Averages , Ratio and Proportion Question 2 - CAT 2024
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:
Step-by-Step Explanation
Let the initial total stock of fruits be S, the number of apples be A, and the number of bananas be B.
- Initial Mangoes
Mangoes = 40% of S = 0.4S
- Combined initial Apples and Bananas
A + B = 60% of S = 0.6S
- Track the fruits sold:
Mangoes sold = 1/2 Ă— 0.4S = 0.2S
Apples sold = 40% of A = 0.4A
Bananas sold = 96
Total Sold = 50% of S = 0.5S
- Formulate the balance equation:
0.2S + 0.4A + 96 = 0.5S âźą 0.4A + 96 = 0.3S
Multiply the entire equation by 10 to clear decimals:
4A + 960 = 3S âźą S = 320 + 4A/3
- Apply integer count constraints:
For S to be a valid integer fruit count, A must be a multiple of 3. Additionally, for the number of apples sold (0.4A = 2A/5) to be a whole number, A must also be a multiple of 5.
Divisibility requirement for A = LCM(3, 5) = 15
Choosing the smallest positive valid value, A = 15:
S = 320 + 4(15)/3 = 320 + 20 = 340
(Checking the remaining constraints: If S = 340, then A + B = 0.6(340) = 204. With A = 15, B = 189, which satisfies the condition of having at least one fruit of each type.)
Correct Answer: 340
đź’ˇ FundaMakers Tip:
Discrete counts require strict integer constraints. Recognizing that the apple sales fraction (2A/5) requires A to be a multiple of 5, while the linear structural formula S = 320 + 4A/3 requires A to be a multiple of 3, reveals that A must be a multiple of 15. Testing the lowest valid value A = 15 gives the answer immediately.