Let the distance from Mohan's home to the station be denoted by $d$ kilometers.

We are given two scenarios:

1. If Mohan travels by bus at a speed of 14.4 km/h, he will be 10 minutes late for the train.
2. If Mohan takes a taxi at a speed of 25.2 km/h, he will be 15 minutes early for the train.

Step 1: Convert minutes to hours

• 10 minutes late = $\frac{10}{60} = \frac{1}{6}$ hours late.
• 15 minutes early = $\frac{15}{60} = \frac{1}{4}$ hours early.

Step 2: Let the scheduled time to catch the train be $t$ hours.

Bus travel:

If Mohan takes the bus, he travels at 14.4 km/h and is 10 minutes late. So, the time taken by bus is $t + \frac{1}{6}$ hours. Using the formula $\text{time} = \frac{\text{distance}}{\text{speed}}$, we have: $\frac{d}{14.4} = t + \frac{1}{6}$

Taxi travel:

If Mohan takes the taxi, he travels at 25.2 km/h and arrives 15 minutes early. So, the time taken by taxi is $t - \frac{1}{4}$ hours. Using the same formula: $\frac{d}{25.2} = t - \frac{1}{4}$

Step 3: Solve the system of equations

We now have two equations:

1. $\frac{d}{14.4} = t + \frac{1}{6}$
2. $\frac{d}{25.2} = t - \frac{1}{4}$

We can solve these two equations to find $d$, the distance from Mohan's home to the station.

Substituting $t$ from one equation into the other:

From the first equation: $t = \frac{d}{14.4} - \frac{1}{6}$

Substitute this value of $t$ into the second equation: $\frac{d}{25.2} = \left( \frac{d}{14.4} - \frac{1}{6} \right) - \frac{1}{4}$

Now, solve for $d$.

First, simplify the equation: $\frac{d}{25.2} = \frac{d}{14.4} - \frac{1}{6} - \frac{1}{4}$

Simplify the right-hand side: $\frac{d}{25.2} = \frac{d}{14.4} - \frac{2}{12} = \frac{d}{14.4} - \frac{1}{3}$

Next, multiply the entire equation by 25.2 to eliminate the denominator on the left side: $d = 25.2 \left( \frac{d}{14.4} - \frac{1}{3} \right)$

Distribute 25.2 on the right side: $d = \frac{25.2d}{14.4} - \frac{25.2}{3}$

Simplify $\frac{25.2}{14.4}$ and $\frac{25.2}{3}$: $d = 1.75d - 8.4$

Rearrange the equation to isolate $d$: $d - 1.75d = -8.4$ $-0.75d = -8.4$

Finally, solve for $d$: $d = \frac{8.4}{0.75} = 11.2$

Conclusion:

The distance from Mohan's home to the station is 11.2 kilometers.

Would you like more details on any part of this solution?

Related Questions:

1. What if the speeds of the bus and taxi were different?
2. How would the result change if Mohan was 5 minutes late instead of 10?
3. How do you solve similar problems involving distance, speed, and time?
4. What happens if the train's departure time changes by a certain amount?
5. How can you verify your result by substituting the values into the original equations?

Tip:

Always make sure to convert time into hours when working with speeds in kilometers per hour to avoid mistakes in calculations.