Question 773516
A fisherman can row against the stream in 20 minutes and return in 15 minutes . Then the speed of the current is?
<pre>
We can't work it unless we know the distance up the river he rowed.
I will assume the distance he rowed is 2 miles.

Since the speed will be in miles per hour we will convert 20 minutes
and 15 minutes to hours: 

{{{20_minutes*(expr((1_hour)/(60_minutes)))}}}{{{""=""}}}{{{20cross(minutes)*(expr((1_hour)/(60cross(minutes))))}}}{{{""=""}}}{{{20/60}}}{{{hour}}}{{{""=""}}}{{{1/3}}}{{{hour}}}

{{{15_minutes*(expr((1_hour)/(60_minutes)))}}}{{{""=""}}}{{{15cross(minutes)*(expr((1_hour)/(60cross(minutes))))}}}{{{""=""}}}{{{15/60}}}{{{hour}}}{{{""=""}}}{{{1/4}}}{{{hour}}}

Let the speed of the current = c
Let his rowing speed (if he were in still water) = r
Then his speed against the current is r-c and
his speed with the current is r+c

We make this chart

                     DISTANCE  =  RATE × TIME
--------------------------------------------
Against the stream      2      =  r-c    {{{1/3}}}
With the stream         2      =  r+c    {{{1/4}}}

2 = (r-c){{{(1/3)}}}
2 = (r+c){{{(1/4)}}}

Clear of fractions:
Multiply both sides of the first equation by 3
Multiply both sides of the first equation by 4

 6 = (r-c)(1)
 8 = (r+c)(1)

 6 = r-c
 8 = r+c

Add the two equations term by term:

14 = 2r      (the c's cancel out)
 7 = r

So his rowing speed is 7 miles per hour.

Substitute 7 for r in

 6 = r-c
 6 = 7-c
 c = 1

So the speed of the current is 1 mile per hour.

Edwin</pre>