Question

A tour bus normally leaves for its destination at 5:00 p.m. for a 200 mile trip. This week however, the bus leaves at 5:40 p.m. To arrive on time, the driver drives 10 miles per hour faster than usual. What is the bus’s usual speed?

Answer 1

The usually speed of the bus is 50 miles/h.

Explanation:

Let the usual speed of the bus be x mile/hour.

We know that

`speed=(distance)/(time)`

`\Rightarrow time = (distance)/(speed)`

The bus travels 200 miles.

To reach its destination it takes time
`=(200)/(x)` h

This week however the bus leaves at 5:40.

The bus late 40 minutes
`=(40)/(60) h`
`= (2)/(3)h`

Now the speed of the bus is = (x+10) miles/h

The new time to reach the destination is
`=(200)/(x+10)` h

According to the problem,

`(200)/(x)-(200)/(x+10)=(2)/(3)`

`\Rightarrow 200[(x+10-x)/(x(x+10))]=(2)/(3)`

`\Rightarrow 200[(10)/(x^2+10x)]=(2)/(3)`

⇒2(x²+10x)=200×10×3

⇒x²+10x = 3000

⇒x²+10x -3000=0

⇒x²+60x-50x-3000=0

⇒x(x+60)-50(x+60)=0

⇒(x+60)(x-50)=0

⇒x= -60,50

∴x=50 [since speed does not negative]

The usually speed of the bus is 50 miles/h.