Answered

Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

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?


Sagot :

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