CAT 2024 Slot 2 QA Question & Solution
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
Solution
(Note: This question repeated in both CAT 2023 and CAT 2024)
Let the initial total stock of all fruits be $S$.
Let the number of apples initially be $a$. Let the number of bananas initially be $b$.
Stock of mangoes = $40%$ of $S = \dfrac{2S}{5}$.
Total fruits sold
= Mangoes sold + Apples sold + Bananas sold
= $\dfrac{2S}{10} + 96 + \dfrac{4a}{10}$.
Given: total fruits sold $= \dfrac{S}{2}$.
So, $$ \dfrac{S}{5} + 96 + \dfrac{2a}{5} = \dfrac{S}{2} $$
Multiply both sides by 10: $$ 2S + 960 + 4a = 5S $$
Rearranging: $$ 960 + 4a = 3S $$
Thus, $$ S = \dfrac{4a + 960}{3} $$
We can write this as: $$ S = \dfrac{4a}{3} + 320 $$
For $S$ to be an integer:
- $\dfrac{4a}{3}$ must be an integer → $a$ must be a multiple of $3$
- Apples sold = $\dfrac{4a}{10}$ must be an integer → $a$ must be a multiple of $5$
Therefore, $a$ must be a multiple of both $3$ and $5$, i.e., a multiple of $15$.
The smallest such value is $a = 15$.
Substitute into $S$: $$ S = \dfrac{4(15)}{3} + 320 = 20 + 320 = 340 $$
Hence, $\boxed{340}$ is the correct answer.