Difference between revisions of "2013 AMC 8 Problems/Problem 5"

Revision as of 02:38, 6 March 2014

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 option (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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Difference between revisions of "2013 AMC 8 Problems/Problem 5"

Revision as of 02:38, 6 March 2014

Problem

Solution

See Also

@@ Line 5: / Line 5: @@
 ==Solution==
+The median here is OBVIOUSLY less than the mean, so option (A) and (B) are out.
 Lining up the numbers (5, 5, 6, 8, 106), we see that the median weight is 6 pounds.