The distance between two stations 'P' and 'Q' is 120 km. The train 'A' starts from station 'P' with `(5)/(6)` of its original speed and reached station 'Q' late by half an hour by its scheduled time. Find the original speed.(in kmph)

Options

A. 54
B. 42
C. 40
D. 48

Answer

To solve the problem step by step, we will use the relationship between distance, speed, and time.

### Step 1: Define Variables
Let the original speed of the train A be \( x \) km/h.

### Step 2: Calculate the Reduced Speed
The train travels at \( \frac{5}{6} \) of its original speed. Therefore, the speed of the train when it starts from station P is:
\[ \text{Speed} = \frac{5}{6} x \text{ km/h} \]

### Step 3: Calculate the Time Taken to Travel the Distance
The distance between stations P and Q is 120 km. The time taken to travel this distance at the reduced speed is given by the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{120}{\frac{5}{6}x} = \frac{120 \times 6}{5x} = \frac{720}{5x} = \frac{144}{x} \text{ hours} \]

### Step 4: Calculate the Scheduled Time
If the train had traveled at its original speed \( x \), the time taken would be:
\[ \text{Scheduled Time} = \frac{120}{x} \text{ hours} \]

### Step 5: Set Up the Equation for Late Arrival
According to the problem, the train arrives half an hour late. Thus, we can set up the equation:
\[ \text{Time taken at reduced speed} = \text{Scheduled Time} + \frac{1}{2} \]
Substituting the expressions we found:
\[ \frac{144}{x} = \frac{120}{x} + \frac{1}{2} \]

### Step 6: Solve the Equation
To eliminate the fractions, we can multiply the entire equation by \( 2x \):
\[ 2x \cdot \frac{144}{x} = 2x \cdot \frac{120}{x} + 2x \cdot \frac{1}{2} \]
This simplifies to:
\[ 288 = 240 + x \]
Now, subtract 240 from both sides:
\[ 288 - 240 = x \]
\[ x = 48 \text{ km/h} \]

### Conclusion
The original speed of the train A is \( 48 \) km/h.

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