.
Going 24 miles downstream in 2 hours, the tugboat has the effective speed = 12 miles per hour.
This speed is the sum of the tugboat in still water (u) and the speed of current (v):
u + v = 12 mph (1)
Going 16 miles upstream in 2 hours, the tugboat has the effective speed = 8 miles per hour.
This speed is the difference of the tugboat in still water and the speed of current:
u - v = 8 mph (2)
Thus you have this system of two equations (1),(2) in two unknowns
u + v = 12 mph (1)
u - v = 8 mph (2)
To solve the system, add two equations (1) and (2) (both sides). You will get
2u = 12 + 8 = 20 ====> u = = 10.
Thus we just found the speed of the tugboat in still water. It is 10 miles per hour.
Then from eq(1), v = 12 - u = 12 - 10 = 2 mph.
Answer. The speed of the tugboat in still water is 10 miles per hour.
The current speed is 2 mph.
Solved.
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For many other similar solved Travel and Distance Upstream and Downstream trip problem see the 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.