AMC 10 · 2010 · #15

Grade 7 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
linear-diophantineoptimizationsystems-of-equations extreme-principle ↑ Prerequisites: systems-of-equations
📏 Medium solution 💡 3 insights
Problem
A math contest has 50 multiple-choice questions. A correct answer scores +4 points, a blank scores 0 points, and a wrong answer scores -1 point. Jesse's total score was 99. What is the largest possible number of questions Jesse answered correctly?

Pick an answer.

(A)
25
(B)
27
(C)
29
(D)
31
(E)
33

AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The three kinds of answers — correct, wrong, and blank — are unknown counts tied together by two facts (they add to 50, and they produce a score of 99). Naming them with letters (Tool #4, Introduce a Variable) turns the word problem into two clean equations. The question asks for a maximum, which is the signal for Tool #14 (Extreme Principle): push the number of correct answers as high as possible while the leftover counts stay non-negative. The key lever is that blanks cannot be negative, which caps how many questions correct and wrong can use up. We finish by checking the winning value against the answer choices (Tool #3, Eliminate Possibilities). No advanced machinery is needed — just setting up and solving one linear inequality.

1STEP 1

Name the three counts

Let c be the correct count, w the wrong count, b the blanks. The three add to 50, and the score says 4c minus w is 99.

c + w + b = 50 4c - w = 99
2STEP 2

Blanks can't be negative

Push c as high as it will go. Blanks can't be negative, so correct plus wrong can never use up more than all 50 questions.

b = 50 - c - w ≥ 0 → c + w ≤ 50
3STEP 3

Replace wrong with correct

Solve the score equation for w: w = 4c - 99. Substituting into that ceiling leaves 5c ≤ 149 — one unknown, one inequality.

w = 4c - 99 → c + (4c - 99) ≤ 50 → 5c - 99 ≤ 50 → 5c ≤ 149
4STEP 4

Take the biggest whole number

Divide by 5: c ≤ 29.8. A count of questions must be a whole number, so the cap is 29.

c ≤ 149/5 = 29.8 → c_max = 29
5STEP 5

Check that 29 really works

Check c = 29: then w = 17 and b = 4, and 4(29) - 17 = 99, so the ceiling is really reached.

c = 29, w = 17, b = 4: 4(29) - 17 = 99 ✓
Answer
29
The bound 5c ≤ 149 gives c ≤ 29.8, and 29 is the largest whole number under that cap — and it is realized exactly by 29 correct, 17 wrong, 4 blank, scoring 4(29) - 17 = 99. Pushing to c = 30 would need w = 4(30) - 99 = 21, so c + w = 51 ＞ 50, which is impossible on a 50-question test. That confirms 29 is both attainable and maximal, matching choice (C).
💡Key takeaway

Name the unknown counts, use 'blanks can't be negative' to build one inequality, and the biggest whole number that fits is your answer.

  • Name the three counts
  • Blanks can't be negative
  • Replace wrong with correct
  • Take the biggest whole number
  • Check that 29 really works

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