2016 AMC 10B Problem 5

We do not know the four ages, and we will not need them. We only need two totals.

1. Use the mean. The mean is $8$ for $4$ cousins, so all four ages add to $4 \cdot 8 = 32$.
2. Use the median. Line the ages up from youngest to oldest. With four ages there is no single middle one, so the median is the average of the 2nd and 3rd ages. That average is $5$, so the two middle ages add to $2 \cdot 5 = 10$.
3. Subtract. The youngest and oldest are everything except the two middle cousins:
$$32 - 10 = 22.$$

(One example: ages $1, 5, 5, 21$. The mean is $32 \div 4 = 8$ and the median is $5$.)

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

The mean tells you the total of all the values. With an even number of values, the median tells you the total of the middle two. The question asks about the values that are left over, so you just subtract one total from the other.

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

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

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

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

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