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once in a lifetime
sun child



Registered: 02/12/15
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Re: mathematical brain teaser thread [Re: Sun King]
#21869557 - 06/28/15 05:21 PM (8 years, 6 months ago) |
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Aha, interesting. . So there is more push on it flying into the wind, than there is when it's flying with the wind, because the difference is so much less when it's flying in the same direction.
-------------------- Innocent, Oldfield & Hegerland Julia Delaney, Bothy Band Rasta Girl, Sister Carol Genesis, Jorma K I Wish You Peace, Lawrence Laughing Do Your Thing, Moondog large . . music garden . . veryall peace them hiStarhouse - main Time Traveler's Guide
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OrgoneConclusion
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Re: mathematical brain teaser thread [Re: Sun King]
#21869567 - 06/28/15 05:22 PM (8 years, 6 months ago) |
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OrgoneConclusion
Blue Fish Group



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Quote:
once in a lifetime said: Aha, interesting. . So there is more push on it flying into the wind, than there is when it's flying with the wind, because the difference is so much less when it's flying in the same direction.
Not exactly.
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once in a lifetime
sun child



Registered: 02/12/15
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What is the answer?
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Sun King



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If the plane crashes it doesn't matter.
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DividedQuantum
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Registered: 12/06/13
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Okay. Here is the official correct answer:
For exactly half of the circular path the wind boosts the airplane's ground speed and for the other half of the path the wind retards it. There is a temptation to suppose that these forces balance each other, with the result that the airplane's time for the entire circle is the same as if there were no wind. This is not the case, because the time during which the plane's speed is boosted is obviously shorter than the time during which it is retarded, with the result that the total time in the wind is greater than if there were no wind.
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once in a lifetime
sun child



Registered: 02/12/15
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Ah, very nice.
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OrgoneConclusion
Blue Fish Group



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A trip without wind will take time T. W = Wind
A trip with the wind will take time T - W. (linguistic not mathematical equation)
A trip against the wind will take time T + W.
One trip is shorter time-wise and the opposite trip is longer time-wise.
Now do the final conclusion.
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Sun King



Registered: 02/15/14
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A windowless room contains three identical light fixtures, each containing an identical light bulb or light globe. Each light is connected to one of three switches outside of the room. Each bulb is switched off at present. You are outside the room, and the door is closed. You have one , and only one, opportunity to flip any of the external switches. After this, you can go into the room and look at the lights, but you may not touch the switches again. How can you tell which switch goes to which light?
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OrgoneConclusion
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Re: mathematical brain teaser thread [Re: Sun King]
#21869627 - 06/28/15 05:33 PM (8 years, 6 months ago) |
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Quote:
Sun King said: If the plane crashes it doesn't matter.
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OrgoneConclusion
Blue Fish Group



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Re: mathematical brain teaser thread [Re: Sun King]
#21869631 - 06/28/15 05:35 PM (8 years, 6 months ago) |
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Quote:
Sun King said: A windowless room contains three identical light fixtures, each containing an identical light bulb or light globe. Each light is connected to one of three switches outside of the room. Each bulb is switched off at present. You are outside the room, and the door is closed. You have one , and only one, opportunity to flip any of the external switches. After this, you can go into the room and look at the lights, but you may not touch the switches again. How can you tell which switch goes to which light?
As a certified electrician I would get a wiring schematic.
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DividedQuantum
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PROBLEM 10
Quote:
Two professors were having drinks in the faculty club bar.
The English professor said, "The poet John William Burgon wrote thoroughly mediocre poems, yet he wrote one of the most marvelous lines in English poetry: 'A rose-red city half as old as time.'"
The mathematician thought for a moment or two, then raised his glass and recited this improved brainteaser:
A rose-red city half as old as Time. / One billion years ago the city's age / Was just two-fifths of what Time's age will be / A billion years from now. Can you compute / How old the crimson city is today?
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once in a lifetime
sun child



Registered: 02/12/15
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hehe
this is one of those algebra problems that I think I may have got wrong after finishing it..
here's what I did.
with x being the age of the city now, and 1 being 1 billion years,
x - 1 = 2/5(x + 1)
x - 1 = 2/5x + 2/5
x = 2/5x + 2/5 + 1
x - 2/5x = 1 2/5
3/5 x = 7/5
3x = 7
x = 7/3
or 2 and 1/3 billion years.
actually I checked it out and I am thinking it's right now.
2 1/3
1 1/3 --- 3 1/3
4/3 - 10/3
is 4 to 10,
and 2 to 5
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DividedQuantum
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Sorry, that is incorrect.
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Sun King



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2A = T A=T/2 A - 1B = 2/5(T+1B) (T/2 - 1B)5/2 = T+1B 1.25T - 2.5B = T +1B .25T = 3.5B T = 14B A = 7B
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DividedQuantum
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Re: mathematical brain teaser thread [Re: Sun King]
#21869996 - 06/28/15 06:59 PM (8 years, 6 months ago) |
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Quote:
Sun King said: 2A = T A=T/2 A - 1B = 2/5(T+1B) (T/2 - 1B)5/2 = T+1B 1.25T - 2.5B = T +1B .25T = 3.5B T = 14B A = 7B
  
Correct!
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DividedQuantum
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PROBLEM 11
Quote:
It is said that the great philosopher Immanuel Kant had such regular habits that the people of Königsberg would adjust their clocks when they saw him stroll past certain landmarks.
One evening Kant discovered that his clock had run down. Evidently his manservant, who had taken the day off, had forgotten to wind it. Kant did not reset the hands because his watch was being repaired and he had no way of knowing the correct time. He walked to the home of his friend Schmidt, a merchant who lived a mile or so away, glancing at the clock in Schmidt's hallway as he entered the house.
After visiting Schmidt for several hours Kant left and walked home along the same route. As always, he walked with a slow, steady gait that had not varied in twenty years. He had no notion of how long this return trip took, since he had never timed himself on this walk. Nevertheless, when Kant entered his house, he immediately set his clock correctly.
How did Kant know the correct time?
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deff
just love everyone



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did he wind his own clock prior to leaving his house, in which case he could tell how long he was gone for total based on the difference in time from when he left to when he got back. then he could subtract the time he spent at his friend's house from this (if he saw the clock at his friend's upon entering and leaving). dividing this difference by two would tell him how long his walk home took, then he could add this to the time of his friend's clock when he left, and set his own clock correctly ?
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DividedQuantum
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Re: mathematical brain teaser thread [Re: deff]
#21870252 - 06/28/15 07:49 PM (8 years, 6 months ago) |
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Quote:
deff said: did he wind his own clock prior to leaving his house, in which case he could tell how long he was gone for total based on the difference in time from when he left to when he got back. then he could subtract the time he spent at his friend's house from this (if he saw the clock at his friend's upon entering and leaving). dividing this difference by two would tell him how long his walk home took, then he could add this to the time of his friend's clock when he left, and set his own clock correctly ?
  
Correct!
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DividedQuantum
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PROBLEM 12
Quote:
Insert mathematical signs anywhere between the digits
1,2,3,4,5,6,7,8,9
to make the expression equal 100.
Please provide two solutions.
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