AMC 10 · 2021 · #5
Grade 6 arithmeticProblem
The quiz scores of a class with k>12 students have a mean of 8. The mean of a collection of 12 of these quiz scores is 14. What is the mean of the remaining quiz scores in terms of k?
Pick an answer.
AMC 10 2021 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.
Toolkit + CCSS Solution
Understand
Restated: A class of $k > 12$ students has quiz mean $8$. A particular group of $12$ students has mean $14$. Find the mean of the remaining $k - 12$ students, expressed in terms of $k$.
Givens: Class size $= k$ with $k > 12$; Mean of the whole class $= 8$, so total class score $= 8k$; A group of $12$ students has mean $14$, so their total $= 14 \times 12 = 168$; Answer choices: (A) $\tfrac{14 - 8}{k - 12}$, (B) $\tfrac{8k - 168}{k - 12}$, (C) $\tfrac{14}{12} - \tfrac{8}{k}$, (D) $\tfrac{14(k-12)}{k^2}$, (E) $\tfrac{14(k-12)}{8k}$
Unknowns: Mean of the remaining $k - 12$ students
Understand
Restated: A class of $k > 12$ students has quiz mean $8$. A particular group of $12$ students has mean $14$. Find the mean of the remaining $k - 12$ students, expressed in terms of $k$.
Givens: Class size $= k$ with $k > 12$; Mean of the whole class $= 8$, so total class score $= 8k$; A group of $12$ students has mean $14$, so their total $= 14 \times 12 = 168$; Answer choices: (A) $\tfrac{14 - 8}{k - 12}$, (B) $\tfrac{8k - 168}{k - 12}$, (C) $\tfrac{14}{12} - \tfrac{8}{k}$, (D) $\tfrac{14(k-12)}{k^2}$, (E) $\tfrac{14(k-12)}{8k}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #8 Analyze the Units, #9 Solve an Easier Related Problem
Tool #7 (Subproblems) breaks the mean into its two ingredients: a sum and a count. Compute each piece for the remaining group, then divide. Tool #8 (Units) keeps "points" and "students" straight so the final ratio has the correct shape "points / students". Tool #9 (Easier Problem) double-checks the algebra by plugging in a friendly number for $k$ (say $k = 13$) — if both the chosen formula and the direct computation agree, the answer is solid.
Execute — Answer: B
6.SP.A.3 Step 1 - Total score of the whole class.
- Mean $\times$ count $=$ sum.
- So total $= 8 \times k = 8k$ points.
💡 Mean is the single number that, multiplied by the count, gives the total — Grade 6 "measure of center summarizes all values".
6.SP.A.3 Step 2 - Total score of the $12$-student group.
- $\text{group total} = 14 \times 12 = 168$ points.
💡 Same mean trick applied to a sub-group — multiply mean by count to get the sum.
6.EE.A.2 Step 3 - Total score of the remaining students.
- Subtract: $\text{remaining total} = 8k - 168$ points.
- The remaining count is $k - 12$ students.
💡 Writing the leftover sum and count as $8k - 168$ and $k - 12$ is Grade 6 "write expressions where letters stand for numbers".
6.EE.A.2 Step 4 - Divide sum by count to get the remaining mean.
- $\dfrac{8k - 168}{k - 12}$.
- This matches choice (B).
💡 Units check: $\tfrac{\text{points}}{\text{students}} = \text{points per student}$ — the right shape for a mean.
6.SP.A.3 Total score of the whole class. Mean $\times$ count $=$ sum. So total $= 8 \time 6.SP.A.3 Total score of the $12$-student group. $\text{group total} = 14 \times 12 = 168$ 6.EE.A.2 Total score of the remaining students. Subtract: $\text{remaining total} = 8k - 6.EE.A.2 Divide sum by count to get the remaining mean. $\dfrac{8k - 168}{k - 12}$. This Review
Reasonableness: Test with $k = 13$ (so only one student remains). Class total $= 8 \times 13 = 104$. Group of $12$ total $= 168$. But $168 > 104$, which means the remaining one student must have $104 - 168 = -64$ points — possible only if scores can be negative. The formula gives $\tfrac{8 \cdot 13 - 168}{13 - 12} = \tfrac{-64}{1} = -64$ — consistent. For $k = 24$: remaining mean $= \tfrac{192 - 168}{12} = \tfrac{24}{12} = 2$, also consistent with eyeballing (the $12$ high scorers pulled the average up, so the rest sits below $8$). Choice (B) survives.
Alternative: Tool #3 (Eliminate Possibilities) plus Tool #9 (Easier Problem). Plug $k = 24$ into each choice and compute: (A) $\tfrac{6}{12} = 0.5$, (B) $\tfrac{24}{12} = 2$, (C) $\tfrac{14}{12} - \tfrac{8}{24} \approx 0.83$, (D) $\tfrac{14 \cdot 12}{576} \approx 0.29$, (E) $\tfrac{14 \cdot 12}{192} = 0.875$. Direct mean computation: class total $192$, group total $168$, so remaining mean $= \tfrac{24}{12} = 2$. Only (B) hits.
CCSS standards used (min grade 6)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Writing the remaining sum as $8k - 168$, the remaining count as $k - 12$, and the final mean as $\tfrac{8k - 168}{k - 12}$.)6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Using "mean $\times$ count $=$ sum" both for the full class ($8 \times k$) and the $12$-student group ($14 \times 12$).)
⭐ This AMC 10 problem only needs Grade 6 "mean $\times$ count $=$ sum" — take the class total ($8k$) minus the $12$-student total ($168$), divide by the leftover $k - 12$, and you get $\tfrac{8k - 168}{k - 12}$.
⭐ This AMC 10 problem only needs Grade 6 "mean $\times$ count $=$ sum" — take the class total ($8k$) minus the $12$-student total ($168$), divide by the leftover $k - 12$, and you get $\tfrac{8k - 168}{k - 12}$.
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