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2010 AMC 10B Problem 15

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2010 AMC 10B Problem 15AMC 10 Step by Step
Name:                             Date: September 13, 2026
Linear Equations & Word Problems · difficulty 3 of 5 · about 3 min · position P11-15

On a 5050 -question multiple choice math contest, students receive 44 points for a correct answer, 00 points for an answer left blank, and −1-1 point for an incorrect answer. Jesse’s total score on the contest was 9999 . What is the maximum number of questions that Jesse could have answered correctly?

  1. A)

    2525

  2. B)

    2727

  3. C)

    2929

  4. D)

    3131

  5. E)

    3333

Answer

C

Key insight

With c correct and w wrong, 4c-w = 99 and c+w <= 50; substituting w = 4c-99 gives c <= 29.

Solution

Let cc be the number correct and ww the number wrong; the remaining 50−c−w≥050 - c - w \ge 0 questions are blank. The score condition is
4c−w=99⟹w=4c−99.4c - w = 99 \quad\Longrightarrow\quad w = 4c - 99.
Substituting into c+w≤50c + w \le 50:
c+4c−99≤50⟹5c≤149⟹c≤29.c + 4c - 99 \le 50 \quad\Longrightarrow\quad 5c \le 149 \quad\Longrightarrow\quad c \le 29.
With c=29c = 29 we get w=17w = 17 and 44 blanks, all nonnegative, and 4⋅29−17=994\cdot 29 - 17 = 99.

The answer is (C) 29\boxed{\textbf{(C)}\ 29}.

Why this works

To maximize the number of correct answers at a fixed score, extra correct answers must be offset by wrong ones, which use up questions. Writing ww in terms of cc turns the question cap into a bound on cc alone; then verify the boundary value is realizable.

Alternative approach

Test the choices from the top: c=33c = 33 needs w=33w = 33, too many questions; c=31c = 31 needs w=25w = 25, total 56>5056 > 50; c=29c = 29 needs w=17w = 17, total 46≤5046 \le 50, so it works.

The trap

Forgetting that wrong answers cost a point and using 4c = 99, or ignoring the 50-question cap and choosing a larger c.

Common mistakes

  • Forgetting that wrong answers cost a point and using 4c = 99, or ignoring the 50-question cap and choosing a larger c.
  • Solving 5c≤1495c \le 149 and rounding up to 3030; c=30c = 30 needs w=21w = 21, but 30+21=5130 + 21 = 51 exceeds 5050.

Techniques

Bound the quantity above/below or estimate to pin it down · Consider the largest/smallest element or boundary case