# 2013 AMC 8 Problems/Problem 5

## Problem

Hammie is in the $6^\text{th}$ grade and weighs 106 pounds. His quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds? $\textbf{(A)}\ \text{median, by 60} \qquad \textbf{(B)}\ \text{median, by 20} \qquad \textbf{(C)}\ \text{average, by 5}$ $$\qquad \textbf{(D)}\ \text{average, by 15} \qquad \textbf{(E)}\ \text{average, by 20}$$

## Solution

The median here is obviously less than the mean, so options $(A)$ and $(B)$ are out.

Lining up the numbers (5, 5, 6, 8, 106), we see that the median weight is 6 pounds.

The average weight of the five kids is $\dfrac{5+5+6+8+106}{5} = \dfrac{130}{5} = 26$.

Therefore, the average weight is bigger, by $26-6 = 20$ pounds, making the answer $\boxed{\textbf{(E)}\ \text{average, by 20}}$.

~ pi_is_3.14

## Video Solution

https://youtu.be/ATZp9dYIM_0 ~savannahsolver

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