AMC 8 · 2000 · #1
Easy mode Grade 3Problem
Picture three people: Aunt Anna, Brianna, and Caitlin.
Aunt Anna is years old. Brianna is half as old as Aunt Anna. Caitlin is years younger than Brianna.
How old is Caitlin?
Pick an answer.
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
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 $2$.
💡 Dividing $42$ by $2$ is a Grade 3 multiplication/division-within-$100$ fact: $2 \times 21 = 42$.
2.OA.A.1 Step 2 - Subproblem 2: find Caitlin's age.
- "$5$ years younger than Brianna" means subtract $5$ from Brianna's $21$.
💡 $21 - 5 = 16$ is a Grade 2 within-$100$ subtraction fact.
3.OA.A.3 Subproblem 1: find Brianna's age. "Half as old as Aunt Anna" means divide Aunt A 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.1Use 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.3Use 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.
⭐ Two clues, two tiny steps: halve $42$, then subtract $5$. AMC 8 #1 only needs Grade 3 arithmetic when you split it into subproblems.