The money collected at a school basketball game was 2,265 from 543 paid admissions. If adult tickets sold at $5.00 each and student tickets sold for $3.50 each, how many tickets of each kind were sold?
Ok, well you need a system of equations.
5x+3.50y=2265
x+y=543
x=543-y
5(543-y)+3.50y=2265
solve for y and substitute in other equation to solve for x
can you take it from here?
I can thank you. For my answer I got 243 adults and 300 students.
Very Good!
To solve this problem, we need to set up a system of equations. Let's assign variables to the unknowns:
Let's call the number of adult tickets sold "a".
Let's call the number of student tickets sold "s".
We can now write two equations based on the given information:
1. The total amount of money collected from adult tickets is $5 multiplied by the number of adult tickets sold: 5a
2. Similarly, the total amount of money collected from student tickets is $3.50 multiplied by the number of student tickets sold: 3.50s
The problem states that the total money collected from all tickets is $2,265. Therefore, we can form the equation:
5a + 3.50s = 2,265 ... (Equation 1)
The problem also states that there were 543 paid admissions in total. Since each admission corresponds to one ticket sold, we can form another equation:
a + s = 543 ... (Equation 2)
We now have a system of two equations with two variables. We can use substitution or elimination methods to solve.
Let's use the substitution method:
From Equation 2, we can isolate a:
a = 543 - s
Now substitute this into Equation 1:
5(543 - s) + 3.50s = 2,265
Distribute the 5:
2,715 - 5s + 3.50s = 2,265
Combine like terms:
-1.50s = -450
Divide both sides by -1.50:
s = 300
Now substitute the value of s back into Equation 2 to solve for a:
a + 300 = 543
a = 543 - 300
a = 243
Therefore, 243 adult tickets were sold, and 300 student tickets were sold.