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

Accepted Answer

340. Let total fruits = $T$. - Mangoes $M = 0.4 T$ - Bananas $B$ and Apples $A$, so $A + B = 0.6 T$ Fruits sold: - Mangoes sold $= 0.2 T$ - Bananas sold $= 96$ - Apples sold $= 0.4 A$ Total sold $= 0.5 T$: $$0.2 T + 96 + 0.4 A = 0.5 T \implies 0.3 T - 0.4 A = 96 \implies 3T - 4A = 960$$ Since $A = 0.6T - B$, substitute $A$: $$3T - 4(0.6T - B) = 960 \implies 0.6T + 4B = 960 \implies 3T + 20B = 4800$$ Since $3T = 4800 - 20B$, $3T$ must be divisible by 20, which implies $T$ is divisible by 20. Also, $A \ge 1 \implies 3T = 960 + 4A \ge 964 \implies T \ge 322$. The smallest multiple of 20 greater than or equal to 322 is $T = 340$. Check for $T = 340$: $$A = \frac{3(340) - 960}{4} = 15 \ge 1$$ $$B = \frac{4800 - 3(340)}{20} = 189 \ge 1$$ $0.4 A = 0.4(15) = 6$ (an integer). Hence, the minimum possible stock size $T = 340$.

Smallest Total Stock of Fruits

Explanation

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