Question 1189570
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2. A boat can go 6 miles downstream in 20 minutes, but the return trip upstream takes 30 minutes . 
Find the rate of the boat in still water and the rate of the current.
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<pre>
Let u represent the rate of the boat in still water and 
let y represent the rate of the current. 


Then  the  effective speed downstream  is  u+v  miles per hour,
while the  effective speed upstream    is  u-v  miles per hour.


    // It is first major point you need understand and use in this sort of problems.



Now, "speed" equation for boat floating downstream is the distance divided by time downstream

    {{{6/((1/3))}}} = u + v        (where 1/3 represents 20 minutes = 1/3 of an hour)

or  18 = u + v.      (1)



Next, "speed" equation for boat floating upstream is the distance divided by time  uptream


    {{{6/((1/2))}}} = u - v        (where 1/2 represents 30 minutes = 1/2 of an hour)

or  12 = u - v.      (2)


    // It is the second major point in solving such problems: you must understand and write these equation automatically !



So, you have this system of equations for u and v

    u + v =  18      (3)
    u - v =  12      (4)


Now add equations (3) and (4) to eliminate "v". You will get

    2u = 18 + 12 = 30.     

Hence,  u = {{{30/2}}} = 15.


Thus you just found the boat' speed in still water. It is 15 miles per hour.


Next, you can find the current rate from equation (3)  

    v = 18 - u = 18 - 15 = 3 miles per hour.


<U>Answer</U>.  The boat' speed in still water is 15 miles per hour.

         The current speed is 3 miles per hour.



Solved.  I advise you to make the check on your own.

         After making the check, you will understand the problem and the solution MUCH better.
</pre>

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It is a typical and standard Upstream and Downstream round trip word problem.


You can find many similar fully solved problems on upstream and downstream round trips with detailed solutions in lessons 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Wind-and-Current-problems.lesson>Wind and Current problems</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/More-problems-on-upstream-and-downstream-round-trips.lesson>More problems on upstream and downstream round trips</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Wind-and-Current-problems-solvable-by-quadratic-equations.lesson>Wind and Current problems solvable by quadratic equations</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Unpowered-raft-moving-downstream-along-a-river.lesson>Unpowered raft floating downstream along a river</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Selected-problems-from-the-archive-on-a-boat-floating-Upstream-and-Downstream.lesson>Selected problems from the archive on the boat floating Upstream and Downstream</A> 

in this site.


Read them attentively and learn how to solve this type of problems once and for all.


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this textbook under the section "<U>Word problems</U>", &nbsp;the topic "<U>Travel and Distance problems</U>".



Save the link to this online textbook together with its description


Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson


to your archive and use it when it is needed.



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