Question 1007654
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A person rowed their boat downstream for 100 miles and they took 2 hours. 
Returning upstream, the trip took 2 hours and 40 minutes.
What is the speed of the water?
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        In the post by @mananth, the solution is produced by a computer code.


        Neither the style of the solution, nor its form of presentation are perfect; 

        they are difficult to read and to understand.


        So, I present here my solution in simple, straightforward and clear, transparent form,

        as it should be done to every school Math problem.



<pre>
Let x be the rate of the boat in still water (in miles per hour)
and y be the rate of the current (in the same units).


Then the effective rate of the boat downstream is x + y
and  the effective rate of the boat   upstream is x - y.


From the problem, the effective rate of the boat downstream is the distance of 100 miles 
divided by the time of 2 hours  {{{100/2}}} = 50 mph.

                  The effective rate of the boat upstream is the distance of 100 miles 
divided by the time of 2 hours and 40 minute, or  2{{{2/3}}} hours, or  {{{8/3}}} hours

    {{{100/((8/3))}}} = {{{300/8}}} = 37.5 mph


So, we have two equations for  'x'  and  'y'

    x + y = 50,      (1)

    x - y = 37.5.    (2)


To find 'y', subtract equations (2) from equation (1).  The terms 'x' and 'x' will cancel each other, and you will get

    2y = 50 - 37.5 = 12.5  --->   y = 12.5/2 = 6.25.


At this point, the solution is complete.


<U>ANSWER</U>.  The rate of the current is 6.25 miles per hour.
</pre>

Solved.


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For a computer code (which @mananth uses to create his solution files), 
there is no difference which style of the solution to produce.


But for a human reader, there is a huge difference what to read and from which source to learn.


So, I created my solution here in order for you see the difference.