Question 1117544
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By algebra....<br>
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).<br>
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:<br>
{{{35(x)+15(50-x) = 1350}}}<br>
I'll let you do the simple algebra from there to find the answer.<br>
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.<br>
So there were 30 advance tickets and 20 same-day tickets.<br>
Check: 30(35)+20(15) = 1050+300 = 1350