AMC 10 · 2019 · #13

Grade 6 arithmetic

Archetype Casework by Position / Vertex Type └ Sub-archetype Digit Place Position

mean-median-mode-rangecaseworklinear-equations-one-var caseworksystematic-enumeration ↑ Prerequisites: mean-median-mode-rangelinear-equations-one-var

📏 Medium solution 💡 2 insights

Problem

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.

Pick an answer.

(A)

-5

(B)

(C)

(D)

$\frac{15}{4}$

(E)

$\frac{35}{4}$

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

Tool #2 (Systematic List): the median can be 6, x, or 8 depending on whether x ≤ 6, 6 ≤ x ≤ 8, or x ≥ 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.

1STEP 1

Write the mean once: the mean of 4,6,8,17,x is $\frac{35+x}{5}$ for every x.

mean =

\frac{4+6+8+17+x}{5}

\frac{35+x}{5}

2STEP 2

By where x lands, the median is 6, x, or 8: 6 if x ≤ 6, x if 6 ≤ x ≤ 8, 8 if x ≥ 8.

median ∈ {6, x, 8} for x ≤ 6, 6 ≤ x ≤ 8, x ≥ 8

3STEP 3

Case A, median 6: $\frac{35+x}{5}$ = 6 → x = -5; since -5 ≤ 6, the list -5,4,6,8,17 has median 6 — valid.

\frac{35+x}{5}

= 6 → x = -5 ( ≤ 6 ✓)

4STEP 4

Case B, median x: $\frac{35+x}{5}$ = x → x = $\frac{35}{4}$ = 8.75; but 6 ≤ x ≤ 8 is required and 8.75 ＞ 8 — invalid.

\frac{35+x}{5}

= x → x =

\frac{35}{4}

= 8.75 ( not ≤ 8 ×)

5STEP 5

Case C, median 8: $\frac{35+x}{5}$ = 8 → x = 5; but x ≥ 8 is required and 5 ＜ 8 — invalid.

\frac{35+x}{5}

= 8 → x = 5 ( not ≥ 8 ×)

6STEP 6

Only Case A survives, so the one valid x is -5 and the sum is -5.

sum = -5

7STEP 7

So -5 is answer (A).

-5 → (A)

Answer

-5

Plug back x = -5: list is -5, 4, 6, 8, 17. Median is the middle value 6. Mean is

\frac{-5+4+6+8+17}{5}

\frac{30}{5}

= 6. Mean = median ✓. The eliminated cases pass a useful sanity check: the Case B candidate x =

\frac{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.

💡Key takeaway

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 = $\frac{35}{4}$ (out of range), x = 5 (out of range). The only legal value is -5, answer (A).

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