SOLUTION: A family drove 1080 miles to their vacation lodge. Bercause of increased traffic denisty, their average speed on the return trip was decreased by 6 miles per hour and the trip too

Question 215307: A family drove 1080 miles to their vacation lodge. Bercause of increased traffic denisty, their average speed on the return trip was decreased by 6 miles per hour and the trip took 2 1/2 hours longer. Determine their average speed on the way to the lodge.
Found 2 solutions by RAY100, ptaylor:
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
Please see problem # 215308

Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt;r=d/t and t=d/r
Let r=average speed to their vacation lodge
time required to get to their vacation lodge=1080/r
r-6=average speed on the return trip
time required for their return trip=1080/(r-6) and we are told that this time is 2.5 hours longer then the time to get to their vacation lodge, so:
(1080/r) + 2.5=1080/(r-6) multiply each term by r(r-6)
1080(r-6)+2.5r(r-6)=1080r get rid of parens
1080r-6480+2.5r^2-15r=1080r subtract 1080r from each side
1080r-1080r-6480+2.5r^2-15r=1080r-1080r collect like terms
2.5r^2-15r-6480=0 divide each term by 2.5
r^2-6r-2592=0 Quadratic in standard form and it can be factored
(r-54)(r+48)=0
r=54 mph----avarage speed to their vacation lodge
and
r=-48 mph---DISREGARD! In this case speeds are positive
CK
time to their vacation lodge=1080/54=20 hours
Time for return trip=1080/48=22.5 hours
22.5-20=2.5
2.5=2.5
Hope this helps---ptaylor
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