Competition · AMC preparation · step 4 of 4
AMC 8 · 2017 · #17
Grade 4 algebraPick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
It is a multiple-choice AMC problem with only five candidate values for G, and each candidate is easy to plug back into the story. Tool #3 (Eliminate Possibilities) is the natural first move on any AMC multiple-choice problem — for every candidate G we can compute the number of chests two different ways (once from Plan A and once from Plan B) and keep only the candidate where the two answers agree. Tool #6 (Guess and Check) is what powers each individual test: guess a G, compute, and see if the two pictures line up. This is faster and more elementary than setting up a system of equations (Tool #13).
Turn both stories into equations
Turn each story into arithmetic: Plan A gives C = G/9 + 2, Plan B gives C = (G-3)/6.
Reading a multi-step word problem and turning each sentence into a small calculation is a Grade 4 multi-step word-problem skill.
4.OA.A.3Eliminate PossibilitiesTest choice A
Try (A) G = 9: Plan A wants 3 chests but Plan B wants 1 — a mismatch, so eliminate (A).
All four checks are just dividing and adding within 100, which is Grade 3 multiplication/division fluency.
3.OA.C.7Guess And CheckTest choice B
Try (B) G = 27: Plan A wants 5 chests but Plan B wants 4 — still a mismatch, so eliminate (B).
Same Grade 3 division facts (27 ÷ 9, 24 ÷ 6) used inside the elimination loop.
3.OA.C.7Guess And CheckTest choice C
Try (C) G = 45: both plans want the same 7 chests, so the counts agree and (C) survives.
Finding the unknown number of chests that makes both equations true is exactly Grade 3 "determine the unknown whole number in a multiplication or division equation."
Forty-five coins fits both stories, because that amount makes Plan A and Plan B call for the very same seven chests.
▸ Why?
Plan A puts 45 coins in stacks of 9, which fills 5 chests, and those 5 filled chests together with the 2 left empty make 7 chests in all.
▸ Why?
Sharing 45 coins into equal stacks of 9 comes to 5 stacks, because 5 stacks of 9 build back up to 45.
▸ Why?
Every chest is either filled or empty with nothing in between, so the 5 filled chests and the 2 empty chests add back to the whole row of chests.
▸ Why?
Plan B puts 6 coins in each of those 7 chests, using 42 coins, and the 3 coins left over bring the total back to 45.
▸ Why?
Seven chests holding 6 coins each is 42 coins, because 6 counted seven times piles up to 42.
▸ Why?
The coins split with no overlap into the 42 placed in chests and the 3 kept back, so those two parts rejoin to 45.
▸ Why?
There is only one real row of chests in the story, so the chest count Plan A points to and the chest count Plan B points to must be the same single number, and only 45 makes both point to 7.
Rule out the rest
(D) G = 63 gives 9 vs 10 and (E) G = 81 gives 11 vs 13 — both mismatch, so only (C) 45 works.
Knocking out the other choices and finishing the multi-step verification is the final Grade 4 word-problem move.
4.OA.A.3Eliminate PossibilitiesThis AMC 8 problem only needs Grade 4 multi-step word-problem skills you already know — just check each answer choice in both stories and keep the one that fits!
- Turn both stories into equations
- Test choice A
- Test choice B
- Test choice C
- Rule out the rest
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