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Math Help - Can this be solved with this info?

  1. #1
    Senior Member Twig's Avatar
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    Can this be solved with this info?

    Hi!

    I read this problem in yesterday and I bothers me that I cant put up an equation on how to solve it.

    A bunch of people were going to ride a rollercoaster.
    A childrens ticket was $15, and an adult ticket was $22.
    Tickets were sold for a total of $349.

    How many children and adults were there?
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  2. #2
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    Let x be the number of children and y be the number of adults. Then
    15x+22y=349
    Since 349\equiv19\mod22, we must require that 15x\equiv19\mod22. Note that 15\times3\equiv1\mod22, so x\equiv3\times19\equiv13\mod22.
    Since 0\leq x\leq23, we must require x=13.
    So x=13 and y=7.
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  3. #3
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    Hello, Twig!

    This can be solved if you're familiar with Modulo Arithmetic.
    But I'll show a primitive algebraic approach.


    A bunch of people were going to ride a rollercoaster.
    A childrens ticket was $15, and an adult ticket was $22.
    Tickets were sold for a total of $349.
    How many children and adults were there?
    Let C = number of children.
    Let A = number of adults.
    Note that A\text{ and }C are nonnegative integers.

    The C children's tickets were $15 each; they cost a total of 15C dollars.
    The A adult tickets were $22 each; they cost a total of 22A dollars.

    The total ticket sales was $349: . 15C + 22A \:=\:349

    Solve for A\!:\;\;A \:=\:\frac{349-15C}{22}

    Then: . A \;=\;\frac{(330 + 19) + (-22C + 7C)}{22} \;=\;\frac{(330-22c) + (19 + 7C)}{22}

    . . A \;=\;\frac{330-22C}{22} + \frac{19C + 7C}{22} \quad\Rightarrow\quad A\;=\;15 - C + \frac{19+7C}{22} .[1]


    Since A is an integer, 19+7C must be a multiple of 22.
    . . 19 + 7C \:=\:22b\;\text{ for some integer }b.

    Solve for C\!:\;\;C \:=\:\frac{22b-19}{7} \;=\;\frac{(21b + b) +(-21 + 2)}{7}

    . . C \;=\;\frac{(21b - 21) + (b + 2)}{7}\quad\Rightarrow\quad C \;=\;3b - 3 + \frac{b+2}{7} .[2]


    Since C is an integer, b+2 must be a multiple of 7.
    The least value of b is: . b = 5

    Substitute into [2]: . C \;=\;3(5) - 3 \frac{5+2}{7}\quad\Rightarrow\quad\boxed{ C \,=\,13}

    Substitute into [1]: . A \;=\;15 - 13 + \frac{19+7(13)}{22} \quad\Rightarrow\quad\boxed{ A \,=\,7}


    There were 13 children and 7 adults.

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  4. #4
    Senior Member Twig's Avatar
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    hey

    Hi!

    I am not familiar with the Modulo Arithemtic.

    I didnt not understand your explanation. I followed up till the "Solve for A".
    Then I dont what you are doing.

    PS: I took this problem from my little brothers math book and he is 15....I cant imagine that they would need this complex math.
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  5. #5
    Senior Member Twig's Avatar
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    I learned a method of solving these types of problems today.
    Called Diofantic eqauations.

    There seem to be many different approaches.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Twig View Post
    I learned a method of solving these types of problems today.
    Called Diofantic eqauations.

    There seem to be many different approaches.
    Diophantine, in English anyway.

    RonL
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  7. #7
    Senior Member Twig's Avatar
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    hi

    Can anyone please explain how "mathsucks" solution works?

    Thanks
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