Central Middle School sold 50 tickets for one night of the school play. Student tickets sold for $2 each, and adult tickets sold for $3 each. They took in $135. How many of each type of ticket did they sell?
Answer (3)
Let's call the number of student tickets they sold s, and the number of adult tickets they sold a. The school sold 50 tickets in all, so a+s=50. For every adult ticket they sold, they made $3, and for every student ticket, they made $2. So the total amount of money they made is 3a+2s. The problem tells us they made $135, so 3a+2s=135. a + s = 50 3a + 2s = 135 This is a system of equations. We will proceed by changing the first equation, solving for a(we could solve for s instead, but I decided to solve for a). What this means is we will subtract s from both sides to get a alone. a + s = 50 a = 50 - s Now we know what a is(in terms of s, that is), so we can plug it into the second equation. 3a + 2s = 135 3(50 - s) + 2s = 135 (Remember to put the parentheses in!) 150 - 3s + 2s = 135 150 - s = 135 -s = -15 s = 15 This means 15 student tickets were sold. Plug this into one of the original equations to figure out how many adult tickets were sold: a + s = 50 a + 15 = 50 a = 35 15 student tickets were sold, and 35 adult tickets were sold.
To solve this problem, set up two equations based on the given information and substitute the value to get the number of tickets sold. ;
Central Middle School sold 15 student tickets and 35 adult tickets. This was found by setting up and solving a system of equations based on the total number of tickets sold and the total revenue. By substituting and simplifying the equations, we arrived at the correct numbers for each type of ticket.
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