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05 Jan 2020, 20:55
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Felixvalentin
Hi Bunuel and ScottTargetTestPrep
If set S was empty, none of the conditions I, II and III must be true, yet the mean of set S would not exceed the mean of any subset of S. Hence, I chose A. I would highly appreciate if you could explain where my flaw is?
Many thanks!
If set S was empty, none of the conditions I, II and III must be true, yet the mean of set S would not exceed the mean of any subset of S. Hence, I chose A. I would highly appreciate if you could explain where my flaw is?
Many thanks!
The mean and the median are concepts which make sense only for non-empty set. If you have a set with no elements, what is the mean of that set? What is the median of that set? Neither the mean nor the median exists for that set; therefore the empty set does not satisfy the requirement that "the mean of set S does not exceed the mean of any subset of set S". Sure, the question could have said "the mean of the non-empty set S ..."; but as a matter of fact, that's not really necessary because as soon as the question is talking about the mean of the set S, you know that set S cannot be empty.
HoneyLemon
25 Jun 2020, 08:21
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If the mean of set S does not exceed mean of any subset of s , which of the following must be true about set s ?
let us consider
set s = { 2,0,15 ,100 200}
now one subset will be s1 = {0,2}
so the mean of s1 will definitely be less than mean of s.
Which is contradicted by the question .
Hence if the mean of set s is less than mean of any subset of s..
All the element in set S must be equal.
If all the elements of set S is equal..Mean and median of set s will be same ..
Hence D is the correct answer
let us consider
set s = { 2,0,15 ,100 200}
now one subset will be s1 = {0,2}
so the mean of s1 will definitely be less than mean of s.
Which is contradicted by the question .
Hence if the mean of set s is less than mean of any subset of s..
All the element in set S must be equal.
If all the elements of set S is equal..Mean and median of set s will be same ..
Hence D is the correct answer
29 Jan 2025, 11:13
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Hi, why have none of the solutions considered Null set as one of the subsets of S? If you consider null set, the 2 and 3 options also break down. Bunuel
Bunuel
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" --> set S can be:
A. \(S=\{x\}\) - S contains only one element (eg {7});
B. \(S=\{x, x, ...\}\) - S contains more than one element and all elements are equal (eg{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>smallest number).
Example: S={3, 5} --> mean of S=4. Pick subset with smallest number s'={3} --> mean of s'=3 --> 3<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.
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" --> set S can be:
A. \(S=\{x\}\) - S contains only one element (eg {7});
B. \(S=\{x, x, ...\}\) - S contains more than one element and all elements are equal (eg{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>smallest number).
Example: S={3, 5} --> mean of S=4. Pick subset with smallest number s'={3} --> mean of s'=3 --> 3<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.
