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22 Oct 2019, 08:48
Kudos
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C
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Difficulty:
5%
(low)
Question Stats:
88% (01:19) correct
12%
(01:34)
wrong
based on 321
sessions
History
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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.
(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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Archit3110
Major Poster
22 Oct 2019, 09:09
Kudos
Bookmarks
Bunuel
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
22 Oct 2019, 11:29
Kudos
Bookmarks
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
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
