Mixed 40% brine and 60% brine. As a result, 200 liters of a solution containing 55% salt were obtained.

Mixed 40% brine and 60% brine. As a result, 200 liters of a solution containing 55% salt were obtained. How many liters of the first solution were added to the mixture?

1. The volume of the first brine is: V1 l;

2. The concentration of salt in this solution: K1 = 40%;

3. The volume of the second solution is equal to: V2 l;

4. In this solution the salt concentration: K2 = 60%;

5. The volume of the resulting solution: Vo = 200 l;

6. Its concentration is equal to: Ko = 55%;

7. We make up the salt balance of all solutions:

K1 * V1 + K2 * V2 = K3 * Vo;

V2 = Vo – V1;

0.4 * V1 + 0.6 * (200 – V1) = 0.55 * 200;

V1 * (0.6 – 0.4) = 200 * (0.6 – 0.55);

V1 = (200 * 0.05) / 0.2 = 10 / 0.2 = 50 l;

V2 = Vo – V1 = 200 – 50 = 150 HP

Answer: 50 liters of the first solution was added to the mixture.