AMC 10 · 2019 · #7

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, and one more real number. Find the sum of all real values of that number making the median of the five equal to their mean.

Pick an answer.

(A)

-5

(B)

(C)

(D)

$\frac{15}{4}$

(E)

$\frac{35}{4}$

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

The mean is written immediately.

mean = (4+6+8+17+x)/5 = (35+x)/5

2STEP 2

Split the median into cases

The median splits into three cases.

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

3STEP 3

Solve the first case

The first case's root lands in range.

(35+x)/5 = 6 → x = -5 ( ≤ 6 ✓)

4STEP 4

Solve the second case

The second lands out of range and is discarded.

(35+x)/5 = x → x = 35/4 = 8.75 ( not ≤ 8 ×)

5STEP 5

Solve the third case

The third is also out of range.

(35+x)/5 = 8 → x = 5 ( not ≥ 8 ×)

6STEP 6

Add the survivors

Only one root survives.

sum = -5

7STEP 7

Match the choice

The sum is negative five.

-5 → (A)

Answer

-5

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.

💡Key takeaway

This AMC 12 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 -5, answer (A).