Question 1167838: Peter can row 11 miles downstream in the same time it takes him to row 7 miles upstream. He rows downstream for 3 hours, then turns and rows back for 4 hours, but finds he is still 5 miles from where he started his trip. What is the speed of the current?
Found 2 solutions by Boreal, Edwin McCravy:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! speed downstream is s+c and (s+c)*t=11
speed upstream is s-c and (s-c)*t=7; so in 4 hours (s-c)*4+5=(s+c)*3
we know that t=11/(s+c) and t=7(s-c)
so that 11s-11c=7s+7c
so that 4s=18c
or 2s=9c
or s=4.5c
Therefore, s-c=3.5 c
and 14c+5=16.5c
and 2.5c=5 and c=2 mph
s then is 9 mph, so that 3 hours downstream is 33 miles and 4 miles upstream is 28 miles. That checks.
11 miles downstream takes 1 hour, and that is the same time it takes him to row 7 miles upstream
The answer is 2 mph.
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website!
He's right but I think I'd do it this way:
Let t be that same time that it takes him to go 11 miles upstream
that it takes him to go downstream 7 miles. So make this chart
distance rate time
downstream 1st time 11 t
upstream 1st time 7 t
downstream 2nd time 3
upstream 2nd time 4
Use rate = distance/time to fill in the rates
distance rate time
downstream 1st time 11 11/t t
upstream 1st time 7 7/t t
downstream 2nd time 11/t 3
upstream 2nd time 7/t 4
Use distance = rate x times to fill in the other two distances:
distance rate time
downstream 1st time 11 11/t t
upstream 1st time 7 7/t t
downstream 2nd time 33/t 11/t 3
upstream 2nd time 28/t 7/t 4
So the two distances are 5 miles apart, so
33/t - 28/t = 5
5/t = 5
5 = 5t
1 = t
So his rate upstream is 11/1 = 11 mph and his rate downstream is 7/1 = 7 mph
The average of those is 9 mph, his speed in still water. So the speed of
the current is how much the current speeds him up going with it, and how much
it slows him down going against it, which is 2 mph.
Edwin
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