A man covered a distance of 18 km in 3 hours, 7.2 km in 2 hours and 15 km in 5 hours. What is the average distance traveled by him per hour?
4.02 km
The problem asks us to find the average distance covered by a man per hour. This is essentially asking for the average speed of the man over the entire journey.
To find the average speed, we need two things:
The average speed is then calculated by dividing the total distance by the total time.
Let's break down the journey into the given segments and sum up the distance and time for each part.
| Segment | Distance (km) | Time (hours) |
|---|---|---|
| 1 | 18 | 3 |
| 2 | 7.2 | 2 |
| 3 | 15 | 5 |
Add the distances covered in each segment:
\(\text{Total Distance} = 18 \text{ km} + 7.2 \text{ km} + 15 \text{ km}\)
\(\text{Total Distance} = 40.2 \text{ km}\)
Add the time taken for each segment:
\(\text{Total Time} = 3 \text{ hours} + 2 \text{ hours} + 5 \text{ hours}\)
\(\text{Total Time} = 10 \text{ hours}\)
Divide the total distance by the total time:
\(\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}\)
\(\text{Average Speed} = \frac{40.2 \text{ km}}{10 \text{ hours}}\)
\(\text{Average Speed} = 4.02 \text{ km/hour}\)
So, the average distance traveled by the man per hour is 4.02 km.
Comparing this result with the given options, we find that 4.02 km matches one of the choices.
| Concept | Formula |
|---|---|
| Average Speed | \(\frac{\text{Total Distance}}{\text{Total Time}}\) |
| Total Distance (multiple segments) | Sum of distances of all segments |
| Total Time (multiple segments) | Sum of time taken for all segments |
The calculation of average speed is a fundamental concept in distance, speed, and time problems. It's important to remember that average speed is not simply the average of the speeds in different segments unless the time taken for each segment is the same.
Understanding the difference between calculating average speed based on distance and time versus just averaging different speeds is crucial for solving these types of problems correctly.
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