AMC 8 · 2000 · #1

Grade 3 arithmetic

Archetype Arithmetic Expression Decomposition

multi-digit-arithmeticfraction-arithmeticlinear-equations-one-var identify-subproblems ↑ Prerequisites: multi-digit-arithmetic

📏 Short solution 💡 2 insights

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Problem

Aunt Anna is $42$ years old. Caitlin is $5$ years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?

$\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 16\qquad\mathrm{(C)}\ 17\qquad\mathrm{(D)}\ 21\qquad\mathrm{(E)}\ 37$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Aunt Anna is $42$. Brianna is half of Aunt Anna's age. Caitlin is $5$ years younger than Brianna. Find Caitlin's age.

Givens: Aunt Anna's age is $42$; Brianna's age is half of Aunt Anna's age; Caitlin's age is Brianna's age minus $5$; Answer choices: (A) $15$, (B) $16$, (C) $17$, (D) $21$, (E) $37$

Unknowns: Caitlin's age

Understand

Restated: Aunt Anna is $42$. Brianna is half of Aunt Anna's age. Caitlin is $5$ years younger than Brianna. Find Caitlin's age.

Givens: Aunt Anna's age is $42$; Brianna's age is half of Aunt Anna's age; Caitlin's age is Brianna's age minus $5$; Answer choices: (A) $15$, (B) $16$, (C) $17$, (D) $21$, (E) $37$

Plan

Primary tool: #7 Identify Subproblems

The question asks about Caitlin, but Caitlin's age is only described through Brianna, and Brianna's age is only described through Aunt Anna. Tool #7 (Identify Subproblems) says: when an unknown sits at the end of a chain of clues, split the work into two tiny subproblems — first find the middle person (Brianna), then use that to find the target (Caitlin). Each subproblem is one operation, so no algebra is needed.

Execute — Answer: B

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

Subproblem 1: find Brianna's age.
"Half as old as Aunt Anna" means divide Aunt Anna's $42$ by $2$.

$$\text{Brianna} = \dfrac{42}{2} = 21$$

💡 Dividing $42$ by $2$ is a Grade 3 multiplication/division-within-$100$ fact: $2 \times 21 = 42$.

#7 Identify Subproblems 2.OA.A.1 Step 2

Subproblem 2: find Caitlin's age.
"$5$ years younger than Brianna" means subtract $5$ from Brianna's $21$.

$$\text{Caitlin} = 21 - 5 = 16 \;\Rightarrow\; \textbf{(B)}$$

💡 $21 - 5 = 16$ is a Grade 2 within-$100$ subtraction fact.

[1] #7 3.OA.A.3 Subproblem 1: find Brianna's age. "Half as old as Aunt Anna" means divide Aunt A

[2] #7 2.OA.A.1 Subproblem 2: find Caitlin's age. "$5$ years younger than Brianna" means subtrac

Review

Reasonableness: Check the chain forward: Aunt Anna $42$, half of that is $21$ (Brianna), five less is $16$ (Caitlin). Each relationship holds. The size also makes sense: Caitlin is younger than Brianna, who is younger than Aunt Anna, and $16 < 21 < 42$. Choice (E) $37$ would be Aunt Anna minus $5$ — a misread. Choice (D) $21$ is Brianna's age, not Caitlin's. The answer is (B) $16$.

Alternative: Tool #3 (Eliminate Possibilities): Brianna is $21$, and Caitlin is younger than Brianna, so Caitlin's age must be less than $21$. That immediately removes (D) $21$ and (E) $37$. From (A) $15$, (B) $16$, (C) $17$, only (B) $16$ makes the gap to Brianna exactly $5$.

CCSS standards used (min grade 3)

2.OA.A.1 Use addition and subtraction within 100 to solve word problems (Subtracting $5$ from $21$ to find Caitlin's age is a Grade 2 word-problem subtraction within $100$.)
3.OA.A.3 Use multiplication and division within 100 to solve word problems (Halving $42$ to get Brianna's age is division within $100$ in a word-problem context.)

⭐ Two clues, two tiny steps: halve $42$, then subtract $5$. AMC 8 #1 only needs Grade 3 arithmetic when you split it into subproblems.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 3)

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