Competition · AMC preparation · step 4 of 4

AMC 8 · 2022 · #5

Grade 2 arithmeticalgebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation
linear-equations-one-varmulti-digit-arithmeticformula-substitution convert-to-algebraidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticlinear-equations-one-var
📏 Short solution 💡 2 insights
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Problem
Anna and Bella share a birthday. Five years ago today, Bella turned 6 and was given a newborn kitten as a present. Right now, the three ages — Anna's, Bella's, and the kitten's — add up to 30. How many years older than Bella is Anna?

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Guess and Check

Two of the three current ages are already fixed once we work backwards five years: Bella is 6 + 5 = 11 and the kitten is 0 + 5 = 5. That leaves Anna as the only unknown. Tool #11 (Work Backwards) carries the two known ages forward five years; Tool #6 (Guess and Check) then lets us try the five answer-choice age differences on top of Bella's 11 to see which makes Anna + 11 + 5 = 30. This keeps the solution arithmetic and avoids reaching for Tool #13 (Algebra), which is overkill for a Grade 2 word problem.

1STEP 1

Move the ages ahead five years

Roll each past age forward five years: Bella is 6 + 5 = 11 today, and the newborn kitten is 0 + 5 = 5 today.

Bella = 6 + 5 = 11, kitten = 0 + 5 = 5
2STEP 2

Add the two known ages

Bella and the kitten together take up 11 + 5 = 16 of the total, leaving the rest of the 30 for Anna.

11 + 5 = 16
3STEP 3

Test each answer choice

Test each choice as the gap d: Anna is 11 + d, so the sum is 27 + d, which hits 30 only when d = 3.

d = 1 → 12 + 11 + 5 = 28; d = 2 → 13 + 11 + 5 = 29; d = 3 → 14 + 11 + 5 = 30; d = 4 → 31; d = 5 → 32
4STEP 4

Read off the age gap

The gap that lands the sum on 30 is d = 3, so Anna is 3 years older than Bella.

Anna - Bella = 14 - 11 = 3 → (C)
Answer
3
All three current ages are whole, positive, and reasonable: Bella 11, kitten 5, Anna 14. Their sum is 14 + 11 + 5 = 30, matching the problem exactly. The gap 3 is also one of the listed choices, so the arithmetic and the multiple-choice structure agree.
💡Key takeaway

This AMC 8 problem only needs Grade 2 add-and-subtract-within-100 word-problem skills you already know!

  • Move the ages ahead five years
  • Add the two known ages
  • Test each answer choice
  • Read off the age gap

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