the mason family went to the amusement park. they spent 56.00$ on tickets. if adults cost 12$ and kids cost 5$ how many adults and children went to the park?
This is guess and check, it doesn’t show how to do the problem.
To solve this problem, let's assume that the number of adults who went to the amusement park is 'A', and the number of children who went to the park is 'C.'
According to the given information, the total amount spent on tickets was $56. We can set up an equation representing the total cost of the tickets:
12A + 5C = 56
We need to determine the values of 'A' and 'C' that satisfy this equation.
One way to approach this is to use trial and error or guess-and-check method. We can start by assuming different values for 'A' and solve for 'C' using the equation. If we find a valid solution, it means that it satisfies the given conditions.
Let's try different values for 'A' and see if we find a valid solution:
Assume A = 1:
12(1) + 5C = 56
12 + 5C = 56
5C = 56 - 12
5C = 44
C = 44 / 5 = 8.8
Since the number of children cannot be a decimal, this combination of values (A = 1, C = 8.8) is not valid. We need to try other values for 'A' and see if we can find a whole number solution for 'C'.
Assume A = 2:
12(2) + 5C = 56
24 + 5C = 56
5C = 56 - 24
5C = 32
C = 32 / 5 = 6.4
Again, this combination of values (A = 2, C = 6.4) is not valid since the number of children cannot be a decimal.
Now, let's try another value for 'A':
Assume A = 3:
12(3) + 5C = 56
36 + 5C = 56
5C = 56 - 36
5C = 20
C = 20 / 5 = 4
This combination of values (A = 3, C = 4) is valid since both 'A' and 'C' are whole numbers, and it satisfies the equation.
Therefore, the Mason family had 3 adults and 4 children who went to the amusement park.
assuming there is one mom and one dad, let the number of kids be x
2(12) + 5x = 56
5x = 32
x = 32/5 , well, that's no good
let the number of adults be x, number of kids be y
12x + 5y = 56
looking for integer solutions,
let y = 4
12x + 20 = 56
x = 3
3 adults, and 4 kids