A man starts off on a journey. He travelled for 16 hours in total. First half of the distance was covered at 40 km/h and the second half at 60 km/h. What was the distance covered?

Options

A. 768 km
B. 576 km
C. 384 km
D. 960 km

Answer

To solve the problem step by step, we need to determine the total distance covered by the man during his journey.

### Step 1: Define the total distance
Let the total distance covered be \( D \). According to the problem, the distance is divided into two halves. Therefore, the first half of the distance is \( \frac{D}{2} \) and the second half is also \( \frac{D}{2} \).

### Step 2: Set up the speeds and times
The first half of the distance is covered at a speed of 40 km/h, and the second half is covered at a speed of 60 km/h. We can denote the time taken to cover the first half as \( T_1 \) and the time taken to cover the second half as \( T_2 \).

Using the formula for time, which is:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
we can express \( T_1 \) and \( T_2 \) as follows:
\[ T_1 = \frac{\frac{D}{2}}{40} = \frac{D}{80} \]
\[ T_2 = \frac{\frac{D}{2}}{60} = \frac{D}{120} \]

### Step 3: Set up the equation for total time
According to the problem, the total time taken for the journey is 16 hours. Therefore, we can write the equation:
\[ T_1 + T_2 = 16 \]
Substituting the expressions for \( T_1 \) and \( T_2 \):
\[ \frac{D}{80} + \frac{D}{120} = 16 \]

### Step 4: Find a common denominator and solve for \( D \)
To solve the equation, we need to find a common denominator for the fractions. The least common multiple of 80 and 120 is 240. Rewriting the equation with a common denominator:
\[ \frac{3D}{240} + \frac{2D}{240} = 16 \]
Combining the fractions gives:
\[ \frac{5D}{240} = 16 \]

### Step 5: Solve for \( D \)
Now, we can solve for \( D \):
\[ 5D = 16 \times 240 \]
Calculating the right side:
\[ 5D = 3840 \]
Now, divide both sides by 5:
\[ D = \frac{3840}{5} = 768 \text{ km} \]

### Conclusion
The total distance covered by the man during his journey is **768 km**.