At a movie theater, tickets for children cost 11.00 each and adult tickets cost 8.75 each. If ticket sales for a group 35 people totaled to 259.00, how many childern were in this group

Answer 1

This is a strange problem. Seemingly a simple 2 equation, 2 unknown algebra problem.

But something goes wrong. Something is odd about this problem.

A+C=35, the total number of tickets sold

8.75A + 11C = $259

Multiply the first equation by 11 to get

11A+11C=385. then subtract the 2nd equation

8.75A+11C=259 to get

2.25A = 385-259= 126, divide by 2.25

A= 126/2.25= 56. That comes out even, no remainder, no partial number, so it looks right, at first.

But check the answer

If A=56, then C=35-56 = -21. That means more adult tickets were purchased than the total number of tickets. It's as if 56 adults bought tickets and 21 children sold tickets to get a grand total sold of 35

Yet, plug in those numbers into the 2nd equation and it works

8.75(56) + 11(-21) = 259

490 -231 = 259

259 = 259

You could graph the two equations and where the two lines intersect would be the solution

A on the y-axis, C on the x-axis, then the point (-21,56) would be where they intersect.

But this may be mixing apples & oranges, as 35 is a pure number, no dollars, while 259 is in dollars. But that's misleading, as we muliply by $11 which makes the units the same.

Looking at it another way. IF A=0 then C= 23+ but not greater than 23 as an integer. to satisfy the 2nd equation, Yet C=35 for the 1st equation. IF C=0, then A = 29+ but not greater than 29 as an integer, yet A=35 to satisfy the 1st equation.

There's no way for Both A and C to be positive, and satisfy both equations.

Answer 2

Hey Cara, we can try to set up some equations to help us out. Lets have A represent the number of adult tickets, and C represent the number of child tickets. Then based on the information given, we know that

A + C = 35. Also, based on the cost of the tickets, we are wondering when 259 = 11C + 8.75A.

Solving this system of equations will give you the answer!