When an article was sold for ₹ 696, percent profit earned was $P\%$. When the same article was sold for ₹ 841, percent profit earned was $(p + 25)\%$. What is the value of $P$?

### Concept & Logic The difference in the selling prices of an article is directly proportional to the difference in the profit percentages, because both profit percentages are calculated on the same constant Cost Price (CP). ### Step-by-Step Solution 1. **Identify the Given Data:** Selling Price 1 ($SP_1$) = ₹ $696$, Profit = $P\%$ Selling Price 2 ($SP_2$) = ₹ $841$, Profit = $(P + 25)\%$ 2. **Calculate Differences:** Difference in Selling Price = $SP_2 - SP_1 = 841 - 696 = 145$ Difference in Profit Percentage = $(P + 25)\% - P\% = 25\%$ 3. **Find the Cost Price (CP):** The $25\%$ difference in profit corresponds exactly to the ₹ $145$ difference in selling price. $$25\% \text{ of } CP = 145$$ $$\frac{25}{100} \times CP = 145$$ $$CP = 145 \times 4 = 580$$ 4. **Calculate Original Profit Percentage ($P$):** In the first case, $SP_1 = 696$. Profit Amount = $SP_1 - CP = 696 - 580 = 116$. $$P = \left(\frac{\text{Profit}}{CP}\right) \times 100$$ $$P = \left(\frac{116}{580}\right) \times 100 = \left(\frac{1}{5}\right) \times 100 = 20$$ ### Exam Strategy & Shortcut Equate the percentage change directly to the value change. $25\% = 145$. $100\% = 145 \times 4 = 580$ (This is the CP). Profit at $696$ is $116$. Fraction = $\frac{116}{580} = \frac{1}{5} = 20\%$. ### Common Pitfall A common pitfall is forming complex algebraic equations for both $SP_1$ and $SP_2$ and trying to substitute, which wastes valuable time instead of simply equating the differences. ### Final Answer Therefore, the correct answer is **20**.