Two race cars are speeding around the racetrack at different speeds. One goes around the track in 2 minutes; the other goes around the track in 1½ minutes. How often does the faster car pass the slower car?

Is there a formula to figure this out?

To determine how often the faster car passes the slower car, we need to calculate the number of laps the slower car completes while the faster car completes one lap.

First, let's find out how many laps each car completes in one minute:

- The slower car completes 1 lap in 2 minutes, so it completes 1/2 lap in 1 minute.
- The faster car completes 1 lap in 1.5 minutes, so it completes 2/3 lap in 1 minute.

To determine how often the faster car overtakes the slower car, we need to find the least common multiple (LCM) of 1/2 and 2/3.

The formula to calculate the LCM is LCM(a, b) = (a * b) / GCD(a, b), where GCD is the greatest common divisor.

In this case, a = 1/2 and b = 2/3.

To calculate the GCD, we need to find the greatest common divisor of the numerators (1 and 2) and the denominators (2 and 3):

- The GCD of 1 and 2 is 1.
- The GCD of 2 and 3 is also 1.

Now we can calculate the LCM:

LCM = (1/2 * 2/3) / (1) = 1/3

The LCM of 1/2 and 2/3 is 1/3.

This means the faster car overtakes the slower car once every 1/3 lap.

In other words, the faster car passes the slower car three times every two laps.

So, the faster car passes the slower car three times.

Yes, there is a formula to figure out how often the faster car passes the slower car. The formula is:

Number of passes = (Total time taken by slower car) / (Time taken by faster car - Time taken by slower car)

In this case, the time taken by the slower car is 2 minutes and the time taken by the faster car is 1.5 minutes.

Number of passes = 2 minutes / (1.5 minutes - 2 minutes)

To simplify this calculation, let's convert both times to seconds.

Number of passes = 120 seconds / (90 seconds - 120 seconds)

Now solve for the number of passes:

Number of passes = 120 seconds / -30 seconds

Number of passes = -4

Therefore, the faster car passes the slower car 4 times.