Hat in the River Problem
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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."
edit: Damn thats not the right answer, hold on
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..
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.
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.
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.
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?
Quote: DocWhen 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:
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.)
I hope nobody will mind that I say that bigfoot's interpreation is that the passenger got on the express train at a stop 10 miles further away from his destination, and changed trains at the usual station. That is certainly not how I interpreted the problem. However, I think if the question is ambiguous then any fault should lie (or is it lay) with the person framing the question. That said, if I were the teacher I would accept bigfoot's solution as well.
Quote: DocSo what's the "other" answer to (b)?
49.090909... MPH.
As a grad student/TA and later as the instructor, my experience was often that, "The answer book is rarely right." I formed the opinion that college solution manuals were often prepared by the professor/author's student assistant who had to prepare hundreds/thousands of solutions in as little time as possible at the lowest rate paid on campus. Result: lots of errors.
Quote: DocAs a grad student/TA and later as the instructor, my experience was often that, "The answer book is rarely right." I formed the opinion that college solution manuals were often prepared by the professor/author's student assistant who had to prepare hundreds/thousands of solutions in as little time as possible at the lowest rate paid on campus. Result: lots of errors.
I wish I could get back all the time I spent beating my head against the wall because of incorrect answers in the "back of the book." In my old high school math books the error rate was quite high. You would think the publisher could pay at least two people to work through all the problems to make sure two heads come up with the same answer.
Back the problem, I think anyone familiar with taking commuter trains would interpret it my way, especially in New York, where trains on the same route run at different speeds. It is also normal for people to go from point A to B over and over, but the time of the commute might change depending on the availability of express trains.
Quote: s2dbakerThe original post is deceptively complicated. Basically, it all boils down to: a hat floats in a river for 20 minutes and travels one mile. How many miles will it travel in an hour?
Except it takes a little figuring to know that it took 20 minutes for it to float that mile.
Quote: DocAs a grad student/TA and later as the instructor, my experience was often that, "The answer book is rarely right." I formed the opinion that college solution manuals were often prepared by the professor/author's student assistant who had to prepare hundreds/thousands of solutions in as little time as possible at the lowest rate paid on campus. Result: lots of errors.
I believe that Steve Wolfram, the creator of Mathematica, faced the same problem when checking his computer program against authoritative sources. His program would often give different answers .... and it turns out it was the sample inputs he had taken from recognized authorities were in the wrong.
