# Child and Adult Ticket Count

## Outline Mathematics

Word Problems

Here's a real life problem taken from the Purplemath site:

Your school is holding a "family friendly" event this weekend. Students have been pre-selling tickets to the event; adult tickets are $5.00, and child tickets (for six-years-old and under) are $2.50.

You consult with your student ticket-sellers, and discover that they have not been keeping track of how many child tickets they have sold. The tickets are identical, until the ticket-seller punches a hole in the ticket, indicating that it is a child ticket. But they don't remember how many holes they've punched. They only know that they've sold 548 tickets for $2460. How many child and adult tickets have been pre-sold?

|Up| |Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

### Solution

Your school is holding a "family friendly" event this weekend. Students have been pre-selling tickets to the event; adult tickets are $5.00, and child tickets (for six-years-old and under) are $2.50.

You consult with your student ticket-sellers, and discover that they have not been keeping track of how many child tickets they have sold. The tickets are identical, until the ticket-seller punches a hole in the ticket, indicating that it is a child ticket. But they don't remember how many holes they've punched. They only know that they've sold 548 tickets for $2460. How many child and adult tickets have been pre-sold?

Let A stand for the number of adult tickets pre-sold, and C stand for the number of child tickets pre-sold. Then A + C = 548,548,2460. Also, since each adult ticket cost $5.00,$2.50,$4.00,$5.00,$6.00, then ($5.00)A stands for the revenue brought in from the adult tickets pre-sold; likewise, $2.50C,$2.50C,$4.00C,$5.00C,$6.00C stands for the revenue brought in from the child tickets. Then ($5.00)A + ($2.50)C = $2460,$548,$2460. What we got is a system of two equations with two unknown quantities A and C:

A + C,A + C,5A + 2.5C = 548

5A + 2.5C,A + C,5A + 2.5C = 2460

There are many ways to solve such a system. One possibility is to express one of the variables from one of the equations and plug the result into the other equation. For example, let's solve the first equation for A:

A = 548 - C,548 + C,548,548 - C

This we substitute into the second equation:

5(548 - C) + 2.5C,5C,548C,2.5C = 2460.

We now have one equation with a single variable. We got to simplify it first:

2740 - 5C + 2.5C = 2460.

Going one step further we obtain

2740 - 2.5C,5C,-2.5C,2.5C = 2460.

And then

2740 - 2460,548,2460 = 2.5C.

From where 2.5C = 280,260,280,300,320 and, finally, C = 112,112,120,122,124.

So that, since A = 548 - C,548 + C,548,548 - C, A = 436,436,446,346,548.

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

63430353 |