AMC 10 · 2002 · #6
Grade 5 arithmeticPick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The correct answer depends on the original number, which is hidden — but we are handed the END of Cindy's wrong procedure (the result 43). Tool #11 (Work Backwards) is built for exactly this: run her two operations in reverse, undoing each with its opposite, to recover the starting number. Tool #7 (Identify Subproblems) then makes the job two clean stages: first recover the number, then feed that number through the correct procedure. The tempting trap is to answer 138 (choice E), which is only the original number — not the corrected result the question actually asks for.
Recover the original number
Undo Cindy's moves in reverse with opposites: 43 × 3 = 129, then 129 + 9 = 138, the original number.
Running a chain of operations backwards with their opposites lands you exactly where you started.
Running a chain of operations backwards with their opposites lands you exactly where you started.
▸ Why?
Each operation is undone by its opposite, so reversing the order recovers the original number.
▸ Why?
Doing the same thing to both sides keeps the statement true at every step of the way back.
Apply the correct procedure
Now run 138 through the correct instructions: 138 - 3 = 135, then 135 ÷ 9 = 15.
Once the starting number is known, the correct result is just a matter of doing the right two operations in order.
5.NBT.B.6Identify SubproblemsMatch to a choice
The corrected result 15 is choice (A); 138 (E) and 43 (C) sit there to catch anyone who stops early.
The question asks for the corrected answer, not the original number or the wrong answer.
4.OA.A.3Identify SubproblemsWhen you know the end result, undo each step with its opposite to get back to the start — then read the question again to answer what it truly asks.
- Recover the original number
- Apply the correct procedure
- Match to a choice
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