There are three types of tickets to enter the amusement park of a particular city. The normal ticket costs R $ 52.00, the child ticket costs R $ 20.00 and the elderly ticket costs R $ 28.00. A group of friends went to the park and paid in total R $ 480.00. In this group, each adult (normal or senior) took two children. How many normal tickets purchased?

3 normal tickets were sold, 3 tickets were sold for the elderly, and 12 tickets were sold for children

Step-by-step 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.