SOLUTION: Ride the peaks. Smith bicycled 45 miles going east from
Durango, and Jones bicycled 70 miles. Jones averaged
5 miles per hour more than Smith, and his trip took one-half
hour lo
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: Ride the peaks. Smith bicycled 45 miles going east from
Durango, and Jones bicycled 70 miles. Jones averaged
5 miles per hour more than Smith, and his trip took one-half
hour lo
Log On
Question 312076: Ride the peaks. Smith bicycled 45 miles going east from
Durango, and Jones bicycled 70 miles. Jones averaged
5 miles per hour more than Smith, and his trip took one-half
hour longer than Smith’s. How fast was each one traveling? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles.
Jones averaged 5 miles per hour more than Smith, and his trip took one-half
hour longer than Smith’s.
How fast was each one traveling?
:
Let s = Smith's speed
then
(s+5) = Jone's speed
:
Write a time equation, Time = dist/speed
:
Jones time - Smiths time = .5 hr - = .5
multiply by s(s+5), results:
70s - 45(s+5) = .5s(s+5)
:
70s - 45s - 225 = .5s^2 + 2.5s
25s - 225 = .5s^2 + 2.5s
combine on the right
0 = .5s^2 + 2.5s - 25s + 225
A quadratic equation
.5s^2 - 22.5s + 225 = 0
Get rid of th decimals, mult by 2
s^2 - 45s + 450 = 0
Factors to
(s-15)(s-30) = 0
Two solutions
s = 15 mph is Smith's time, then Jones time = 20 mph
and
s = 30 mph is Smith's time, then Jones time = 35
:
The slower time would make more sense for bicycle
Check by finding the times
70/20 = 3.5
45/15 = 3.0
------------
differ: .5h
