Is mobile tips a bus from his home to the station she will travel at 14.4 km per hour to reach 10 minutes late for the train if he takes a taxi he will travel at 25.2 km per hour to reach 15 minute early for the train how far is the station from Mohan at home

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.

This problem explores how to solve a distance-speed-time problem where Mohan travels by bus at 14.4 km/h and taxi at 25.2 km/h, arriving either 10 minutes late or 15 minutes early for a train. Using algebra and system of equations, the distance to the station is calculated as 11.2 km. Suitable for students in Grades 9-11.

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