.
A boat moves 7 kilometers upstream in the same amount of time it moves 20 kilometers downstream.
If the rate of the current is 5 kilometers per hour, find the rate of the boat in still water.
~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let "u" be the rate of the boat in still water, in .
Then your governing equation ("time equation") is
= .
The left side is the time to travel 7 kilometers upstream.
The right side is the time to travel 20 kilometers downstream.
To solve the equation, multiply both sides by (u-5)*(u+5). You will get
7(u+5) = 20(u-5), or
7u + 35 = 20u - 100, or
35 + 100 = 20u - 7u, or
135 = 13u ---> u = .
Answer. The rate of the boat in still water is .
Unfortunately, the person who invented these input numbers, select them too curve to get an integer number as the answer.
Perhaps, this person didn't solve the problem himself.
It is a typical "upstream and downstream Travel and Distance" word problem.
See the lessons
- Wind and Current problems
- More problems on upstream and downstream round trips
- Selected problems from the archive on the boat floating Upstream and Downstream
in this site.
You will find the detailed solutions of many similar problems there.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this textbook under the section "Word problems", the topic "Travel and Distance problems".
If after reading my solution you still have a question "why the governing equation has this form ?",
then read the lessons above. They contain the detailed answer to this question.