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Both 1 and 2 are not sufficient to find an exact median value. E should be the correct answer.
1) All numbers can be 3 or 6(30 6's or 30 3's) which give 3 and 6 as median respectively.
2) 3*30 = 90. Which leaves 14, this amount can be distributed in various ways resulting in different median values.
Using 1 and 2 by using the approach in 2 will again result in different median values.

tricky one
i agree insufficient info in both stmts
we know tht the numbers are between 3-6 and the total is 104 but we dont know the frequency of the numbers
when you map it take for example
16 6's + remainder of 8 (distribute that either as 3,5 or 4,4)
15 5's +remainder of 4
so two possibilities already pop up
therefore we cant ditermine what can be done
cheers
(correct me wherevr i may be wrong)

310.What is the median of 30 numbers? Each of the numbers is greater more or equal to 3. 1) Each number <=6 2) The sum of the numbers is 104

I could be entirely wrong on this one, but 3 will be the median number if 3 is listed more often than any other number in the series.
if we have 16 3's, then the sum is 48
that means that the remaining 14 values need to add up to 56. You can only get that to work by plugging 14 fours. In this case, 3 is the median. If your set will include 4s, or 5s, that you have to have even more 3s so that the total stays at 104. Hence I selected b as the answer.

From the sum 104 compare to 3*30=90 we can conclude that there must be a lot of 3s. The larger the other numbers are, the more 3s we can have; the smaller the other numbers are, the less 3s we can have.

So we can assume the other numbers are 4.
3x +4(30-x)=104
x=120-104=16

In other words, we will have at least 16 3s. And therfore the median of these 30 numbers would 3.

Therefore (B) is the answer.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
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what if I had 10 3's and 20 3.7's or
9 3's and 21 77/21's which is greater than 3, yet we get diff answers.
the stem doesn't say the numbers are integers.