SOLUTION: Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one same-day ticket is $55. For one performance, 30 adv

Algebra ->  Equations -> SOLUTION: Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one same-day ticket is $55. For one performance, 30 adv      Log On


   

Question 1189913: Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one same-day ticket is $55. For one performance, 30 advance tickets and 40 same-day tickets were sold. The total amount paid for the tickets was $1900. What was the price of each kind of ticket?
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
a = number of advance tickets.
s = number of same day tickiets.

equations that need to be solved simultaneously are:
a + s = 55
30a + 40s = 1900

multiply both sides of the first equation by 30 and leave the second equatin as is to get:
30a + 30s = 1650
30a + 40s = 1900

subtract the first equation from the second to get:
10s = 250
solve for s to get:
s = 25
that makes a = 30 because a + s = 55

30a + 40s becomes 30 * 30 + 40 * 25 = 900 + 1000 = 1900
this confirms the values of a and s are correct.

the price of an advance ticket was 30 and the price of a same day ticket was 25.
that's your solution.

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.


            It can be solved using arithmetic only,  without  Algebra. 


Since the combined cost of one advance ticket and one some-day tickets is $55,

30 pairs of such combined tickets cost  30*55 = 16750 dollars.


It means that remaining 40-30 = 10 same day tickets cost 1900-1650 = 250  dollars.


Hence, one same-day ticket costs 250/10 = 25 dollars.


It means that each advanced ticket costs 55-25 = 30 dollars.

Solved  (mentally).