Math CentralQuandaries & Queries


Question from Stephanie, a student:
beth paddled a canoe to a picnic spot in the river in 2 hours.
she traveled up the stream back to the original spot in three hours.
the current was two miles per hour.
What was her rate of speed?


Which way is the current?

What is the ratio of her speed on the way there to her speed on the way back?

Let her speed relative to the water be X miles per hour.
What was her speed relative to the land on the way there?
What was it on the way back?
Set the ratio of these equal to the value you found above and solve for X

Good Hunting!

(By the way, at the 2008 Olympics the winning time for single person racing canoes over 1 km was nearly 4 minutes. That's about 9.3 mph. Is the problem realistic?)


Hi Stephanie.

You need to break this up into speed components. That means that the overall speed is made up of Beth paddling and also the speed of the current (first it works for her, then it works against her). On both trips, she goes the same distance (from the original to the picnic spot).

So let's use some variables for algebra:

Let B = Beth's padding speed in still water,

So Beth's total speed going downstream is B + 2, because she gets a 2 mph "boost" from the current. But her speed coming back is B - 2, because the current is working against her.

You know that when you multiply speed by time you get distance, right? For example, 3 hours at 60 mph = 180 miles. So that means that 2(B + 2) is the distance she went downstream, because it took two hours. Coming back, she went 3(B - 2) because it took 3 hours at the speed B - 2 to come back.

But you know she went the same distance there and back, of course!

So that means these two expressions are equal to each other, because they represent the distance travelled!


2(B + 2) = 3(B - 2)

Solve for B, that's Beth's padding speed, which is what the question wants from you.

Stephen La Rocque.

PS: Beth is a very fast paddler!!


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