AMC 8 · 2025 · #9
Grade 6 arithmeticProblem
Ningli looks at the 6 pairs of numbers directly across from each other on a clock. She takes the average of each pair of numbers. What is the average of the resulting 6 numbers?
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: On a standard clock face with numbers $1$ through $12$, Ningli pairs up the six numbers that sit directly opposite each other (a number and the one $6$ hours across from it). For each of those $6$ pairs she computes the average of the two numbers, getting $6$ new numbers. We need the average of those $6$ averages.
Givens: A standard clock face shows the numbers $1, 2, 3, \dots, 12$; There are exactly $6$ pairs of opposite numbers (each number is paired with the one $6$ positions away); For each pair, the "average" is $\dfrac{a + b}{2}$; Answer choices: (A) $5$, (B) $6.5$, (C) $8$, (D) $9.5$, (E) $12$
Unknowns: The average of the $6$ pair-averages
Understand
Restated: On a standard clock face with numbers $1$ through $12$, Ningli pairs up the six numbers that sit directly opposite each other (a number and the one $6$ hours across from it). For each of those $6$ pairs she computes the average of the two numbers, getting $6$ new numbers. We need the average of those $6$ averages.
Givens: A standard clock face shows the numbers $1, 2, 3, \dots, 12$; There are exactly $6$ pairs of opposite numbers (each number is paired with the one $6$ positions away); For each pair, the "average" is $\dfrac{a + b}{2}$; Answer choices: (A) $5$, (B) $6.5$, (C) $8$, (D) $9.5$, (E) $12$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #5 Look for a Pattern, #1 Draw a Diagram
There are only $6$ pairs, so the most direct attack is Tool #2 (Systematic List): list the pairs in order $(1,7), (2,8), \dots, (6,12)$, compute each average, and then average those $6$ numbers. While doing the list we can use Tool #5 (Pattern): the six pair-averages turn out to be $4, 5, 6, 7, 8, 9$ — a perfectly regular arithmetic run, whose average is just the midpoint $\tfrac{4+9}{2}$. Tool #1 (Diagram) shows why every pair must sum to $1+7 = 13$, $2+8 = 10$, $3+9 = 12$ … wait, they do NOT all share the same sum, but they do all share the same average of $\tfrac{a+b}{2}$ that creeps up by $1$ each time — exactly the pattern we exploit.
Execute — Answer: B
4.OA.C.5 Step 1 - Identify the $6$ opposite-number pairs by walking around the clock in order.
- "Opposite" means $6$ positions apart, so we pair each number $k$ from $1$ to $6$ with $k+6$.
💡 Generating opposite pairs by the rule "$k$ pairs with $k+6$" is a Grade 4 "follow a given rule to make a pattern" move.
3.OA.C.7 Step 2 - Compute the average of each pair as $\dfrac{a+b}{2}$.
- We add the two numbers and divide by $2$, recording the results in order.
💡 Each pair sum is at most $18$ and dividing by $2$ is fluent Grade 3 multiplication/division within $100$.
6.SP.B.5 Step 3 - Look at the resulting list of six pair-averages: $4, 5, 6, 7, 8, 9$.
- They form a consecutive run of integers (an arithmetic sequence with common difference $1$).
- For any such evenly spaced list, the average is the midpoint, i.e.
- the average of the first and last terms.
💡 Recognizing that the average of an evenly spaced list equals its midpoint is a Grade 6 "summarize a numerical data set with a measure of center" insight.
5.NF.B.3 Step 4 Double-check arithmetically by summing all six values and dividing by $6$.
💡 Reading $\tfrac{39}{6}$ as "$39$ divided by $6$" and simplifying to $6.5$ is Grade 5 "fraction as division".
4.OA.C.5 Identify the $6$ opposite-number pairs by walking around the clock in order. "Op 3.OA.C.7 Compute the average of each pair as $\dfrac{a+b}{2}$. We add the two numbers and 6.SP.B.5 Look at the resulting list of six pair-averages: $4, 5, 6, 7, 8, 9$. They form a 5.NF.B.3 Double-check arithmetically by summing all six values and dividing by $6$. Review
Reasonableness: The clock numbers $1$ through $12$ have overall average $\tfrac{1+2+\dots+12}{12} = \tfrac{78}{12} = 6.5$. Since every clock number appears in exactly one pair and an average $\tfrac{a+b}{2}$ weights each of its two members equally, averaging the six pair-averages is the same as averaging all twelve numbers. So the answer must be $6.5$ — matching (B). Magnitude is plausible: $6.5$ sits right between $1$ and $12$, exactly where a balanced average should land.
Alternative: Tool #16 (Change Focus): instead of computing six pair-averages and re-averaging, notice that $\text{average of pair-averages} = \dfrac{\sum_{i=1}^{6}\frac{a_i+b_i}{2}}{6} = \dfrac{(a_1+b_1)+\dots+(a_6+b_6)}{12} = \dfrac{1+2+\dots+12}{12} = \dfrac{78}{12} = 6.5$. The two layers of averaging collapse into one overall average — no per-pair computation needed.
CCSS standards used (min grade 6)
4.OA.C.5Generate a number or shape pattern following a given rule (Pairing each clock number $k$ from $1$ to $6$ with $k+6$ to enumerate all six opposite pairs.)3.OA.C.7Fluently multiply and divide within 100 (Adding each pair (sums at most $18$) and dividing by $2$ to get the six pair-averages $4, 5, 6, 7, 8, 9$.)6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Recognizing that the mean (average) of the evenly spaced list $4, 5, 6, 7, 8, 9$ is the midpoint $\tfrac{4+9}{2} = 6.5$, the formal Grade 6 "measure of center" concept.)5.NF.B.3Interpret a fraction as division of the numerator by the denominator (Reading $\tfrac{39}{6}$ as "$39 \div 6 = 6.5$" to confirm the average arithmetically.)
⭐ This AMC 8 problem only needs Grade 6 "average is the center of the data" reasoning you already know — and once you see the pair-averages line up as $4, 5, 6, 7, 8, 9$, the answer is just the middle: $6.5$!
⭐ This AMC 8 problem only needs Grade 6 "average is the center of the data" reasoning you already know — and once you see the pair-averages line up as $4, 5, 6, 7, 8, 9$, the answer is just the middle: $6.5$!
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