Moving at ` (5 ) /(6 ) ` of its usual speed, a train is 10 minutes "late. Its usual time to cover the joumey is:

Options

A. 40 min
B. 50 min
C. 35 min
D. 55 min

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step-by-Step Solution:

1. **Understand the Problem**:
- The train is moving at \( \frac{5}{6} \) of its usual speed and is 10 minutes late.
- We need to find the usual time it takes to cover the journey.

2. **Define Variables**:
- Let the usual speed of the train be \( x \) (in distance units per hour).
- Let the usual time to cover the journey be \( t \) (in hours).
- The distance covered can be expressed as \( d = x \cdot t \).

3. **Calculate Distance at Reduced Speed**:
- When the train moves at \( \frac{5}{6} \) of its usual speed, its speed becomes \( \frac{5}{6}x \).
- The time taken at this reduced speed is \( t + \frac{10}{60} \) hours (since 10 minutes is \( \frac{10}{60} \) hours).
- Therefore, the distance can also be expressed as:
\[ d = \left(\frac{5}{6}x\right) \cdot \left(t + \frac{1}{6}\right) \]

4. **Set the Distances Equal**:
- Since the distance is the same in both scenarios, we can set the two expressions for distance equal to each other:
\[ x \cdot t = \left(\frac{5}{6}x\right) \cdot \left(t + \frac{1}{6}\right) \]

5. **Cancel Out \( x \)**:
- Assuming \( x \neq 0 \), we can divide both sides by \( x \):
\[ t = \frac{5}{6} \cdot \left(t + \frac{1}{6}\right) \]

6. **Distribute and Rearrange**:
- Distributing \( \frac{5}{6} \):
\[ t = \frac{5}{6}t + \frac{5}{36} \]
- Rearranging gives:
\[ t - \frac{5}{6}t = \frac{5}{36} \]
\[ \frac{1}{6}t = \frac{5}{36} \]

7. **Solve for \( t \)**:
- Multiply both sides by 6 to isolate \( t \):
\[ t = 6 \cdot \frac{5}{36} = \frac{30}{36} = \frac{5}{6} \text{ hours} \]

8. **Convert Hours to Minutes**:
- To convert \( \frac{5}{6} \) hours into minutes:
\[ t = \frac{5}{6} \times 60 = 50 \text{ minutes} \]

### Final Answer:
The usual time to cover the journey is **50 minutes**.