Question

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

Answer 1

Correct Answer: 340

Let the total number of fruits be $T$. 

Then:

• Number of mangoes = $40\%$ of $T = 0.4T$
• Number of bananas = $B$ (say)
• Number of apples = $T - 0.4T - B = 0.6T - B$

 

The fruit seller sells:

• Half of the mangoes = $\dfrac{1}{2} \times 0.4T = 0.2T$
• 96 bananas
• $40\%$ of the apples = $0.4 \times (0.6T - B) = 0.24T - 0.4B$

 

Total fruits sold = $0.2T + 96 + 0.24T - 0.4B = 0.44T - 0.4B + 96$

According to the question, this is $50\%$ of the total fruits, i.e. $0.5T$.

So, $$0.44T - 0.4B + 96 = 0.5T$$

Rearranging the equation: $$0.44T - 0.4B = 0.5T - 96$$ $$-0.4B = 0.06T - 96$$ $$0.4B = 96 - 0.06T$$ $$B = \dfrac{96 - 0.06T}{0.4} = 240 - 0.15T$$

But $B$ must be less than or equal to $0.6T$ (since apples = $0.6T - B \geq 1$). So, $$B \leq 0.6T - 1$$

Let's substitute $B = 0.6T - 1$ into the equation: $$0.4B = 0.4(0.6T - 1) = 0.24T - 0.4$$

Plug into original equation: $$0.44T - (0.24T - 0.4) = 96$$ $$0.2T + 0.4 = 96$$ $$0.2T = 95.6$$ $$T = \dfrac{95.6}{0.2} = 478$$

So the smallest integer value of $T$ that satisfies all conditions is: $\boxed{478}$