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Difference between revisions of "2025 AMC 8 Problems/Problem 14"
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The median of the list is <math>7</math>, so the mean of the new list will be <math>7 \cdot 2 = 14</math>. Since there will be <math>6</math> numbers in the new list, the sum  of the <math>6</math> numbers will be <math>14 \cdot 6 = 84</math>. Therefore, <math>2+6+7+7+28+N = 84 \rightarrow N = \boxed{\text{(E)\ 34}}</math>
 
The median of the list is <math>7</math>, so the mean of the new list will be <math>7 \cdot 2 = 14</math>. Since there will be <math>6</math> numbers in the new list, the sum  of the <math>6</math> numbers will be <math>14 \cdot 6 = 84</math>. Therefore, <math>2+6+7+7+28+N = 84 \rightarrow N = \boxed{\text{(E)\ 34}}</math>
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~Soupboy0

Revision as of 21:54, 29 January 2025

A number $N$ is inserted into the list $2$, $6$, $7$, $7$, $28$. The mean is now twice as great as the median. What is $N$?

$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 20\qquad \textbf{(D)}\ 28\qquad \textbf{(E)}\ 34$

Solution

The median of the list is $7$, so the mean of the new list will be $7 \cdot 2 = 14$. Since there will be $6$ numbers in the new list, the sum of the $6$ numbers will be $14 \cdot 6 = 84$. Therefore, $2+6+7+7+28+N = 84 \rightarrow N = \boxed{\text{(E)\ 34}}$

~Soupboy0

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