Ticket Sales System: Solve for Adult & Child Attendance
What This Problem Teaches
- Setting up systems of linear equations from real-world scenarios involving two unknowns
- Translating "total quantity" and "total value" constraints into mathematical equations
- Using elimination method to solve systems efficiently, especially with decimal coefficients
- Converting decimal problems to whole numbers to reduce calculation errors
- Verifying solutions by checking both original conditions independently
Visualizing the Problem
Let's organize the given information to see the structure clearly:
| Ticket Type | Price Each | Number Sold | Total Revenue |
|---|---|---|---|
| Adult | $17.95 | a (unknown) | 17.95a |
| Children | $12.95 | c (unknown) | 12.95c |
| Totals | — | 500 people | $7,355 |
This gives us two constraints: a + c = 500 (total people) and 17.95a + 12.95c = 7355 (total revenue).
Solution: Method 1 — The Elimination Approach
Step 1 — Define variables
Let a = number of adult tickets sold and c = number of children's tickets sold.
Step 2 — Set up the system of equations
From the problem constraints:
17.95a + 12.95c = 7355 (total revenue)
Step 3 — Eliminate decimals by multiplying by 100
To avoid decimal arithmetic, multiply the second equation by 100:
1795a + 1295c = 735500
Step 4 — Use elimination to solve for one variable
Multiply the first equation by -1295 to eliminate c:
1795a + 1295c = 735500
Add the equations:
a = 176
Step 5 — Find the other variable
Substitute a = 176 into the first equation:
c = 324
Solution: Method 2 — The Substitution Strategy
Step 1 — Solve the simpler equation for one variable
From a + c = 500, we get c = 500 - a.
Step 2 — Substitute into the revenue equation
Replace c in the revenue equation:
17.95a + 6475 - 12.95a = 7355
5a + 6475 = 7355
5a = 880
a = 176
Step 3 — Find the second variable
Using c = 500 - a:
Verification
Let's check both conditions:
Total people: 176 + 324 = 500 ✓
Total revenue:
Children revenue: 324 × $12.95 = $4,195.80
Total: $3,159.20 + $4,195.80 = $7,355.00 ✓
Both conditions are satisfied, confirming our solution is correct.
Watch Out For These
Writing
17.95 + 12.95 = 7355 instead of 17.95a + 12.95c = 7355. Remember that total revenue comes from multiplying each price by the number of tickets sold at that price.
Trying to work directly with
17.95a + 12.95c = 7355 often leads to arithmetic mistakes. Converting to whole numbers by multiplying by 100 makes the calculations much cleaner and more reliable.
Getting an answer like 400 adult tickets and 100 children's tickets. This would mean 80% adult attendance, which might seem high. While mathematically possible, the actual answer (176 adults, 324 children) shows children's tickets made up the majority—which makes sense given the lower price point.
How to Spot This Problem Type
Look for these tell-tale signs that you're dealing with a ticket sales system problem:
- "Two different prices" and "total revenue" — signals you need both quantity and value equations
- "Total number of people/tickets" combined with "total money collected"
- Phrases like "They made $X from Y people" where you need to find how many of each type
- "Adult tickets cost... children's tickets cost..." — classic setup phrasing
- Any problem where you have two unknowns and two pieces of total information (total count + total value)
The Pattern Behind This
This is a classic mixture problem involving two components with different unit values. The general pattern is:
P₁x + P₂y = T₂ (total value)
Where x and y are the quantities of each type, P₁ and P₂ are the unit prices, T₁ is the total quantity, and T₂ is the total value.
Real Applications
- Theater and venue management: Box offices use these calculations daily to analyze sales mix and plan capacity for different seating sections
- Event planning: Conference organizers track regular vs. student registration to forecast revenue and plan accordingly
- Transportation systems: Airlines, buses, and trains analyze passenger mix between different fare classes to optimize pricing strategies
What If?
Let a = adult tickets, c = children's tickets:a + c = 50019.95a + 14.95c = 7355
Multiply the second equation by 100:1995a + 1495c = 735500
Multiply first equation by -1495:-1495a - 1495c = -747500
Add to get: 500a = -12000
So a = -24
This problem has no solution! The negative answer tells us it's impossible to collect exactly $7,355 from 500 people with these higher prices. The minimum revenue would be 500 × $14.95 = $7,475, which exceeds our target.
Let a = adult, s = senior, c = children's tickets.
Given: c = 2s (twice as many children's as senior)
a + s + c = 50017.95a + 15.95s + 12.95c = 7355
Substitute c = 2s:a + s + 2s = 500 → a + 3s = 50017.95a + 15.95s + 12.95(2s) = 7355 → 17.95a + 41.85s = 7355
From first equation: a = 500 - 3s
Substitute: 17.95(500 - 3s) + 41.85s = 73558975 - 53.85s + 41.85s = 7355-12s = -1620s = 135
s = 135 senior ticketsc = 2s = 270 children's ticketsa = 500 - 3(135) = 95 adult tickets
Total: 95 + 135 + 270 = 500 ✓
Revenue: 95($17.95) + 135($15.95) + 270($12.95) = $1,705.25 + $2,153.25 + $3,496.50 = $7,355 ✓
Let c = price of children's tickets
Then c + 5 = price of adult tickets
180(c + 5) + 320c = 7355180c + 900 + 320c = 7355500c + 900 = 7355
500c = 6455c = 12.91
Adult price = c + 5 = 12.91 + 5 = 17.91
Children's tickets: $12.91, Adult tickets: $17.91
Check: 180($17.91) + 320($12.91) = $3,223.80 + $4,131.20 = $7,355 ✓
a + c = 50018a + 13c = 7355
From first equation: c = 500 - a
Substitute: 18a + 13(500 - a) = 735518a + 6500 - 13a = 73555a = 855a = 171
c = 500 - 171 = 329
Total: 171 + 329 = 500 ✓
Revenue: 171($18) + 329($13) = $3,078 + $4,277 = $7,355 ✓
This problem has a clean integer solution! This shows that not all ticket problems lead to fractional answers. When the numbers are chosen carefully, real-world constraints (whole numbers of tickets) are naturally satisfied. Problems with messy decimals often indicate pricing or attendance figures that wouldn't occur in practice.
Frequently Asked Questions
2026-09-18