Example

Solving a Ticket Mixture Problem: Student and Adult Tickets

Apply the seven-step problem-solving strategy and the total-value model to a ticket mixture problem where each ticket type has a whole-dollar value and the relationship between ticket counts involves both multiplication and subtraction.

Problem: At a school concert, the total value of tickets sold was $1,506. Student tickets sold for $6 each and adult tickets sold for $9 each. The number of adult tickets sold was five less than three times the number of student tickets sold. How many student tickets and how many adult tickets were sold?

  1. Read the problem and identify the types involved: student tickets (worth $6 each) and adult tickets (worth $9 each). The total value of all tickets is $1,506.
  2. Identify what to find: the number of student tickets and the number of adult tickets.
  3. Name the unknowns using a single variable. Let ss = the number of student tickets. The phrase "five less than three times" combines multiplication by 33 with subtracting 55, so the number of adult tickets is 3s53s - 5. Organize in a table:
TypeNumberValue ($)Total Value ($)
Studentss66s6s
Adult3s53s - 599(3s5)9(3s - 5)
1,506
  1. Translate into an equation by adding the total values and setting the sum equal to the overall total:

6s+9(3s5)=15066s + 9(3s - 5) = 1506

  1. Solve the equation:
  • Distribute 99: 6s+27s45=15066s + 27s - 45 = 1506
  • Combine like terms: 33s45=150633s - 45 = 1506
  • Add 4545 to both sides: 33s=155133s = 1551
  • Divide both sides by 3333: s=47s = 47

Find the number of adult tickets: 3(47)5=1415=1363(47) - 5 = 141 - 5 = 136.

  1. Check: 476=28247 \cdot 6 = 282 and 1369=1,224136 \cdot 9 = 1{,}224. Adding: 282+1,224=1,506282 + 1{,}224 = 1{,}506 \checkmark
  2. Answer: They sold 47 student tickets and 136 adult tickets.

This example demonstrates that ticket and stamp problems follow the same total-value model used for coin problems — each type of ticket has a fixed value, just as each type of coin does. Here the relationship "five less than three times" translates to 3s53s - 5, and distributing the $9 ticket price across 3s53s - 5 produces whole-number coefficients (27s27s and 45-45) that are easier to work with than the decimal coefficients typical of coin problems. After distribution, combining 6s+27s6s + 27s into 33s33s and solving the resulting two-step equation yields the answer.

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Updated 2026-04-21

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