AMC 10 · 2023 · #10

Grade 6 arithmetic
mean-median-mode-rangelinear-equations-two-varsystems-of-equations convert-to-algebraguess-and-checkidentify-subproblems ↑ Prerequisites: mean-median-mode-rangelinear-equations-one-var
📏 Medium solution 💡 2 insights
Problem
Maureen has taken n quizzes with current mean M (so her current total is n · M). If she scores 11 on the next quiz the mean rises to M + 1; if she scores 11 on each of the next three quizzes the mean rises to M + 2. Find M.

Pick an answer.

(A)
4
(B)
5
(C)
6
(D)
7
(E)
8

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Guess and Check

Five small whole-number choices (4, 5, 6, 7, 8) and a clean integer relationship make Tool #6 (Guess and Check) the fastest path: for each candidate M, find the n forced by the first condition, then test whether the second condition holds. Tool #3 (Eliminate Possibilities) catches us when a candidate forces a non-positive-integer n. Tool #13 (Convert to Algebra) becomes the verification path — set up both conditions as equations and solve the 2 × 2 linear system to confirm.

1STEP 1

Translate the first clue with the mean definition; expanding and canceling nM leaves the clean relation n + M = 10.

nM + 11 = (n+1)(M+1) = nM + n + M + 1 → n + M = 10
2STEP 2

Sweep the five choices: each candidate M reads off n = 10 - M, and every one gives a positive integer n.

M & n = 10-M & n valid? ; 4 & 6 & ✓ ; 5 & 5 & ✓ ; 6 & 4 & ✓ ; 7 & 3 & ✓ ; 8 & 2 & ✓
3STEP 3

Set up the second clue: three more 11's give total nM + 33 on n + 3 quizzes, which must equal (n+3)(M+2).

check: nM + 33 ?= (n+3)(M+2)
4STEP 4

Run the check on all five: only M = 7 (with n = 3) gives nM + 33 = 54 = (n+3)(M+2); every other candidate misses.

M & n & nM+33 & (n+3)(M+2) & match? ; 4 & 6 & 57 & 9 · 6 = 54 & × ; 5 & 5 & 58 & 8 · 7 = 56 & × ; 6 & 4 & 57 & 7 · 8 = 56 & × ; 7 & 3 & 54 & 6 · 9 = 54 & ✓ ; 8 & 2 & 49 & 5 · 10 = 50 & ×
5STEP 5

Only M = 7 survives both clues, with n = 3 quizzes taken — matching choice (D).

M = 7 → (D)
Answer
7
Verify directly with the winning numbers. Maureen has taken 3 quizzes with mean 7, so her total is 21. After one more quiz scoring 11, her total is 32 on 4 quizzes, mean 324\frac{32}{4} = 8 = 7 + 1 ✓. After three more quizzes of 11, her total is 21 + 33 = 54 on 6 quizzes, mean 546\frac{54}{6} = 9 = 7 + 2 ✓. Both conditions match exactly. Magnitude check: M = 7 is comfortably below 11, which is necessary for an extra 11 to pull the mean up.
💡Key takeaway

This AMC 10 problem only needs Grade 6 understanding of mean you already know — "total = number of quizzes × mean" turns each clue into one equation, and trying each answer choice locates M = 7 in under a minute.