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Sim (Guest)
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Sep-23-01, 07:41 PM (EST) |
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""cool" problem"
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I need some help finding a cool math problem. It's a problem that have a obvious answer, but it isn't the correct answer. It's something you need to think about and then get the answer...HELP! thanx! |
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Whymme
guest
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Oct-09-01, 09:40 PM (EST) |
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5. "RE: "cool" problem"
In response to message #3
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>>You drove half the way from A to be at the speed of 20 mi/h. Then you realized that at this rate you'd be late to the evening services. So you sped up an covered the other half of the way at 30 mi/h. What was your average speed? >>25
WRONG! And exactly the kind of trap that is set by this puzzle.
Imagine that the distance from A to B is 120 miles. You drive the first sixty miles at a speed of 20 m/h, the second half at 30 m/h. This means that in total you drive 3+2, or 5 hours. Dividing 120 by 5 gives you the answer that you have an average speed of 24 m/h. Not really a problem: Harvey always takes the train to work. Normally, the ride takes an hour and 20 minutes, but last Wednesday, Harvey made the trip in only 80 minutes. Can you explain why?
Whymme |
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WhyNotMe
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Oct-11-01, 07:59 AM (EST) |
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6. "RE: "cool" problem"
In response to message #5
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How about this one on average speed? I came from city A to city B at a constant speed of 50 mph. Now I am going back from city B to city A. At what speed should I be if I want the average speed to be 100 mph for the round trip? |
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Bryan Wagstaff
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Oct-06-01, 05:35 AM (EST) |
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2. "RE:"
In response to message #0
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This isn't really a 'problem', more of a statement that can be proved in a number theory class. It looks cool, though: e ^ (pi * i) + 1 = 0. It uses several natually occuring constants (e, pi, i), and 1,0. Proof of this is a little beyond an e-mail, but you can plug it into your fancy calculator and see that it is actually true. bryan. |
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WhyNotMe
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Oct-11-01, 07:59 AM (EST) |
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7. "RE: "cool" problem"
In response to message #0
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A problem posed by Richard I. Hess in the Pi Mu Epsilon Journal. On a 1x1x2 rectangular box, "distance" is defined as the shortest straight line connecting two points after the box is opened and flattened out. 1. If you start from one corner of the box, what is the longest distance you can cover on the box? (It is not the square root of 8, i.e. the corner on the "opposite" side.) 2. What is the longest distance you can cover on the box if you can start anywhere on the box? (It is not 3.) |
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WhyNotMe
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Oct-11-01, 00:09 AM (EST) |
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11. "RE: "cool" problem"
In response to message #9
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Answer 1: sqrt(2.75^2+0.75^2)=2.85044 {a^b means a to the b power} Answer 2: sqrt(16-4 sqrt(3))=3.01197For years, the answer to part one was thought to be the "opposite" corner. The Japanese mathematician Kotani found the point is actually on the 1x1 square at 1/4 away from the "opposite" corner. Part two is trickier. Check out the Pi Mu Epsilon Journal, Fall 1999 issue, pp 58-62 for full detail. |
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Robyn
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Dec-17-01, 08:58 PM (EST) |
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13. "RE: "cool" problem"
In response to message #0
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How much faster should one go to get there twenty percent sooner? The answer? Five times faster.e.g. It takes you twenty minutes to walk to a friends house. Your walking pace is 2 mph. So if you left at 5:00 P.M., you would arrive at 5:20 P.M. . If you wanted to arrive 20% sooner, you'd have to arrive at 5:04. Assuming you left at 5:00, your speed would be 10 mph. A brisk walking pace indeed! What this shows is that the relationship between how fast you go and how soon you get there is reciprical. |
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alexb
Charter Member
792 posts |
Dec-17-01, 09:03 PM (EST) |
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14. "RE: "cool" problem"
In response to message #13
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>How much faster should one go to get there twenty percent >sooner? Meaning, I gather, in twenty percent less time, right? >The answer? Five times faster. Hmm. >e.g. >It takes you twenty minutes to walk to a friends house. Your >walking >pace is 2 mph. So if you left at 5:00 P.M., you would arrive >at 5:20 P.M. . >If you wanted to arrive 20% sooner, you'd have to arrive at >5:04. 20% of 20 min is 4 min. I.e. one should spend 4 mintues less on the road and arrive at 5:16 P.M. |
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NJZ
Member since Mar-23-02
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Aug-16-02, 06:04 PM (EST) |
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16. "RE:"
In response to message #0
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Along those same lines is one of my favorit's - determine which step in the following proof leads to the fallacy that 1 = 2 (using beginning algebra): 1) Let a = b 2) a^2 = ab 3) a^2 + a^2 = a^2 + ab 4) 2a^2 = a^2 + ab 5) 2a^2 - 2ab = a^2 + ab - 2ab 6) 2a^2 - 2ab = a^2 - ab 7) 2(a^2 - ab) = 1(a^2 - ab) 8) 2 = 1 Another false proof that 2 = 1 (using complex numbers): 1) -1/1 = 1/-1 2) (-1/1)^.5 = (1/-1)^.5 3) (-1^.5)/(1^.5) = (1^.5)/(-1^.5) 4) i/1 = 1/i 5) i / 2 = 1 / (2i) 6) i/2 + 3/(2i) = 1/(2i) + 3/(2i) 7) i (i/2 + 3/(2i) ) = i ( 1/(2i) + 3/(2i) ) 8) (i^2)/2 + (3i)/(2i) = i/(2i) + (3i)/(2i) 9) (-1)/2 + 3/2 = 1/2 + 3/2 10 1 = 2 To give credit where credit is due, these and other classic fallacies can be found at https://www.math.toronto.edu/mathnet/falseProofs/fallacies.html. NJZ |
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