.
One way distance is 30 miles.
Downstream travel took 2 hours.
Upstream travel took 2.5 hours.
The effective speed downstream was = 15 miles per hour. (Speed = ).
The effective speed upstream was = 12 miles per hour. (Speed = ).
The effective speed downstream is u + v, where u is the speed of the boat in still water and v is the current speed.
The effective speed upstream is u - v.
Thus you have this system of two equations:
u + v = 15, (1)
u - v = 12. (2)
Add the equations (both sides). You will get
2u = 15 + 12 = 27 ====> u = = 13.5.
Thus we just found the speed of the boat in still water u = 13.5 mph.
Then from eq(1), v = 15 - 13.5 = 1.5 mph.
Answer. The effective speed of the boat downstream is 15 mph.
The effective speed of the boat upstream is 12 mph.
The speed of the boat in still water is 13.5 mph.
The speed of the current is 1.5 mph.
Solved. All the question are answered.
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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.