and a 50% solution was taken from kg, then, according to the condition of the problem, we can compose a system of linear equations:

14 * x / 100 + 50 * y / 100 = 22 * (x + y + 10) / 100,

14 * x / 100 + 50 * y / 100 + 50 * 10/100 = 32 * (x + y + 10) / 100.

From the second equation we get:

14 * x + 50 * y + 500 = 32 * x + 32 * y + 320,

18 * x – 18 * y = 180,

x – y = 10,

y = x – 10.

Let’s solve the first equation of the system:

14 * x + 50 * y = 22 * x + 22 * y + 220,

28 * y – 8 * x = 220,

28 * (x – 10) – 8 * x = 220,

28 * x – 280 – 8 * x = 220,

20 * x = 500,

x = 25 (kg) – the mass of a 14% solution.