BTGmoderatorDC wrote:A movie theater sold 120 tickets to the matinee showing of a popular children's movie, and 150 tickets to the evening showing. The theater sold the same number of adult tickets for each show, but for the evening show was just 20 children's tickets short of selling twice as many children's tickets as it did for the matinee. How many children's tickets were sold to the evening show?
A. 50
B. 60
C. 70
D. 80
E. 90
OA D
Source: Veritas Prep
The theater sold the same number of adult tickets for each show
Let A = # of adult tickets sold for MATINEE show
So,
A = # of adult tickets sold for EVENING show
The evening show was just 20 children's tickets short of selling twice as many children's tickets as it did for the matinee
In other words, if they had sold 20 extra children's tickets for the evening show, then the number of children's tickets for the evening show would have been TWICE the number of children's tickets for the matinee show
So, we can write: (# of EVENING show children's tickets) + 20 = 2(# of MATINEE show children's tickets)
Another way to write this is: (# of EVENING show children's tickets) = 2(# of MATINEE show children's tickets) - 20
Let
C = # of children's tickets sold for MATINEE show
So,
2C - 20 = # of children's tickets sold for EVENING show
A movie theater sold 120 tickets to the matinee showing of a popular children's movie, and 150 tickets to the evening showing
We can write the following:
A + C = 120
A + (2C - 20) = 150
Rewrite as:
A + 2C = 170
A + C = 120
Subtract
bottom equation from
top equation to get: C = 50
So, 50 children's tickets were sold for the MATINEE show.
How many children's tickets were sold to the evening show?
We already know that
2C - 20 = # of children's tickets sold for EVENING show
Since C = 50, we can replace C with 50 to get:
2(50) - 20 = # of children's tickets sold for EVENING show
Evaluate:
80 = # of children's tickets sold for EVENING show
Answer: D
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
