June 12th, 2012 at 4:14:24 PM
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Joe puts his canoe in the river and starts paddling upstream. After a mile his hat falls in the river. Ten minutes later he realizes his hat is missing and immediately paddles downstream to retrieve it. He catches up to it at the same place he launched his canoe in the first place.

What is the speed of the river current?

I would suggest posting your answers openly. However, if you wish to post a solution, please hide it in "spoiler tags."

What is the speed of the river current?

I would suggest posting your answers openly. However, if you wish to post a solution, please hide it in "spoiler tags."

nice day

It's not whether you win or lose; it's whether or not you had a good bet.

June 12th, 2012 at 4:18:32 PM
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6 mph, assuming he paddles consistently.

edit: Damn thats not the right answer, hold on

edit: Damn thats not the right answer, hold on

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June 12th, 2012 at 4:21:17 PM
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3 mph

June 12th, 2012 at 4:26:21 PM
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Not enough information.

How fast does he paddle? Also, one assumes he paddles faster to catch up to his hat, so his upstream speed will not be the same.

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June 12th, 2012 at 4:31:15 PM
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I agree with Chester Dog: 3mph.

June 12th, 2012 at 4:33:55 PM
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damnit you are right. it must be 3 mph

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June 12th, 2012 at 4:49:34 PM
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I confirm I get 3 mph too. I have an easy solution and a hard solution...

If Joe paddles upstream and them downstream the same amount of time in both directions then his effort paddling in both directions an equal amount of time would cancel out. In other words, he would end up in the same place if he floated the whole time.

We know he paddled upstream for ten minutes. So, after paddling downstream for another ten minutes he would end up in the same spot as he would had he floated for 20 minutes. We also know from the problem that in this same 20 minutes the hat floated one mile. So if the hat travels a mile in 20 minutes then it would travel three miles in an hour, for a speed of 3 m.ph..

We know he paddled upstream for ten minutes. So, after paddling downstream for another ten minutes he would end up in the same spot as he would had he floated for 20 minutes. We also know from the problem that in this same 20 minutes the hat floated one mile. So if the hat travels a mile in 20 minutes then it would travel three miles in an hour, for a speed of 3 m.ph..

Let:

p=paddling speed in still water.

c=current of the water.

d=distance Joe paddled upstream since his hat fell out of the canoe.

Recall that distance = rate * time. Consider the unknown distance Joe went upstream since his hat fell out:

(1) d=(p-c)*(1/6)

This is because his net speed is the paddling speed less the current speed. We know he paddled without his hat upstream for 1/6 of an hour.

We’re also given that it took the same time for the hat to float downriver a mile as it took for Joe to travel d miles upstream and 1+d miles downstream. Let’s set up an equation, balancing for time.

(2) Time for hat to float one mile downstream = Time for Joe to travel d miles upstream + Time for Joe to travel 1+d times downstream.

Time for hat to float one mile downstream = one mile/rate of current = 1/c.

Time for Joe to travel d miles upstream = 1/6 (in hours), as provided in the question.

Time for Joe to travel 1+d times downstream = distance Joe traveled downstream / rate Joe traveled downstream = [1 + (1/6)*(p-c) ]/ (p+c).

The distance is the one mile the hat floated in the river + the (1/6)*(p-c) from equation (1). Joe’s speed going downstream is p+c, which is the sum of his paddling speed and current speed.

Now we’re ready to solve for equation (2)

1/c = (1/6) + [1 + (1/6)*(p-c) ]/ (p+c)

1/c = (1/6) + (6+p-c)/(6p+6c)

6/c = 1 + (6+p-c)/(p+c) Multiplying both sides by 6.

6/c = (p+c)/(p+c) + (6+p-c)/(p+c) Finding a common denominator

6/c = (6+2p)/(p+c)

6p + 6c = 6c + 2pc

6p = 2pc

3=c

So the current is 3 miles per hour.

p=paddling speed in still water.

c=current of the water.

d=distance Joe paddled upstream since his hat fell out of the canoe.

Recall that distance = rate * time. Consider the unknown distance Joe went upstream since his hat fell out:

(1) d=(p-c)*(1/6)

This is because his net speed is the paddling speed less the current speed. We know he paddled without his hat upstream for 1/6 of an hour.

We’re also given that it took the same time for the hat to float downriver a mile as it took for Joe to travel d miles upstream and 1+d miles downstream. Let’s set up an equation, balancing for time.

(2) Time for hat to float one mile downstream = Time for Joe to travel d miles upstream + Time for Joe to travel 1+d times downstream.

Time for hat to float one mile downstream = one mile/rate of current = 1/c.

Time for Joe to travel d miles upstream = 1/6 (in hours), as provided in the question.

Time for Joe to travel 1+d times downstream = distance Joe traveled downstream / rate Joe traveled downstream = [1 + (1/6)*(p-c) ]/ (p+c).

The distance is the one mile the hat floated in the river + the (1/6)*(p-c) from equation (1). Joe’s speed going downstream is p+c, which is the sum of his paddling speed and current speed.

Now we’re ready to solve for equation (2)

1/c = (1/6) + [1 + (1/6)*(p-c) ]/ (p+c)

1/c = (1/6) + (6+p-c)/(6p+6c)

6/c = 1 + (6+p-c)/(p+c) Multiplying both sides by 6.

6/c = (p+c)/(p+c) + (6+p-c)/(p+c) Finding a common denominator

6/c = (6+2p)/(p+c)

6p + 6c = 6c + 2pc

6p = 2pc

3=c

So the current is 3 miles per hour.

It's not whether you win or lose; it's whether or not you had a good bet.

June 12th, 2012 at 4:51:58 PM
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I like these problems, even if I end up looking ignorant for posting the wrong answer!

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June 12th, 2012 at 5:16:26 PM
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Let "X" equal the size of the hat.

Let "Y" be the pretty girl at the craps table who is wearing the "Algebra, Go Solve Your Own Proofs" tee-shirt.

Let "Z" be the size of your bankroll.

Take bankroll and stunning girl to dinner, drinks and post-prandial enjoyment of better things than people who paddle along not realizing they are hatless.

Let "Y" be the pretty girl at the craps table who is wearing the "Algebra, Go Solve Your Own Proofs" tee-shirt.

Let "Z" be the size of your bankroll.

Take bankroll and stunning girl to dinner, drinks and post-prandial enjoyment of better things than people who paddle along not realizing they are hatless.

June 12th, 2012 at 9:45:23 PM
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Quote:FleaStiffLet "Y" be the pretty girl at the craps table who is wearing the "Algebra, Go Solve Your Own Proofs" tee-shirt.

I'd rather go canoeing hatless, listening to "Men Without Hats," than suffer through dinner with any woman wearing that shirt. However, if she took it off, I could see setting aside a math problem for a few minutes.

It's not whether you win or lose; it's whether or not you had a good bet.