Adult tickets for a play cost $13 and child tickets cost $3. If there were 20 people at a performance and the theater collected $200 from ticket sales, how many children attended the play?
Set up a system of equations to model the situation.
Let a be the number of adult tickets sold, and let c be the number of child tickets sold.
See what you get, and I'll help you.
a+c=20??
That's one of your equations, but a system of equations is two equations.
Find the other using the ticket pricing.
13a+3c=200??
Yes, very good. Now, we are trying to find the number of children who attended the play (c).
We have these equations:
a + c = 20
13a + 3c = 200
The easiest way to solve this is substitution. Solve the first equation for a, and then plug that into the second equation. You will then have all c's in the equation and will be able to solve it like any other equation.
a=20-c
13(20-c)+3c = 200
260 -13c +3c = 200
60=10c
6=c
The jurassic zoo charges $14 for each adult admission and $10 for each child . The total bill for the 100 people from a school trip was $1,0. How many adults and how many children went to the zoo?
the tickets for a dance recital cost $5.00 for adults and $2.00 for children. if the total number of tick sold was 295 and the total amount collected was 1,200, how many adult tickets were sold
To solve this problem, we can set up a system of equations. Let's denote the number of adult tickets sold as "A" and the number of child tickets sold as "C".
From the given information, we know that the cost of an adult ticket is $13 and the cost of a child ticket is $3. We also know that there were 20 people at the performance and the theater collected $200 from ticket sales.
The first equation represents the total number of people at the performance:
A + C = 20
The second equation represents the total amount collected from ticket sales:
13A + 3C = 200
Now, let's solve this system of equations to find the number of children who attended the play.
We can start by multiplying the first equation by 3 to eliminate C when we add the equations together:
3A + 3C = 60
Now we can add the equations:
13A + 3C + 3A + 3C = 200 + 60
16A + 6C = 260
Next, divide the entire equation by 2 to simplify it further:
8A + 3C = 130
Now we can solve this equation for C by performing isolation:
3C = 130 - 8A
C = (130 - 8A) / 3
Since C represents the number of child tickets sold, it needs to be an integer. We can test different values for A (the number of adult tickets sold) and see which one gives us an integer value for C.
Let's try different values for A, starting from 1 and incrementing by 1 until we find an integer value for C:
For A = 1, C = (130 - 8) / 3 = 122 / 3 = Not an integer
For A = 2, C = (130 - 16) / 3 = 114 / 3 = Not an integer
For A = 3, C = (130 - 24) / 3 = 106 / 3 = Not an integer
For A = 4, C = (130 - 32) / 3 = 98 / 3 = Not an integer
For A = 5, C = (130 - 40) / 3 = 90 / 3 = 30
By trying different values for A, we find that the number of adult tickets sold is 5 and the number of child tickets sold is 30. Therefore, 30 children attended the play.