Difference between revisions of "2025 AMC 8 Problems/Problem 14"

Revision as of 20:53, 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, $2+6+7+7+28+N = 84 \rightarrow N = \boxed{\text{(E)\ 34}}$

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-The 2025 AMC 8 is not held yet. Please do not post false problems.
+A number <math>N</math> is inserted into the list <math>2</math>, <math>6</math>, <math>7</math>, <math>7</math>, <math>28</math>. The mean is now twice as great as the median. What is <math>N</math>?
+<math>\textbf{(A)}\ 7\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 20\qquad \textbf{(D)}\ 28\qquad \textbf{(E)}\ 34</math>
+==Solution==
+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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Difference between revisions of "2025 AMC 8 Problems/Problem 14"

Revision as of 20:53, 29 January 2025

Solution