Question 1152852
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Here is a fast way to solve this and a large number of similar problems if a formal algebraic solution is not required:<br>
(1) 164 tickets all at $10 would cost $1640.
(2) The actual total cost of the tickets was $1945; that is $305 more than the figure from (1).
(3) The difference between the costs of the two kinds of tickets is $2.50.
(4) The number of more expensive tickets is the $305 total from (2), divided by the $2.50 difference from (3): 305/2.5 = 610/5 = 1220/10 = 122.<br>