Moving 6/7 of its usual speed a train is 10 min late. Find its usual time to cover the journey

Options

A. 25 min
B. 15 min
C. 35 min
D. 60 min

Answer

To solve the problem, we need to find the usual time taken by the train to cover the journey. Let's denote the following:

- Let the usual speed of the train be \( S \) (in distance units per time unit).
- Let the usual time taken to cover the journey be \( T \) (in time units).
- Let the distance covered by the train be \( D \).

From the relationship between speed, distance, and time, we have:

\[ D = S \times T \]

When the train moves at \( \frac{6}{7} \) of its usual speed, its new speed becomes:

\[ \text{New Speed} = \frac{6}{7} S \]

The time taken to cover the same distance \( D \) at this new speed is given by:

\[ \text{New Time} = \frac{D}{\text{New Speed}} = \frac{D}{\frac{6}{7} S} = \frac{7D}{6S} \]

Now, we know that the train is 10 minutes late when moving at this new speed. This means:

\[ \text{New Time} = T + 10 \text{ minutes} \]

We can set up the equation:

\[ \frac{7D}{6S} = T + 10 \]

Now, substituting \( D \) from the first equation \( D = S \times T \) into the equation, we get:

\[ \frac{7(S \times T)}{6S} = T + 10 \]

This simplifies to:

\[ \frac{7T}{6} = T + 10 \]

Now, we will eliminate \( T \) from the right side:

\[ \frac{7T}{6} - T = 10 \]

To combine the terms on the left, we can rewrite \( T \) as \( \frac{6T}{6} \):

\[ \frac{7T}{6} - \frac{6T}{6} = 10 \]

This simplifies to:

\[ \frac{T}{6} = 10 \]

Now, multiplying both sides by 6 gives us:

\[ T = 60 \text{ minutes} \]

Thus, the usual time taken by the train to cover the journey is **60 minutes**.