Question 1142438
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This is, in essence, a mixture problem.  You are mixing tickets at $6 each and tickets at $8 each to get tickets averaging some price in between.<br>
Here is a solution method without algebra which, if you understand it, will get you to the answer to mixture problems like this much faster than the usual algebraic method.<br>
(1) The average ticket price in this example is $180/24 = $7.50.
(2) Use any of a number of different methods to determine that $7.50 is 3/4 of the way from $6 to $8. (perhaps picture the three numbers on a number line...?)
(3) That means 3/4 of the tickets must be the more expensive $8 tickets.<br>
ANSWER: 3/4 of the 24 tickets, or 18 tickets, were adult tickets; the other 1/4, or 6 tickets, were children's tickets.