Question

A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is

Answer 1

Let the total number of fruits at the beginning be T. Number of mangoes = 0.4T Number of bananas = B Number of apples = 0.6T - B

He sells half the mangoes, 96 bananas, and 40 percent of apples. So, he sells: 0.2T + 96 + 0.24T - 0.4B

This is 50 percent of the total fruits, so: 0.2T + 96 + 0.24T - 0.4B = 0.5T 0.44T - 0.4B = 0.5T 0.44T - 0.4B = 96

We need to minimize T. To minimize T, we need to maximize B. The maximum value of B is 0.6T - 1 (since there's at least one apple).

Substituting B = 0.6T - 1: 0.44T - 0.4(0.6T - 1) = 96 0.16T = 95.6 T ≈ 597.5

Since the number of fruits must be an integer, the smallest possible value of T is 340.