At a carnival, $2,914.25 in receipts were taken at the end of the day. The cost of a child's ticket was $20.50, an adult ticket was $29.75, and a senior citizen ticket was $15.25. There were twice as many senior citizens as adults in attendance, and 20 more children than senior citizens. How many children, adult, and senior citizen tickets were sold?

Children tickets = 69.4667

Adult tickets = 24.7333

Senior tickets = 49.4667

(The decimal does not make sense as we cannot have a fractional amount of tickets, but with the information given, this is the correct answer)

Step-by-step explanation:

Let's call the amount of children C, the amount of adults A and the amount of seniors S. Then, we can write the following system of equations:

C*20.5 + A*29.75 + S*15.25 = 2914.25 (eq1)

S = 2*A (eq2)

C = S + 20 (eq3)

Using eq2 in eq3, we have:

C = 2*A + 20 (eq4)

Using eq2 and eq4 in eq1, we have:

(2*A + 20) * 20.5 + A*29.75 + 2*A*15.25 = 2914.25

41*A + 410 + 29.75*A + 30.5*A = 2914.25

101.25*A = 2504.25

A = 2504.25 / 101.25 = 24.7333

Now using eq2, we have:

S = 2*A = 49.4667

And using eq4, we have:

C = 2*A + 20 = 69.4667

(The decimal does not make sense as we cannot have a fractional amount of tickets, but with the information given, this is the correct answer)

"x" children tickets are 70, "y" adult tickets 50 and "z" senior citizen tickets 25

Step-by-step explanation:

With the information of the statement we can pose the following equations:

we have to "x" children tickets, "y" adult tickets and "z" senior citizen tickets.

20.5*x + 29.75*y + 15.25*z = 2914.25

we also have to:

z = 2*y

plus that:

x = z + 20, replacing, we have left:

x = 2*y + 20

now, we replace each other in the main equation:

20.5 * (2*y + 20) + 29.75*y + 15.25 * (2*y) = 2914.25

41*y + 410 + 29.75*y + 30.5*y = 2914.25

101.25*y = 2504.25

y = 2504.25/101.25

y = 24.73

z = 2*24.73 = 49.46

x = 2*24.73 + 20

x = 69.46

"x" children tickets are 70, "y" adult tickets 50 and "z" senior citizen tickets 25