A coding contest awards points for solving problems as follows. For up to 12 problems, a competitor earns 50 points...
GMAT Algebra : (Alg) Questions
A coding contest awards points for solving problems as follows. For up to 12 problems, a competitor earns 50 points for the first problem solved and 20 points for each additional problem solved. Which of the following equations gives the total points \(\mathrm{P}\) for solving \(\mathrm{n}\) problems, where \(\mathrm{n}\) is a positive integer and \(\mathrm{n \leq 12}\)?
1. TRANSLATE the problem information
- Given information:
- First problem solved: 50 points
- Each additional problem: 20 points
- Need equation for total points P when solving n problems
• This tells us we have a base amount plus a variable amount
2. INFER the mathematical structure
- Key insight: "additional problems" means problems beyond the first one
- If we solve n problems total, then we have:
- 1 first problem worth 50 points
- \(\mathrm{(n-1)}\) additional problems worth 20 points each
• Total points = Points from first problem + Points from additional problems
3. Set up the equation
\(\mathrm{P = 50 + 20(n-1)}\)
4. SIMPLIFY the expression
\(\mathrm{P = 50 + 20(n-1)}\)
\(\mathrm{P = 50 + 20n - 20}\)
\(\mathrm{P = 20n + 30}\)
5. Verify with test cases
- When \(\mathrm{n = 1}\): \(\mathrm{P = 20(1) + 30 = 50}\) ✓
- When \(\mathrm{n = 2}\): \(\mathrm{P = 20(2) + 30 = 70 = 50 + 20}\) ✓
Answer: D
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skill: Misinterpreting what "additional" means in the context
Students often think every problem is worth the same amount and set up \(\mathrm{P = 50n}\) (thinking all n problems are worth 50 points each) or \(\mathrm{P = 20n}\) (thinking all problems are worth 20 points each). They miss that the first problem has special scoring.
This may lead them to eliminate all the given choices and guess, or misread the problem entirely.
Second Most Common Error:
Poor INFER reasoning: Setting up the wrong expression for additional problems
Students correctly identify that the first problem is special but write \(\mathrm{P = 50 + 20n}\) instead of \(\mathrm{P = 50 + 20(n-1)}\). They count all n problems as "additional" rather than recognizing that only \(\mathrm{(n-1)}\) are additional beyond the first.
This may lead them to select Choice C (\(\mathrm{P = 20n + 50}\)) after incorrectly rearranging their wrong equation.
The Bottom Line:
The trickiest part is recognizing that "additional" creates an off-by-one situation - when you solve n problems, only \(\mathrm{(n-1)}\) of them count as "additional" beyond the first special one.