Kala drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 7 hours. When Kala drove home, there was no traffic and the trip only took 5 hours. If her average rate was 18 miles per hour faster on the trip home, how far away does Kala live from the mountains?
@TheSmartOne
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@satellite73
again it is distance is rate times time the the rate going is say \(R\) then the rate returning is \(R+18\)
so one expression for the distance is \[D=7R\] another is \[D=5(R+18)\]
I'm not very good with the distance, rate and time equations
since evidently the distances are the same, you can set them equal and solve \[7R=5(R+18)\]
it is really one idea distance is rate times time \[D=RT\] or \[T=\frac{D}{R}\] or \[R=\frac{D}{T}\]
i am sure you can solve \[7R=5(R+18)\] right?
Yes. Thank you!
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