In an alloy of a certain mass, copper, aluminum and manganese are contained in a ratio of 17: 4: 1.

In an alloy of a certain mass, copper, aluminum and manganese are contained in a ratio of 17: 4: 1. In the second alloy of the same mass, in the ratio of 181: 16: 3, one of the metals contains 87.3 kilograms more. What is the mass of the alloy?

Let us denote by x the content of manganese in the first alloy, and by y in the second alloy.

Then: in 1 alloy of manganese there will be x kg, aluminum 4x kg., And copper 17x kg. Only 22 kg.

In the 2nd alloy of manganese there will be 3 kg, aluminum 16 kg, copper 181 kg. Only 200 kg.

Their masses are the same, therefore 22x = 200y; 11x = 100y.

It is known that in alloy 2 one of the metals is 87.3 kg more. Let’s determine what kind of metal it is.

1) Let this metal be manganese, then from the equation we express 3y = x + 87.3; x = 3y – 87.3; and substitute 11 * (3y – 87.3) = 100y, 33y – 960.3 = 100y into the equation.

We get that y <0, this does not suit us, because manganese is put in the alloy.

2) Now suppose this metal is aluminum.

Then 16y = 4x + 87.3; x = 4y – 21.825.

11 (4y – 21.825) = 100y, 44y – 240.075 = 100y, here also y <0, which means this metal is not suitable either.

3) It remains to check – copper.

181y = 17x + 87.3; x = (181y-87.3) / 17, 11/17 * (181y – 87.3) = 100y.

11 * 181y – 11 * 87.3 = 1700y, 1991y – 1700y = 960.3, y = 960.3 / 291 = 3.3 kg.

Then the mass of the alloy will be: 100 * y = 330 kg.