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

Updated On: Nov 12, 2025
  • 100
  • 240
  • 340
  • None of Above
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The Correct Option is C

Solution and Explanation

Let the initial stock of all fruits be denoted by \( S \). Let the number of bananas be \( b \) and the number of apples be \( a \).

Stock of Mangoes \( = 40\% \) of \( S = \frac{2S}{5} \) 

Total number of fruits sold = Mangoes Sold + Apples Sold + Bananas Sold

\( = \frac{2S}{10} + 96 + \frac{4a}{10} = \frac{S}{2} \) (Given)

\( \Rightarrow \frac{S}{5} + 96 + \frac{2a}{5} = \frac{S}{2} \)

Multiply both sides by 10 to eliminate denominators:

\( 2S + 960 + 4a = 5S \)

\( \Rightarrow 3S = 4a + 960 \)

\( \Rightarrow S = \frac{4a + 960}{3} = \frac{4a}{3} + 320 \)

For \( S \) to be an integer, \( a \) must be a multiple of 3. Also, from the term \( \frac{4a}{10} \), \( a \) must be a multiple of 5.

Hence, the smallest value of \( a \) that satisfies both conditions (LCM of 3 and 5) is \( a = 15 \)

Substitute into the formula for \( S \):

\( S = \frac{4 \times 15}{3} + 320 = 20 + 320 = 340 \)

Correct answer is (C): 340

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