A person covers a distance of 300 km partly by train and partly by car. If he travels 60 km by train and rest by car it takes him 4 hours to cover that distance and if he covers 100 km by train and rest by car it takes him 4 hours 10 minutes to cover that distance. Find out the speed of train and car.

Answer

To solve the problem, we need to find the speeds of the train and the car based on the information given. Let's denote the speed of the train as \( T \) km/h and the speed of the car as \( C \) km/h.

### Step 1: Set up the equations based on the first scenario
In the first scenario, the person travels 60 km by train and the remaining distance (300 km - 60 km = 240 km) by car. The total time taken is 4 hours.

The time taken to travel by train is:
\[ \text{Time by train} = \frac{60}{T} \]

The time taken to travel by car is:
\[ \text{Time by car} = \frac{240}{C} \]

According to the problem, the total time is 4 hours:
\[ \frac{60}{T} + \frac{240}{C} = 4 \quad \text{(Equation 1)} \]

### Step 2: Set up the equations based on the second scenario
In the second scenario, the person travels 100 km by train and the remaining distance (300 km - 100 km = 200 km) by car. The total time taken is 4 hours and 10 minutes, which is \( \frac{25}{6} \) hours.

The time taken to travel by train is:
\[ \text{Time by train} = \frac{100}{T} \]

The time taken to travel by car is:
\[ \text{Time by car} = \frac{200}{C} \]

According to the problem, the total time is \( \frac{25}{6} \) hours:
\[ \frac{100}{T} + \frac{200}{C} = \frac{25}{6} \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
Now we have two equations:
1. \( \frac{60}{T} + \frac{240}{C} = 4 \)
2. \( \frac{100}{T} + \frac{200}{C} = \frac{25}{6} \)

We can solve these equations simultaneously.

From Equation 1, we can express \( \frac{240}{C} \) in terms of \( T \):
\[ \frac{240}{C} = 4 - \frac{60}{T} \]
\[ \frac{240}{C} = \frac{4T - 60}{T} \]
\[ C = \frac{240T}{4T - 60} \quad \text{(Equation 3)} \]

Now, substitute Equation 3 into Equation 2:
\[ \frac{100}{T} + \frac{200}{\left(\frac{240T}{4T - 60}\right)} = \frac{25}{6} \]

Simplifying the second term:
\[ \frac{200(4T - 60)}{240T} = \frac{800T - 12000}{240T} = \frac{800}{240} - \frac{12000}{240T} = \frac{10}{3} - \frac{50}{T} \]

Now, substituting this back into Equation 2:
\[ \frac{100}{T} + \left(\frac{10}{3} - \frac{50}{T}\right) = \frac{25}{6} \]

Combine the terms:
\[ \left(\frac{100 - 50}{T}\right) + \frac{10}{3} = \frac{25}{6} \]
\[ \frac{50}{T} + \frac{10}{3} = \frac{25}{6} \]

Now, convert \( \frac{10}{3} \) to have a common denominator of 6:
\[ \frac{10}{3} = \frac{20}{6} \]
So:
\[ \frac{50}{T} + \frac{20}{6} = \frac{25}{6} \]
Subtract \( \frac{20}{6} \) from both sides:
\[ \frac{50}{T} = \frac{25}{6} - \frac{20}{6} = \frac{5}{6} \]

### Step 4: Solve for \( T \)
Cross-multiply to solve for \( T \):
\[ 50 \cdot 6 = 5T \implies 300 = 5T \implies T = 60 \text{ km/h} \]

### Step 5: Find \( C \) using \( T \)
Substituting \( T \) back into Equation 3 to find \( C \):
\[ C = \frac{240 \cdot 60}{4 \cdot 60 - 60} = \frac{14400}{240 - 60} = \frac{14400}{180} = 80 \text{ km/h} \]

### Final Answer
The speed of the train is \( 60 \) km/h and the speed of the car is \( 80 \) km/h.