SOLUTION: You ride your bike to campus a distance of 5 miles and return home on the same route. Going to campus, you average 9 mph faster than on your return trip home. If the round trip tak
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Question 241922: You ride your bike to campus a distance of 5 miles and return home on the same route. Going to campus, you average 9 mph faster than on your return trip home. If the round trip takes one hour and ten minutes, what is your average rate on the return trip? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! You ride your bike to campus a distance of 5 miles and return home on the same route.
Going to campus, you average 9 mph faster than on your return trip home.
If the round trip takes one hour and ten minutes, what is your average rate on the return trip?
:
let s = speed on the return trip
then
(s+9) = speed to the campus
:
Write a time equation: Time = dist/speed
Change time to a fraction 1 = hrs
: + =
Multiply equation by 6s(s+9), results
6(s+9)*5 + 6s(5) = s(s+9)*7
:
5(6s+54) + 30s = 7(s^2 + 9s)
:
30s + 270 + 30s = 7s^2 + 63s
Arrange as a quadratic equation
0 = 7s^2 + 63s - 60s - 270
:
7s^2 + 3s - 270 = 0
Factors to:
(7s + 45)(s - 6) = 0
The positive solution
s = 6 mph on the return trip
;
;
Check this by finding the times
5/6 + 5/15 = 1.667; which is 1 hr 10 min
