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TchrWill
Advanced Member

USA
79 Posts

Posted - 04/03/2013 :  12:08:31  Show Profile  Reply with Quote
Ah come on guys. There must be a few of you out there that really enjoy the challenge of a good math problem. My first challenging word problem was given to me by my supervisor at my first job as an aero-engineer. It hooked me and I have been collecting word problems ever since. Admittedly, most of them were acquired after I retired but the 84 year old brain still enjoys the challenge of any new ones. If you have any favorites that you might have encountered over the years, I would appreciate your sending them to me or posting them in this forum. Who knows, you might really get into it as I have for the past 23 retired years.


Just for laughs, I offer you the first word problem that I encountered 63 years ago followed by another one for you to tackle, if you desire.

One of the oldest and most popular problems in the field of recreational mathematics involves the plight of a fly traveling back and forth between two trains approaching one another on the same track. The usual initial reaction is that a solution is obtainable only with some level of higher mathematics. While it is true that the answer can be derived by summing an infinite geometric series, it is also derivable by means of two simple observations and calculations. The following is a typical version of the problem. The long and short solutions are given below.
Two trains are traveling toward one another on the same track. One train is traveling at a speed of 40mph while the other is traveling at a speed of 60mph. At the point in time when the two trains are 200 miles apart, a fly sitting on the nose of one train starts to fly down the track toward the other train. When the fly reaches the other train, it instantaneously turns around and starts flying back toward the first train. The fly continues in this manner, flying back and forth, until the two trains crash into one another and the fly is crushed. If, during this cyclic flight path of the fly, it travels at the constant speed of 80mph, how much distance will the fly have covered in his back and forth journey?


A boy, a girl, and a trained dog, go for a 10 mile walk. The boy and girl can walk at 2 MPH and the dog can trot at 4 MPH. They also have a bicycle which only one of them can use at a time (including the smart dog!). Ruff! When riding the bike, the boy and girl can each peddle at 12 MPH while the dog can pedal at 16 MPH. Ruff! They all leave at the same time and arrive at their destination "at the same time." Ruff! What is the shortest time in which all three can simultaneously complete the trip? Ruff Ruff! (Good luck!) [Hint-Don't bark up the wrong tree.]

Enjoy.
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Ultraglide
Advanced Member

Canada
299 Posts

Posted - 04/04/2013 :  14:30:00  Show Profile  Reply with Quote
Problem 1:
The distance between the trains is decreasing at 100 mph so they will meet in 2 hours. Since the fly is flying for two hours at 80 mph (a mighty fast fly) it will travel 160 miles.

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TchrWill
Advanced Member

USA
79 Posts

Posted - 04/04/2013 :  15:19:54  Show Profile  Reply with Quote
Right on the money ultraglide. I was reasonably certain that most, if not all, having read this would know the short answer. The guy that gave me the problem 63 years ago had given it to one of his professors who then went down the calculus path. It does fool some people.

Thanks for taking the time to respond.

Bill
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Ultraglide
Advanced Member

Canada
299 Posts

Posted - 04/04/2013 :  17:39:29  Show Profile  Reply with Quote
I have a couple of questions about the second problem.
Do they all cover exactly 10 miles and do any of them stop (other than to possibly mount or dismount the bicycle - which we can consider zero time)?
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Admin
Forum Admin

USA
635 Posts

Posted - 04/04/2013 :  17:56:08  Show Profile  Reply with Quote
quote:
Originally posted by TchrWill

Right on the money ultraglide. I was reasonably certain that most, if not all, having read this would know the short answer. The guy that gave me the problem 63 years ago had given it to one of his professors who then went down the calculus path. It does fool some people.

Thanks for taking the time to respond.

Bill



Thanks for posting these interesting problems, Bill!

All the best,
Gisele
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TchrWill
Advanced Member

USA
79 Posts

Posted - 04/04/2013 :  20:55:16  Show Profile  Reply with Quote
quote:
Originally posted by Ultraglide

I have a couple of questions about the second problem.
Do they all cover exactly 10 miles and do any of them stop (other than to possibly mount or dismount the bicycle - which we can consider zero time)?



Wow! It has been so long ago that I tackled this one that I can't be certain. As best as I can remember, without pawing over the solution again, any one of them can reverse direction in order to pick up another. This implies that the total distance traveled by anyone of them could be more than 10 miles. As implied, either or all could be traveling backwards at one point or another.


Thanks for taking a shot at it.

Bill
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someguy
Advanced Member

Canada
143 Posts

Posted - 04/21/2013 :  06:28:03  Show Profile  Reply with Quote
I haven't cracked this one yet. All I have done so far is compute the time taken for two different methods of getting everyone to the destination at the same time. One when they always move forward with the boy and girl riding the bike for an equal amount of time, and one where the dog takes the bike backwards sometimes. I have not convinced myself as to whether or not there is a better way.

Anybody have any ideas?
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