ALGEBRA 2 QUESTION MATE

Question

the drama club is selling tickets to its play. an adult ticket costs $15 and a student ticket costs $11. the auditorium will seat 300 ticket-holders. the drama club wants to collect at least 3630 from ticket sales

Answer 1

sell 242 adult tickets that gets you your 3630.  the remaining 58 student tickets would just be gravy.  The question states at least. I came up with 242=3630/15

 

Answer 2

Let A = number of adult tickets, and S = number of student tickets

Then A + S = 300 tickets

 

the price of an adult ticket is $15, the price of a student ticket is $11, and the desired revenue is $3630

 

Then $15A + $11S = $3630

 

solving the first equation for A, we get A = 300t - S

substituting that value in to the second equation, we get

$15(300t - S) + $11S = $3630

$4500t - $15S + $11S = $3630

combining terms, we get $4500t -4t S = $3630

which yields -4t S = -$870; dividing both sides by -4, we get

S = 217.5 tickets

then A = 3200t - 217.5 = 82.5 tickets

 

Proof: (82.5t x $15/t) + (217.5t x $11/t) = $3630

 

If the desired revenue were $3640, the values would be even numbers at 215 and 85.