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A theater charges $80 for premium seats and $60 for standard seats. For a particular show, a total of 21...

GMAT Algebra : (Alg) Questions

Source: Prism
Algebra
Systems of 2 linear equations in 2 variables
MEDIUM
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Notes
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A theater charges \(\$80\) for premium seats and \(\$60\) for standard seats. For a particular show, a total of \(21\) tickets were sold, and the revenue from premium seats was \(\$280\) more than the revenue from standard seats. How many standard seats were sold?

  1. 7
  2. 10
  3. 11
  4. 14
A
7
B
10
C
11
D
14
Solution

1. TRANSLATE the problem information into mathematical language

  • Given information:
    • Premium seats cost $80 each, standard seats cost $60 each
    • Total tickets sold = 21
    • Premium seat revenue was $280 more than standard seat revenue
  • What we need to find: Number of standard seats sold

2. TRANSLATE the relationships into equations

Let \(\mathrm{s}\) = number of standard seats and \(\mathrm{p}\) = number of premium seats

  • From total tickets: \(\mathrm{p + s = 21}\)
  • From revenue relationship: Revenue from premium = Revenue from standard + 280
    So: \(\mathrm{80p = 60s + 280}\)

3. INFER the solution strategy

Since we have two equations with two unknowns, this is a system of equations. We can use substitution since the first equation easily gives us \(\mathrm{p}\) in terms of \(\mathrm{s}\).


4. SIMPLIFY by substitution and algebraic manipulation

  • From \(\mathrm{p + s = 21}\), we get: \(\mathrm{p = 21 - s}\)
  • Substitute into the revenue equation:
    \(\mathrm{80(21 - s) = 60s + 280}\)
  • Expand: \(\mathrm{1680 - 80s = 60s + 280}\)
  • Collect like terms: \(\mathrm{1680 - 280 = 60s + 80s}\)
  • Simplify: \(\mathrm{1400 = 140s}\)
  • Solve: \(\mathrm{s = 10}\)

Answer: B (10)




Why Students Usually Falter on This Problem


Most Common Error Path:

Weak TRANSLATE skill: Students often struggle to correctly interpret "revenue from premium seats was $280 more than revenue from standard seats" and write it as \(\mathrm{80p + 280 = 60s}\) instead of \(\mathrm{80p = 60s + 280}\).

This reverses the relationship and leads to:
\(\mathrm{80(21 - s) + 280 = 60s}\)
\(\mathrm{1680 - 80s + 280 = 60s}\)
\(\mathrm{1960 = 140s}\)
\(\mathrm{s = 14}\)

This may lead them to select Choice D (14).


Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up the equations but make algebraic errors, particularly when distributing or combining like terms. Common mistakes include sign errors or forgetting to distribute properly.

This leads to confusion and potentially guessing among the remaining answer choices.


The Bottom Line:

This problem requires careful reading to properly translate the revenue relationship, then systematic algebraic manipulation. The key insight is recognizing that "A was $280 more than B" means \(\mathrm{A = B + 280}\), not \(\mathrm{A + 280 = B}\).

Answer Choices Explained
A
7
B
10
C
11
D
14
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