One solution contains 20% (by volume) salt, and the second contains 70% salt.

One solution contains 20% (by volume) salt, and the second contains 70% salt. How many liters of the first and second solutions do you need to take to make 100 liters of 50% brine?

Let’s take the required amount of a 20% solution for xl, and for yl – a 70% solution. Then x + y = 100 l. The salt content in a 20% solution will be 0.2x, in 70% – 0.7y, and in the resulting 50% – 0.5 * 100 l or 0.2x + 0.7y. Let’s compose and solve the system of equations:
x + y = 100,
0.2x + 0.7y = 0.5 * 100;

x = 100-y,
2x + 7y = 500;

x = 100-y,
2 (100-y) + 7y = 500;

x = 100-y
200-2y + 7y = 500

x = 100-y,
5y = 300;

x = 100-y,
y = 60;

x = 40,
y = 60.
Answer: to obtain 100 liters of a 50% solution, you need to take 40 liters of a 20% solution and 60 liters of a 70% solution.