Question 1155419
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Here are two (very similar) quick and easy non-algebraic methods for solving the problem (if an algebraic solution is not required!).<br>
Method 1....<br>
(1) 320 tickets all at $10 would bring in $3200; the actual total was $4200, which is $1000 greater.
(2) Each regular ticket costs $5 more than a student ticket.  The number of regular tickets sold, in order to make that additional $1000, is 1000/5 = 200.<br>
ANSWER: 200 regular tickets; 120 student tickets.<br>
CHECK: 200(15)+120(10) = 3000+1200 = 4200<br>
Method 2....<br>
(1) 320 tickets at $10 would bring in $3200; 320 all at $15 would bring in $4800.
(2) The actual total of $4200 is 1000/1600 = 5/8 of the way from $3200 to $4800, so 5/8 of the tickets were the more expensive tickets.<br>
ANSWER: 5/8 of the 320 tickets, or 200, were the $15 tickets; the other 120 were the less expensive.<br>