Hello Bunuel,

What if : set S = {1,2,3,4} ; Mean=2.5
subset of S ={1,2} ; Mean =1.5

Then option 2 " All elements in set S are equal" does not fit in.
Please let me where am i going wrong.

Thanks,
Saba

Option 1, Set S contains only one element means that there is only one elements in S: {X}. This option is not necessarily true. Please re-read the solution.

What is S= 1,2,3,4,5
Mean=3
Median=3
Subset(4,5); Mean =4.5>3
So, Statment 3: False

Hello !
For case III, let's take an example where the median of set S equals the mean of set S : set S = {1,2,3}, mean=median=2. In that case, {1} is a subset of set S and 2>1. Conclusion: the mean of set S DOES exceed mean of a subset of set S
How is III still correct ?

Hi Bunuel,

"If the mean of set S does not exceed mean of any subset of set S''

Does it have to be true that Mean of S and Mean of any subset of S is equal? What if set S has negative numbers?

For Ex:- S= {3,4,5,6,-7} and subset of S is {3,4,5,6}. Here mean does not exceed but is lower that its subset.

In above example none of the three statement holds true? Could you please advise.

Thank you.

I don't agree with the explanation. Statement III.
If median of set = mean of set
eg. mean {0,1,2} = median {0,1,2} = 1
here, mean exceeds the subset of set which is {2}.

Only II should be the corret answer?

Official Solution:


If the mean of set \(S\) does not exceed mean of any subset of set \(S\), which of the following must be true about set \(S\)?

I. Set \(S\) contains only one element

II. All elements in set \(S\) are equal

III. The median of set \(S\) equals the mean of set \(S\)


A. none of the three qualities is necessary
B. II only
C. III only
D. II and III only
E. I, II, and III


"The mean of set \(S\) does not exceed mean of ANY subset of set \(S\)" implies that set \(S\) could be:

A. \(S=\{x\}\) - \(S\) contains only one element (e.g. {7 });

B. \(S=\{x, x, ...\}\) - \(S\) contains more than one element and all elements are equal (e.g. {7,7,7,7 }).

Why is that? Because if set \(S\) contains two (or more) different elements, then we can always consider the subset with smallest number and the mean of this subset (mean of subset=smallest number) will be less than mean of entire set (mean of full set\(\gt\)smallest number).

Example: if \(S=\{3, 5\}\), then mean of \(S=4\). Pick subset with the smallest number: \(s'=\{3\}\), mean of \(s'=3\). As we see \(3 \lt 4\).

Now let's consider the statements:

I. Set \(S\) contains only one element - not always true, we can have scenario \(B\) too (\(S=\{x, x, ...\}\));

II. All elements in set \(S\) are equal - true for both \(A\) and \(B\) scenarios, hence always true;

III. The median of set \(S\) equals the mean of set \(S\) - - true for both \(A\) and \(B\) scenarios, hence always true.

So statements II and III are always true.


Answer: D