Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2025 AMC 8 Problems/Problem 14"
(→‎Solution)
Line 8: Line 8:
  
 
~Soupboy0
 
~Soupboy0
 +
 +
==Vide Solution 1 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=jTTcscvcQmI

Revision as of 22:16, 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2025_AMC_8_Problems/Problem_14&oldid=241109"