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A boat takes 2/3 as much time to travel downstream as to its return. If the rate of the river's current is 8 kph,
what is the rate of the boat in still water?
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Let "x" be the the rate of the boat in still water.
Then its effective rate downstream is (x+8) km/h,
while the effective rate upstream is (x-8) km/h.
The "time equation" is
= , (<<<---=== it says "time downstream = 2/3 time upstream")
where D is one way distance.
You can cancel the factor D. After that, multiply both sides by 3*(x-8)*(x+8) to get
3*(x-8) = 2*(x+8),
3x - 24 = 2x + 16,
3x - 2x = 16 + 24 ====> x = 40.
Answer. the rate of the boat in still water is 40 km/h.
Solved.
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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.