A group of visited paid a total of $380 entrance tickets for a zoo. The entrance ticket for a child cost $4. The entrance ticket for an adult cost twice as much. If there were 10 more adult the childrens, how many adult were there?.
My workout 1 chlid -$4. 1 adult -$4×2twice= $8.
10 more adult $8 ×10 = $80.
x is the number of childs, and y is the number of adults. Then
4x+8y = 380
y = x + 10
Solve the system
4x + 8(x+10) = 4x + 8x + 80 =380 =>
12x = 300,
then x = 300/12 = 25.
y = x + 10 = 35
35 adults
To solve this problem, let's assign variables to the unknowns.
Let's say the number of children is represented by C, and the number of adults is represented by A.
We know that the entrance ticket for a child costs $4, so the total cost for all the children is 4 * C.
We also know that the entrance ticket for an adult costs twice as much, so the total cost for all the adults is 8 * A.
According to the problem, the total cost for all the entrance tickets is $380, so we can set up the equation:
4 * C + 8 * A = 380
We are also given that there were 10 more adults than children, so we can write another equation:
A = C + 10
Now we have a system of two equations:
1) 4 * C + 8 * A = 380
2) A = C + 10
To solve for the number of adults, we can substitute equation 2) into equation 1):
4 * C + 8 * (C + 10) = 380
Now we can simplify and solve for C:
4 * C + 8 * C + 80 = 380
12 * C = 300
C = 25
So there were 25 children in the group. To find the number of adults, we can substitute this value back into equation 2):
A = 25 + 10
A = 35
Therefore, there were 35 adults in the group.