Question

Two buses leave towns 405 miles apart at the same time and travel toward each other. One bus travels 15 faster than the other. If they meet in 3 hours,hwhat is the rate of each bus?

Answer 1

Let x be the distance traveled by one of the buses (let's call it Bus A) and y be the rate of the other bus (let's call it Bus B).

Let's remember the relation that exist between the speed (rate), the distance and the time:

`V=(d)/(t)`

We know that it takes 3 hours for Bus A to travel x miles at a speed of y+15 (it is 15 faster than Bus B). This expressed as an equation is:

`y+15=(x)/(3)`

We also know that it takes 3 hours for Bus B to travel 405-x miles at a speed of y:

`y=(405-x)/(3)`

Solve the obtained system of equations by equalization:

`\begin{gathered} y+15=(x)/(3) \\ x=3(y+15) \\ y=(405-x)/(3) \\ x=405-3y \end{gathered}`
`\begin{gathered} x=x \\ 3y+45=405-3y \\ 6y=360 \\ y=(360)/(6) \\ y=60 \end{gathered}`

The rate of Bus B is 60 and the rate of Bus A is 75 (Bus A is 15 faster than Bus B).