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



Let the total number of fruits in the stock at the beginning of the day be denoted as \( x \). The fruit seller has mangoes, bananas, and apples, with the number of mangoes representing 40% of the total stock. Thus, the number of mangoes at the start is \( 0.4x \).

Let:
- \( m = 0.4x \) be the number of mangoes,
- \( b \) be the number of bananas,
- \( a \) be the number of apples.

Since \( x = m + b + a \), we know that:

\[
x = 0.4x + b + a
\]

This simplifies to:

\[
x - 0.4x = b + a
\]

\[
0.6x = b + a
\]

Now, the fruit seller sells half of the mangoes, 96 bananas, and 40% of the apples. So, the number of mangoes sold is:

\[
\frac{m}{2} = \frac{0.4x}{2} = 0.2x
\]

The number of apples sold is:

\[
0.4a
\]

At the end of the day, the total number of fruits sold is half of the original stock, i.e., \( \frac{x}{2} \). Therefore, the total fruits sold is the sum of the mangoes, bananas, and apples sold:

\[
0.2x + 96 + 0.4a = \frac{x}{2}
\]

Simplifying this equation:

\[
0.2x + 96 + 0.4a = \frac{x}{2}
\]

Multiply through by 10 to eliminate the decimals:

\[
2x + 960 + 4a = 5x
\]

Rearranging the equation:

\[
960 + 4a = 3x
\]

Thus, we have the relationship:

\[
x = \frac{960 + 4a}{3}
\]

Since \( x \) must be an integer, \( 960 + 4a \) must be divisible by 3. To find the smallest value of \( a \), we check the smallest values of \( a \) such that \( 960 + 4a \) is divisible by 3.

First, check \( a = 0 \):

\[
960 + 4(0) = 960 \quad (\text{Not divisible by 3})
\]

Next, check \( a = 3 \):

\[
960 + 4(3) = 960 + 12 = 972 \quad (\text{Divisible by 3})
\]

Thus, the smallest possible value of \( a \) is 3. Substituting this into the equation for \( x \):

\[
x = \frac{960 + 4(3)}{3} = \frac{972}{3} = 324
\]

Therefore, the smallest possible total number of fruits at the beginning of the day is \( \boxed{324} \).

Would you like further details or clarifications? Here are some related questions for practice:

1. If 60% of the mangoes were sold instead of 50%, what would the new total stock be?
2. How many apples were there at the beginning of the day?
3. What percentage of the total stock was bananas?
4. If 20 more bananas were sold, how many fruits would have been sold in total?
5. What is the ratio of bananas to apples in the original stock?

**Tip**: When working with percentages, it's helpful to express them as fractions to simplify calculations.

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

Discover how to find the smallest possible total number of fruits in a store's stock using algebra and percentage concepts. This problem involves solving linear equations with percentages, covering key algebraic principles for students in Grades 9-12.

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