Hi Kimberley,

I can help get you started.

Suppose the speed of the stream is $s$ (in whatever units you are using) and $b$ is the speed of the beaver in still water. Then the speed of the beaver downstream, with the current, is $b + s$ and the speed of the beaver upstream, against the current is $b - s.$

Do you see how to complete the problem?

Write back if you need more assistance,
Penny

Kimberly wrote back

Don't we still need more variables in order to figure it out?? I've been on this for two hours and still no idea!

Going downstream the beaver swims 4 times as fast as he does swimming upstream, thus

\[b + s = 4(b - s).\]

What is the relationship between $b$ and $s?$

Penny

Ok, we are getting more confused... are we going to be able to come up with an exact answer? The curriculum we are using is asking for and exact answer rounded to the nearest hundredth??

Simplify the expression

\[b + s = 4(b - s)\]

to find a relationship between $b$ and $s.$ What did you get?

Penny

-3b=-5s ?????
But i have no variables to input for b or s??

The question is "How many times faster is its speed in still water than the speed of the stream?"

The speed in still water is $b$ and the speed of the stream is $s$ so the question is " Find the number $k$ so that $b = k \times s.$ What is $k?$"

Penny