An alloy of copper and tin (bronze), containing 12 kg less tin than copper, was alloyed with 4 kg of pure copper.

An alloy of copper and tin (bronze), containing 12 kg less tin than copper, was alloyed with 4 kg of pure copper. As a result, the copper content in the alloy increased by 2.5%. How much tin could be in the alloy?

1. Mass of tin in the original alloy: Mo kg;
2. The mass of copper in it: Mm = (Mo + 12) kg;
3. Copper was added to the alloy: M = 4 kg;
4. The copper content increased by: K = 2.5%;
5. Let’s compose the balance for copper:
Ko – Kc = 0.025;
(Mm + M) / (Mm + Mo + M) – Mm / (Mm + Mo) = 0.025;
(Mo + 12 + 4) / (Mo + 12 + Mo + 4) – (Mo + 12) / (Mo + 12 + Mo) = 0.025;
(Mo + 16) / (2 * (Mo + 8)) – (Mo + 12) / (2 * (Mo + 6)) = 0.025;
(Mo + 16) * (Mo + 6) – (Mo + 12) * (Mo + 8) = 0.05 * (Mo + 6) * (Mo + 8);
Mo² + 22 * Mo + 96 – Mo² – 20 * Mo – 96 = 0.05 * Mo² + 0.7 * Mo + 2.4;
Mo² – 26 * Mo + 48 = 0;
Mo1,2 = 13 + -√ (13² – 48) = 13 + – 11;
Mo1 = 13 – 11 = 2 kg (does not satisfy the condition of the problem K> 2.5%);
Mo = 13 + 11 = 24 kg.
Answer: there is 24 kg of tin in the alloy.