πŸ‡ΊπŸ‡Έ AMC 8 ⇄ switch contest
2022 AMC 8

Problem 5

Problem 5 · 2022 AMC 8 Stretch
Logic & Word Problems ageswork-backwardsum-constraint

Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned 6 years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is 30. How many years older than Bella is Anna?

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Answer: C — 3 years older.
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Hint 1 of 2
The two ages you can actually pin down are Bella's and the kitten's — nail those today, and the total does the rest.
Still stuck? Show hint 2 →
Hint 2 of 2
Bella was 6 five years ago, so she's 11; the kitten was newborn, so it's 5. Subtract both from 30 to uncover Anna's age, then compare.
Show solution
Approach: pin down the ages you can know first, then let the total reveal the unknown
  1. Insight: the kitten is the easy clue, not a distraction — “newborn five years ago” means it's exactly 5 today. And Bella, 6 five years ago, is 11 today. Both ages are now fixed.
  2. Anna is the only mystery, and the total 30 hands her to us: Anna = 30 − 11 (Bella) − 5 (kitten) = 14.
  3. Anna − Bella = 14 − 11 = 3 years.
  4. Sanity check: 14 + 11 + 5 = 30. ✓
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