AMC 8 · 2025 · #14
Grade 6 arithmeticProblem
A number N is inserted into the list 2, 6, 7, 7, 28. The mean is now twice as great as the median. What is N?
Pick an answer.
AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Start with the list $2, 6, 7, 7, 28$. We insert one more number $N$ so the new six-number list has a mean that is exactly twice the new median. Which of the five answer choices works?
Givens: Original list: $2, 6, 7, 7, 28$ (already sorted, sum $= 50$); Exactly one new value $N$ is inserted, giving a list of $6$ numbers; New mean $= 2 \times $ new median; Answer choices: (A) $7$, (B) $14$, (C) $20$, (D) $28$, (E) $34$
Unknowns: The value of $N$
Understand
Restated: Start with the list $2, 6, 7, 7, 28$. We insert one more number $N$ so the new six-number list has a mean that is exactly twice the new median. Which of the five answer choices works?
Givens: Original list: $2, 6, 7, 7, 28$ (already sorted, sum $= 50$); Exactly one new value $N$ is inserted, giving a list of $6$ numbers; New mean $= 2 \times $ new median; Answer choices: (A) $7$, (B) $14$, (C) $20$, (D) $28$, (E) $34$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #6 Guess and Check, #7 Identify Subproblems
This is a multiple-choice problem with only five candidates, so Tool #3 (Eliminate) is the natural first move. Tool #7 (Identify Subproblems) splits the work into two clean pieces — "find the new median" and "find the new mean" — and once we see that the median is locked at $7$ for every choice (since $N \geq 7$), Tool #6 (Guess and Check) just plugs each candidate into the mean formula and keeps the one that lands on $2 \times 7 = 14$. No algebra is required, though algebra would also work.
Execute — Answer: E
3.NBT.A.2 Step 1 - Sort the original list so we can think about the median.
- The list $2, 6, 7, 7, 28$ is already in order, and its sum is $2 + 6 + 7 + 7 + 28 = 50$.
- We will reuse this sum when computing the new mean.
💡 Adding five small whole numbers is the Grade 3 fluent-addition skill, and getting the sum out of the way now keeps later steps simple.
6.SP.A.3 Step 2 - Find the median of the new six-number list.
- Every answer choice satisfies $N \geq 7$, so when $N$ is dropped into the sorted list, the first four entries stay $2, 6, 7, 7$.
- With six numbers, the median is the average of the $3$rd and $4$th entries, both equal to $7$.
💡 Eliminating possibilities for where $N$ lands shows the middle pair is forced to be $(7, 7)$, so the median is a single fixed number.
4.OA.A.1 Step 3 - Translate the condition "mean is twice the median" into a required new mean.
- Since the new median is $7$, the new mean must equal $2 \times 7 = 14$.
💡 "Twice as great" is a Grade 4 multiplicative-comparison sentence: one quantity is $2$ times the other.
6.SP.B.5 Step 4 - Use the definition of the mean.
- The new list has $6$ numbers whose sum is $50 + N$, so the new mean is $(50 + N) / 6$.
- Set that equal to $14$ and check each answer choice $N \in \{7, 14, 20, 28, 34\}$.
💡 Plugging each choice into the sum-divided-by-count formula is straight Grade 6 "summarize a data set with the mean" reasoning.
4.OA.A.3 Step 5 - Solve for $N$ by subtracting $50$ from both sides.
- Only $N = 34$ makes the mean exactly $14$ — choices $7, 14, 20, 28$ give means of $\tfrac{57}{6}, \tfrac{64}{6}, \tfrac{70}{6}, \tfrac{78}{6}$, none equal to $14$.
💡 A one-step subtraction inside a multi-step word problem is squarely Grade 4 multi-step-word-problem work.
3.NBT.A.2 Sort the original list so we can think about the median. The list $2, 6, 7, 7, 2 6.SP.A.3 Find the median of the new six-number list. Every answer choice satisfies $N \ge 4.OA.A.1 Translate the condition "mean is twice the median" into a required new mean. Sin 6.SP.B.5 Use the definition of the mean. The new list has $6$ numbers whose sum is $50 + 4.OA.A.3 Solve for $N$ by subtracting $50$ from both sides. Only $N = 34$ makes the mean Review
Reasonableness: Sanity check: with $N = 34$, the new list is $2, 6, 7, 7, 28, 34$. Middle two are still $7$ and $7$, so the median is $7$. The sum is $50 + 34 = 84$, the mean is $84 / 6 = 14$, and $14 = 2 \times 7$. Everything matches. Also, the answer $34$ is the largest choice, which makes sense — to pull the mean far above the median, we need a value that is significantly larger than $7$.
Alternative: Tool #13 (Convert to Algebra) gives the same answer in one line: let $S = 50$ be the original sum. The new median is locked at $7$ (because $N \geq 7$ for every choice), so $\tfrac{S + N}{6} = 2 \cdot 7$ gives $N = 84 - 50 = 34$. Tool #3 + #6 is shown above because it stays closer to a $5$th-grader's instinct of "try the choices and see".
CCSS standards used (min grade 6)
3.NBT.A.2Fluently add and subtract within 1000 (Adding the five original entries $2 + 6 + 7 + 7 + 28 = 50$ and later doing $84 - 50 = 34$.)4.OA.A.1Interpret a multiplication equation as a comparison (Reading "the mean is twice as great as the median" as the multiplicative comparison $\text{mean} = 2 \times \text{median}$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Chaining the addition, multiplication, and subtraction steps that turn the conditions into the single equation $50 + N = 84$ and solve for $N$.)6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Treating the median of the six-number list as a single summary value ($= 7$) that the mean condition must match.)6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Computing the mean of the new six-number list as $\tfrac{\text{sum}}{\text{count}} = \tfrac{50 + N}{6}$ — the core Grade 6 mean-of-a-data-set formula.)
⭐ This AMC 8 problem only needs Grade 6 mean and median ideas — and the trick that the median stays at 7 — which you already know!
⭐ This AMC 8 problem only needs Grade 6 mean and median ideas — and the trick that the median stays at 7 — which you already know!
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