Competition · AMC preparation · step 4 of 4
AMC 8 · 2000 · #1
Grade 3 arithmeticPick an answer.
AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Find Brianna's age
Subproblem 1: halve Aunt Anna's 42 to get Brianna's age, 21.
Dividing 42 by 2 is a Grade 3 multiplication/division-within-100 fact: 2 × 21 = 42.
Brianna is 21 years old, because she is half as old as the 42-year-old Aunt Anna.
▸ Why?
"Half as old as Aunt Anna" means splitting her 42 years into two equal parts, which is the division 42 ÷ 2.
▸ Why?
That division gives 21 because division reverses multiplication, and two groups of 21 rebuild 42 (2 × 21 = 42).
Find Caitlin's age
Subproblem 2: subtract 5 from Brianna's 21 to get Caitlin's age, 16.
21 - 5 = 16 is a Grade 2 within-100 subtraction fact.
2.OA.A.1Identify SubproblemsTwo clues, two tiny steps: halve 42, then subtract 5. AMC 8 #1 only needs Grade 3 arithmetic when you split it into subproblems.
- Find Brianna's age
- Find Caitlin's age
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