Asked by Anne B.

At an ordinary rate, a man can row the distance from A to B, about 15 km, in 5 hours less time than it takes him to return. Could he double his rate, his time to B would only be one hour less than his time to A. What is his usual rate in still water? What is the rate of the stream?
Answered by Steve
Since time=distance/speed, if his speed is s and the current's is c, then

15/(s+c) = 15/(s-c)-5
15/(2s+c) = 15/(2s-c)-1

Now just solve for s and c.
Answered by Anne B.
My working equations were 15/(s+c) = 5 - 15/(s-c) and 15/(2s+c) = 15/(2s-c) - 1

Whichever I use, I can't seem to get the right values of s and c. My answers are not in the choices given.
Answered by Reiny
I don't think Steve is on line right now, so I will try.

Did you notice the the right side of both of Steve's equations are the opposite of yours?
Steve's solution would result in
s = 4
c = 2

Yours give inadmissible results of
s = appr -14 and c = 16.7

Your translation of the given data into equations is incorrect.
Answered by Anne B.
Oh yeah, I see that.
Well, I guess I'm a bit confused with the less thingy. Anyway, could you please explain how you got 4&2?
Thank you.
Answered by Anne B.
Nevermind, I got it. Thanks for all the help!
Answered by Gail A.
How were you able to obtain the answer? My equation is similar to what steve formulated but im stuck in the problem thanks for the help.
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