2025 AMC 8 Problems/Problem 14
Revision as of 00:09, 30 January 2025 by Hydromathgod (talk | contribs)
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