# School Play and Ticket problem.

I may be missing something obvious here, but cant seem to see what. Can anyone give me some insight on how to solve this.

100 tickets are sold for a school play.
Tickets for a child cost £1.50 each.
Tickets for an adult cost £2.50 each.
If the organisation had charged £1 for each child's ticket, and £5 for each adults ticket, they would have made £10 more.
How many children and how many adults came to the play?

• Write two equations in two unknowns, then solve them. I recommend your two unknowns be the number of child tickets, and the number of adult tickets. Mar 26 '14 at 17:54

Let's call the number of child tickets $x$ and number of adult tickets $y$.

Then we know that $$x+y = 100 \text{ (Number of tickets sold)}$$ $$1.50x + 2.50y = x + 5y - 10 \text{ (The difference in pricing)}$$

This is a linear system of equations:

$$x+y=100 \\ 0.5x - 2.5y = -10$$

It can be solved using a method of your choice.

I just copped onto what i was missing. The 100 tickets total was what i kept forgetting.

I let child tickets $=x$ and adult tickets $=y$

Two simultaneous equations were $$1.50x+2.50y=z\\1x+5y=z-10$$ Where $z$ is the total money taken in. Then I let $x+y=100 \Rightarrow y=100-x$

Subbed the y in, got two equations in two unknowns and hey presto! $$x=80\\y=20$$