SOLUTION: Sandy bicycled 45 miles going east from Cambridge and Janie, 70 miles. Janie averaged 5 miles per hour more than Sandy and her trip took 1/2 hour longer than Sandy's. How fast was
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: Sandy bicycled 45 miles going east from Cambridge and Janie, 70 miles. Janie averaged 5 miles per hour more than Sandy and her trip took 1/2 hour longer than Sandy's. How fast was
Log On
Question 193910: Sandy bicycled 45 miles going east from Cambridge and Janie, 70 miles. Janie averaged 5 miles per hour more than Sandy and her trip took 1/2 hour longer than Sandy's. How fast was each one traveling? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Sandy bicycled 45 miles going east from Cambridge and Janie, 70 miles. Janie averaged 5 miles per hour more than Sandy and her trip took 1/2 hour longer than Sandy's. How fast was each one traveling?
:
I assume this means that Sandy traveled 45 miles and Janie traveled 70 miles.
:
let s = Sandy's speed
then
(s+5) = Janie's speed
:
Write a time equation; time = dist/speed
:
Janie's travel time = Sandy's travel time + .5 hrs = + .5
Multiply equation by s(s+5) to get rid of the denominators, results:
70s = 45(s+5) + .5s(s+5)
:
70s = 45s + 225 + .5s^2 + 2.5s
:
0 = .5s^2 + 2.5s + 45s - 70s + 225
A quadratic equation
.5s^2 - 22.5s + 225 = 0
Multiply equation by 2 to eliminate the decimals
s^2 - 45s + 450 = 0
Factor
(s - 15)(s - 30) = 0
Two solutions for Sandy's speed
s = 15
and
s = 30
:
:
Check using both solutions by finding the times;
Sandy: 15 mph; Janie: 20 mph
70/20 = 3.5 hr
45/15 = 3.0 hr; a half hr difference
;
Sandy: 30 mph; Janie: 35 mph
70/35 = 2 hrs
45/30 = 1.5 hrs; a half hr difference here too
