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Two buses left town a for town b at the same time. the speed of one bus was 10 mph greater than the speed of the other. in 3 1/2 hours, the faster bus reached town b, but the other bus still had to travel 1/6 of the distance between towns a and b. find the speed of the buses and the distance between towns a and b.

User Altan
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Final answer:

The speed of the slower bus is 40 mph, the speed of the faster bus is 50 mph, and the distance between towns A and B is 180 miles.

Step-by-step explanation:

We can solve this problem by setting up a system of equations. Let's call the speed of the slower bus 'x' mph. Then the speed of the faster bus would be 'x + 10' mph, since it is 10 mph greater. We know that the faster bus reached town B in 3 1/2 hours, and that the slower bus still had to travel 1/6 of the distance between towns A and B. Let's say the total distance between the towns is 'd' miles.

To find the speed of the buses and the distance between the towns, we can set up the following equations:

3 1/2(x + 10) = d

3 1/2x + 10 = 5/6d

By multiplying both sides of the second equation by 2/3, we get:

7/2x + 20/3 = 10/3d

Now we can solve the system of equations to find the values of 'x' and 'd'.

First, multiply the first equation by 2 to get rid of the fraction:

7x + 20 = 10d

Next, substitute the value of 'd' in terms of 'x' from the second equation into the first equation:

7x + 20 = 10(2/3x + 20/3)

Expand and solve for 'x':

7x + 20 = (20/3)x + (200/3)

(11/3)x = 140/3

x = 40

Now substitute 'x' back into the second equation to find 'd':

3 1/2(40) + 10 = 5/6d

140 + 10 = 5/6d

150 = 5/6d

d = 180

So the speed of the slower bus is 40 mph, the speed of the faster bus is 50 mph, and the distance between towns A and B is 180 miles.

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