AMC 8 · 2022 · #5
Grade 2 arithmeticalgebraPick an answer.
AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Roll each past age forward five years: Bella is 6 + 5 = 11 today, and the newborn kitten is 0 + 5 = 5 today.
Adding 5 to small ages to roll the clock forward is a Grade 1 add-within-20 word-problem move.
1.OA.A.1Work BackwardsBella and the kitten together take up 11 + 5 = 16 of the total, leaving the rest of the 30 for Anna.
Combining the two known ages into one running total is a one-step "put together" word problem within 100.
2.OA.A.1Work BackwardsTest each choice as the gap d: Anna is 11 + d, so the sum is 27 + d, which hits 30 only when d = 3.
Plugging each candidate gap into a one-step addition check is exactly the Grade 2 "add/subtract within 100" word-problem skill.
2.OA.A.1Guess And CheckThe gap that lands the sum on 30 is d = 3, so Anna is 3 years older than Bella.
Finding the difference of two small whole numbers is a Grade 1 subtract-within-20 skill.
1.OA.A.1Guess And CheckThis AMC 8 problem only needs Grade 2 add-and-subtract-within-100 word-problem skills you already know!
More like this
Same archetype — closest grade level first.
