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
Explanation:
Let total number of fruits at the beginning of the day = 5x ∴ Number of mangoes = 40% of 5x = 2x
Let the number of apples be '5a', hence the number of bananas = 5x - 5a.
⇒ He sells 50% of 2x mangoes, 96 bananas and 40% of 5a apples i.e., x + 96 + 2a fruits.
Also, he sold 50% of his fruits ⇒ x + 96 + 2a = 50% of 5x = 2.5x ⇒ 1.5x = 96 + 2a ⇒ x = 64 + 4a/3 ⇒ 5x = 320 + 20a/3
To minimise 5x, we need to minimise 2aa/3.
Minimum integral value of 20a/3 will be 20 when a = 3.
∴ 5x = 320 + 20 = 340
Hence, 340.
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