Sure, let's solve this problem step by step.

1. First, let's convert the time difference into hours since the speed is given in km/h. 6 minutes late is 6/60 = 0.1 hours and 4 minutes early is 4/60 = 0.0667 hours.

2. Let's denote the distance to school as D (in km) and the time it should take to get to school as T (in hours).

3. From the problem, we can write down two equations based on the relationship speed = distance/time.

For the first scenario (18km/h speed), we have:
D = 18(T + 0.1)

For the second scenario (20km/h speed), we have:
D = 20(T - 0.0667)

4. Since both equations are equal to D, we can set them equal to each other and solve for T:

18(T + 0.1) = 20(T - 0.0667)

Simplifying this equation gives: T = 0.567 hours

5. Now, we can substitute T back into one of the original equations to find D. Let's use the first one:

D = 18(T + 0.1) = 18(0.567 + 0.1) = 12 km

So, the distance to school is 12 km.

Question

Sure, let's solve this problem step by step.

1. First, let's convert the time difference into hours since the speed is given in km/h. 6 minutes late is 6/60 = 0.1 hours and 4 minutes early is 4/60 = 0.0667 hours.

2. Let's denote the distance to school as D (in km) and the time it should take to get to school as T (in hours).

3. From the problem, we can write down two equations based on the relationship speed = distance/time.

For the first scenario (18km/h speed), we have: 
   D = 18(T + 0.1)

For the second scenario (20km/h speed), we have: 
   D = 20(T - 0.0667)

4. Since both equations are equal to D, we can set them equal to each other and solve for T:

18(T + 0.1) = 20(T - 0.0667)

Simplifying this equation gives: T = 0.567 hours

5. Now, we can substitute T back into one of the original equations to find D. Let's use the first one:

D = 18(T + 0.1) = 18(0.567 + 0.1) = 12 km

So, the distance to school is 12 km.

Knowee AI · Accepted Answer