Question 1179660
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A motorboat traveling with the current went 42 mi in 3.5 h. Traveling against the current, the boat went 18 mi in 3 h. 
Find the rate of the boat in calm water and the rate of the current.
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<pre>
Let x be the rate of the motorboat 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 motorboat downstream is x + y
and  the effective rate of the motorboat   upstream is x - y.


From the problem, the effective rate of the motorboat downstream is the distance of 42 miles 
divided by the time of 3.5 hours  {{{42/3.5}}} = 12 mph.

                  The effective rate of the motorboat upstream is the distance of 18 miles 
divided by the time of 3 hours  {{{18/3}}} = 6 mph.


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

    x + y = 12,    (1)

    x - y =  6.    (2)


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

    2x = 12 + 6 = 18  --->   x = 18/2 = 9.

Now from equation (1)

     v = 12 - u = 12 - 9 = 3.


<U>ANSWER</U>.  The rate of the motorboat in still water is 9 mph.  The rate of the current is 3 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

        42/(x+y) = 3.5   --->   divide by 3.5  12/(x+y) = 1  --->   x+y = 12,

    while in my solution I simply write for the effective rate 

        x + y = 42/3.5 = 12.


    Second loop in the @mananth post is writing

                       18/(x-y) = 3   --->   divide by 3  6/(x-y) = 1  --->   x-y = 6,

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

        x - y = 18/3 = 6.
</pre>

It is why I present 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.