AMC 10 · 2021 · #19

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

mean-median-mode-rangesystems-of-equationslinear-equations-two-vardecimal-arithmetic convert-to-algebraidentify-subproblems ↑ Prerequisites: mean-median-mode-rangesystems-of-equations

📏 Medium solution 💡 3 insights

Problem

Suppose that $S$ is a finite set of positive integers. If the greatest integer in $S$ is removed from $S$ , then the average value (arithmetic mean) of the integers remaining is $32$ . If the least integer in $S$ is also removed, then the average value of the integers remaining is $35$ . If the greatest integer is then returned to the set, the average value of the integers rises to $40$ . The greatest integer in the original set $S$ is $72$ greater than the least integer in $S$ . What is the average value of all the integers in the set $S$ ?

Pick an answer.

(A)

36.2

(B)

36.4

(C)

36.6

(D)

36.8

(E)

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.

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Toolkit + CCSS Solution

Understand

Restated: $S$ is a finite set of positive integers. Three averaging conditions are given: remove the greatest integer $G$ and the average drops to $32$; remove $L$ (the least) as well and the average becomes $35$; put $G$ back (so $L$ is the only one removed) and the average is $40$. Also, $G - L = 72$. Find the average of all integers in the original set $S$.

Givens: $\dfrac{\Sigma - G}{n - 1} = 32$ (drop greatest); $\dfrac{\Sigma - G - L}{n - 2} = 35$ (drop greatest and least); $\dfrac{\Sigma - L}{n - 1} = 40$ (drop least only); $G - L = 72$; Choices: (A) $36.2$, (B) $36.4$, (C) $36.6$, (D) $36.8$, (E) $37$

Unknowns: $\dfrac{\Sigma}{n}$ — the average of all integers in $S$

Understand

Plan

Primary tool: #13 Convert to Algebra

Secondary: #7 Identify Subproblems

Tool #13 (Algebra) — the four sentences are tailor-made to be turned into four equations in $n$, $\Sigma$, $G$, $L$. Once written, the system isn't messy: the two "average over $n-1$" lines differ only in whether $G$ or $L$ is removed, so subtracting them isolates $G - L = 8(n - 1)$ and uses condition (4) to get $n$ in one step. Tool #7 (Identify Subproblems) sequences the unknowns: first $n$, then $L$ (from the $n - 2$ equation), then $\Sigma$ (back into one of the $n - 1$ equations). Finally divide $\Sigma$ by $n$ for the answer. Each sub-step is just careful arithmetic — no need for complex algebra moves.

Execute — Answer: D

#13 Convert to Algebra 6.EE.B.7 Step 1

Let $n = $ number of integers in $S$ and $\Sigma = $ total sum.
Translate the four English sentences into four algebraic equations.

$$\begin{aligned}(1)\ \Sigma - G &= 32(n-1)\\(2)\ \Sigma - G - L &= 35(n-2)\\(3)\ \Sigma - L &= 40(n-1)\\(4)\ G - L &= 72\end{aligned}$$

💡 "Average $=$ sum $\div$ count" turns each averaging condition into a single linear equation.

#7 Identify Subproblems 6.EE.B.7 Step 2

Subtract equation (1) from equation (3): the $\Sigma$'s cancel, leaving $G - L$ on the left and $8(n-1)$ on the right.

$$(3) - (1):\ G - L = (40 - 32)(n - 1) = 8(n - 1)$$

💡 Two "$n-1$" averages differ only by which extreme is removed; their gap in sum is exactly $G - L$.

#13 Convert to Algebra 6.EE.B.7 Step 3

Combine with equation (4): $72 = 8(n - 1)$, so $n - 1 = 9$ and $n = 10$.
So the original set has $10$ integers.

$$72 = 8(n - 1) \Rightarrow n - 1 = 9 \Rightarrow n = 10$$

💡 Divide both sides by $8$ to get the count instantly.

#7 Identify Subproblems 6.EE.B.7 Step 4

Plug $n = 10$ into (1) and (2).
From (1): $\Sigma - G = 32 \cdot 9 = 288$.
From (2): $\Sigma - G - L = 35 \cdot 8 = 280$.
Subtracting gives $L = 288 - 280 = 8$.

$$\Sigma - G = 288,\quad \Sigma - G - L = 280 \Rightarrow L = 8$$

💡 Once $n$ is known, two equations differing only in $L$ pin down $L$ in one subtraction.

#13 Convert to Algebra 6.EE.B.7 Step 5

Plug $L = 8$ into equation (3): $\Sigma - 8 = 40 \cdot 9 = 360$, so $\Sigma = 368$.

$$\Sigma = 40 \cdot 9 + L = 360 + 8 = 368$$

💡 Once $L$ and $n - 1$ are known, the full sum is one addition.

#13 Convert to Algebra 5.NBT.B.7 Step 6

Average of all of $S$: $\dfrac{\Sigma}{n} = \dfrac{368}{10} = 36.8$.
This matches choice (D).

$$\dfrac{\Sigma}{n} = \dfrac{368}{10} = 36.8\ \Rightarrow\ (D)$$

💡 Divide by $10$ — shift the decimal one place.

[1] #13 6.EE.B.7 Let $n = $ number of integers in $S$ and $\Sigma = $ total sum. Translate the fo

[2] #7 6.EE.B.7 Subtract equation (1) from equation (3): the $\Sigma$'s cancel, leaving $G - L$

[3] #13 6.EE.B.7 Combine with equation (4): $72 = 8(n - 1)$, so $n - 1 = 9$ and $n = 10$. So the

[4] #7 6.EE.B.7 Plug $n = 10$ into (1) and (2). From (1): $\Sigma - G = 32 \cdot 9 = 288$. From

[5] #13 6.EE.B.7 Plug $L = 8$ into equation (3): $\Sigma - 8 = 40 \cdot 9 = 360$, so $\Sigma = 36

[6] #13 5.NBT.B.7 Average of all of $S$: $\dfrac{\Sigma}{n} = \dfrac{368}{10} = 36.8$. This matche

Review

Reasonableness: Verify with $G$: equation (4) gives $G = L + 72 = 8 + 72 = 80$. Check equation (1): $\Sigma - G = 368 - 80 = 288 = 32 \cdot 9$ ✓. Check equation (2): $\Sigma - G - L = 368 - 80 - 8 = 280 = 35 \cdot 8$ ✓. Check equation (3): $\Sigma - L = 368 - 8 = 360 = 40 \cdot 9$ ✓. All four conditions hold, and the answer $36.8$ sits squarely between $32$ (lowest reported avg) and $40$ (highest) — exactly the kind of number a balanced average should produce.

Alternative: Tool #6 (Guess and Check) on the answer choices — each option corresponds to a specific $\Sigma = 10 \cdot 36.X$, and the $G - L = 72$ + average $= 32$ over $n - 1$ relation can be cross-tested. But with four clean equations and four unknowns, direct algebra is faster than back-substituting five candidates.

CCSS standards used (min grade 6)

6.EE.B.7 Solve real-world problems by writing and solving equations of the form $px = q$ (Translating the four averaging / gap conditions into four linear equations in $n, \Sigma, G, L$ and solving by elimination and substitution.)
5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Computing the final average $368 \div 10 = 36.8$.)

⭐ This AMC 10 problem only needs Grade 6 "sum $=$ average $\times$ count" equations you already know! Subtract two $n - 1$ averages to get $G - L = 8(n-1) = 72$, so $n = 10$. Then $\Sigma = 368$ and the answer is $\dfrac{368}{10} = 36.8$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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