Tickets for next month's production of The Sea Gull at the local commu

­Although official solution too talks of 11 tickets costing 101, I would take it with a pinch of salt.
All tickets are not of same value, so 33 tickets and 44 tickets would surely have different combination. Say 33 tickets consisted of 22 A and 11 C while 44 tickets consisted of 20A and 24 C. In that case the difference would be 13 of C and -2 of A. 
But if 44 tickets had more of each of A and C, say 28A and 16C, then you will have 6 of A and 5 of C adding to 11.

Anyways, I would make equations and then work on number properties:
Yesterday: Let there be x of A and 33-x of C => x*A + (33-x) * C = 323.
(1) Surely exactly one of xA or (33-x)C would be odd to get odd+even = odd.
If x is odd, then 33-x is even and vice versa. Surely one of A and C will be odd.
Noe, there is only one odd option, so one of A or C would be 11.
(2) If we take all to be of C value, we will get the least sum, of tickets => 33*10 = 330, which is greater than given sum 323.
Hence, C is less than 10. Only possibility 6 or 8. Also, A will be 11 for sure 

Just from yesterday sale, we know A=11 and C is either 8 or 6.
Although, you can substitute (11,8) and (11,6) here itself to check which is correct, let use today's sale to ease our calculations. After all, efforts have been made to add the details to the question.

Today:  Let there be y of A and 44-y of C => y*A + (44-y) * C = 424.

Subtract the two equations to get (y-x)*A + 11*C -(y-x)*C = 101
(A-C) (y-x) = 101-11C
(11,8): (11-8)(y-x) = 101-11*8 ........ 3(y-x) = 13 ...... y-x = \(\frac{13}{3}\), Not an integer.
(11,6): (11-6)(y-x) = 101-11*6 ........ 5(y-x) = 35 ...... y-x = 7, Possible. Correct

Solution: A=11 and C=6

If we go for typical algebra for this question it will take over 6 minutes I feel. After the first few basic algebra steps , we realise that 10, 11 or 12. And that A has to be odd number (which will lead to A being 11). Then on can substitute 6 and 8 in the basic algebra equation to get the answer as 6 . A tough question

@­HarshaBujji, there is a concern about your POE: x is the difference of adults between 2 days, x could be positive or negative

 

­Hi KarishmaB
I couldn't understand the yellow part in the above explanation. How can we conclude that C can be 6/8 and A can be 10/11/12?
One thing i could figure out while solving was that, ''11'' will be one of the values, because total = 323 is an odd no.

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When you sell 33 tickets and get $323 for it, your average price per ticket is 323/33 = 9.8 (approximately)

If your tickets are actually priced at 2 different price points, then the average must be somewhere between those price points.
Say you are selling tickets of $8 and $10. The average will be between 8 and 10. 
Say you are selling tickets of $6 and $8, the average will be between 6 and 8. 

If your average is 9.8, one ticket price will be less than 9.8 and the other will be more than 9.8.
So C will be 6 or 8 (only two options less than 9.8) and A will be 10/11/12 (three options more than 9.8).
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