First, let's calculate the total amount of sausages that will be consumed:
- 1.7 parents x $3 = $5.10 per family
- 2.3 children x $2 = $4.60 per family
- Total sausages per family = $5.10 + $4.60 = $9.70
Now, let's calculate the total amount of sausages consumed by all 120 families:
- $9.70 x 120 = $1164
Adding the cleaning cost of $300, the total cost of the event is:
- $1164 + $300 = $1464
Let's now calculate the number of adults and children attending the event:
- 1.7 parents x 120 families = 204 adults
- 2.3 children x 120 families = 276 children
Since each adult ticket costs twice as much as a child ticket, let's assume the child ticket price is X.
Therefore, the adult ticket price would be 2X.
The total revenue from ticket sales would be:
- 204 adults x 2X + 276 children x X = $1464
Solving the equation:
- 408X + 276X = $1464
- 684X = $1464
- X = $2.14
Therefore, the child ticket price is $2.14 and the adult ticket price is $4.28.
The primary school should charge $4.28 for adult tickets and $2.14 for child tickets to cover the cost of the event and break even.
A primary school is planning to hold a fete, but needs help in figuring out how much to
charge for entry. They intend to have an ‘all you can eat’ sausage sizzle, and they think
each adult will eat about $3 worth of sausages, and each child $2 worth. The cost of
cleaning after the event will be $300. They want an adult ticket to cost twice as much as
a child ticket, and they expect that, on average, 1.7 parents and 2.3 children will come
from each family. How much will they need to charge for adult and child tickets if they
are expecting 120 families to attend and they don’t want to make a profit or a loss?
(Round to the nearest cent.)
1 answer