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.
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