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2016 AMC 10B Problem 5

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2016 AMC 10B Problem 5AMC 10 Step by Step
Name:                             Date: September 13, 2026
Statistics & Data · difficulty 1 of 5 · about 1 min · position P1-10

The mean age of Amanda's 44 cousins is 88 , and their median age is 55 . What is the sum of the ages of Amanda's youngest and oldest cousins?

  1. A)

    1313

  2. B)

    1616

  3. C)

    1919

  4. D)

    2222

  5. E)

    2525

Answer

D

Key insight

Mean 8 makes the four ages total 32; median 5 makes the two middle ages total 10; the youngest and oldest are what is left: 22.

Solution

Let the ages in increasing order be a≤b≤c≤da \le b \le c \le d.

The mean is 88, so a+b+c+d=4⋅8=32a + b + c + d = 4 \cdot 8 = 32.

With four values the median is the average of the two middle ones, so b+c2=5\frac{b + c}{2} = 5, giving b+c=10b + c = 10.

Subtracting, a+d=32−10=22a + d = 32 - 10 = 22.

The answer is (D) 22\boxed{\textbf{(D)}\ 22}.

Why this works

The mean controls the total and the median controls the middle pair; the question asks for the complement of the middle pair, so the two facts subtract directly. You never need to know the ages themselves. For an even-sized data set, always translate "median" into "the two middle values add to twice the median."

The trap

Trying to determine the four individual ages, which are not determined, instead of subtracting the middle pair's total from the overall total.

Common mistakes

  • Trying to determine the four individual ages, which are not determined, instead of subtracting the middle pair's total from the overall total.
  • Treating the median 55 as a single middle age and subtracting only 55 from 3232, giving 2727, or subtracting it twice from a wrong total.

Techniques

Set up the equation/formula and compute; no special trick needed