SOLUTION: linda is going to visit a freind who lives in san luis obispo, which is 300 miles away. on the return trip, she was able to average 20 mi/hr faster, and it took her one hour and a

Question 453381: linda is going to visit a freind who lives in san luis obispo, which is 300 miles away. on the return trip, she was able to average 20 mi/hr faster, and it took her one hour and a quarter less time. what was her average speed on the way down, and what was her average speed on the way back.
Thanks alot!
Answer by pedjajov(51) (Show Source): You can put this solution on YOUR website!
On her way TO her friend Linda drove with:
r - speed
t - time it took her to her friend
d - 300mi, distance she passed
:
rt=d -> rt=300 -> r=300/t
:
On her way back Linda drove with:
r+20 - 20mi/h faster than on her way to friend
t-5/4 - she used 5/4 of the hour less time to travel
d - distance was the same 300 miles
:
(r+20)(t-5/4)=300
rt-5r/4+20t-25=300, multiply by 4
4rt-5r+80t-100=1200, substitute rt with 300 and r with 300/t
4*300-5*300/t+80t-100=1200, subtract 1200 from both sides
1200-1500/t+80t-100-1200=0, combine numbers
80t-1500/t-100=0, multiply by t
80t^2-100t-1500=0, divide by 20
4t^2-5t-75=0, solve using quadratic formula
:

or
:
Only acceptable solution is positive number so t=5 hours on her way down.
:
Her speed on the way down was 300/5 = 60mi/h.
Her speed on the way back was 60+20 = 80mi/h.

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