AMC 10 · 2019 · #13
Grade 6 arithmeticProblem
What is the sum of all real numbers x for which the median of the numbers 4,6,8,17, and x is equal to the mean of those five numbers?
Pick an answer.
AMC 10 2019 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: Consider the five numbers $4, 6, 8, 17, x$ for some real number $x$. Find the sum of all real $x$ that make the median of these five numbers equal to their mean.
Givens: Four fixed numbers: $4, 6, 8, 17$; A fifth number $x$, real and unknown; Mean of the five numbers $= \dfrac{4+6+8+17+x}{5} = \dfrac{35+x}{5}$; Median of the five numbers depends on where $x$ lands in the sorted order; Answer choices: $-5,\, 0,\, 5,\, \tfrac{15}{4},\, \tfrac{35}{4}$
Unknowns: The sum of every valid value of $x$
Understand
Restated: Consider the five numbers $4, 6, 8, 17, x$ for some real number $x$. Find the sum of all real $x$ that make the median of these five numbers equal to their mean.
Givens: Four fixed numbers: $4, 6, 8, 17$; A fifth number $x$, real and unknown; Mean of the five numbers $= \dfrac{4+6+8+17+x}{5} = \dfrac{35+x}{5}$; Median of the five numbers depends on where $x$ lands in the sorted order; Answer choices: $-5,\, 0,\, 5,\, \tfrac{15}{4},\, \tfrac{35}{4}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
Tool #2 (Systematic List): the median can be $6$, $x$, or $8$ depending on whether $x \le 6$, $6 \le x \le 8$, or $x \ge 8$ — exactly three cases to check, no more. Tool #7 (Subproblems): solve each case's mean = median equation separately, then test the answer against its own case condition. Tool #3 eliminates the two cases whose solutions violate the case condition; only one $x$ survives, and the sum is that one value. Algebra is just one line per case — no need for a heavier tool.
Execute — Answer: A
6.SP.B.5 Step 1 - Write the mean once.
- The mean of $4,6,8,17,x$ is $\dfrac{35 + x}{5}$ for every $x$.
💡 Grade 6 data: mean is just (sum of values) / (count).
6.SP.A.3 Step 2 - List the three cases for the median by where $x$ falls in the sorted order of $\{4,6,8,17\}$.
- Case A: $x \le 6$, sorted order starts with $x$ (or $4$) then $4$ (or $x$), then $6, 8, 17$ — median $= 6$.
- Case B: $6 \le x \le 8$, sorted order is $4,6,x,8,17$ — median $= x$.
- Case C: $x \ge 8$, sorted order ends with $x$ at or after $8$, so median $= 8$.
💡 Grade 6 statistics: the median of $5$ numbers is the $3$rd one after sorting.
6.EE.B.7 Step 3 - Case A — median $= 6$.
- Set mean equal to median: $\dfrac{35+x}{5} = 6 \Rightarrow 35 + x = 30 \Rightarrow x = -5$.
- Check $x = -5$ against the case rule $x \le 6$: $-5 \le 6$ ✓.
- Sorted list is $-5, 4, 6, 8, 17$ with median $6$ — valid solution.
💡 Grade 6 one-step equations: a single multiply-by-$5$ unlocks $x$.
6.EE.B.7 Step 4 - Case B — median $= x$.
- Set mean equal to median: $\dfrac{35+x}{5} = x \Rightarrow 35 + x = 5x \Rightarrow x = \dfrac{35}{4} = 8.75$.
- Check against the case rule $6 \le x \le 8$: $8.75 > 8$ — violates the case.
- Invalid.
💡 Grade 6: solving gives $x$, but the case demands $x \le 8$ — fails.
6.EE.B.7 Step 5 - Case C — median $= 8$.
- Set mean equal to median: $\dfrac{35+x}{5} = 8 \Rightarrow 35 + x = 40 \Rightarrow x = 5$.
- Check against the case rule $x \ge 8$: $5 < 8$ — violates.
- Invalid.
💡 Grade 6: again, the candidate $x$ leaves its own case range — eliminate.
6.NS.C.5 Step 6 - Only Case A survives, with $x = -5$.
- The sum of all valid $x$ is $-5$.
💡 Grade 6 negative numbers: one valid value means the "sum" is just that single value.
6.NS.C.6 Step 7 Match $-5$ to the choices.
💡 Grade 6 number line: $-5$ is the negative answer choice.
6.SP.B.5 Write the mean once. The mean of $4,6,8,17,x$ is $\dfrac{35 + x}{5}$ for every $ 6.SP.A.3 List the three cases for the median by where $x$ falls in the sorted order of $\ 6.EE.B.7 Case A — median $= 6$. Set mean equal to median: $\dfrac{35+x}{5} = 6 \Rightarro 6.EE.B.7 Case B — median $= x$. Set mean equal to median: $\dfrac{35+x}{5} = x \Rightarro 6.EE.B.7 Case C — median $= 8$. Set mean equal to median: $\dfrac{35+x}{5} = 8 \Rightarro 6.NS.C.5 Only Case A survives, with $x = -5$. The sum of all valid $x$ is $-5$. 6.NS.C.6 Match $-5$ to the choices. Review
Reasonableness: Plug back $x = -5$: list is $-5, 4, 6, 8, 17$. Median is the middle value $6$. Mean is $(-5+4+6+8+17)/5 = 30/5 = 6$. Mean $=$ median ✓. The eliminated cases pass a useful sanity check: the Case B candidate $x = 35/4 = 8.75$ would sit between $8$ and $17$, not between $6$ and $8$ as required, and the Case C candidate $x = 5$ would sit between $4$ and $6$, not above $8$ as required. So the elimination is correct.
Alternative: Tool #1 (Diagram) — draw the number line $4, 6, 8, 17$ and sketch where the median sits as $x$ slides from $-\infty$ to $+\infty$. The median is a piecewise function: flat at $6$ for $x \le 6$, equal to $x$ on $[6,8]$, flat at $8$ for $x \ge 8$. The mean $\tfrac{35+x}{5}$ is a single straight line. Geometrically you are intersecting one line with three line segments; only the leftmost segment $y = 6$ catches the line, at $x = -5$. Same answer $(A)$.
CCSS standards used (min grade 6)
6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Identifying the median as the $3$rd value of the sorted list of $5$ numbers.)6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Computing the mean $(35+x)/5$ of the five numbers.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving mean = median in each of the three cases for $x$.)6.NS.C.5Understand that positive and negative numbers describe quantities (Accepting a negative value $x = -5$ as a legitimate real solution.)6.NS.C.6Understand a rational number as a point on the number line (Locating $x = -5$ to the left of all fixed values on the number line.)
⭐ This AMC 10 problem only needs Grade 6 mean-vs-median reasoning you already know! The median is either $6$, $x$, or $8$ depending on where $x$ lands. Setting mean = median in each case gives $x = -5$ (valid), $x = 35/4$ (out of range), $x = 5$ (out of range). The only legal value is $\mathbf{-5}$, answer $(A)$.
⭐ This AMC 10 problem only needs Grade 6 mean-vs-median reasoning you already know! The median is either $6$, $x$, or $8$ depending on where $x$ lands. Setting mean = median in each case gives $x = -5$ (valid), $x = 35/4$ (out of range), $x = 5$ (out of range). The only legal value is $\mathbf{-5}$, answer $(A)$.
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