Question 1177332
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A kayak can travel 24 miles downstream in 2 ​hours, while it would take 12 hours to make the same trip upstream. 
Find the speed of the kayak in still​ water, as well as the speed of the current. 
Let k represent the speed of the kayak in still​ water, and let c represent the speed of the current.
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<pre>
Let k be the rate of the kayak in still water (in miles per hour)
and c be the rate of the current (in the same units).


Then the effective rate of the kayak downstream is k + c
and  the effective rate of the kayak   upstream is k - c.


From the problem, the effective rate of the kayak downstream is the distance of 24 miles 
divided by the time of 2 hours  {{{24/2}}} = 12 mph.

                  The effective rate of the plane upstream is the distance of 24 miles 
divided by the time of 12 hours  {{{24/12}}} = 2 mph.


So, we have two equations to find 'k' and 'c'

    k + c = 12,    (1)

    k - c =  2.    (2)


To solve, add equations (1) and (2).  The terms 'c' and '-c' will cancel each other, and you will get

    2k = 12 + 2 = 14  --->   k = 14/2 = 7.

Now from equation (1)

     v = 12 - u = 12 - 7 = 5.


<U>ANSWER</U>.  The rate of the kayak in still water is 7 mph.  The rate of the current is 5 mph km/h.
</pre>

Solved.


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This solution produces the same answer as in the post by @mananth, but has an advantage
that it does not contain excessive calculations that the solution by @mananth has.


We, the tutors, write here our solutions not only to get certain numerical answer.
We write to teach - and, in particular, to teach solving in a right style.


This style solving presented in my post, is straightforward with no logical loops.
The solution presented in the post by @mananth has two logical loops.


<pre>
    One loop in the @mananth post is writing

        24/(k+c) = 2   --->   2k + 2c = 24  --->  /2  --->  k+c = 12,

    while in my solution I simply write for the effective rate 

        k + c = 24/2 = 12.


    Second loop in the @mananth post is writing

                       24/(x-y) = 12   --->   12x - 12c = 24  --->  /12  --->  k-c = 2,

    while in my solution I simply write for the effective rate upstream

        u - v = 24/12 = 2.
</pre>

It is why I presented my solution here and why I think it is better than the solution by @mananth:
- because it teaches students to present their arguments in a straightforward way, without logical zigzags.


@mananth repeats his construction of solution with no change for all similar problems on floating
with and against the current simply because his COMPUTER CODE is written this way.
But this way is not pedagogically optimal - in opposite, it is pedagogically imperfect.