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Thread: Riddle me this

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  1. #1
    ATK Member Vicious Horizon's Avatar
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    Re: Riddle me this

    I have two strings.

    I name one 'Fred', the other 'George'

    I take fred and light both ends at the same time.

    With the thread burning at either end, when the two ends eventually meet, then this means 30 minutes has passed. (The time is halved as it's being burnt on either end.)

    I take George, and at the same time as Fred, I burn only /one/ end of George.

    30 minutes passes and Fred is gone, I then light the other end of George.

    Now George is roughly 50% of its original size, meaning there is roughly 30 minutes left if I leave it as it is.

    Imagine if it's easier that George was always at half size.

    now I light the opposite end of George, and wait until the string has been burnt completely.

    This gives me 45 minutes, as 30 of these minutes was signified by 100% of Fred, and 50% of George (we assume).

    100% (1 Hour) + 50% (George's remaining time) is equal to 30 minutes.

    30 / 2 = 15, 30 + 15 = 45.

    15 = 50% of George.



    George may not be 50% in actual size, but 50% of the hour that George contains is gone. This isn't about length, this is about time.





    I think I won?
    Take that ya one-eyed, bomb-lobbin', cactus eatin', pot bellied, thug fat jigglin-chicken whoopin' big, back-stabbin lob-armed creepy spastic bloody, blind-eyed pashy little twitchy pickle-headed rocke- hoppin, potato-poppin' phony two-faced stealthy mutant bastard!


  2. #2
    Yellow and mellow Calneon's Avatar
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    Re: Riddle me this

    Quote Originally Posted by Vicious Horizon View Post
    I have two strings.

    I name one 'Fred', the other 'George'

    I take fred and light both ends at the same time.

    With the thread burning at either end, when the two ends eventually meet, then this means 30 minutes has passed. (The time is halved as it's being burnt on either end.)

    I take George, and at the same time as Fred, I burn only /one/ end of George.

    30 minutes passes and Fred is gone, I then light the other end of George.

    Now George is roughly 50% of its original size, meaning there is roughly 30 minutes left if I leave it as it is.

    Imagine if it's easier that George was always at half size.

    now I light the opposite end of George, and wait until the string has been burnt completely.

    This gives me 45 minutes, as 30 of these minutes was signified by 100% of Fred, and 50% of George (we assume).

    100% (1 Hour) + 50% (George's remaining time) is equal to 30 minutes.

    30 / 2 = 15, 30 + 15 = 45.

    15 = 50% of George.



    George may not be 50% in actual size, but 50% of the hour that George contains is gone. This isn't about length, this is about time.





    I think I won?
    You won the googling contest yeah.

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