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

Latest revision as of 18:15, 31 January 2025

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 1

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

~Soupboy0

Solution 2

Since the average right now is 10, and the median is 7, we see that N must be larger than 10, which means that the median of the 6 resulting numbers should be 7, making the mean of these 14. We can do 2 + 6 + 7 + 7 + 28 + N = 14 * 6 = 84. 50 + N = 84, so N = $\boxed{\text{(E)\ 34}}$

~Sigmacuber

Solution 3

We try out every option by inserting each number into the list. After trying, we get $\boxed{\text{(E)\ 34}}$

Note that this is very time-consuming and it is not the most practical solution.

~codegirl2013

Video Solution 1 by SpreadTheMathLove

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

Video Solution 2 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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 {{AMC8 box|year=2025|num-b=13|num-a=15}}
 {{MAA Notice}}
-[[Introductory Algebra Problems]]
+[[Category:Introductory Algebra Problems]]

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