child's fair ticket is $5 and adult fair ticket are $5. 1050 people attend the fair. fair collects $7350. how many of each type of ticket were sold?
sorry.
child is $5 and adult is $8
Try this:
x = no. of children attending the fair
y = no. of adults attending the fair
so, 5x + 8y = 7350 (eqn 1)
x + y = 1050 (eqn 2)
x = 1050 - y
Substt. the above in eqn 1,
8y + 5(1050-y) = 7350
8y + 5250 - 5y = 7350
3y = 7350 - 5250
y = 2100/3 = 700
Substt. the above in eqn-2
x = 1050 - 700 = 350
Therefore, 350 children and 700 adults attended the fair.
To solve this problem, we can use a system of equations. Let's assume that the number of children attending the fair is represented by "x" and the number of adults attending the fair is represented by "y". We are given two pieces of information:
1. The cost of a child's ticket is $5, so the total revenue from the children's tickets is 5x.
2. The cost of an adult's ticket is $8, so the total revenue from the adults' tickets is 8y.
The total revenue collected from all the tickets is $7350. So, we can set up the equation:
5x + 8y = 7350 ... (equation 1)
We are also given that 1050 people attended the fair, so we can set up the equation:
x + y = 1050 ... (equation 2)
Now, we need to solve this system of equations using substitution or elimination. Let's solve it using substitution.
We can rearrange equation 2 to solve for x:
x = 1050 - y
Now we can substitute this value of x in equation 1:
5(1050 - y) + 8y = 7350
Simplifying this equation, we get:
5250 - 5y + 8y = 7350
3y = 7350 - 5250
3y = 2100
y = 2100/3
y = 700
Substituting this value of y back into equation 2, we can find the value of x:
x + 700 = 1050
x = 1050 - 700
x = 350
Therefore, the number of children attending the fair is 350, and the number of adults attending the fair is 700.