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Julie and Eric row their boat (at a constant speed) 63 miles downstream for 7 hours, helped by the current.
Rowing at the same rate, the trip back against the current takes 9 hours. Find the rate of the current.
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Let u be the rate of the boat in a still water (at no current), in miles per hour.
Let v be the rate of the stream.
Then the rate rowing downstream is (u+v) relative to the river banks.
From the other side, this rate is = 9 mph, according to the condition,
It gives you your first equation
u + v = 9.
The rate rowing upstream is (u-v) relative to the river banks.
From the other side, this rate is = 7 mph, according to the condition,
It gives you your second equation
u - v = 9.
Combining equations (1) and (2) together, you get the system of two equations
u + v = 9, (1)
u - v = 7. (2)
To solve the system, add the equations (1) and (2). You will get
2u = 9 + 7 = 16 ====> u = = 8.
Thus you found that the rate of the boat in the still water is 8 miles per hour.
Then from (1), v = 9 - u = 9 - 8 = 1 mph.
Answer. The rate of the current is 1 mile per hour.
Solved.
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".