AMC 10 · 2023 · #10
Grade 6 arithmeticProblem
Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an on the next quiz, her mean will increase by . If she scores an on each of the next three quizzes, her mean will increase by . What is the mean of her quiz scores currently?
Pick an answer.
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
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) $8$
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
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.
💡 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.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.
💡 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.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.
💡 Same definition-of-mean check, now applied to the "three more quizzes" scenario. Grade 6 statistics.
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.
💡 Plug-in arithmetic on five small candidates — Grade 3 multi-step calculation, fast in your head.
6.EE.B.6 Step 5 - Only $M = 7$ passes both clues, with $n = 3$ quizzes taken so far.
- That matches choice (D).
💡 One candidate survives the second clue; eliminate the rest — Grade 6 "solve for the value that satisfies all conditions".
6.SP.B.5 Translate the first clue into a number sentence. Current total is $nM$; one more 6.EE.B.6 Now sweep the five answer choices. For each candidate $M$, the first clue gives 6.SP.B.5 Test the second clue. After three more quizzes of $11$, the new total is $nM + 3 3.OA.D.8 Run the check on each candidate. The left side is $nM + 33$, the right side is $ 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.5Summarize 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.6Use 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.8Solve 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.
⭐ 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.