Quote:bigfoot66I like these problems, even if I end up looking ignorant for posting the wrong answer!

You mean "talking through your hat"? LOL

Quote:WizardJoe 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."Have a nice day.

Quote:WizardI confirm I get 3 mph too. I have an easy solution and a hard solution...

I still maintain the original question did not contain enough information.

v the speed of the current,

and c the speed of the canoe without regard to the current

The hat floats 1 mile in (1/6 + t) hours, so v = 1 / (1/6 + t).

Joe travels at speed (v - c) for 1/6 hours, then speed (v + c) for t hours; his total distance is 1 mile.

1/6 (v - c) + t (v + c) = 1

(1/6 + t) v + (t - 1/6) c = 1

We know v = 1 / (1/6 + t), so (1/6 + t) v = 1:

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

(t - 1/6) c = 0

This means that c = 0 or t = 1/6

If c = 0, then Joe would have been alongside the hat the entire time, and not "caught up to it"

Therefore, t = 1/6, and v = 1 / (1/6 + t) = 1 / (1/3) = 3 mph.

Quote:ewjones080I hate these f***ing problems. They're my achilles heal in mathematics.

When I was an undergraduate, I came home one weekend, and my mother told me that there had been a big flap. My younger brother in elementary school had missed a question on a math test and didn't understand why his answer was wrong. He gave the problem to our father who shared it with the group of engineers he worked with. They agreed with my brother's answer, but the teacher still disagreed. I read the problem and immediately recognized why there was disagreement.

Of course, I don't remember the exact problem, but it went something like this:

Quote:A man is traveling by train. First, he takes an express train for 60 miles, with that segment requiring one hour. He switches to a local train and travels 20 miles farther, with the second segment requiring an additional 40 minutes.

(a) What was his average speed for the trip?

(b) If he had ridden the express train for 10 more miles before he changed trains, what would have been his average speed?

Neglect any time involved in changing from one train to the other and treat each train as if it travels at a constant speed.

You may answer (a) and (b), but my additional question is this: Why was there such disagreement on the answer to part (b)? (No, it was not just someone's incompetence at arithmetic.)

Quote:DocA man is traveling by train. First, he takes an express train for 60 miles, with that segment requiring one hour. He switches to a local train and travels 20 miles farther, with the second segment requiring an additional 40 minutes.

(a) What was his average speed for the trip?

(b) If he had ridden the express train for 10 more miles before he changed trains, what would have been his average speed?

A) 48 MPH

B) 53.333... MPH

I'll wait for more answers to pour in, but off hand I don't see any justification for any other answer to B.

It takes 70 minutes to travel the first 70 minutes, and 20 minutes to travel the last 10. So, overall it took 90 minutes to travel 80 miles. That is a speed of 80/90 =0.888... miles per minute, or 53.333 miles per hour.

By the way, I like the picture in Aye's spoiler three posts above. Definitely a good canoeing hat.

Quote:WizardI'll wait for more answers to pour in, but off hand I don't see any justification for any other answer to B.

Other than a minor typo in your spoiler, I'd go along with that answer and solution. At least for the problem as you interpreted it.

Is there another way to interpret the question and get a different answer?