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Thread: Puzzle involving multiple unknowns

  1. #1
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    Puzzle involving multiple unknowns

    Hey, I've got this puzzle that I'm trying to work out. It seems simple, but I can't seem to form a coherent equation to build in order to solve the puzzle.

    The puzzle is this.
    A group of friends are going on a flight, and are going to split the costs between them, so that the total comes to 6000 dollars. At the last minute, 3 of the friends can't make it, and everyone else has to pitch in an additional 100 dollars each to cover the costs. How much are the price of the tickets per person?

    Any help would be great. Thanks.
    Last edited by topsquark; Apr 7th 2018 at 07:27 AM. Reason: Tweeked dollar signs
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  2. #2
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    Re: Puzzle involving multiple unknowns

    Do you mean 3 people in the group didn't have enough money to pay for their tickets, and the rest people should add 100 for every of them?
    Last edited by Fox333; Apr 7th 2018 at 03:41 AM.
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    Re: Puzzle involving multiple unknowns

    There is not enough information to answer the question unambiguously. Indeed, there may be 5 people in the group, then one ticket costs 1200, and two people should pay extra 150. Or there may be 6 people in the group, then one ticket costs 1000, and three people should pay extra 100. Both answers don't contradict your conditions. Perhaps, they are other possible answers.
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  4. #4
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    Re: Puzzle involving multiple unknowns

    a group of $n$ friends: each pays $\displaystyle \frac{6000}{n}$

    $n-3$ friends: each pays $\displaystyle \frac{6000}{n-3}$

    the difference must be $100$ so

    $\displaystyle \frac{6000}{n-3}-\frac{6000}{n}=100$
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  5. #5
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    Re: Puzzle involving multiple unknowns

    Sorry, I misunderstood. Since the text was corrected, it's become clearer.

    Let x be the price of one ticket and y the amount of people in the group. Then, xy=6000. It's going to be our first equation. In fact, there were only y-3 people who paid, and each of them paid not only x dollars for him/herself, but also extra 100 dollars. Totally, they paid 6000 dollars. Thus, (x+100)(y-3)=6000
    So, we have a system of two equations.

    xy=6000
    (x+100)(y-3)=6000

    It has two possible solutions for x, but only one positive solution x=400
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