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:

• A. 345
• B. 340
• C. 448
• D. 225
• E. 196

Hint:

Set up equations based on percentages and the condition that all quantities must be integers. Use the divisibility constraints to find the minimum value.

Answer 1

The Correct Option is B

Step 1: Let total fruits initially = $T$. Mangoes = $0.4T$. Bananas = $B$, Apples = $A$. So $0.4T + B + A = T \implies B + A = 0.6T$.
Step 2: Sold: Mangoes sold = $0.5 \times 0.4T = 0.2T$. Bananas sold = 96. Apples sold = $0.4A$.
Step 3: Total sold = $0.2T + 96 + 0.4A = 0.5T$ (50% of fruits sol(d).
Step 4: $0.4A = 0.5T - 0.2T - 96 = 0.3T - 96 \implies A = \frac{0.3T - 96}{0.4} = 0.75T - 240$.
Step 5: From $B + A = 0.6T$, we get $B = 0.6T - A = 0.6T - (0.75T - 240) = 240 - 0.15T$.
Step 6: Since $B \ge 1$, $240 - 0.15T \ge 1 \implies 0.15T \le 239 \implies T \le 1593.33$. Since $A \ge 1$, $0.75T - 240 \ge 1 \implies 0.75T \ge 241 \implies T \ge 321.33$. Since $0.4T$ must be integer (mangoes count), $T$ must be multiple of 5. Also $B$ and $A$ must be integers.
Step 7: $B = 240 - 0.15T$ must be integer, so $T$ must be multiple of 20. Also $A = 0.75T - 240$ must be integer.
Step 8: Minimum $T$ satisfying $T \ge 322$ and multiple of 20 is $T = 340$. Check: $T=340$, mangoes = 136, $A = 0.75 \times 340 - 240 = 255 - 240 = 15$, $B = 240 - 0.15 \times 340 = 240 - 51 = 189$. All integers, at least 1.
Step 9: Final Answer: The smallest possible total is 340.