An alloy of gold and silver weighs 50 g. It contains 80% gold. How much gold should be added to the alloy so that percentage of gold is increased to 90?

### Concept & Logic This is an **Alloy & Mixture** problem based on a single constant component. Since only gold is added to the mixture, the absolute mass of silver remains constant before and after the addition. $$ \text{Initial Mass of Silver} = \text{Final Mass of Silver} $$ ### Step-by-Step Solution * **Given:** * Initial alloy weight = $50$ g * Initial gold percentage = $80\%$ * Final gold percentage required = $90\%$ * **Calculation:** * Since the initial gold is $80\%$, the initial silver is $100\% - 80\% = 20\%$. * Absolute mass of silver = $20\%$ of $50 = 10$ g. * In the new alloy, the gold percentage must be $90\%$, making the new silver percentage $100\% - 90\% = 10\%$. * Let the total weight of the new alloy be $x$. * Since the mass of silver has not changed: $10\%$ of $x = 10$ g. * $\frac{10}{100} \times x = 10 \implies x = 100$ g. * The new total weight is $100$ g. The initial total weight was $50$ g. * Amount of gold added = $100 - 50 = 50$ g. ### Exam Strategy & Shortcut **Constant Component Shortcut:** The Silver percentage changes from $20\%$ to $10\%$. Since the silver amount is strictly constant, the total mixture must be inversely proportional to the silver percentage. Total Weight Ratio (Old : New) = $10 : 20 = 1 : 2$. Old weight ($1$ unit) = $50$ g. New weight ($2$ units) = $100$ g. Added amount = $100 - 50 = 50$ g. ### Common Pitfall Students often try to set up complex algebraic equations directly with the added gold (e.g., $0.80(50) + x = 0.90(50 + x)$). While mathematically correct, it is slower and prone to algebra errors. Using the constant component (silver) is much safer and faster. ### Final Answer **Therefore, the correct answer is 50 g.**