AMC 10 · 2016 · #5

Grade 6 arithmetic
mean-median-mode-range identify-subproblems ↑ Prerequisites: mean-median-mode-range
📏 Short solution 💡 2 insights
Problem
Four cousins have a mean age of 8 and a median age of 5. Find the sum of the youngest cousin's age and the oldest cousin's age.

Pick an answer.

(A)
13
(B)
16
(C)
19
(D)
22
(E)
25

AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

We are never asked for the four ages one by one, only for the two ends added together. Line the ages up youngest to oldest: the total splits into the two ends plus the two in the middle. Tool #16 (Change Focus / Count the Complement) says the two ends are just the total minus the middle pair, so we never need the individual ages. Tool #7 (Identify Subproblems) breaks the work into two easy pieces — get the total from the mean, get the middle pair from the median — and Tool #4 (Introduce a Variable) lets us treat each of those pairs as one quantity.

1STEP 1

Total age from the mean

Total = mean × count, so the four ages add up to 8 × 4 = 32.

total = mean × count = 8 × 4 = 32
2STEP 2

Middle pair from the median

The median of four ages is the average of the two middle ones, so the middle pair sums to 2 × 5 = 10.

two middle ages = 2 × median = 2 × 5 = 10
3STEP 3

Ends are the total minus the middle

The two ends are the total minus the middle pair: 32 - 10 = 22, which is (D).

youngest + oldest = 32 - 10 = 22 = (D)
Answer
22
Check with a concrete set of ages that fits: 3, 4, 6, 19 has mean (3+4+6+19)/4 = 32/4 = 8 and median (4+6)/2 = 5, and the ends give 3 + 19 = 22. The two middle ages are small (around 5) while the total is 32, so the ends must carry most of the total — a sum of 22 for the two ends fits, and it matches (D).
💡Key takeaway

Mean times the count gives the total; the rest you don't need to find one by one — just subtract the part you already know.

  • Total age from the mean
  • Middle pair from the median
  • Ends are the total minus the middle