Formulate the following problems as a pair of equations, and hence find their solutions:
Roohi travels 300km to her home partly by train and partly by bus. She taken 4 hours if she travels 60km by train and the remaining by bus. If she travels 100km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Answer

To solve the problem, we will formulate a pair of linear equations based on the information given and then solve those equations to find the speeds of the train and the bus.

### Step 1: Define Variables
Let:
- \( x \) = speed of the train (in km/h)
- \( y \) = speed of the bus (in km/h)

### Step 2: Formulate the First Equation
According to the problem, Roohi travels 60 km by train and the remaining distance by bus, which is \( 300 - 60 = 240 \) km. The total time taken for this journey is 4 hours.

Using the formula for time, \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \), we can write the equation for the first scenario:
\[ \frac{60}{x} + \frac{240}{y} = 4 \]

### Step 3: Formulate the Second Equation
In the second scenario, Roohi travels 100 km by train and the remaining distance by bus, which is \( 300 - 100 = 200 \) km. The total time taken for this journey is 4 hours and 10 minutes, which is \( 4 + \frac{10}{60} = \frac{25}{6} \) hours.

Using the same formula for time, we can write the equation for the second scenario:
\[ \frac{100}{x} + \frac{200}{y} = \frac{25}{6} \]

### Step 4: Simplify the Equations
Now we have the following two equations:
1. \( \frac{60}{x} + \frac{240}{y} = 4 \) (Equation 1)
2. \( \frac{100}{x} + \frac{200}{y} = \frac{25}{6} \) (Equation 2)

To eliminate the fractions, we can multiply both equations by \( 6xy \):
1. \( 6y \cdot 60 + 6x \cdot 240 = 24xy \)
\[ 360y + 1440x = 24xy \quad \text{(Multiply by 6)} \]
Rearranging gives:
\[ 24xy - 360y - 1440x = 0 \quad \text{(Equation 1)} \]

2. \( 6y \cdot 100 + 6x \cdot 200 = 25xy \)
\[ 600y + 1200x = 25xy \quad \text{(Multiply by 6)} \]
Rearranging gives:
\[ 25xy - 600y - 1200x = 0 \quad \text{(Equation 2)} \]

### Step 5: Solve the System of Equations
Now we have two equations:
1. \( 24xy - 360y - 1440x = 0 \)
2. \( 25xy - 600y - 1200x = 0 \)

We can solve these equations simultaneously. Let's express \( y \) from the first equation:
\[ y = \frac{1440x}{24x - 360} \]

Substituting this expression for \( y \) into the second equation:
\[ 25x \left(\frac{1440x}{24x - 360}\right) - 600\left(\frac{1440x}{24x - 360}\right) - 1200x = 0 \]

This will lead to a quadratic equation in terms of \( x \). After solving for \( x \), we can substitute back to find \( y \).

### Step 6: Calculate Values
After solving the equations, we find:
- \( x = 60 \) km/h (speed of the train)
- \( y = 80 \) km/h (speed of the bus)

### Final Answer
The speed of the train is **60 km/h** and the speed of the bus is **80 km/h**.

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