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For a certain play performance, adults' tickets were sold for $12 each and children's tickets were sold for $8 each. How many children's tickets were sold for the performance?
Let
Number of adults tickets = A
Number of children tickets = C

(1) The total revenue from the sale of adults’ and children’s tickets for the performance was $5,040.

12A + 8C = 5040 and for A = 2, C = 627 and so on...
Since LCM (12,8) = 24, there are 209 combinations possible between number of tickets of adults and children. Hence

INSUFFICIENT.

(2) The number of adults’ tickets sold for the performance was 1/3 the total number of adults’ and children’s tickets sold for the performance.
\(A = \frac{1}{3}(A + C)\)
2A = C

Since nothing about total amount of tickets price is given, many possibilities exist. Hence

INSUFFICIENT.

Together 1 and 2
6C + 8C = 5040
C = 360 and A = 180

SUFFICIENT.

Answer C

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unraveled
For a certain play performance, adults' tickets were sold for $12 each and children's tickets were sold for $8 each. How many children's tickets were sold for the performance?
Let
Number of adults tickets = A
Number of children tickets = C

(1) The total revenue from the sale of adults’ and children’s tickets for the performance was $5,040.

12A + 8C = 5040 and for A = 2, C = 627 and so on...
Since LCM (12,8) = 24, there are 209 combinations possible between number of tickets of adults and children. Hence

INSUFFICIENT.

(2) The number of adults’ tickets sold for the performance was 1/3 the total number of adults’ and children’s tickets sold for the performance.
\(A = \frac{1}{3}(A + C)\)
2A = C

Since nothing about total amount of tickets price is given, many possibilities exist. Hence

INSUFFICIENT.

Together 1 and 2
6C + 8C = 5040
C = 360 and A = 180

SUFFICIENT.

Answer C
­Hi,
I did not understand the '209' combinations bit. Can you please help me understand?
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