Difference between revisions of "2025 AMC 8 Problems/Problem 14"
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− | 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 = <math>\boxed{\text{(E)\ 34}}</math> | + | 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 | ~Sigmacuber |
Latest revision as of 20:59, 15 August 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 , and the median is
, we see that
must be larger than
, which means that the median of the 6 resulting numbers should be
, making the mean of these
. We can do
.
, so
=
~Sigmacuber
Solution 3
We try out every option by inserting each number into the list. After trying out each number, we get
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 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 twice of the median, which is still
. Therefore, only choice left is
~ 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
Video Solution(Quick, fast, easy!)
~MC
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.