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: Sep 23, 2024
  • 100
  • 240
  • 340
  • None of Above
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The Correct Option is C

Solution and Explanation

Let initial stock of all the fruits is SS and we have bb and aa mangoes initially.
Stock of Mangoes =40%= 40\% of SS =2S5= \frac {2S}{5}
Total number of fruits sold = Mangoes Sold + Apples Sold + Bananas Sold
=2S10+96+4a10=S2= \frac {2S}{10} + 96 + \frac {4a}{10}= \frac S2 (Given)

S5+96+2a5=S2⇒ \frac S5 +96+\frac {2a}{5}= \frac S2

S=4a+9603⇒ S=\frac {4a+960}{3}
S=4a3+320⇒ S=\frac {4a}{3}+320
aa has to be multiple of 33 for the above term to be an integer but aa has to be multiple of 55 for 4a10\frac {4a}{10} to be an integer.
⇒ Smallest value of aa satisfying both conditions is 1515.
4a3+320⇒ \frac {4a}{3}+320

4×153+320⇒ \frac {4 \times 15}{3}+320

20+320⇒ 20+320
340⇒340

So, the correct option is (C):340(C): 340

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