At a local fair, the admission fees are $1.50 for children, $3 for college students, and $5 for adults. On Saturday, 1500 tickets were sold and brought in $5500. The number of college student tickets sold was 100 less than the number of children's tickets sold.

Respuesta :

Answer:

Number of children tickets = 400

Number of students tickets = 300

Number of adults tickets = 800

Step-by-step explanation:

Let the number of children tickets = [tex]x[/tex]

Let the number of students tickets = [tex]y[/tex]

Let the number of adults tickets = [tex]z[/tex]

Total number of tickets sold = 1500

Writing the equation:

[tex]x+y+z=1500[/tex] .... (1)

Price for each children ticket = $1.50

Price for each student ticket = $3

Price for each adult ticket = $5

Total money brought in = $5500

[tex]1.5x+3y+5z = 5500[/tex] ..... (2)

Number of student tickets sold was 100 lesser than number of children's tickets.

i.e.

[tex]y = x -100[/tex] ..... (3)

Putting value of [tex]y[/tex] from (3) in equation (1) and equation (2):

[tex]x+x-100+z=1500\\\Rightarrow 2x+z=1600 .... (4)[/tex]

[tex]1.5x+3(x-100)+5z=5500\\\Rightarrow 4.5x+5z=5800 ..... (5)[/tex]

Multiplying equation (4) by 5 and subtracting equation (5) from it:

[tex]5.5x=2200\\\Rightarrow x = 400[/tex]

By equation (3):

[tex]y = 400 - 100 = 300[/tex]

By equation (1):

[tex]z = 1500 - 400 - 300 = 800[/tex]

Therefore, the answer is:

Number of children tickets = 400

Number of students tickets = 300

Number of adults tickets = 800

400 children tickets, 300 college ticket and 800 adult tickets were sold.

Let x represent the number of children tickets, y represent the number of college tickets and z represent the number of adult tickets.

Since 1500 tickets were sold, hence:

x + y + z = 1500     (1)

They brought in $5500, hence:

1.5x + 3y + 5z = 5500   (2)

Also:

y = x - 100

-x + y = -100     (3)

Solving equations 1, 2 and 3 simultaneously gives:

x = 400, y = 300, z = 800

400 children tickets, 300 college ticket and 800 adult tickets were sold.

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