Given:

• Total distance $= 450$ km.
• Two scenarios for traveling:
  1. Travels 240 km by train and the rest (210 km) by car in 8 hours 40 minutes.
  2. Travels 180 km by train and the rest (270 km) by car in 9 hours (20 minutes more).

Let's break down the information and solve for the car's speed step-by-step.

Step 1: Convert the time into hours

1. First scenario: 8 hours 40 minutes = $8 + \frac{40}{60} = 8 + \frac{2}{3} = \frac{26}{3}$ hours.
2. Second scenario: 20 minutes more than 8 hours 40 minutes = 9 hours.

Step 2: Define variables

Let:

• $x$ = speed of the train (in km/h).
• $y$ = speed of the car (in km/h).

Step 3: Set up equations for each scenario

First scenario:

• Distance by train = 240 km
• Distance by car = 210 km
• Total time = $\frac{240}{x} + \frac{210}{y} = \frac{26}{3}$ hours

This gives the equation: $\frac{240}{x} + \frac{210}{y} = \frac{26}{3}$

Second scenario:

• Distance by train = 180 km
• Distance by car = 270 km
• Total time = 9 hours

This gives the equation: $\frac{180}{x} + \frac{270}{y} = 9$

Step 4: Solve the system of equations

From the equations:

1. $\frac{240}{x} + \frac{210}{y} = \frac{26}{3}$
2. $\frac{180}{x} + \frac{270}{y} = 9$

We'll solve these step-by-step:

Equation 1: Multiply through by 3 to clear the fraction

$3 \times \frac{240}{x} + 3 \times \frac{210}{y} = 26$ $\frac{720}{x} + \frac{630}{y} = 26$

Equation 2: Simplify as is

$\frac{180}{x} + \frac{270}{y} = 9$

We now have:

1. $\frac{720}{x} + \frac{630}{y} = 26$
2. $\frac{180}{x} + \frac{270}{y} = 9$

Step 5: Solve these equations

Multiply the second equation by 4 to align the coefficients of $\frac{180}{x}$ with $\frac{720}{x}$: $4 \times \left(\frac{180}{x} + \frac{270}{y} = 9\right)$ $\frac{720}{x} + \frac{1080}{y} = 36$

Now we have the system:

1. $\frac{720}{x} + \frac{630}{y} = 26$
2. $\frac{720}{x} + \frac{1080}{y} = 36$

Subtract the first equation from the second: $\left(\frac{720}{x} + \frac{1080}{y}\right) - \left(\frac{720}{x} + \frac{630}{y}\right) = 36 - 26$ $\frac{450}{y} = 10$

Solve for $y$: $y = \frac{450}{10} = 45$

Final Answer

The speed of the car is 45 km/h.

Would you like more details or have any questions?

Relative Questions

1. How would the solution change if the travel times were different?
2. Can you explain why we needed to set up two separate equations for this problem?
3. How can we find the speed of the train using this information?
4. What are the implications if the total distance traveled changes?
5. How would this problem change if the car's speed varied on different parts of the trip?

Tip

When dealing with time, always convert minutes to hours for consistency in equations.