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At a college production of a​ play, 420 tickets were sold. The ticket prices were​ $8, $10, and​ $12, and the total income from ticket sales was ​$3900. How many tickets of each type were sold if the combined number of​ $8 and​ $10 tickets sold was 5 times the number of​ $12 tickets​ sold?

 Apr 6, 2020
 #2
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Let  x  represent the number of $8.00 tickets sold

       y  represent the number of $10.00 tickets sold

       z  represent the number of $12.00 tickets sold

 

Since the combined number of $8.00 tickets and $10.00 tickets sold was 5 times the number of $12.00 tickets sold:

      x + y  =  5z   --->     z  =  (x + y)/5

 

Value of the tickets sold:  8x + 10y + 12(x + y)/5  =  3900

(multiply by 5)                  40x + 50y + 12x + 12y  =  19 500

                                                            52x + 62y  =  19 500

 

Number of tickets sold:  x + y + (x + y)/5  =  420

(multiply by 5)                 5x + 5y + x + y  =  2100

                                                    6x + 6y  =  2100

(divide by 6)                                     x + y  =  350

 

Combining the two equations:  52x + 62y  =  19500

                                                     x +     y  =  350

 

(multiplying the bottom equation by  -52:

                                     52x + 62y  =   19500  

                                   -52x -  52y  =   -18200

(adding down)                       10y  =  1300

                                                 y  =  130

 

x + y   = 350   --->   x + 130  =  350   --->   x  =  220

 

z  =  (x + y)/5   --->   z  =  (220 + 130) / 5   --->   z  =  70

 Apr 6, 2020

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