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2005 AMC 10B Problem 6

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2005 AMC 10B Problem 6AMC 10 Step by Step
Name:                             Date: September 13, 2026
Ratios, Percents & Averages · difficulty 2 of 5 · about 2 min · position P1-10

At the beginning of the school year, Lisa's goal was to earn an AA on at least 80%80\% of her 5050 quizzes for the year. She earned an AA on 2222 of the first 3030 quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an AA ?

  1. A)

    11

  2. B)

    22

  3. C)

    33

  4. D)

    44

  5. E)

    55

Answer

B

Key insight

The goal is 40 A's out of 50; she has 22, so 18 of the last 20 quizzes must be A's, leaving at most 2 non-A's.

Solution

Her target is 80%80\% of 5050 quizzes, which is 0.8⋅50=400.8 \cdot 50 = 40 A's for the year.

She already has 2222 A's, so she still needs 40−22=1840 - 22 = 18 more.

There are 50−30=2050 - 30 = 20 quizzes left. If 1818 of them must be A's, then at most 20−18=220 - 18 = 2 can be below an A.

The answer is (B) 2\boxed{\textbf{(B)}\ 2}.

Why this works

Percent goals become concrete once converted to a count of the whole. Then the problem is pure bookkeeping: needed minus achieved gives what is still required, and remaining minus required gives the slack. Always ask "percent of what?" before computing.

The trap

Applying 80% to the remaining 20 quizzes (16 A's, 4 misses) instead of to all 50.

Common mistakes

  • Applying 80% to the remaining 20 quizzes (16 A's, 4 misses) instead of to all 50.
  • Reporting 1818 (A's still needed) or mixing up "at most" with "at least."

Techniques

Set up the equation/formula and compute; no special trick needed