One alloy is 40% copper and another is 55% copper. Melting them together gives a third alloy that is 45% copper. If the first alloy weighs 100 kg, what is the mass of the third alloy in kilograms?

Step-by-step solution

1. 1

   Choose the quantity that is conserved. Percentages do not add, but mass of copper does: the copper in the third alloy is exactly the copper from the first plus the copper from the second. Set up the balance on copper, not on percentages.

2. 2

   Name the unknown carefully. Let $m$ be the mass of the second alloy in kg. Then the third alloy has mass $100 + m$, because nothing is lost in melting. Reading the question as "$m$ = mass of the third alloy" is the usual trap — the question asks for $100 + m$ at the end, not for $m$.

3. 3

   Write the copper balance.

   $0.40(100) + 0.55m = 0.45(100 + m)$

4. 4

   Solve for $m$.

   $40 + 0.55m = 45 + 0.45m \;\Longrightarrow\; 0.10m = 5 \;\Longrightarrow\; m = 50$

5. 5

   Answer the question that was asked. The third alloy is the sum of both inputs:

   $100 + 50 = 150 \text{ kg}$

6. 6

   Verify the copper content. Copper in: $0.40(100) + 0.55(50) = 40 + 27.5 = 67.5$ kg. Copper required: $0.45(150) = 67.5$ kg ✓. A useful cross-check is the lever rule: $45\%$ sits $5$ points above $40\%$ and $10$ points below $55\%$, so the masses must be in the inverse ratio $10 : 5 = 2 : 1$, giving $100 : 50$ ✓.