From the two alloys containing silver, a third alloy was obtained. The mass of the first alloy is 50 g, it contains

From the two alloys containing silver, a third alloy was obtained. The mass of the first alloy is 50 g, it contains 60% pure silver, and the second alloy contains 80% pure silver. The third alloy contains 64% pure silver. What is the mass of the second alloy?

We take the mass of the second alloy as x, then:
80 * x / 100 = 0.8 * x g – the amount of silver in the second alloy.
The amount of silver in the first:
60 * 50/100 = 30 g,
and in the resulting:
64 (50 + x) / 100 g.
Since the sum of the masses of silver in the first two alloys is equal to the mass in the resulting one, we have the equation:
0.8x + 30 = 64 (50 + x) / 100
80x + 3000 = 3200 + 64x
80x-64x = 3200-3000
16x = 200
x = 200/16 = 12.5 g.