Difference between revisions of "2025 AMC 8 Problems/Problem 14"
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~Soupboy0 | ~Soupboy0 | ||
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| + | ==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}}$ | ||
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| + | ~HydroMathGod | ||
==Vide Solution 1 by SpreadTheMathLove== | ==Vide Solution 1 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=jTTcscvcQmI | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
Revision as of 00:09, 30 January 2025
A number 
is inserted into the list 
, 
, 
, , 
. The mean is now twice as great as the median. What is ?
Solution
The median of the list is , so the mean of the new list will be 
. Since there will be numbers in the new list, the sum of the numbers will be 
. Therefore, 
~Soupboy0
Solution 2 (Using answer choices)
We could use answer choices to solve this problem. The sum of the 
numbers is 
. If you add to the list, 
is not divisible by , therefore it will not work. Same thing applies to 
and 
. The only possible choices left are and 
. Now you check . You see that doesn't work because 
and 
is not divisible by . Therefore, only choice left is \boxed{\text{(E)\ 34}}$
~HydroMathGod
