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 ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> 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       Log On


   

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) About Me  (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) About Me  (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) About Me  (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