.
Let v be the rate of current, in miles per hour.
Notice that one way distance is 5 miles.
Then Sandra's rate rowing downstream is (6+v) miles per hour, and the time rowing 6 miles downstream is hours.
Sandra's rate rowing upstream is (6-v) miles per hour, and the time rowing 6 miles upstream is hours.
The time equation is
+ = 3 hours.
To solve it, multiply both sides by (6+v)*(6-v). You will get
5*(6-v) + 5*(6+v) = 3*(6^2-v^2),
60 = 108 - 3v^2 ====> 3v^2 = 108 - 60 = 48 ====> v^2 = = 16 ====> v = 4 miles per hour.
Answer. The rate of the current is 4 mph.
Solved.
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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
- Wind and Current problems
- More problems on upstream and downstream round trips
- Wind and Current problems solvable by quadratic equations
- Unpowered raft floating downstream along a river
- Selected problems from the archive on the boat floating Upstream and Downstream
in this site.
Read them attentively and learn how to solve this type of problems once and for all.
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".
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.