A petting zoo sells two types of tickets. The standard ticket, for admission only, costs $5. The premium ticket, which...
GMAT Algebra : (Alg) Questions
A petting zoo sells two types of tickets. The standard ticket, for admission only, costs \(\$5\). The premium ticket, which includes admission and food to give to the animals, costs \(\$12\). One Saturday, the petting zoo sold a total of \(250\) tickets and collected a total of \(\$2,300\) from ticket sales. Which of the following systems of equations can be used to find the number of standard tickets, \(\mathrm{s}\), and premium tickets, \(\mathrm{p}\), sold on that Saturday?
\(\mathrm{s + p = 250}\)
\(\mathrm{5s + 12p = 2,300}\)
\(\mathrm{s + p = 250}\)
\(\mathrm{12s + 5p = 2,300}\)
\(\mathrm{5s + 12p = 250}\)
\(\mathrm{s + p = 2,300}\)
\(\mathrm{12s + 5p = 250}\)
\(\mathrm{s + p = 2,300}\)
1. TRANSLATE the problem information
- Given information:
- Standard ticket: $5 (admission only)
- Premium ticket: $12 (admission + food)
- Total tickets sold: 250
- Total revenue collected: $2,300
- Variables: s = standard tickets, p = premium tickets
2. INFER the approach
- We have two unknowns (s and p), so we need two equations
- The problem gives us two different constraints to work with:
- A constraint about total quantity (tickets)
- A constraint about total value (money)
3. TRANSLATE the first constraint
- "Total of 250 tickets sold" means:
\(\mathrm{s + p = 250}\)
4. TRANSLATE the second constraint
- "Collected $2,300 from ticket sales" means:
- s standard tickets × $5 each = \(\mathrm{5s}\) dollars
- p premium tickets × $12 each = \(\mathrm{12p}\) dollars
- Total: \(\mathrm{5s + 12p = 2,300}\)
5. Match to answer choices
- Our system:
\(\mathrm{s + p = 250}\)
\(\mathrm{5s + 12p = 2,300}\)
- This exactly matches Choice A
Answer: A
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE reasoning: Students mix up which cost goes with which ticket type
They might think "standard sounds fancier, so it should cost more" or rush through reading and accidentally swap the $5 and $12 costs. This creates the wrong second equation: \(\mathrm{12s + 5p = 2,300}\).
This may lead them to select Choice B (\(\mathrm{s + p = 250}\), \(\mathrm{12s + 5p = 2,300}\)) or Choice D (\(\mathrm{12s + 5p = 250}\), \(\mathrm{s + p = 2,300}\)).
Second Most Common Error:
Poor TRANSLATE execution: Students mix up what equals what in the equations
They might think the money equation should equal 250 and the ticket equation should equal 2,300, completely reversing the setup. This fundamental misunderstanding of what each constraint represents creates entirely wrong equations.
This may lead them to select Choice C (\(\mathrm{5s + 12p = 250}\), \(\mathrm{s + p = 2,300}\)).
The Bottom Line:
Success requires careful, systematic translation of each constraint separately. The key insight is recognizing that "total tickets" and "total revenue" represent completely different types of quantities that need separate equations.
\(\mathrm{s + p = 250}\)
\(\mathrm{5s + 12p = 2,300}\)
\(\mathrm{s + p = 250}\)
\(\mathrm{12s + 5p = 2,300}\)
\(\mathrm{5s + 12p = 250}\)
\(\mathrm{s + p = 2,300}\)
\(\mathrm{12s + 5p = 250}\)
\(\mathrm{s + p = 2,300}\)