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.

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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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 =

Let the number of students tickets =

Let the number of adults tickets =

Total number of tickets sold = 1500

Writing the equation:

.... (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

..... (2)

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

i.e.

..... (3)

Putting value of from (3) in equation (1) and equation (2):

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

By equation (3):

By equation (1):

Therefore, the answer is:

Number of children tickets = 400

Number of students tickets = 300

Number of adults tickets = 800

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step-by-step explanation:

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