The average speed (correct to one place of decimal) of a train running at the rate of 30 km/h during the first 100 km at 40 km/h during the second 100 km at 50 km/h during the third 100 km is

Options

A. 38.1 km/h
B. 38.2 kmh
C. 38.3 kmh
D. 38.5 km/h

Answer

To solve the problem of finding the average speed of a train that runs different segments of a journey at different speeds, we can follow these steps:

### Step 1: Identify the distances and speeds
The train travels three segments of 100 km each, with the following speeds:
- First 100 km at 30 km/h
- Second 100 km at 40 km/h
- Third 100 km at 50 km/h

### Step 2: Calculate the time taken for each segment
To find the time taken for each segment, we use the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]

1. **Time for the first segment (T1)**:
\[ T1 = \frac{100 \text{ km}}{30 \text{ km/h}} = \frac{100}{30} = \frac{10}{3} \text{ hours} \]

2. **Time for the second segment (T2)**:
\[ T2 = \frac{100 \text{ km}}{40 \text{ km/h}} = \frac{100}{40} = \frac{5}{2} \text{ hours} \]

3. **Time for the third segment (T3)**:
\[ T3 = \frac{100 \text{ km}}{50 \text{ km/h}} = \frac{100}{50} = 2 \text{ hours} \]

### Step 3: Calculate the total time taken
Now, we add the times for all three segments:
\[ \text{Total Time} = T1 + T2 + T3 = \frac{10}{3} + \frac{5}{2} + 2 \]

To add these fractions, we need a common denominator. The least common multiple of 3, 2, and 1 (for 2) is 6.

Converting each term:
- \(T1 = \frac{10}{3} = \frac{20}{6}\)
- \(T2 = \frac{5}{2} = \frac{15}{6}\)
- \(T3 = 2 = \frac{12}{6}\)

Now, we can add them:
\[ \text{Total Time} = \frac{20}{6} + \frac{15}{6} + \frac{12}{6} = \frac{20 + 15 + 12}{6} = \frac{47}{6} \text{ hours} \]

### Step 4: Calculate the total distance
The total distance traveled by the train is:
\[ \text{Total Distance} = 100 + 100 + 100 = 300 \text{ km} \]

### Step 5: Calculate the average speed
The average speed is calculated using the formula:
\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{300 \text{ km}}{\frac{47}{6} \text{ hours}} = 300 \times \frac{6}{47} = \frac{1800}{47} \text{ km/h} \]

### Step 6: Calculate the numerical value
Now we need to calculate \( \frac{1800}{47} \):
\[ \frac{1800}{47} \approx 38.297 \]

### Step 7: Round off to one decimal place
Rounding \( 38.297 \) to one decimal place gives us:
\[ \text{Average Speed} \approx 38.3 \text{ km/h} \]

### Final Answer
Thus, the average speed of the train is **38.3 km/h**.

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