5 kg of tea and 8 kg of sugar together cost ₹ 172. The price of tea has risen by 20% and that of sugar by 10%. Hence the same quantities of tea and sugar now cost ₹ 199.20. What is the original price of tea per kg?

### Concept & Formula When tackling itemized cost problems with distinct percentage increases, you can either model the new total cost using multipliers ($1.2$, $1.1$) or model strictly the **difference** (the price increase) to form a much simpler linear equation. ### Step-by-Step Solution * Let the original price of tea per kg be $T$ and sugar per kg be $S$. * Write the initial cost equation: $$5T + 8S = 172 \quad \text{--- (Eq 1)}$$ * Write the new cost equation based on the percentage increases: $$5(1.2T) + 8(1.1S) = 199.20$$ $$6T + 8.8S = 199.20 \quad \text{--- (Eq 2)}$$ * To solve for $T$, eliminate $S$. Multiply (Eq 1) by $1.1$ to match the $S$ coefficients: $$5.5T + 8.8S = 189.20 \quad \text{--- (Eq 3)}$$ * Subtract (Eq 3) from (Eq 2): $$(6T - 5.5T) + (8.8S - 8.8S) = 199.20 - 189.20$$ $$0.5T = 10$$ * Solve for $T$: $$T = \frac{10}{0.5} = 20$$ ### Exam Strategy & Shortcut Instead of working with the full new prices, analyze just the **increase in cost**. Total cost increase = $199.20 - 172 = 27.20$. This increase comes directly from a 20% bump on tea total and a 10% bump on sugar total: $$0.2(5T) + 0.1(8S) = 27.20$$ $$T + 0.8S = 27.20$$ We know from the original scenario that $5T + 8S = 172$. Dividing that entire equation by 10 gives: $$0.5T + 0.8S = 17.2$$ Subtracting this from the increase equation kills the $S$ term instantly: $$0.5T = 10 \implies T = 20$$. This method minimizes large decimal arithmetic under pressure. ### Common Pitfall Trying to isolate variables using substitution (e.g., $S = \frac{172 - 5T}{8}$) right from the start. This creates messy fractions and increases the chance of arithmetic errors compared to the elimination or cost-increase method. ### Final Answer **Therefore, the correct answer is ₹ 20.**