AMC 8 · 2000 · #1

Grade 3 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

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

📏 Short solution 💡 2 insights

📘 View easy version →

Problem

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

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy 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.

1STEP 1

Subproblem 1: halve Aunt Anna's 42 to get Brianna's age, 21.

Brianna =

\frac{42}{2}

= 21

2STEP 2

Subproblem 2: subtract 5 from Brianna's 21 to get Caitlin's age, 16.

Caitlin = 21 - 5 = 16 → (B)

Answer

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.

💡Key takeaway

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

More like this