2025 AMC 8 Problems/Problem 14: Difference between revisions

Latest revision as of 19:59, 15 August 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 out each number, we get $\boxed{\text{(E)\ 34}}$

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

$~$ ~codegirl2013 ~ Minor edit by KangarooPrecise

Solution 4

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 $34$ . Now you check $28$ . You see that $28$ 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

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/VP7g-s8akMY?si=z0ZzRRMAMp9LYd1V&t=1442 ~hsnacademy

Video Solution 1 by SpreadTheMathLove

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

Video Solution 2 by Thinking Feet

https://youtu.be/PKMpTS6b988

Video Solution(Quick, fast, easy!)

https://youtu.be/fdG7EDW_7xk

~MC

@@ Line 1: / Line 1: @@
+== Problem ==
 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==
+== Solution 1 ==
+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 are <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 \implies N = \boxed{\text{(E)\ 34}}</math>
+~Soupboy0
+==Solution 2==
+Since the average right now is <math>10</math>, and the median is <math>7</math>, we see that <math>N</math> must be larger than <math>10</math>, which means that the median of the 6 resulting numbers should be <math>7</math>, making the mean of these <math>14</math>. We can do <math>2 + 6 + 7 + 7 + 28 + N = 14 * 6 = 84</math>. <math>50 + N = 84</math>, so <math>N</math> = <math>\boxed{\text{(E)\ 34}}</math>
+~Sigmacuber
+== Solution 3 ==
+We try out every option by inserting each number into the list. After trying out each number, we get <math>\boxed{\text{(E)\ 34}}</math>
+Note that this is very time-consuming and it is not the most practical solution.
+<math>~</math>
+~codegirl2013
+~ Minor edit by KangarooPrecise
+== Solution 4 ==
+We could use answer choices to solve this problem. The sum of the <math>5</math> numbers is <math>50</math>. If you add <math>7</math> to the list, <math>57</math> is not divisible by <math>6</math>, therefore it will not work. Same thing applies to <math>14</math> and <math>20</math>. The only possible choices left are <math>28</math> and <math>34</math>. Now you check <math>28</math>. You see that <math>28</math> doesn't work because <math>(28+50) \div 6 = 13</math> and <math>13</math> is not twice of the median, which is still <math>7</math>. Therefore, only choice left is <math>\boxed{\text{(E)\ 34}}</math>
+~ HydroMathGod
+==Video Solution (A Clever Explanation You’ll Get Instantly)==
+https://youtu.be/VP7g-s8akMY?si=z0ZzRRMAMp9LYd1V&t=1442
+~hsnacademy
+== Video Solution 1 by SpreadTheMathLove ==
+https://www.youtube.com/watch?v=jTTcscvcQmI
+== Video Solution 2 by Thinking Feet ==
+https://youtu.be/PKMpTS6b988
+==Video Solution(Quick, fast, easy!)==
+https://youtu.be/fdG7EDW_7xk
+~MC
+== See Also ==
-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>
+{{AMC8 box|year=2025|num-b=13|num-a=15}}
+{{MAA Notice}}
+[[Category:Introductory Algebra Problems]]

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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