1Cademy - Solving a Ticket Mixture Problem: Childrens and Adult Tickets with a Known Total

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Expressing Unknowns Using a Known Total

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Solving a Ticket Mixture Problem: Children's and Adult Tickets with a Known Total

Apply the seven-step problem-solving strategy, the total-value model, and the known-total technique to a ticket mixture problem where the total number of tickets sold is given rather than a phrase relating the two counts.

Problem: Galen sold $810$ tickets for his church's carnival for a total of $2,820. Children's tickets cost $3 each and adult tickets cost $5 each. How many children's tickets and how many adult tickets did he sell?

Read the problem and identify the types involved: children's tickets (worth $3 each) and adult tickets (worth $5 each). The total number of tickets is $810$ and the total revenue is $2,820.
Identify what to find: the number of children's tickets and the number of adult tickets.
Name the unknowns using a single variable. Because the total ticket count is known, use the known-total technique. Let $c$ = the number of children's tickets. Then the number of adult tickets is $810 - c$ . Organize in a table:

Type	Number	Value ($)	Total Value ($)
Children	$c$	3	$3c$
Adult	$810 - c$	5	$5(810 - c)$
			2,820

Translate into an equation by adding the total values and setting the sum equal to the overall total:

$3c + 5(810 - c) = 2{,}820$

Solve the equation:

Distribute $5$ : $3c + 4{,}050 - 5c = 2{,}820$
Combine like terms: $-2c + 4{,}050 = 2{,}820$
Subtract $4{,}050$ from both sides: $-2c = -1{,}230$
Divide both sides by $-2$ : $c = 615$

Find the number of adult tickets: $810 - 615 = 195$ .

Check: $615 \cdot 3 = 1{,}845$ and $195 \cdot 5 = 975$ . Adding: $1{,}845 + 975 = 2{,}820$ $\checkmark$
Answer: Galen sold $615$ children's tickets and $195$ adult tickets.

This example differs from ticket problems where the two counts are related by a phrase like "five less than three times." Here, the known total of $810$ tickets directly links the two unknowns through subtraction: if $c$ tickets are children's, then $810 - c$ must be adult. Distributing $5$ across $(810 - c)$ produces the constant $4{,}050$ and the negative term $-5c$ . After combining $3c$ and $-5c$ into $-2c$ , the resulting equation involves dividing both sides by a negative number, which is a common source of errors for students.

Updated 2026-04-21

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OpenStax Elementary Algebra

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