AMC 10 · 2023 · #10

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

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 is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Maureen has taken $n$ quizzes with current mean $M$ (so her current total is $n \cdot M$). 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 $M$.

Givens: Current mean is $M$ after $n$ quizzes; total so far is $nM$; One more $11$ pushes the mean to $M+1$ (now $n+1$ quizzes, total $nM + 11$); Three more $11$'s push the mean to $M+2$ (now $n+3$ quizzes, total $nM + 33$); Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$

Unknowns: The current mean $M$

Understand

Plan

Primary tool: #6 Guess and Check

Secondary: #3 Eliminate Possibilities, #13 Convert to Algebra

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 $2 \times 2$ linear system to confirm.

Execute — Answer: D

#13 Convert to Algebra 6.SP.B.5 Step 1

Translate the first clue into a number sentence.
Current total is $nM$; one more $11$ makes the new total $nM + 11$ on $n+1$ quizzes, and the new mean is $M+1$.
Multiply out and cancel $nM$ from both sides.

$$nM + 11 = (n+1)(M+1) = nM + n + M + 1 \;\Rightarrow\; n + M = 10$$

💡 The definition of mean — total ÷ count — turns the first clue into one simple sentence: $n + M = 10$. Grade 6 "summarize a data set with mean" thinking.

#6 Guess and Check 6.EE.B.6 Step 2

Now sweep the five answer choices.
For each candidate $M$, the first clue gives $n = 10 - M$.
We only need $n$ to be a positive integer.

$$\begin{array}{c|c|c} M & n = 10-M & n \text{ valid?} \\ \hline 4 & 6 & \checkmark \\ 5 & 5 & \checkmark \\ 6 & 4 & \checkmark \\ 7 & 3 & \checkmark \\ 8 & 2 & \checkmark \end{array}$$

💡 Each guess picks a candidate $M$ and reads $n$ off the first clue — Grade 6 "use a variable to record an unknown" worked one number at a time.

#6 Guess and Check 6.SP.B.5 Step 3

Test the second clue.
After three more quizzes of $11$, the new total is $nM + 33$ on $n + 3$ quizzes; the new mean must equal $M + 2$.
Equivalently the total must equal $(n+3)(M+2)$.
Compute both sides for each $(M, n)$ pair.

$$\text{check: } nM + 33 \stackrel{?}{=} (n+3)(M+2)$$

💡 Same definition-of-mean check, now applied to the "three more quizzes" scenario. Grade 6 statistics.

#6 Guess and Check 3.OA.D.8 Step 4

Run the check on each candidate.
The left side is $nM + 33$, the right side is $(n+3)(M+2)$.
For $M = 7, n = 3$: $nM + 33 = 21 + 33 = 54$ and $(n+3)(M+2) = 6 \cdot 9 = 54$.
Match.
All other candidates fail.

$$\begin{array}{c|c|c|c|c} M & n & nM+33 & (n+3)(M+2) & \text{match?} \\ \hline 4 & 6 & 57 & 9 \cdot 6 = 54 & \times \\ 5 & 5 & 58 & 8 \cdot 7 = 56 & \times \\ 6 & 4 & 57 & 7 \cdot 8 = 56 & \times \\ 7 & 3 & 54 & 6 \cdot 9 = 54 & \checkmark \\ 8 & 2 & 49 & 5 \cdot 10 = 50 & \times \end{array}$$

💡 Plug-in arithmetic on five small candidates — Grade 3 multi-step calculation, fast in your head.

#3 Eliminate Possibilities 6.EE.B.6 Step 5

Only $M = 7$ passes both clues, with $n = 3$ quizzes taken so far.
That matches choice (D).

$$M = 7 \;\Rightarrow\; \textbf{(D)}$$

💡 One candidate survives the second clue; eliminate the rest — Grade 6 "solve for the value that satisfies all conditions".

[1] #13 6.SP.B.5 Translate the first clue into a number sentence. Current total is $nM$; one more

[2] #6 6.EE.B.6 Now sweep the five answer choices. For each candidate $M$, the first clue gives

[3] #6 6.SP.B.5 Test the second clue. After three more quizzes of $11$, the new total is $nM + 3

[4] #6 3.OA.D.8 Run the check on each candidate. The left side is $nM + 33$, the right side is $

[5] #3 6.EE.B.6 Only $M = 7$ passes both clues, with $n = 3$ quizzes taken so far. That matches

Review

Reasonableness: 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 $\tfrac{32}{4} = 8 = 7 + 1$ ✓. After three more quizzes of $11$, her total is $21 + 33 = 54$ on $6$ quizzes, mean $\tfrac{54}{6} = 9 = 7 + 2$ ✓. Both conditions match exactly. Magnitude check: $M = 7$ is comfortably below $11$, which is necessary for an extra $11$ to pull the mean up.

Alternative: Tool #13 (Convert to Algebra) gives the same answer without testing each choice. Expand the second condition: $nM + 33 = (n+3)(M+2) = nM + 2n + 3M + 6$, so $2n + 3M = 27$. Combined with $n + M = 10$ from the first condition, subtract $2 \cdot (n+M) = 20$ from $2n + 3M = 27$ to get $M = 7$, then $n = 3$. Cleanest algebra, identical answer (D).

CCSS standards used (min grade 6)

6.SP.B.5 Summarize numerical data sets in relation to their context (Using the definition of mean (total ÷ count) to translate each scenario into an equation between total and (new count) × (new mean).)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Introducing $n$ (number of quizzes so far) and $M$ (current mean) and reading off $n = 10 - M$ from the first clue to test each candidate $M$.)
3.OA.D.8 Solve two-step word problems using the four operations (Plug-in arithmetic checking $nM + 33 \stackrel{?}{=} (n+3)(M+2)$ on each candidate — small whole-number multi-step calculations.)

⭐ 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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