Competition · AMC preparation · step 4 of 4
AMC 8 · 2012 · #11
Grade 6 arithmeticPick an answer.
AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Three conditions must all hold at once: mean, median, and unique mode are equal. The cleanest path is Tool #7 (Identify Subproblems) — pin down the mode first (it constrains itself), then use it to force the mean. Because 6 already appears twice and no other number repeats, the mode is forced to be 6 — that single observation collapses the problem. Tool #3 (Eliminate Possibilities) is a natural double-check: scan the five answer choices and drop any that would create a tied mode or shift the mean off 6. Tool #6 (Guess and Check) backs up the arithmetic by plugging the winning x back into the list.
Find the mode
Only 6 already repeats (twice) while every other value appears once, so whatever x turns out to be, the unique mode is 6.
Splitting the three conditions and tackling the mode first is the Tool #7 move — one subproblem locks in a number we can use everywhere else.
6.SP.B.5Identify SubproblemsUse the mean to solve for x
The mean is 6 too, so the seven numbers total 7 × 6 = 42; subtract the six knowns (31) to get x = 11.
The mean is just (sum)/(count). Multiplying both sides by the count turns the average condition into a simple sum-to-42 subproblem.
The seven numbers must add up to 42.
▸ Why?
The mean is fixed at 6, and the mean equals the total divided by the count of seven, so the total must be 6 × 7 = 42.
▸ Why?
The single shared center value the problem demands can only be 6.
▸ Why?
Among the six known numbers only 6 repeats — twice, against one copy each of 3, 4, 5, and 7 — so 6 is the single most-frequent value, the unique mode.
▸ Why?
Pairing 6's two copies against any other value's one copy leaves a 6 unmatched, showing 6's count is the larger one.
▸ Why?
The problem forces the mean, median, and mode to be one and the same number, so since that mode is 6, the mean is 6 as well.
▸ Why?
Multiplying the mean back by the count of seven reverses the division that defines the mean, returning the total.
Rule out the other choices
Check each choice: 5 and 7 tie 6 (mode not unique), 6 and 12 miss the mean — only x = 11 survives.
Tool #3 (Eliminate) on a multiple-choice problem: knock out anything that fails a stated condition. Four choices die fast.
6.SP.B.5Eliminate PossibilitiesVerify the median
With x = 11 the sorted list is 3, 4, 5, 6, 6, 7, 11, whose middle value is 6 — median matches the mean and mode.
Tool #6 (Guess and Check) finishes the job — plug the candidate back into the original setup to make sure every requirement (not just the one you used) is satisfied. Answer: (D) 11.
6.SP.B.5Guess And CheckOnce you spot that the mode has to be 6, the rest is a Grade 6 mean problem — sum equals count times average.
- Find the mode
- Use the mean to solve for x
- Rule out the other choices
- Verify the median
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