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 makes 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. What is the smallest possible total number of fruits in the stock at the beginning of the day?

Answer 1

Correct Answer: 960

Step 1: Let the total number of fruits at the beginning of the day be \( T \)

Since mangoes make up \(40\%\) of the total stock:
\(\text{Mangoes} = 0.4 \times T.\)
The remaining \(60\%\) of the stock consists of bananas and apples:
\(\text{Bananas} + \text{Apples} = 0.6 \cdot T.\)

Step 2: Fruits sold during the day
During the day:

• Mangoes sold = \(\frac{1}{2} \cdot \text{Mangoes} = \frac{1}{2} \cdot 0.4 \cdot T = 0.2 \cdot T\)
• Bananas sold = 96.
• Apples sold = \(40\%\) of the apples. Let the apples be \( A \). Then:
   \(\text{Apples sold} = 0.4 \cdot A.\)

Step 3: Total fruits sold

At the end of the day, \(50\%\) of the total fruits were sold. Thus:
\(\text{Total fruits sold} = 0.5 \cdot T.\)

Substitute the components:
\(\text{Total fruits sold} = (\text{Mangoes sold}) + (\text{Bananas sold}) + (\text{Apples sold}).\)

Substitute the values:
\(0.5 \cdot T = 0.2 \cdot T + 96 + 0.4 \cdot A. \quad \text{(Equation 1)}\)

Step 4: Express apples in terms of \( T \)

From the stock composition:
\(A + \text{Bananas} = 0.6 \cdot T.\)

Let bananas \( B = 96 \). Substitute:
\(A + 96 = 0.6 \cdot T.\)

Solve for \( A \):
\(A = 0.6 \cdot T - 96. \quad \text{(Equation 2)}\)

Step 5: Substitute \( A \) into Equation 1

Substitute \( A = 0.6 \times T - 96 \) into Equation 1:
\(0.5 \cdot T = 0.2 \cdot T + 96 + 0.4 \cdot (0.6 \cdot T - 96).\)

Simplify:
\(0.5 \cdot T = 0.2 \cdot T + 96 + 0.4 \cdot 0.6 \cdot T - 0.4 \cdot 96.\)

\(0.5 \cdot T = 0.2 \cdot T + 96 + 0.24 \cdot T - 38.4.\)

Step 6: Combine terms

Combine terms:
\(0.5 \cdot T = 0.44 \cdot T + 57.6.\)

Simplify further:
\(0.5 \cdot T - 0.44 \cdot T = 57.6.\)

\(0.06 \cdot T = 57.6.\)

Solve for \( T \):
\(T = \frac{57.6}{0.06} = 960.\)

Final Answer
The smallest possible total number of fruits in the stock at the beginning of the day is: \(\boxed{960}.\)