Two men, both with rowing speeds of 5 mph in still water, start out rowing on a river at the same time in the morning. Man A rows upstream from point A to point B, while man B rows downstream from point B to point A. They pass each other at noon on the river and continue rowing. Man B, rowing downstream, reaches point A at 4 in the afternoon. Man A, rowing upstream, reaches point B at 9 in the evening.
How fast is the river current?
Just fast enough to get the results as they were stated. Sorry…I have a young relative visiting and that’s her answer. She wanted to see it in print.
Too fast for my speed. I ended up with a quadratic equation with imaginary roots. I assume algebra is required to solve this one, unless I missed something.
Maybe the only imagined they passed each other then 
Only linear algebra, nothing more.
Current speed = 1 mph.
You are correct !!!
let D1 be distance from A to point at which boats meet
let D2 be distance from B to point at which boats meets
let Vdown be speed of boat going downstream
let Vup be speed up boat going upstream
let X be speed of current
let T1 be elapsed time between departure and meeting at noon
For first part of the trip (until noon):
T1 = D1/Vup = D2/Vdown
For second part of trip ( after meeting at noon until finish):
D1 = Vdown x 4 hrs
D2 = Vup x 9 hrs
Combining above eqns:
4Vdown/Vup = 9Vup/Vdown
Cross multiply:
4Vdown(squared) = 9Vup(squared)
Since still water speed of both boats is 5 mph, relative to fixed points A and B,
Vup = 5-X
Vdown = 5+x
Substitute into the equatioin with the Vsquareds:
4(5+x)(squared) = 9(5-x)(squared)
expand and solve for x.
or just trial and error.
x=1