Answer:
80 adult tickets were sold
Explanation:
If you subtract the extra 20 adult tickets from ticket numbers and revenue, you get ...
equal numbers of $4 and $8 tickets were sold for a total of $720 collected
Since that is some number of ticket pairs worth $12 per pair, there must have been ...
$720/12 = 60 pairs of $4 and $8 tickets
The number of adult tickets is 20 more than this, so is
60 + 20 = 80 . . . . adult tickets sold
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If you feel better with an equation, rather than numerical reasoning, you can let n represent the number of adult tickets. Then n-20 is the number of student tickets and total revenue is ...
8n +4(n-20) = 880
12n -80 = 880 . . . . . simplify
12n = 960 . . . . . . . . . add 80
n = 80 . . . . . . . . . . . . divide by 12
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Comment on the different solutions
Using n for the number of adult tickets, we effectively chose a scenario where we added 20 student tickets to bring the total to $960 for some number of $12 pairs of tickets. That number is 80, the number of adult tickets.
In the "word solution" given at first, we effectively solved for the number of student tickets, then added 20 to get the number of adult tickets. The corresponding equation would be ...
4s +8(s+20) = 880 . . . s = number of student tickets sold
12s = 720 . . . . . . . . . subtract 160, the price of the 20 extra adult tickets