Admission to a zoo costs $10 for adults and $6 for children. A group of 29 people attending the zoo paid a total of $222 in admission fees.
(Part a) Write a system of equations to represent the situation. Let a represent the number of adult admissions, and let c represent the number of child admissions.
(Part B) Solve the system you wrote in part (a) using the substitution method. Show your work.
(Part C) Interpret your solution in the context of the problem.
consider the people and the money:
a+c = 29
10a + 6c = 222
Now crank 'er out...
I need the answers
Part a:
The cost of an adult admission is $10, so the total cost for adult admissions is 10a.
The cost of a child admission is $6, so the total cost for child admissions is 6c.
The total number of people attending the zoo is 29, so we have the equation: a + c = 29.
The total amount paid in admission fees is $222, so we have the equation: 10a + 6c = 222.
Part b:
We have the system of equations:
a + c = 29
10a + 6c = 222
From the first equation, we can solve for a in terms of c: a = 29 - c.
Substituting this into the second equation, we have:
10(29 - c) + 6c = 222
Distributing and simplifying, we get:
290 - 10c + 6c = 222
-4c = -68
c = 17
Substituting this value of c into the equation a + c = 29, we have:
a + 17 = 29
a = 29 - 17
a = 12
Therefore, there were 12 adult admissions and 17 child admissions.
Part c:
The solution to the system of equations is a = 12 and c = 17. This means that there were 12 adults and 17 children in the group of 29 people attending the zoo.
This solution tells us that the total cost of adult admissions was 12 * $10 = $120, and the total cost of child admissions was 17 * $6 = $102. The sum of these costs is $120 + $102 = $222, which matches the total amount paid in admission fees.
So, the solution is consistent with the problem statement and represents a valid scenario in which the given conditions are satisfied.
Part a:
Let a represent the number of adult admissions
Let c represent the number of child admissions
The total number of people attending the zoo is 29, so we can set up the equation:
a + c = 29 (Equation 1)
The total amount paid in admission fees is $222, which can be represented by the equation:
10a + 6c = 222 (Equation 2)
Part b:
To solve the system of equations using the substitution method, we can solve one equation for one variable and substitute it into the other equation.
From Equation 1, we can solve for a in terms of c:
a = 29 - c
Now we can substitute this value of a into Equation 2:
10(29 - c) + 6c = 222
Distribute the 10:
290 - 10c + 6c = 222
Combine like terms:
290 - 4c = 222
Subtract 290 from both sides:
-4c = -68
Divide by -4:
c = 17
Now we can substitute the value of c back into Equation 1 to find the value of a:
a + 17 = 29
Subtract 17 from both sides:
a = 12
So we have a = 12 and c = 17 as the solution to the system of equations.
Part c:
The solution to the system of equations tells us that there are 12 adult admissions and 17 child admissions in the group of 29 people. This means that there are 12 adults and 17 children in the group.
The adults pay $10 each for admission, so the total revenue from adult admissions is $10 * 12 = $120.
The children pay $6 each for admission, so the total revenue from child admissions is $6 * 17 = $102.
The total revenue from admission fees is $120 + $102 = $222, which matches the total amount paid by the group.
Therefore, the solution is consistent with the problem, and it means that there were 12 adults and 17 children in the group attending the zoo.