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:

Options

  1. 340
  2. 240
  3. 300
  4. 540

Correct Answer

340

Explanation

Let total initial fruits = N. Mangoes = 40% of N = (2/5)N (must be integer), apples = A, bananas = B. After sales: mangoes sold = (1/2)M = (1/5)N, bananas sold = 96, apples sold = 40% of A = (2/5)A. Total sold = 50% of N = (1/2)N. Equation for total sold: (1/5)N + 96 + (2/5)A = (1/2)N Multiply by 10: 2N + 960 + 4A = 5N ⇒ 3N = 4A + 960 ⇒ A = (3N − 960)/4. Constraints: • N is a multiple of 5 (so M = 2/5 N is integer). • A must be a positive integer and a multiple of 5 (so that 40% of A is an integer). • M must be even (so half of mangoes is an integer). • B = N − M − A ≥ 96. Search for the smallest N meeting these conditions gives: N = 340, with M = 136, A = 15, B = 189. Check: • Mango sold = 68, bananas sold = 96, apples sold = 6 → total sold = 170 = 50% of 340. • All counts are integers and each fruit type ≥ 1. Therefore the smallest possible initial total number of fruits is 340

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