Concert fundraising: checking whether the target can be reached at all
The Grade 6 class must raise $150,000 towards a national schools' mathematics competition trip. They are holding a concert in the school hall, which seats 240. Adult tickets sell at $800 and children's tickets at $500, and each ticket is for one seat. The hall costs $25,000 to rent for the evening, the sound system $12,400, and printing tickets and programmes costs $8,600. Refreshments are donated. So far they have sold 186 tickets: 74 adult and 112 children's. A parent promises to buy a block of 20 adult tickets if the class reaches 90% of its target by the night before the concert. A teacher estimates that walk-in sales on the night will be about 10% of seats and suggests cutting the children's price to $400 to fill the hall.
1. Work out the profit made so far if no more tickets are sold, and the amount still needed to reach the target.
Read the worked answer
Costs (fixed, whatever the ticket sales): 25,000 + 12,400 + 8,600 = $46,000
Money taken so far:
- Adults: 74 × $800 = $59,200
- Children: 112 × $500 = $56,000
- Total: 59,200 + 56,000 = $115,200
Check the ticket count: 74 + 112 = 186 seats sold, so 240 − 186 = 54 seats remain.
Profit so far: 115,200 − 46,000 = $69,200
Amount still needed: 150,000 − 69,200 = $80,800
The costs are already committed, so the whole ticket income from here on is extra profit - but each of the 54 remaining seats can bring in at most $800.
2. Can the class reach $150,000 using the remaining 54 seats, with or without the parent's 20-ticket block? Show the best case.
Read the worked answer
Best case with all 54 remaining seats sold at the adult price: 54 × $800 = $43,200 New profit = 69,200 + 43,200 = $112,400 - still $37,600 short.
Realistic case with the parent's block: The block is 20 adult tickets, leaving 34 seats for walk-ins.
- 20 × $800 = $16,000
- 34 × $500 = $17,000
- Extra income = $33,000 New profit = 69,200 + 33,000 = $102,200 - $47,800 short.
Does the parent's condition ever trigger? 90% of $150,000 = $135,000. Even filling every seat at the adult price gives $112,400, so the class never reaches 90% and the block is in practice unavailable.
Finally, cutting children's tickets to $400 lowers income: on 34 seats that is 34 × $100 = $3,400 less. Filling the hall at a lower price makes the shortfall bigger. The plan as written cannot reach the target.
3. A committee member claims: 'Even if we fill all 240 seats at the adult price, one concert cannot raise $150,000.' Test this claim and recommend what the class should change.
Read the worked answer
Test the best possible one-concert outcome:
- 240 seats × $800 (every seat at the highest price) = $192,000
- Less costs of $46,000 = $146,000
$146,000 is $4,000 below the $150,000 target. The claim is correct: at these prices and costs, one concert falls short even in an impossible best case where every child's ticket becomes an adult's ticket.
What the numbers say must change:
- Cut the fixed costs. The $25,000 hall is the biggest item; a free school hall would give 146,000 + 25,000 = $171,000, comfortably over target.
- Raise the price. To gain $4,000 from 240 seats needs about $17 more per seat, e.g. adult $850.
- Run a second performance, which reuses the printing and sound system but adds hall rent.
Lowering the children's price is the wrong lever - it moves the class further from the target. The real problem is that 240 seats at a top price of $800 barely exceed the target before costs; the class should check the maximum possible income before promising a target.