Question 1156500

A tour bus normally leaves for its destination at 5:00 p.m. for a 378 mile trip. This week however, the bus leaves at 6:00 p.m. To arrive on time, the driver drives 9 miles per hour faster than usual. What is the bus` usual speed? 

Let bus` usual  be x mph

Distance of the trip = 378 miles
Normal time taken

Time taken = 378/x 

Speed increased by 9  mph

speed = (x+9) mph

Time taken = 378/(x+9)

Difference in time = 1 hour ( started one hour late

{{{ 378/x - 378/(x+9) = 1}}}

(378(x+9)-378x)/(x(x+9)) = 1}}}


378x + 3402 -378x = x^2+9x

x^2+9x -3402=0

(x+63)(x-54)=0


x=-63 or 54 . Ignore negative value

normal speed = 54 mph