Anonymous asked in Entertainment & MusicJokes & Riddles ยท 8 years ago

Two complicated math riddles I don't know how to solve?

1. In front of you are several long fuses. You know they burn for exactly one hour after you light them at one end. The entire fuse does no necessarily burn at a constant speed. For example, it might take five minutes to burn through half the fuse and fifty-five minutes to burn the other half. With your lighter and using these fuses, how can you measure exactly three-quarters of an hour of time?

2. Diophantus's youth lasted 1/6 of his life. He had his first beard in the next 1/12 of his life. At the end of the following 1/7 Diophantus got married. Five years from then his son was born. His son lived exactly 1/12 of Diophantus's life. Diophantus died 4 years after the death of his son... How long did Diophantus live?

I would much rather you tell me how to do these than just giving me the answers. Thank you for any help!


His son lived 1/2 of Diophantus' life! My mistake.

2 Answers

  • 8 years ago
    Best Answer

    1. The key here is recognizing that you can burn a fuse at both ends and also burn two fuses at the same time. Burning a fuse at both ends means that it'll burn twice as fast. That means instead of burning one hour, by burning on both ends, it'll burn only half an hour.

    You can solve this problem by burning one fuse on one end, and another on both ends. When the one burning on both ends finishes, then you know that 30 minutes have passed. At this time, you know that the fuse only burning on one end will burn for exactly 30 more minutes. Light the other end and voila, it'll finish burning in 15 minutes instead of 30. That means you get exactly three-quarters of an hour when the second fuse finish burning.

    2. This is just algebra. The variable you're trying to solve is the number of years Diophantus lived. Let's call this X. The other variable is the number of years his son lived. Let's call this Y.

    You know that Y = 1/2 X.

    Then you can figure out the length of time the son lived that overlapped with X.

    You know the son is born 1/6X + 1/12X + 1/7X + 5 (which is 11/28X + 5)

    You know the son died 1/2X later.

    When the son is born, Diophantus has (17/28X - 5) years remaining in his life.

    You also know that the remainder of his life is really just his son's life plus 4 years.

    So you have two expressions basically describing the remainder of Diophantus's life. They are equal because they describe the same thing. Therefore:

    17/28X - 5 = Y + 4

    But you know that Y = 1/2X, so now you have two equations and two variables and you can solve for X. So 51/84X = 42/84X + 9

    X would be 84.

  • 8 years ago

    1. At the same time, light one fuse at both ends and one fuse from one end.

    It will take the first fuse 30 minutes to completely burn. Once this has burned, light the other fuse at the other end. This will have 30 minutes left, but burning from both ends it will take 15 minutes.

    30 minutes for the first fuse to completely burn from both ends, + 15 minutes for the 2nd fuse to finish burning from both ends = 45 minutes

    This one just takes some insight as to how to get a total of 45 minutes (3/4 of an hour) by being able to burn a fuse from both ends, which will take half as long

    2. let x be Diophantus' lifetime

    1/6 x + 1/12 x + 1/7 x + 5 + 1/2 x + 4 = x

    common denominator between 6 , 12 , and 7 is 84 (this is a big hint!)

    14x / 84 + 7x / 84 + 12x / 84 + 5 + 42x / 84 + 4 = x

    75x / 84 + 9 = x

    9 = 9x / 84

    x = 84

    so his youth lasted 84 / 6 = 14 years

    first beard 84 / 12 = 7 years

    next 84/7 = 12 years

    5 years

    son lived 84 / 2 = 42 years

    4 years

    the total of these is 84 years, Diophantus' lifetime

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