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

Answer:

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

Step-by-step explanation:

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

We know that

$speed=\frac{distance}{time}$

$\Rightarrow time = \frac{distance}{speed}$

The bus travels 200 miles.

To reach its destination it takes time $=\frac{200}{x}$ h

This week however the bus leaves at 5:40.

The bus late 40 minutes $=\frac{40}{60} h$ $= \frac{2}{3}h$

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

The new time to reach the destination is $=\frac{200}{x+10}$ h

According to the problem,

$\frac{200}{x}-\frac{200}{x+10}=\frac{2}{3}$

$\Rightarrow 200[\frac{x+10-x}{x(x+10)}]=\frac{2}{3}$

$\Rightarrow 200[\frac{10}{x^2+10x}]=\frac{2}{3}$

⇒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.