AMC 10 · 2022 · #8
Grade 6 arithmeticProblem
A data set consists of 6 (not distinct) positive integers: 1, 7, 5, 2, 5, and X. The average (arithmetic mean) of the 6 numbers equals a value in the data set. What is the sum of all positive values of X?
Pick an answer.
AMC 10 2022 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: A data set has six positive integers $\{1, 7, 5, 2, 5, X\}$. The arithmetic mean of the six numbers equals one of the values that already appears in the data set. Find every positive integer $X$ that makes this happen, and add those $X$-values together.
Givens: Five fixed values: $1, 7, 5, 2, 5$; Sixth value: a positive integer $X$; The mean equals some value in $\{1, 2, 5, 7, X\}$ (the distinct entries); Answer choices: (A) $10$, (B) $26$, (C) $32$, (D) $36$, (E) $40$
Unknowns: All valid positive-integer values of $X$; Their sum
Understand
Restated: A data set has six positive integers $\{1, 7, 5, 2, 5, X\}$. The arithmetic mean of the six numbers equals one of the values that already appears in the data set. Find every positive integer $X$ that makes this happen, and add those $X$-values together.
Givens: Five fixed values: $1, 7, 5, 2, 5$; Sixth value: a positive integer $X$; The mean equals some value in $\{1, 2, 5, 7, X\}$ (the distinct entries); Answer choices: (A) $10$, (B) $26$, (C) $32$, (D) $36$, (E) $40$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #6 Guess and Check, #11 Work Backwards
The phrase "equals a value in the data set" gives a small finite list of cases: the mean can be $1, 2, 5, 7,$ or $X$. Tool #2 (Systematic List) walks through every case in order so nothing is missed. Within each case Tool #11 (Work Backwards) finds $X$ — we know the mean, so we run the average formula backwards to recover the missing piece. Tool #6 (Guess and Check) is the verification: plug the candidate $X$ back into the full data set and confirm the recomputed mean really does land in the set.
Execute — Answer: D
6.SP.B.5 Step 1 - Add the five fixed numbers so the mean can be written cleanly.
- The total of the known values is $1+7+5+2+5=20$.
- So the mean of all six is $\frac{20+X}{6}$.
💡 The arithmetic mean is the total divided by the count — a Grade 6 statistics move.
6.SP.A.3 Step 2 - List every distinct value already in the set: $\{1, 2, 5, 7\}$.
- Plus the wildcard $X$.
- Five cases to check, in order: mean = $1, 2, 5, 7, X$.
💡 An ordered list of cases guarantees nothing is overlooked and nothing counted twice.
6.EE.B.7 Step 3 - Case mean=$1$: $\frac{20+X}{6}=1$ gives $X=-14$.
- Case mean=$2$: $\frac{20+X}{6}=2$ gives $X=-8$.
- Both are not positive — reject.
- The total $20$ alone is already $\frac{20}{6}>3$, so the average can never drop to $1$ or $2$ with a positive $X$.
💡 Run the average formula backwards: $X = 6 \cdot \text{mean} - 20$. If the result is not a positive integer, discard.
6.EE.B.7 Step 4 - Case mean=$5$: $\frac{20+X}{6}=5$ gives $X=10$.
- Positive integer — keep.
- Verify: $\{1,7,5,2,5,10\}$, sum $30$, mean $5$, and $5$ is in the set ✓.
💡 Plug $X=10$ back to confirm the mean really equals $5$, a value already present.
6.EE.B.7 Step 5 - Case mean=$7$: $\frac{20+X}{6}=7$ gives $X=22$.
- Positive integer — keep.
- Verify: $\{1,7,5,2,5,22\}$, sum $42$, mean $7$, and $7$ is in the set ✓.
💡 Same backwards move — the mean fixes $X$ uniquely.
6.EE.B.7 Step 6 - Case mean=$X$: $\frac{20+X}{6}=X$ gives $20+X=6X$, so $5X=20$, $X=4$.
- Positive integer — keep.
- Verify: $\{1,7,5,2,5,4\}$, sum $24$, mean $4$, and $X=4$ is by construction the value in the set ✓.
💡 Setting the mean equal to $X$ itself is a self-consistent loop, solved by a one-step equation.
4.NBT.B.4 Step 7 - The valid positive values are $X \in \{4, 10, 22\}$.
- Add them: $4 + 10 + 22 = 36$, which is choice (D).
💡 Three two-digit numbers — straightforward Grade 4 addition.
6.SP.B.5 Add the five fixed numbers so the mean can be written cleanly. The total of the 6.SP.A.3 List every distinct value already in the set: $\{1, 2, 5, 7\}$. Plus the wildcar 6.EE.B.7 Case mean=$1$: $\frac{20+X}{6}=1$ gives $X=-14$. Case mean=$2$: $\frac{20+X}{6}= 6.EE.B.7 Case mean=$5$: $\frac{20+X}{6}=5$ gives $X=10$. Positive integer — keep. Verify: 6.EE.B.7 Case mean=$7$: $\frac{20+X}{6}=7$ gives $X=22$. Positive integer — keep. Verify: 6.EE.B.7 Case mean=$X$: $\frac{20+X}{6}=X$ gives $20+X=6X$, so $5X=20$, $X=4$. Positive i 4.NBT.B.4 The valid positive values are $X \in \{4, 10, 22\}$. Add them: $4 + 10 + 22 = 36 Review
Reasonableness: Each surviving $X$ has been independently verified by recomputing the mean of the full six-element set and confirming the value lives in $\{1,2,5,7,X\}$. The cases mean=$1$ and mean=$2$ were correctly rejected because the five fixed numbers already total $20$, forcing $X$ negative. The sum $36$ matches choice (D); the nearby choices $32$ and $40$ would correspond to forgetting one of $\{4, 22\}$ or including a spurious case.
Alternative: Tool #6 (pure Guess and Check) on the answer choices: try $X=4$ (works), $X=10$ (works), $X=22$ (works). Their sum is $36$, matching (D). This is slower because it gives no guarantee of completeness — without the case analysis you have not proven you found every valid $X$. The systematic-list path proves the list of $X$-values is exhaustive.
CCSS standards used (min grade 6)
6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Writing the arithmetic mean of the six-element set as $\frac{20+X}{6}$.)6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Listing the distinct values $\{1,2,5,7,X\}$ as the only candidates the mean could equal.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving $\frac{20+X}{6} = \text{value}$ for $X$ in each case (including the self-referential case mean$=X$).)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the surviving values $4 + 10 + 22 = 36$ for the final answer.)
⭐ This AMC 10 problem only needs Grade 6 mean and one-step equations you already know — write the average as $\frac{20+X}{6}$, try each in-set value for it, keep the cases that give a positive integer, and add the survivors $4+10+22=36$.
⭐ This AMC 10 problem only needs Grade 6 mean and one-step equations you already know — write the average as $\frac{20+X}{6}$, try each in-set value for it, keep the cases that give a positive integer, and add the survivors $4+10+22=36$.
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