# To the solution containing 40% salt was added some of the solution containing 80%

**To the solution containing 40% salt was added some of the solution containing 80% salt, and 2 kg of the solution containing 52% salt was obtained. How much of each of the stock solutions were mixed?**

Let x kg be the mass of the first solution. Since it contains 40% salt (40% = 0.4), the weight of the salt in it will be 0.4x.

Since you get 2 kg of a new solution, the mass of the second solution will be (2 – x) kg. Since it contains 80% salt (80% = 0.8), the weight of one salt in it will be 0.8 * (2 – x).

The mass of the resulting solution is 2 kg, and the salt concentration is 52% (52% = 0.52), then the pure salt in it is 0.52 * 2 = 1.04.

We compose and solve the equation:

0.4x + 0.8 (2 – x) = 1.04.

0.4x + 1.6 – 0.8x = 1.04.

-0.4x = 1.04 – 1.6.

-0.4x = -0.56.

x = -0.56: (-0.4).

x = 1.4 (kg) is the mass of the first solution.

2 – x = 2 – 1.4 = 0.6 (kg) is the mass of the second solution.