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Difference between revisions of "2025 AMC 8 Problems/Problem 14"
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==Solution 2 (Using answer choices)==
 
==Solution 2 (Using answer choices)==
  
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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 divisible by <math>2</math>. Therefore, only choice left is \boxed{\text{(E)\ 34}}$
+
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 divisible by <math>2</math>. Therefore, only choice left is <math>\boxed{\text{(E)\ 34}}</math>
  
 
~HydroMathGod
 
~HydroMathGod

Revision as of 00:10, 30 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

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 $34$. Now you check $28$. You see that $28$ doesn't work because $(28+50) \div 6 = 13$ and $13$ is not divisible by $2$. Therefore, only choice left is $\boxed{\text{(E)\ 34}}$

~HydroMathGod

Vide Solution 1 by SpreadTheMathLove

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

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