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For a certain play performance, adults' tickets were sold for $12 each

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Bunuel
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For a certain play performance, adults' tickets were sold for $12 each [#permalink] New post   22 Oct 2019, 08:48
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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?


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

(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.
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Re: For a certain play performance, adults' tickets were sold for $12 each [#permalink] New post   22 Oct 2019, 09:09
Bunuel wrote:
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?


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

(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.


given
12x+8y
#1
The total revenue from the sale of adults’ and children’s tickets for the performance was $5,040.
12x+8y=5040
3x+2y=1260
x&y unknown
insufficent
#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.
x=1/3(x+y)
2x=y
again insufficient
from 1&2
7x=1260
x=180
y=360
sufficient
IMO C
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Re: For a certain play performance, adults' tickets were sold for $12 each [#permalink] New post   22 Oct 2019, 11:29
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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Re: For a certain play performance, adults' tickets were sold for $12 each [#permalink] New post   23 Oct 2019, 03:14
lnm87 wrote:
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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Re: For a certain play performance, adults' tickets were sold for $12 each [#permalink] New post   27 Jul 2022, 00:09
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Re: For a certain play performance, adults' tickets were sold for $12 each [#permalink]
27 Jul 2022, 00:09
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