AMC 8 · 2022 · #16

Grade 4 arithmeticalgebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

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

📏 Short solution 💡 3 insights

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Problem

Four numbers are written in a row. The average of the first two is $21,$ the average of the middle two is $26,$ and the average of the last two is $30.$ What is the average of the first and last of the numbers?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2022 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: Four numbers sit in a row. Call them $a, b, c, d$. We are told three two-number averages: $\tfrac{a+b}{2} = 21$, $\tfrac{b+c}{2} = 26$, and $\tfrac{c+d}{2} = 30$. The question asks for the average of the first and the last numbers, $\tfrac{a+d}{2}$.

Givens: Average of the first two numbers $\tfrac{a+b}{2} = 21$; Average of the middle two numbers $\tfrac{b+c}{2} = 26$; Average of the last two numbers $\tfrac{c+d}{2} = 30$; Answer choices: (A) $24$, (B) $25$, (C) $26$, (D) $27$, (E) $28$

Unknowns: The average of the first and last numbers, $\tfrac{a+d}{2}$

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #7 Identify Subproblems

Trying to solve for $a, b, c, d$ individually is hopeless — three equations, four unknowns. But we do not need the individual values; we only need $a+d$. Tool #16 (Change Focus / Complement) is the key insight: inside the whole sum $a+b+c+d$, the pair $a+d$ is exactly the *complement* of the middle pair $b+c$. If we know the whole and we know the middle, the outside is forced. Tool #7 (Identify Subproblems) breaks the work into three clean subproblems: (1) turn each given average into a pair sum by multiplying by $2$, (2) build the whole-sum from the first and last pair sums, (3) subtract the middle pair sum, then halve to get the desired average. No algebra is needed — only the definition of average and four-operation arithmetic.

Execute — Answer: B

#7 Identify Subproblems 3.OA.A.3 Step 1

Turn each average into a pair sum.
The average of two numbers is their sum divided by $2$, so each given average doubles to give the pair's sum.

$a+b = 21 \times 2 = 42$, $\;b+c = 26 \times 2 = 52$, $\;c+d = 30 \times 2 = 60$

💡 Reading "the average of two numbers is $21$" as "the two numbers add to $42$" is a Grade 3 multiplication word-problem move.

#7 Identify Subproblems 4.OA.A.3 Step 2

Add the first pair sum and the last pair sum.
The numbers $a, b, c, d$ each appear exactly once, so this gives the total of all four numbers.

$(a+b) + (c+d) = 42 + 60 = 102$, so $a+b+c+d = 102$

💡 Combining pair sums whose terms don't overlap is the Tool #7 "add the subproblem answers" move — pure multi-step arithmetic.

#16 Change Focus / Count the Complement 4.OA.A.3 Step 3

Use the complement viewpoint.
Inside the whole sum $a+b+c+d = 102$, the pair $b+c$ is the middle, and $a+d$ is what's left — the *complement* of the middle within the whole.
Since the middle pair sums to $52$, the outer pair must make up the rest.

$$a+d = (a+b+c+d) - (b+c) = 102 - 52 = 50$$

💡 When you know the whole and one part, the other part is just whole $-$ part — that's exactly the Tool #16 complement trick, dressed as Grade 4 subtraction.

#7 Identify Subproblems 3.OA.A.3 Step 4

Convert the pair sum $a+d = 50$ back into the desired average by dividing by $2$.

$$\tfrac{a+d}{2} = \tfrac{50}{2} = 25 \;\Rightarrow\; \textbf{(B)}$$

💡 Same definition as Step 1, in reverse: "sum is $50$, so the average of the two numbers is $25$" — Grade 3 division.

[1] #7 3.OA.A.3 Turn each average into a pair sum. The average of two numbers is their sum divid

[2] #7 4.OA.A.3 Add the first pair sum and the last pair sum. The numbers $a, b, c, d$ each appe

[3] #16 4.OA.A.3 Use the complement viewpoint. Inside the whole sum $a+b+c+d = 102$, the pair $b+

[4] #7 3.OA.A.3 Convert the pair sum $a+d = 50$ back into the desired average by dividing by $2$

Review

Reasonableness: The three given averages are $21, 26, 30$ — increasing as we slide the window of two from the left end to the right end of the row. So the numbers themselves trend upward, and we should expect the average of the leftmost and rightmost numbers ($a$ and $d$) to sit somewhere near the middle of that $21$-to-$30$ range. The answer $25$ lands right in that band — closer to the lower end, which fits because $a$ (small end) drags the average down. A concrete check: $a = 16, b = 26, c = 26, d = 34$ satisfies all three given averages and gives $\tfrac{a+d}{2} = \tfrac{16+34}{2} = 25$. ✓

Alternative: Tool #6 (Guess and Check) by picking convenient values. Set $b = 26$ (the middle average is a natural anchor). Then $a = 42 - 26 = 16$, $c = 52 - 26 = 26$, $d = 60 - 26 = 34$. Average of first and last: $\tfrac{16+34}{2} = 25$. Any other choice of $b$ shifts $a$ down and $d$ up by the same amount, so the average $\tfrac{a+d}{2}$ stays at $25$ — a nice invariant the problem is quietly built around.

CCSS standards used (min grade 4)

3.OA.A.3 Solve multiplication and division word problems within 100 (Translating between "average of two numbers" and "sum of two numbers" by multiplying or dividing by $2$ — once to go from each given average to a pair sum, and once at the end to go from the pair sum $50$ back to the average $25$.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Adding the first and last pair sums to get the whole ($42 + 60 = 102$) and then subtracting the middle pair sum ($102 - 52 = 50$) to isolate $a+d$ — multi-step whole-number arithmetic.)

⭐ This AMC 8 problem only needs Grade 4 multi-step arithmetic — add the outer pair sums, subtract the middle pair sum — that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 4)

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