SOLUTION: At a local Brownsville play production, 420 tickets were sold. The ticket prices varied on the seating arrangements and cost $8, $10, or $12. The total income from ticket sales rea

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Question 1179222: At a local Brownsville play production, 420 tickets were sold. The ticket prices varied on the seating arrangements and cost $8, $10, or $12. The total income from ticket sales reached $3920. If the combined number of $8 and $10 priced tickets sold was 5 times the number of $12 tickets sold, how many tickets of each type were sold?
Found 3 solutions by josgarithmetic, MathLover1, ikleyn:
Answer by josgarithmetic(39623)   (Show Source): You can put this solution on YOUR website!
PRICES         COUNTS           COSTS
  8              e               8e
 10            420-e-t          10(420-e-t)
 12              t              12t
               420               3920



First equation gives .

Second equation simplifies to .
Substitution gives .

Quantity of $10 tickets by difference, .
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!

At a local Brownsville play production, tickets were sold.
let's ticket that cost $ be , ticket that cost $ be , and ticket that cost $ be

.........eq.1

if the total income from ticket sales reached $, we have
.....eq.2
If the combined number of $ and $ priced tickets sold was 5 times the number of $tickets, we have

....eq.3

go to
.........eq.1, substitute from eq.3




go to
....eq.3
..........solve for
..............eq.1)

go to
.....eq.2, substitute and
.......solve for






go to

..............eq.1), substitute



sold, how many tickets of each type were sold?
tickets that cost $ were sold,
tickets that cost $ were sold,
and tickets that cost $ were sold

Answer by ikleyn(52832)   (Show Source): You can put this solution on YOUR website!
.
At a local Brownsville play production, 420 tickets were sold.
The ticket prices varied on the seating arrangements and cost $8, $10, or $12.
The total income from ticket sales reached $3920.
If the combined number of $8 and $10 priced tickets sold was 5 times the number of $12 tickets sold,
how many tickets of each type were sold?
~~~~~~~~~~~~ 

Let x be the number of the $8 tickets, and

let y be the number of the $10 tickets.


Then, from the condition, the number  (x+y)  is  5/6  of 420, i.e. x+y = 350,

      and the number of those who bought the $12 tickets was the rest  1/6 of 420, i.e. 70 persons.


Now we have the system of 2 (two) equations in two unknowns


     x +   y         =  350     (1)    

    8x + 10y + 12*70 = 3920     (2)    (total revenue)


We simplify this system to this EQUIVALENT strandard form


      x +   y        =  350     (3)  

     8x + 10y        = 3080     (4)  


Multiply equation (3) by 10 (both sides) and then subtract from it equation (4).  You will get then


    10x - 8x         = 3500 - 3080

       2x            =  420

        x            = 420/2 = 210.


Thus 210 persons bought $8 tickets;  hence,  350-210 = 140 bought $10 tickets;  and the rest  70 persons bought $12 tickets.

Solved.


////////////


By the way,  the problem' setup,  presented in the post by @josgarithmetic,

                        IS TOTALLY WRONG,


            so for your safety,  you better ignore it . . .



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