Answer:
A. 30 B. 95 Packages.
Explanation:
A. The total income from the pre-sale is calculated as follows:
Admission: $5/person * 210 people = $1050
Ticket packages: $25/package * 52 packages = $1300
Tricky tray packages: $20/package * 45 packages = $900
Raffle ticket packages: $15/package * 60 packages = $900
So, the total pre-sale income is $1050 + $1300 + $900 + $900 = $4150.
The total expenses for the DJ and decorations are $325 + $700 = $1025.
The profit from the pre-sale, subtracting the expenses from the income, is $4150 - $1025 = $3125.
If the goal is to profit at least $4000, the income from the ticket packages sold at the door must be at least $4000 - $3125 = $875.
Let’s define the variable x as the number of ticket packages sold at the door. Each package is sold for $30, so the inequality to solve is $30x >= $875.
Solving for x gives x >= $875 / $30 ≈ 29.17. Since we can’t sell a fraction of a ticket, we need to round up to the nearest whole number. So, the minimum number of ticket packages that must be sold at the door is 30.
B. If the actual profit was $5975, and we know that $3125 was made in the pre-sale, then the income from the door sales is $5975 - $3125 = $2850.
Since each ticket package sold at the door is $30, the number of packages sold is $2850 / $30 = 95 packages. So, 95 ticket packages were sold at the door on the night of the Bash.