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 me break this down using simple variables and logic that's easy to follow.

Let:

• $M$ = mangoes at start
• $B$ = bananas at start
• $A$ = apples at start
• Total fruits at start = $M + B + A$

Key facts:

• Mangoes = $40\%$ of total stock
• He sells: half the mangoes + $96$ bananas + $40\%$ of apples
• Total sold = $50\%$ of all fruits

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Since mangoes are $40\%$ of total stock:

$M = 0.4 \times (M + B + A)$

Why do we multiply by $0.4$? Because $40\% = \frac{40}{100} = 0.4$

Solving this:

$M = 0.4M + 0.4B + 0.4A$

$M - 0.4M = 0.4B + 0.4A$

$0.6M = 0.4(B + A)$

Let's get rid of decimals by multiplying both sides by $10$:

$6M = 4(B + A)$

$3M = 2(B + A)$

$3M = 2B + 2A$

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Total sold = $50\%$ of initial stock:

$\frac{M}{2} + 96 + 0.4A = 0.5(M + B + A)$

The left side is what he sold, the right side is $50\%$ of everything.

Expanding the right side:

$\frac{M}{2} + 96 + 0.4A = 0.5M + 0.5B + 0.5A$

Notice that $0.5M = \frac{M}{2}$, so they cancel out:

$96 + 0.4A = 0.5B + 0.5A$

$96 = 0.5B + 0.5A - 0.4A$

$96 = 0.5B + 0.1A$

Multiply by $10$ to remove decimals:

$960 = 5B + A$

$A = 960 - 5B$

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From the first equation:

$M = \frac{2B + 2A}{3}$

Substituting $A = 960 - 5B$:

$M = \frac{2B + 2(960 - 5B)}{3} = \frac{2B + 1920 - 10B}{3} = \frac{1920 - 8B}{3}$

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Total = $M + B + A$

$= \frac{1920 - 8B}{3} + B + (960 - 5B)$

$= \frac{1920 - 8B + 3B + 2880 - 15B}{3}$

$= \frac{4800 - 20B}{3}$

To minimize this total, we need $B$ to be as large as possible!

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We need:

1. $A \geq 1$ (at least one apple): $960 - 5B \geq 1 \rightarrow B \leq 191.8 \rightarrow B \leq 191$
2. $M \geq 1$ (at least one mango): Need to check when we find $B$
3. $B \geq 96$ (must sell $96$ bananas, so must have at least $96$)
4. $M$ must be a whole number: $(1920 - 8B)$ must be divisible by $3$

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For $M$ to be whole: $1920 - 8B \equiv 0 \pmod{3}$

Since $1920 \equiv 0 \pmod{3}$ and $8 \equiv 2 \pmod{3}$:

We need $2B \equiv 0 \pmod{3}$, which means $B \equiv 0 \pmod{3}$

So $B$ must be divisible by $3$.

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Largest $B \leq 191$ that's divisible by $3$: $B = 189$

Let's check:

• $A = 960 - 5(189) = 960 - 945 = 15$
• $M = \frac{1920 - 8(189)}{3} = \frac{1920 - 1512}{3} = \frac{408}{3} = 136$

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Total = $M + B + A = 136 + 189 + 15 = 340$

Answer: $340$