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Difference between revisions of "2025 AMC 8 Problems/Problem 14"
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==Vide Solution 1 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=jTTcscvcQmI

Revision as of 23:16, 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

Vide Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

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