2013 AMC 8 Problems/Problem 5: Difference between revisions

Revision as of 00:10, 20 January 2015

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 $\boxed{\textbf{(E)}\ \text{average, by 20}}$ .

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AJHSME/AMC 8 Problems and Solutions

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 ==Solution==
-The median here is obviously less than the mean, so option (A) and (B) are out.
+The median here is obviously less than the mean, so options <math>(A)</math> and <math>(B)</math> are out.
 Lining up the numbers (5, 5, 6, 8, 106), we see that the median weight is 6 pounds.

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2013 AMC 8 Problems/Problem 5: Difference between revisions

Revision as of 00:10, 20 January 2015

Problem

Solution

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