AMC 10 · 2021 · #5

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

mean-median-mode-rangefraction-arithmeticformula-substitution identify-subproblemsdimensional-analysis ↑ Prerequisites: mean-median-mode-range

📏 Medium solution 💡 1 insight

Problem

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.

Pick an answer.

(A)

$~\frac{14-8}{k-12}$

(B)

$~\frac{8k-168}{k-12}$

(C)

$~\frac{14}{12} - \frac{8}{k}$

(D)

$~\frac{14(k-12)}{k^2}$

(E)

$~\frac{14(k-12)}{8k}$

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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.

1STEP 1

Mean × count = sum, so the whole class totals 8k points.

class total = 8 × k = 8k

2STEP 2

Same rule on the group of 12: their total is 168 points.

group total = 14 × 12 = 168

3STEP 3

Subtract: the leftover sum is 8k - 168 over k - 12 students.

remaining total = 8k - 168, remaining count = k - 12

4STEP 4

Divide sum by count: the remaining mean is $\frac{8k - 168}{k - 12}$ — choice (B).

remaining mean =

\frac{8k - 168}{k - 12}

→ (B)

Answer

\frac{8k-168}{k-12}

Test with k = 13 (so only one student remains). Class total = 8 × 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

\frac{8 · 13 - 168}{13 - 12}

\frac{-64}{1}

= -64 — consistent. For k = 24: remaining mean =

\frac{192 - 168}{12}

\frac{24}{12}

= 2, also consistent with eyeballing (the 12 high scorers pulled the average up, so the rest sits below 8). Choice (B) survives.

💡Key takeaway

This AMC 10 problem only needs Grade 6 "mean × count = sum" — take the class total (8k) minus the 12-student total (168), divide by the leftover k - 12, and you get $\frac{8k - 168}{k - 12}$ .

More like this