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**Q24.** 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:

Let's solve this problem step by step.

### Step 1: Define Variables
Let \( T \) be the total number of fruits at the beginning of the day.
Let \( M \) be the number of mangoes, \( B \) be the number of bananas, and \( A \) be the number of apples at the beginning of the day.

Given:
- \( M = 0.4T \) (since mangoes make up 40% of the total stock)
- The seller sells \( \frac{M}{2} \) mangoes.
- The seller sells 96 bananas.
- The seller sells \( 0.4A \) apples.

### Step 2: Calculate the Fruits Sold
The total number of fruits sold:
\[
\text{Total fruits sold} = \frac{M}{2} + 96 + 0.4A
\]

We know that at the end of the day, the seller has sold 50% of his fruits, so:
\[
\frac{T}{2} = \frac{M}{2} + 96 + 0.4A
\]

Substitute \( M = 0.4T \) into the equation:
\[
\frac{T}{2} = \frac{0.4T}{2} + 96 + 0.4A
\]
\[
\frac{T}{2} = 0.2T + 96 + 0.4A
\]

### Step 3: Simplify the Equation
Subtract \( 0.2T \) from both sides:
\[
\frac{T}{2} - 0.2T = 96 + 0.4A
\]
\[
0.3T = 96 + 0.4A
\]

### Step 4: Solve for \( A \) and \( T \)
Solve for \( A \):
\[
0.4A = 0.3T - 96
\]
\[
A = \frac{0.3T - 96}{0.4}
\]
\[
A = 0.75T - 240
\]

Since \( A \) must be a positive integer, \( 0.75T - 240 \) must also be positive. Therefore, \( T \) must satisfy:
\[
0.75T > 240
\]
\[
T > 320
\]

The smallest integer value for \( T \) that satisfies this inequality is \( T = 321 \).

### Step 5: Check Consistency
For \( T = 321 \):
- \( M = 0.4 \times 321 = 128.4 \), but since the number of fruits must be an integer, \( T = 321 \) is not valid.

Let's try \( T = 400 \) (next logical step to check for consistency):
- \( M = 0.4 \times 400 = 160 \)
- \( A = 0.75 \times 400 - 240 = 60 \)

Now check:
\[
0.3 \times 400 = 96 + 0.4 \times 60
\]
\[
120 = 96 + 24
\]
This satisfies the equation, so the minimum possible total number of fruits at the beginning of the day is **400**.

### Final Answer
The smallest possible total number of fruits in the stock at the beginning of the day is **400**.

Would you like more details or have any questions?

Here are five related questions:
1. How would the problem change if the percentage of mangoes was different?
2. What if the seller sold a different percentage of apples?
3. How can the problem be modified to involve more types of fruits?
4. What if the total fruits sold were 60% instead of 50%?
5. Can you solve a similar problem where the fruit percentages are given for different fruits?

**Tip:** When solving percentage problems, always ensure that the final quantities are reasonable and consistent with the conditions of the problem.

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:

Learn how to calculate the minimum total number of fruits at the beginning of the day based on percentages and sales conditions. This problem involves key algebraic concepts and percentage calculations, suitable for students in Grades 9-10.

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