Difference between revisions of "2025 AMC 8 Problems/Problem 14"
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| − | ==Problem== | + | == Problem == |
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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>? | 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> | <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 | + | 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 | ~Soupboy0 | ||
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~Sigmacuber | ~Sigmacuber | ||
| − | ==Solution 3== | + | == Solution 3 == |
We try out every option by inserting each number into the list. After trying, we get <math>\boxed{\text{(E)\ 34}}</math> | We try out every option by inserting each number into the list. After trying, we get <math>\boxed{\text{(E)\ 34}}</math> | ||
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~codegirl2013 | ~codegirl2013 | ||
| − | == | + | == Video Solution 1 by SpreadTheMathLove == |
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https://www.youtube.com/watch?v=jTTcscvcQmI | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
| − | ==Video Solution by Thinking Feet== | + | == Video Solution 2 by Thinking Feet == |
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https://youtu.be/PKMpTS6b988 | https://youtu.be/PKMpTS6b988 | ||
| − | ==See Also== | + | == See Also == |
| + | |||
{{AMC8 box|year=2025|num-b=13|num-a=15}} | {{AMC8 box|year=2025|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | [[Introductory Algebra Problems]] | ||
Revision as of 18:15, 31 January 2025
Contents
Problem
A number 
is inserted into the list 
, 
, 
, , 
. The mean is now twice as great as the median. What is ?
Solution 1
The median of the list is , so the mean of the new list will be 
. Since there are numbers in the new list, the sum of the numbers will be 
. Therefore, 
~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 = 
~Sigmacuber
Solution 3
We try out every option by inserting each number into the list. After trying, we get
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
See Also
| 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. 
