A man buys an equal number of oranges at two rates: 4 a rupee and 5 a rupee. He mixes them and sells the mixture at 4 a rupee. What is his overall percentage result (gain or loss)?

Introduction / Context:This is a mixture pricing problem with equal counts. Compute total cost from the two purchase rates, then the selling revenue at a single mixed rate, and finally compare for gain or loss percentage on total cost.

Given Data / Assumptions:

• Let n be the number bought at each rate.
• Cost1 per orange = ₹0.25 (4 a rupee); Cost2 per orange = ₹0.20 (5 a rupee).
• Selling price per orange = ₹0.25 (4 a rupee).

Concept / Approach:Total cost = n(0.25) + n(0.20) = 0.45n. Total oranges = 2n. Total revenue at 0.25 each = 0.50n. Profit% = (Revenue − Cost)/Cost * 100.

Step-by-Step Solution:Total cost = 0.45nTotal revenue = 0.50nProfit = 0.05nProfit% = (0.05n) / (0.45n) * 100 = 11.111...% = 11 1/9 % gain

Verification / Alternative check:Choose n = 20. Cost = 9; revenue = 10; profit = 1; profit% = 1/9 * 100 = 11.11%.

Why Other Options Are Wrong:Loss options are invalid since revenue exceeds cost; 12 1/2% is higher than the exact fraction 1/9.

Common Pitfalls:Using an average of rates instead of computing exact totals; forgetting counts are equal (not costs).

Final Answer:11 1/9 % gain