how do i solve this math problem?

A theater has a seating capacity of 597 and charges $2 for children, $4 for students, and $6 for adults. At a certain screening with full attendance, there were half as many adults as children and students combined. The receipts totaled $2406. How many children attended the show?

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  • 9 years ago

    Hey,

    Let me help you in solving this problem.

    Let there be 'x' children and 'y' students

    Total seating capacity = 597

    So number of adults = 597 - ( no of children + no. of students)

    = 597 - (x + y)

    Also we are given that number of adults = (x + y)/2

    Or, (x + y)/2 = 597 - (x + y)

    => On solving this equation we can get that:

    x + y = 398

    So number of adults = (x + y)/2 = 199

    Also we are given that:

    2x + 4y + 6(199) = 2406

    2x + 4y + 1194 = 2406

    2x + 4y = 2406 - 1194 = 1012

    Or by simplifying: x + 2y = 506

    Also from the earlier equations we know:

    x + y = 398

    So we can solve the given two equation to find the value of 'x' or the number of children:

    x + 2y = 506

    x + y = 398

    --------------------

    - - - SUBTRACT

    ----------------------

    0 + y = 108

    So y = 108

    Substitute 108 for y in x + y = 398 to find x:

    Or, x = 398 - 108

    = 290

    So, 290 children attended the show.

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  • 9 years ago

    First consider the attendance: 597 = C+S+A

    2A = C+S, since there are 1/2 as many adults...

    So, 597 = C+S+A = 2A+A = 3A. Hence A = 199, and C+S = 398 which implies C = 398-S

    Looking at cost, 2406 = 2C+4S+6A = 2C+4S+6*(199) = 2C+4S+1194.

    Hence, 1212 = 2C+4S.

    Since C = 398-S from above, we have 1212 = 2*(398-S)+4S = 796-2S+4S = 796+2S.

    That's a pretty darn good start, so I'm going to let you finish it. Good luck.

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