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: Understanding the Concept:
This problem uses percentage-based linear equations. We need to find the smallest integer total stock \( T \) that satisfies all conditions.
Step 2: Key Formula or Approach:
Let total stock = \( T \). Mangoes \( M = 0.4T \). Bananas = \( B \), Apples = \( A \).
Sold fruits = \( 0.5M + 96 + 0.4A = 0.5T \).
Constraint: \( M + B + A = T \).
Step 3: Detailed Explanation:
1. Substitute \( M \) in the sold equation:
\( 0.5(0.4T) + 96 + 0.4A = 0.5T \implies 0.2T + 96 + 0.4A = 0.5T \).
\( 96 + 0.4A = 0.3T \).
Multiply by 10: \( 960 + 4A = 3T \).
2. From \( M + B + A = T \), substitute \( M = 0.4T \):
\( B + A = 0.6T \).
3. We have \( 3T - 4A = 960 \). For \( T \) to be an integer, \( A \) must be a multiple of 3 because \( 3T \) and 960 are multiples of 3.
4. Test smallest options for \( T \):
If \( T = 340 \): \( 3(340) - 960 = 4A \implies 1020 - 960 = 60 \implies A = 15 \).
If \( A = 15 \), then \( B = 0.6(340) - 15 = 204 - 15 = 189 \).
Since \( B>96 \) (as he sold 96), this is a valid smallest case.
Step 4: Final Answer:
The smallest possible total number of fruits is 340.