A man walks on a straight road form his home to a market 2.5 km away with speed of 5 `(km)/(hr)`. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 `(km)/(hr)`. The average speed of the man over the intervel of time 0 to 40 min is equal to

Options

A. `5(km)/(hr)`
B. `(25)/(4)(km)/(hr)`
C. `(30)/(4)(km)/(hr)`
D. `(45)/(8)(km)/(hr)`

Explanation

To solve this problem, we need to determine the average speed of the man over the given time interval. The average speed is defined as the total distance traveled divided by the total time taken.

### Step-by-Step Solution:

1. **Calculate the time taken to reach the market:**
- Distance to the market, \( D_1 = 2.5 \) km
- Speed to the market, \( V_1 = 5 \) km/hr
- Time taken to reach the market, \( T_1 = \frac{D_1}{V_1} = \frac{2.5}{5} = 0.5 \) hours

2. **Convert the time to minutes:**
- \( 0.5 \) hours = \( 0.5 \times 60 = 30 \) minutes

3. **Determine the remaining time for the return journey:**
- Total time given, \( T_{total} = 40 \) minutes
- Time taken to return home, \( T_2 = T_{total} - T_1 = 40 - 30 = 10 \) minutes

4. **Convert the return time to hours:**
- \( T_2 = 10 \) minutes = \( \frac{10}{60} = \frac{1}{6} \) hours

5. **Calculate the distance traveled on the return journey:**
- Speed on the return journey, \( V_2 = 7.5 \) km/hr
- Distance traveled on the return journey, \( D_2 = V_2 \times T_2 = 7.5 \times \frac{1}{6} = 1.25 \) km

6. **Calculate the total distance traveled:**
- Total distance, \( D_{total} = D_1 + D_2 = 2.5 + 1.25 = 3.75 \) km

7. **Convert the total time to hours:**
- Total time in hours, \( T_{total} = \frac{40}{60} = \frac{2}{3} \) hours

8. **Calculate the average speed:**
- Average speed, \( V_{avg} = \frac{D_{total}}{T_{total}} = \frac{3.75}{\frac{2}{3}} = 3.75 \times \frac{3}{2} = 5.625 \) km/hr

Therefore, the average speed of the man over the interval of time 0 to 40 minutes is \( 5.625 \) km/hr.