SOLUTION: Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost $35 and same-day tickets cost $15 . For one performance, there were 50

Question 1117544: Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost
$35
and same-day tickets cost
$15
. For one performance, there were
50
tickets sold in all, and the total amount paid for them was
$1350
.How many tickets of each type were sold?

Found 2 solutions by josmiceli, greenestamps:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = number of advance tickets sold
Let = number of same-day tickets sold
---------------------------------------------
(1)
(2)
-------------------------------
(1)
Subtract (1) from (2)
(2)
(1)
-------------------------------

and
(1)
(1)
----------------------------
30 advance tickets were sold
20 same-day tickets were sold

Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

By algebra....

Let x be the number of advance tickets sold.
Then since the total number of tickets was 50, the number of same-day tickets is (50-x).

Then write and solve the equation that says x tickets at $35 each plus (50-x) tickets at $15 each makes a total of $1350:

I'll let you do the simple algebra from there to find the answer.

By logical reasoning and a bit of mental arithmetic....
If all 50 tickets had been same-day tickets, the total cost would have been 50*15 = 750; the actual total was 1350-750=600 more than that.
The difference between the cost of an advance ticket and a same-day ticket is 35-15=20.
To make up the "extra" 600, the number of advance tickets had to be 600/20 = 30.

So there were 30 advance tickets and 20 same-day tickets.

Check: 30(35)+20(15) = 1050+300 = 1350
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