At a local Brownsville play production, 560 tickets were sold. The ticket prices varied on the seating arrangements and cost $8, $10, or $12. The total income from ticket sales reached $5320. If the combined number of $8 and $10 priced tickets sold was 7 times the number of $12 tickets sold, how many tickets of each type were sold?

Number of $8 tickets sold

Number of $10 tickets sold

Number of $12 tickets sold

Number of $8 tickets sold  = 210

Number of $10 tickets sold  = 280

Number of $12 tickets sold  = 70

Step-by-step explanation:

The number of ticket sold is 560  tickets .

Total number of ticket sold = 560 tickets

The prices varies as $8, $10 and $12.

The total income from the ticket sold = $5320

Let

number of  $8 ticket sold = a

number of  $10 ticket sold = b

number of  $12 ticket sold = c

a + b = 7c

Therefore,

8a + 10b + 12c = 5320

a + b + c = 560

7c + c = 560

8c = 560

divide both sides by 8

c = 560/8

c = 70

Insert the value of c in the equations below

8a + 10b + 12c = 5320

8a + 10b + 12(70) = 5320

8a + 10b + 840 = 5320

8a + 10b = 5320 - 840

8a + 10b = 4480(i)

a + b + c = 560

a + b + 70 = 560

a + b = 560 - 70

a + b = 490(ii)

8a + 10b = 4480(i)

a + b = 490(ii)

a = 490 - b

insert the value of a in equation(i)

8(490 - b) + 10b = 4480

3920 - 8b + 10b = 4480

3920 + 2b = 4480

2b = 4480 - 3920

2b = 560

b = 560/2

b = 280

insert the value of b in equation (ii)

a + b = 490(ii)

a + 280 = 490

a = 490 - 280

a = 210

Number of $8 tickets sold  = 210

Number of $10 tickets sold  = 280

Number of $12 tickets sold  = 70

1. 3

2. 4

3. 1

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answer: for each 3 girls there is 2 boys . 3: 2

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