We are given that a car can travel 200 miles in the same time it takes a bus to travel 160 miles, and the bus's speed is 10 miles per hour slower than the car's speed.

Let's define variables to represent the rates:

• Let $r$ represent the rate (speed) of the car in miles per hour.
• Then, the rate of the bus will be $r - 10$ miles per hour.

Step 1: Relating Time

Since both the car and bus travel for the same amount of time, we use the formula for time: $\text{Time} = \frac{\text{Distance}}{\text{Rate}}$

• The time for the car is: $\frac{200}{r}$
• The time for the bus is: $\frac{160}{r - 10}$

These two times are equal, so: $\frac{200}{r} = \frac{160}{r - 10}$

Step 2: Solve the Equation

To solve for $r$, first cross-multiply: $200(r - 10) = 160r$

Expand both sides: $200r - 2000 = 160r$

Now, isolate $r$: $200r - 160r = 2000$ $40r = 2000$ $r = \frac{2000}{40} = 50$

Step 3: Calculate the Rate of the Bus

Since $r = 50$, the rate of the car is 50 miles per hour. The bus's rate is $50 - 10 = 40$ miles per hour.

Final Answer:

• The rate of the car is 50 miles per hour.
• The rate of the bus is 40 miles per hour.

Would you like further details or have any other questions?

Here are five follow-up questions based on this concept:

1. If the car traveled 300 miles in the same time, what would the new speed of the bus be?
2. How much time did both the car and bus take to travel their respective distances?
3. If the speed of the car increased by 10 miles per hour, how much faster would it reach 200 miles?
4. What would be the speed of the bus if the car's speed was reduced to 40 miles per hour?
5. How far would the bus travel in 3 hours at its current speed?

Tip: Always set up a relationship between distances and speeds when dealing with problems involving time and rate comparisons.