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Jim is taking a canoe in the Grand River. The speed of the current is 3 miles per hour.
Jim can canoe 4 miles upstream in the same time that it would take him to canoe 10 miles downstream.
What is the speed of Jim’s canoe in still water?
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Let "u" be the unknown the speed of Jim’s canoe in still water.
Then the speed canoeing downstream is (u+3) miles per hour (the speed relative to the bank of the river),
while the speed canoeing upstream is (u-3) miles per hour.
The time Jim spends canoeing 4 miles upstream is 4/(u-3) hours.
The time Jim spends canoeing 10 miles downstream is 10/(u+3) hours.
According to the condition, these amounts of time are the same.
It gives you an equation
= .
To solve it, multiply both sides by (u-3)*(u+3). You will get
4*(u+3) = 10*(u-3),
4u + 12 = 10u - 30,
12 + 30 = 10u - 4u,
6u = 42 ---> u = = 7.
Answer. The speed Jim canoes in still water is 7 miles per hour.
Ignore writing by "josgarithmetic". It is WRONG.
To see more similar solved problems, look into 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.
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".