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 us choose the total number of fruits as $5x$ (instead of just $x$) because this makes percentage calculations much cleaner.

Since mangoes make up 40% of the stock, and 40% = $\tfrac{2}{5}$, if our total is $5x$, then:

Mangoes = $\tfrac{2}{5} \times 5x = 2x$ (a nice clean number!)

So at the beginning of the day:

Total fruits = $5x$

Mangoes = $2x$

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Since Total = Mangoes + Bananas + Apples:

$5x = 2x + \text{Bananas} + \text{Apples}$

Therefore: $\text{Bananas} + \text{Apples} = 3x$

Now, let's say Apples = $5a$ (We're choosing $5a$ because apples are sold at 40% = $\tfrac{2}{5}$, making calculations clean)

Then: Bananas = $3x - 5a$

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Mangoes sold = Half of $2x = x$

Bananas sold = 96 (given)

Apples sold = 40% of $5a = \tfrac{2}{5} \times 5a = 2a$

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Total fruits sold = Mangoes sold + Bananas sold + Apples sold

$x + 96 + 2a = 50\%$ of $5x$

$x + 96 + 2a = 0.5 \times 5x = 2.5x$

$96 + 2a = 2.5x - x = 1.5x$

$192 + 4a = 3x$

Therefore: $3x = 4a + 192$ ... (This is our key equation!)

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Mangoes ≥ 1: Since mangoes = $2x$, we need $x ≥ 1$

Apples ≥ 1: Since apples = $5a$, we need $a ≥ 1$

Bananas constraint: This is the tricky one!

We need bananas ≥ 1: So $3x - 5a ≥ 1$

More importantly: We need enough bananas to sell 96, so $3x - 5a ≥ 96$

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From $3x = 4a + 192$, we get $x = \tfrac{4a + 192}{3}$

For $x$ to be a whole number, $(4a + 192)$ must be divisible by 3.

Since $192 = 3 \times 64$, we need $4a$ to be divisible by 3.

Since 4 and 3 have no common factors (they're coprime), we need $a$ itself to be divisible by 3.

The smallest positive value where $a$ is divisible by 3 is $a = 3$.

When $a = 3$:

$x = \tfrac{4(3) + 192}{3} = \tfrac{12 + 192}{3} = \tfrac{204}{3} = 68$

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With $a = 3$ and $x = 68$:

Total fruits = $5x = 5(68) = 340$

Mangoes = $2x = 136$

Apples = $5a = 15$

Bananas = $3x - 5a = 204 - 15 = 189$

Checking our constraints:

All fruit types ≥ 1

Enough bananas to sell 96: $189 > 96$

Checking the main condition:

Fruits sold = $68 + 96 + 6 = 170$

50% of total = $0.5 \times 340 = 170$

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The smallest possible total number of fruits is 340.