Question

A tour bus normally leaves for its destination at 5:00 p.m. for a 220 mile trip. This week however, the bus leaves at 5:20 p.m. To arrive on time, the driver drives 5 miles per hour faster than normal. What is the normal speed of the bus?

Answer 1

Usual Week

let the normal speed of the bus is V (in miles per hour)

so time taken for the bus = distnace/speed = 220/V

This week

speed = V+5

time taken = 220/(V+5)

time taken this week is 20 min (1/3 hour) less than usual trip

220/(V+5) = 220/V - 1/3

220/V - 220/(V+5) = 1/3

5/V(V+5) = 1/660

V(V+5) = 660*5 = 10*11*6*5 = 55*60

so V = 55 miles per hour

(the other is to solve quadrativc equation V^2 + 5V - 330 = 0)

so answer is 55 miles per hour.