A piece of copper and zinc containing 9kg of zinc was alloyed with 3kg of copper.

A piece of copper and zinc containing 9kg of zinc was alloyed with 3kg of copper. The new alloy contains 10% more copper than the original. How many kg of copper was there in the original alloy?

Let the mass of copper in the alloy be x, then the whole alloy is x + 9. The mass of the new alloy is x + 9 + 3 or x + 12.

The copper content in the old alloy is x / (x + 9), and in the new one (x + 3) / (x +12). Let’s make the equation:

((x + 3) / (x + 12)) – x / (x + 9) = 10/100

(x + 3) (x + 9) – x * (x + 12) = 0.1 * (x + 9) * (x + 12)

X ^ 2 + 12x + 27 – x ^ 2 – 12x = 0.1 (x ^ 2 + 21x + 108)

X ^ 2 + 21x + 108 = 270

X ^ 2 + 21 – 162 = 0

D = 21 ^ 2 + 4 * 16 ^ 2 = 441 + 648 = 332

X = (-21 + 33) / 2 = 6 or x = (-21 – 33) / 2 = – 27 – does not satisfy the conditions of the problem.

Answer: 6 kilograms of copper was in the original alloy.