.
Rowing with the current, a canoeist paddled 18 mi in 3 h. Against the current, the canoeist could paddle only 6 mi in the same amount of time.
Find the rate of the canoeist in calm water and the rate of the current.
~~~~~~~~~~~~~~~~~~~~~~~
Your equations are
(1) = u + v, ("Rowing with the current, a canoeist paddled 18 mi in 3 h")
(2) = u - v. ("Against the current, the canoeist could paddle only 6 mi in the same amount of time.")
Here u is the rate of the canoeist in calm water and v is the rate of the current.
Simplify equations (1) and (2). You will get
u + v = 6, (3)
u - v = 3. (4)
Add equations (3) and (4). You will get
2u = 6 + 3 = 9 ---> u = = 4.5.
Thus you just found the rate of the canoeist in calm water. It is 4.5 mph.
Now find v from (3):
v = 6 - u = 6 - 4.5 = 1.5 mph.
Answer. The rate of the canoeist in calm water is 4.5 mph.
The rate of the current is 1.5 mph.
It is a typical Upstream and Downstream Travel and Distance problem.
You can find similar fully solved similar problems on upstream and downstream trips with detailed solutions in 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
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".