2025 AMC 8 Problems/Problem 14

Problem

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

Solution 2 (Using answer choices)

We could use answer choices to solve this problem. The sum of the $5$ numbers is $50$ . If you add $7$ to the list, $57$ is not divisible by $6$ , therefore it will not work. Same thing applies to $14$ and $20$ . The only possible choices left are $28$ and $34$ . You now check $28$ . $28$ doesn't work because $(28+50) \div 6 = 13$ and $13$ is not twice of the median, which is still $7$ . Therefore, only choice left is $\boxed{\text{(E)\ 34}}$

~HydroMathGod

Vide Solution 1 by SpreadTheMathLove

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

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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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