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

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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) About Me  (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
:
t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
t+=+%28-%28-5%29+%2B-+sqrt%28+%28-5%29%5E2-4%2A4%2A%28-75%29+%29%29%2F%282%2A4%29+
t+=+%285+%2B-+sqrt%28+25%2B1200%29%29%2F8+
t+=+%285+%2B-+35%29%2F8
t+=+%285%2B35%29%2F8+=+5 or t=%285-35%29%2F8+=+3.75
:
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.