Help please !!! At a local Brownsville play production, 420 tickets were sold. The ticket prices varied on the seating arrangements an

Question

Help please !!! At a local Brownsville play production, 420 tickets were sold. The ticket prices varied on the seating
arrangements and cost $8, $10, or $12. The total income from ticket sales reached $3920. If the
combined number of $8 and $10 priced tickets sold was 5 times the number of $12 tickets sold, how
many tickets of each type were sold?

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bonexptip 2 months 2021-08-02T07:55:41+00:00 1 Answers 3 views 0

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    2021-08-02T07:56:46+00:00

    Answer:

    n = 12$    x = 8$   y = 10$     where n, x, and y are number of tickets

    12 n + 8 x + 10 y = 3920     and n + x + y = 420

    12n + 8 (x + y) + 2 y = 3920

    12 n + 8 (5 n) + 2 y = 3920        since 5 (x + y) = n

    52 n + 2 y = 3920   or  y = 1960 – 26 n

    Also, n + x + y = 420   or n + 5 n = 420   since x + y = 5 n

    n = 70    so 70 of the $12 were sold

    And since y = 1960 – 26 n     we have y = 140 tickets

    Now 12 * 70 + 8 x + 140 * 10 = 3920

    This gives x = 210 tickets

    Check:   210 + 140 + 70 = 420 tickets

    Also, 12 * 70 + 210 * 8 + 140 * 10 = 3920

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