Answer:
3 normal tickets were sold, 3 tickets were sold for the elderly, and 12 tickets were sold for children
Explanation:
Let x = normal ticket number
Let y = child ticket number
Let z = elderly ticket number
The total amount spent is 480, so we can create this equation:
52 * x + 20 * y + 28 * z = 480
They also tell us that each adult took two children, so:
y = 2 * (x + z)
We have two equations and three unknowns, therefore we must simplify and do trial and error. First, let's substitute "y" in terms of "x" and "z" in the first equation, then simplify:
52 * x + 20 * y + 28 * z = 480
52 * x + 20 * 2 * (x + z) + 28 * z = 480
52 * x + 40 * (x + z) + 28 * z = 480
52 * x + 40 * x + 40 * z + 28 * z = 480
92 * x + 68 * z = 480
23 * x + 17 * z = 120
Let's solve for x in terms of e:
23 * x = 120-17 * z
x = (120-17 * z) / 23
We know that (120 - 17 * z) has to be a multiple of 23 for "x" to be an integer. So let's look at testing some values of "z" to make valid values for "x":
120-17 (0) = 120 (not multiple of 23)
120-17 (1) = 103 (not multiple of 23)
120-17 (2) = 86 (not multiple of 23)
120-17 (3) = 69 (multiple of 23)
120-17 (4) = 52 (not multiple of 23)
120-17 (5) = 35 (not multiple of 23)
120-17 (6) = 18 (not multiple of 23)
120-17 (7) = 1 (not multiple of 23)
After that, we get negative values for "x", so here there is only one possible value for "x", so we know that z = 3:
Now we can solve for "x" and "y":
x = (120-17 * 3) / 23
x = (120-51) / 23
x = 69/23
x = 3
Now for and:
y = 2 * (x + y)
y = 2 (3 + 3)
y = 12
Therefore, 3 normal tickets were sold, 3 tickets were sold for the elderly, and 12 tickets were sold for children.