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D01-18

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D01-18  [#permalink]

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New post 16 Sep 2014, 00:12
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Re D01-18  [#permalink]

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New post 16 Sep 2014, 00:12
Official Solution:


Statement 1: If the mean and median of the set are positive, the standard deviation could be any. The set could have elements {1, 1, 1} or {1, 2, 3} or {10, 20, 30, 40, 50}. In each case, the standard deviation is not the same. So not sufficient..

Statement 2: If difference between any elements of the set is equal, then the set has to have same elements because the number of elements is greater than 2. So standard deviation is 0. Sufficient.


Answer: B
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Re: D01-18  [#permalink]

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New post 02 Dec 2014, 12:20
Bunuel wrote:
Official Solution:


Statement 1: If the mean and median of the set is positive, the standard deviation could be any. The set could have elements {1, 1, 1} or {1, 2, 3} or {10, 20, 30, 40, 50}. In each case, the standard deviation isn’t the same. So NSF.

Statement 2: If difference between any elements of the set is equal, then the set has to have same elements because the number of elements is greater than 2. So standard deviation is 0. Sufficient.


Answer: B


Statement 2: If difference between any elements of the set is equal, then the set has to have same elements because the number of elements is greater than 2. So standard deviation is 0. Sufficient.


I'm confused why they have to be the same elements because the number of elements is greater than 2.. If the difference is equal, can't it just be {1,3,5..} or {1,5,9..} which means SD can be anything..

Also, can you explain difference between elements and numbers in this case? This may be adding to my confusion.
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Re: D01-18  [#permalink]

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New post 03 Dec 2014, 04:06
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codeblue wrote:
Bunuel wrote:
Official Solution:


Statement 1: If the mean and median of the set is positive, the standard deviation could be any. The set could have elements {1, 1, 1} or {1, 2, 3} or {10, 20, 30, 40, 50}. In each case, the standard deviation isn’t the same. So NSF.

Statement 2: If difference between any elements of the set is equal, then the set has to have same elements because the number of elements is greater than 2. So standard deviation is 0. Sufficient.


Answer: B


Statement 2: If difference between any elements of the set is equal, then the set has to have same elements because the number of elements is greater than 2. So standard deviation is 0. Sufficient.


I'm confused why they have to be the same elements because the number of elements is greater than 2.. If the difference is equal, can't it just be {1,3,5..} or {1,5,9..} which means SD can be anything..

Also, can you explain difference between elements and numbers in this case? This may be adding to my confusion.


Second statement says that the difference between ANY two elements of the set is equal. If the set does not have all the elements equal, for example, if the set is {1, 3, 5}, then the difference between ANY two elements of the set won't be equal: 3-1=2 but 5-1=4. Hence the set must have same elements.

As for your other question: element of a set and number of a set are the same thing - member of a set.
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Re: D01-18  [#permalink]

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New post 18 Jan 2015, 14:11
Given statement 1, can we conclude Set S is evenly spaced?
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Re: D01-18  [#permalink]

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New post 27 Mar 2015, 11:05
Statement mentions that mean and median are equal not positive as mentioned in the answer explanation.
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New post 27 Mar 2015, 11:05
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Statement mentions that mean and median are equal not positive as mentioned in the answer explanation.
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Re: D01-18  [#permalink]

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New post 05 Jul 2015, 18:04
Bunuel wrote:
Official Solution:


Statement 1: If the mean and median of the set is positive, the standard deviation could be any. The set could have elements {1, 1, 1} or {1, 2, 3} or {10, 20, 30, 40, 50}. In each case, the standard deviation is not the same. So NSF.

Statement 2: If difference between any elements of the set is equal, then the set has to have same elements because the number of elements is greater than 2. So standard deviation is 0. Sufficient.


Answer: B


Suppose n=2
Then what will happen for Statement ii)

Consider a case {1,2}

2-1=1 and 1-2 =-1 .
The difference is not the same correct ??
So in a way I am disproving the given statement and this approach is incorrect.

If Set {1,1}

difference is always 0. Therefore SD is 0.

Are there other possibilities or other insights to this ? I want to understand the n=2 case better.
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Re: D01-18  [#permalink]

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New post 06 Jul 2015, 01:05
anurag356 wrote:
Bunuel wrote:
Official Solution:


Statement 1: If the mean and median of the set is positive, the standard deviation could be any. The set could have elements {1, 1, 1} or {1, 2, 3} or {10, 20, 30, 40, 50}. In each case, the standard deviation is not the same. So NSF.

Statement 2: If difference between any elements of the set is equal, then the set has to have same elements because the number of elements is greater than 2. So standard deviation is 0. Sufficient.


Answer: B


Suppose n=2
Then what will happen for Statement ii)

Consider a case {1,2}

2-1=1 and 1-2 =-1 .
The difference is not the same correct ??
So in a way I am disproving the given statement and this approach is incorrect.

If Set {1,1}

difference is always 0. Therefore SD is 0.

Are there other possibilities or other insights to this ? I want to understand the n=2 case better.


Question says that n > 2, why are you considering n = 2 there?
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Re: D01-18  [#permalink]

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New post 06 Jul 2015, 02:29
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Bunuel wrote:
anurag356 wrote:
Bunuel wrote:
Official Solution:


Statement 1: If the mean and median of the set is positive, the standard deviation could be any. The set could have elements {1, 1, 1} or {1, 2, 3} or {10, 20, 30, 40, 50}. In each case, the standard deviation is not the same. So NSF.

Statement 2: If difference between any elements of the set is equal, then the set has to have same elements because the number of elements is greater than 2. So standard deviation is 0. Sufficient.


Answer: B


Suppose n=2
Then what will happen for Statement ii)

Consider a case {1,2}

2-1=1 and 1-2 =-1 .
The difference is not the same correct ??
So in a way I am disproving the given statement and this approach is incorrect.

If Set {1,1}

difference is always 0. Therefore SD is 0.

Are there other possibilities or other insights to this ? I want to understand the n=2 case better.


Question says that n > 2, why are you considering n = 2 there?



Its true that N>2 is given.

But if in the exam no such condition was given then I ll have to consider 2 elements in a set as well.In that case what will happen is what Im trying to understand. So that I can be prepared.

To understand things better im asking this case.
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Re: D01-18  [#permalink]

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New post 12 Aug 2015, 19:51
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Official Solution:


Statement 1: If the mean and median of the set is positive, the standard deviation could be any. The set could have elements {1, 1, 1} or {1, 2, 3} or {10, 20, 30, 40, 50}. In each case, the standard deviation is not the same. So NSF.

Statement 2: If difference between any elements of the set is equal, then the set has to have same elements because the number of elements is greater than 2. So standard deviation is 0. Sufficient.


Answer: B
Statement mentions that mean and median are equal not positive as mentioned in the answer explanation.
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D01-18  [#permalink]

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New post 04 Jul 2017, 23:09
Even if the constraint x>2 was not given, option ii would still be sufficient right?
I am not sure how x>2 helps the second option.
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New post 04 Jul 2017, 23:34
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New post 07 Aug 2018, 06:58
Bunuel , I have one doubt, It might be stupid but still I have it
When we say 'Set' --> doesn't it mean a collection of distinct objects. So how come , we are taking Set as {1,1,1} as per second statement.
Doesn't it violate basic definition of Set?
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Re: D01-18   [#permalink] 07 Aug 2018, 06:58
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