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805+ (Hard)|   Math Related|         
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gullyboy09
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Tickets for next month's production of The Sea Gull at the local commu 11 Jul 2026, 01:10
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No doubt why you are an expert! Most importantly, the technique you used is so efficient that many questions can be easily doable now. I was trying to solve 4 equations with possible values of A and C. I knew there must be some easy approach but couldn't think of it. Thanks!
KarishmaB
the sale of 33 tickets generated a total revenue of $323,

From 33 tickets revenue is $323 which means that the average revenue per ticket is 9.something.
Hence, cost C should be 6/8 and cost A should be 10/11/12.

Next, to generate a revenue of $323, an odd amount, we need one price to be odd. Except 11, all other 4 options are even.
Hence A must be $11.
ANSWER

Next, assume all 33 tickets were for $11. The revenue then would be $363. But the actual revenue is $40 less (it is $323) because some of these tickets were cheaper.
If some of these tickets were for $6 i.e. $5 less than $11, then it would explain the reduced revenue of $40 (8 tickets were cheaper).
But if some of these tickets were for $8 i.e. $3 less than $11, then we cannot reduce the revenue by $40. We will not get an integer number of tickets and hence C cannot be $8.
Hence C is $6.
ANSWER

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