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 makes 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:

Question

Accepted Answer

1. **Define Variables**:
   - Let the number of mangoes at the beginning be $$ 4x $$, which is 40% of the total stock.
   - The total number of fruits is $$ 10x $$ (as $$ 4x $$ constitutes 40%).
   - Let the number of bananas be $$ y $$, and the number of apples be $$ 6x - y $$ (as bananas and apples together make up $$ 60\% $$).

2. **Selling Conditions**:
   - The seller sells:
     - Half of the mangoes: $$ 0.5 	imes 4x = 2x $$,
     - $$ 96 $$ bananas,
     - $$ 40\% $$ of apples: $$ 0.4 	imes (6x - y) = 2.4x - 0.4y $$.
   - Total fruits sold = $$ 50\% $$ of total fruits = $$ 0.5 	imes 10x = 5x $$.

3. **Formulate Equation**:
   From the total fruits sold:
   $$
   2x + 96 + (2.4x - 0.4y) = 5x
   $$
   Simplify:
   $$
   4.4x - 0.4y + 96 = 5x
   $$
   $$
   0.6x = 0.4y + 96
   $$
   $$
   x = \frac{2}{3}y + 160
   $$

4. **Constraints**:
   - Since $$ y \geq 96 $$ (bananas cannot be less than those sold),
   - Apples ($$ 6x - y $$) must be non-zero:
     $$
     (6x - y) > 0
     $$
     Substituting $$ x = \frac{2}{3}y + 160 $$:
     $$
     (6(\frac{2}{3}y + 160) - y) > 0
     $$
     Simplify:
     $$
     (4y + 960 - y) > 0
     $$
     $$
     y < 192
     $$

5. **Minimize Total Fruits**:
   To minimize $$ 10x $$, minimize $$ x = \frac{2}{3}y + 160 $$. This happens when $$ y $$ is maximized ($$ y = 189 $$, as it must be a multiple of $$ 3 $$).

6. **Calculate Minimum Total Fruits**:
   Substituting $$ y = 189 $$:
   $$
   x = \frac{2}{3}(189) + 160 = 126 + 160 = 34
   $$
   Total fruits = $$ 10x = 10(34) = 340$$.

Thus, the smallest possible total number of fruits is **340**.