SOLUTION: Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost
$15
and same-day tickets cost
$20
. For one performance, there were
55
Algebra.Com
Question 1160489: Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost
$15
and same-day tickets cost
$20
. For one performance, there were
55
tickets sold in all, and the total amount paid for them was
$900
. How many tickets of each type were sold?
Found 3 solutions by solver91311, Boreal, greenestamps:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Solve the 2X2 system for 
and 
.
John
My calculator said it, I believe it, that settles it
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
A+S=55. number equation
15A+20S=900. money equation
-15A-15S=-825
5S=75
S=15 same day or $300
A=40 advance or $600
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
A quick mental solution method for problems like this, if a formal algebraic solution method is not required....
(1) 55 tickets all at $20 each would have brought in a total of 55*$20 = $1100; that is $200 more than the actual total.
(2) The difference between the cost of one of each type of ticket is $5.
(3) The $200 difference is because the number of $15 tickets was $200/$5 = 40.
So there were 40 advance tickets sold, which means 15 same-day tickets.
ANSWER: 40 advance tickets; 15 same-day tickets
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