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: Define initial fruit quantities.

Let \( T \) be the total number of fruits initially. Mangoes are \(40\%\) of the total:

\(\text{Mangoes} = 0.4 \times T.\)

The remaining \(60\%\) are bananas and apples:

\(\text{Bananas} + \text{Apples} = 0.6 \cdot T.\)

Step 2: Calculate fruits sold.

During the day:

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

Step 3: Determine total fruits sold.

At the end of the day, \(50\%\) of the total fruits were sold:

\(\text{Total fruits sold} = 0.5 \cdot T.\)

The total fruits sold is the sum of each type sold:

\(\text{Total fruits sold} = (\text{Mangoes sold}) + (\text{Bananas sold}) + (\text{Apples sold}).\)

Substituting the values:

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

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

From the initial stock composition, we know:

\(\text{Apples} + \text{Bananas} = 0.6 \cdot T.\)

Let \( B \) be the initial number of bananas. We are given that 96 bananas were sold, which means \( B \) must be at least 96. For the purpose of relating \( A \) to \( T \), let's use the equation from Step 2: \( \text{Bananas} + A = 0.6 \cdot T \). If we assume the number of bananas sold (96) is the total number of bananas, then:

\(96 + A = 0.6 \cdot T.\)

Solving for \( A \):

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

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

Substitute the expression for \( A \) from Equation 2 into Equation 1:

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

Simplify the equation:

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

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

Step 6: Combine like terms to solve for \( T \).

Group the \( T \) terms and the constant terms:

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

\(0.5 \cdot T = 0.44 \cdot T + 57.6.\)

Isolate \( T \):

\(0.5 \cdot T - 0.44 \cdot T = 57.6.\)

\(0.06 \cdot T = 57.6.\)

Solve for \( T \):

\(T = \frac{57.6}{0.06}.\)

\(T = 960.\)

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