SOLUTION: Sorry I seem to be really bad at word problems and I can never get how to set them up and I always get thrown off I don't understand how to do this step by step Landon can cli

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Question 1008903: Sorry I seem to be really bad at word problems and I can never get how to set them up and I always get thrown off I don't understand how to do this
step by step
Landon can climb a certain hill at a rate that is 0.5mph slower than his rate coming down the hill. If it takes him 2 hours to climb the hill and 110 minutes to come down the hill, what is the rate coming down ?
Answer by ptaylor(2198) About Me  (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=rate coming down hill
Then r-0.5=rate going up hill
We can't deal in both hours and minutes, sooo:
110 minutes =110/60 or 11/6 hours
Going up the hill:
d(up)=(r-0.5)*2
d(down)=r*(11/6)
We know that d(up)=d(down)
(r-0.5)*2=r*(11/6)
2r-1=11r/6 multiply each side by 6
12r-6=11r
12r-11r=6
r=6 mph
CK
d(up)=(6-0.5)*2=(5.5)*2=11 mi
d(down)=6*(11/6)=11 mi
Hope this helps---ptaylor