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M16-23

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Intern
Joined: 06 Apr 2016
Posts: 23
GMAT 1: 540 Q42 V23

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24 Aug 2017, 11:14
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
Math Expert
Joined: 02 Sep 2009
Posts: 50627

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24 Aug 2017, 11:25
1
saba@4010 wrote:
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

"The mean of set $$S$$ does not exceed mean of ANY subset of set $$S$$".

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Joined: 06 Apr 2016
Posts: 23
GMAT 1: 540 Q42 V23

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24 Aug 2017, 18:53
Bunuel wrote:
saba@4010 wrote:
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

"The mean of set $$S$$ does not exceed mean of ANY subset of set $$S$$".

Thanks Bunuel ,understood !

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Joined: 12 Feb 2015
Posts: 58
Location: India
GPA: 3.84

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25 Aug 2017, 01:47
Just need to confirm one thing.S={1,2,3} then can the subset be like A={1,1,}
Math Expert
Joined: 02 Sep 2009
Posts: 50627

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25 Aug 2017, 04:50
himanshukamra2711 wrote:
Just need to confirm one thing.S={1,2,3} then can the subset be like A={1,1,}

No, you don't have three 1's in S. Subsets of {1, 2, 3} are:
{1, 2, 3}
{1, 2}
{1, 3}
{2, 3}
{1}
{2}
{3}
{} - an empty set.
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Joined: 14 May 2017
Posts: 48

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29 Aug 2017, 05:04
Bunuel wrote:
ManSab wrote:
1. Set S contains only one element.
Does it mean count of element or value of element?
My understanding is that set S has only one element by count not by value... therefore {X,X...} is not possible. Please help understand.

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.

Hi,

I am confused with second option. If i didn't misunderstood the question, it state that mean of Set S should be less than mean of its subset.

Lets Consider:

S =(2,4,6) - mean =4
S'= (4,6) - mean = 5

Thanks,
Arpit
Math Expert
Joined: 02 Sep 2009
Posts: 50627

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29 Aug 2017, 05:07
NeverGiveUp- Arpit wrote:
Bunuel wrote:
ManSab wrote:
1. Set S contains only one element.
Does it mean count of element or value of element?
My understanding is that set S has only one element by count not by value... therefore {X,X...} is not possible. Please help understand.

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.

Hi,

I am confused with second option. If i didn't misunderstood the question, it state that mean of Set S should be less than mean of its subset.

Lets Consider:

S =(2,4,6) - mean =4
S'= (4,6) - mean = 5

Thanks,
Arpit

I'd advice to re-read the stem and the solution carefully and go through the discussion once more.
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Joined: 14 May 2017
Posts: 48

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29 Aug 2017, 05:56
[quote="Bunuel"

I'd advice to re-read the stem and the solution carefully and go through the discussion once more.[/quote]

Got it. Thanks mate.

I overlooked the word 'Any '.

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Intern
Joined: 07 Oct 2015
Posts: 7

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18 Sep 2017, 23:48
What is S= 1,2,3,4,5
Mean=3
Median=3
Subset(4,5); Mean =4.5>3
So, Statment 3: False
Math Expert
Joined: 02 Sep 2009
Posts: 50627

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18 Sep 2017, 23:50
3ksnikhil wrote:
What is S= 1,2,3,4,5
Mean=3
Median=3
Subset(4,5); Mean =4.5>3
So, Statment 3: False

"The mean of set $$S$$ does not exceed mean of ANY subset of set $$S$$".

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Joined: 19 Jun 2017
Posts: 4

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22 Dec 2017, 14:18
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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 50627

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23 Dec 2017, 00:14
mahagmat wrote:
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 ?

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Joined: 22 Aug 2016
Posts: 6

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20 Jun 2018, 23:03
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.
Math Expert
Joined: 02 Sep 2009
Posts: 50627

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20 Jun 2018, 23:16
hrishipatil72 wrote:
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.

The mean of {3, 4, 5, 6, -7} is 2.2.
The mean of one of the subset of the above set, {3, -7}, is 2.

So, your example is not valid.

The stem says: "The mean of set $$S$$ does not exceed mean of ANY subset of set $$S$$".
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Joined: 05 Mar 2018
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10 Aug 2018, 02:47
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?
Math Expert
Joined: 02 Sep 2009
Posts: 50627

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10 Aug 2018, 02:59
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?

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Joined: 10 Dec 2017
Posts: 22
Location: India

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21 Sep 2018, 00:47
As the question stem says "what must be true"?

let's consider S=(1,2,3,4,5)
Subset of S say S'=(3,4,5)

If the mean of set S does not exceed mean of any subset of set S(Given)
Mean of the Set= 3
Mean of the subset= 4
Does not exceed means, either the mean of the set is smaller than that of its subset or the mean of the set is equal to that of its subset.
3<4
so the consideration of the numbers is correct.

Here, option 3 is valid, but Could you guys please tell me why the second option is correct as the question specifically given "must be true"????

Thanks,
Intern
Joined: 17 Aug 2017
Posts: 1

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11 Oct 2018, 00:26
Hi Bunel

I am bit confused between statement 1 and 2
statement 1 cannot always be true because stmt 2 is a possibility
we can say the same to stmt 2 also right it cannot always be true stmt 1 is also a possibility

and what if there was an option that says both statment 1& 3

Bunuel wrote:
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.

Re: M16-23 &nbs [#permalink] 11 Oct 2018, 00:26

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