Keturah264
Answered

Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

A motorboat takes 5 hours to travel 200 km going upstream. The return trip takes 4 hours going downstream. What is the rate of the boat in still water and what is the rate of the current?

Sagot :

arnfrs
rate of motorboat: b
rate of current: c
So the rate the boat travels upstream is $b - c$, and the rate it travels downstream is $b + c$

[tex]d = rt[/tex]
[tex]200 = (b - c)\cdot 5[/tex]
[tex]200 = (b + c)\cdot 4[/tex]
[tex]b - c = 40[/tex]
[tex]b + c = 50[/tex]

Adding these equations, we get:
[tex]2b = 90[/tex]
[tex]b = 45[/tex]

So
[tex]c = 50 - 45 = 5[/tex]

Therefore the rate of the boat is 45kph, and the rate of the current is 5kph

Hope this helps :)



Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.