# 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

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 Click here to see ALL problems on Travel Word Problems 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(2048)   (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