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M20-35

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M20-35  [#permalink]

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New post 16 Sep 2014, 01:09
2
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A
B
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D
E

Difficulty:

  65% (hard)

Question Stats:

58% (01:03) correct 42% (01:24) wrong based on 170 sessions

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Re M20-35  [#permalink]

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New post 16 Sep 2014, 01:09
Official Solution:


Note that the median of a set with odd number of elements is just the middle element, when arranged in ascending/descending order.

(1) The median of set \(\{x, -1, 1, 3, -x\}\) is 0. The median of this set of 5 (odd) elements must be the middle term, hence \(x=0\). Sufficient.

(2) The median of set \(\{x, -1, 1, 3, -x\}\) is \(\frac{x}{2}\). It could be that \(x=0\) (in this case the set is \(\{-1, 0, 0, 1, 3\}\)) as well as it could be that \(x=2\) (in this case the set is \(\{-2, -1, 1, 2, 3\}\)). Not sufficient.


Answer: A
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Re: M20-35  [#permalink]

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New post 28 Jul 2015, 10:05
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Bunuel wrote:
Official Solution:


Note that the median of a set with odd number of elements is just the middle element, when arranged in ascending/descending order.

(1) The median of set \(\{x, -1, 1, 3, -x\}\) is 0. The median of this set of 5 (odd) elements must be the middle term, hence \(x=0\). Sufficient.

(2) The median of set \(\{x, -1, 1, 3, -x\}\) is \(\frac{x}{2}\). It could be that \(x=0\) (in this case the set is \(\{-1, 0, 0, 1, 3\}\)) as well as it could be that \(x=2\) (in this case the set is \(\{-2, -1, 1, 2, 3\}\)). Not sufficient.


Answer: A


Hi Bunuel,
At first i correctly thought that x must be 0 as {\(-1, 0, 0, 1, 3\)} is possible but than I eliminated stmt 1 as 0 can not be written as -0 ( as mentioned -x)
is there a typo in the Q or can we assume both -x and x to be 0
thanks.
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Re: M20-35  [#permalink]

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New post 29 Jul 2015, 01:18
Ankur9 wrote:
Bunuel wrote:
Official Solution:


Note that the median of a set with odd number of elements is just the middle element, when arranged in ascending/descending order.

(1) The median of set \(\{x, -1, 1, 3, -x\}\) is 0. The median of this set of 5 (odd) elements must be the middle term, hence \(x=0\). Sufficient.

(2) The median of set \(\{x, -1, 1, 3, -x\}\) is \(\frac{x}{2}\). It could be that \(x=0\) (in this case the set is \(\{-1, 0, 0, 1, 3\}\)) as well as it could be that \(x=2\) (in this case the set is \(\{-2, -1, 1, 2, 3\}\)). Not sufficient.


Answer: A


Hi Bunuel,
At first i correctly thought that x must be 0 as {\(-1, 0, 0, 1, 3\)} is possible but than I eliminated stmt 1 as 0 can not be written as -0 ( as mentioned -x)
is there a typo in the Q or can we assume both -x and x to be 0
thanks.


-0 = 0, so nothing wrong with it.
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Re: M20-35  [#permalink]

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New post 13 Oct 2017, 00:58
Hi,

what if x is negative eg ( -4,-1,1,3,4) median here is 0. isnt st1 insufficient?
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Re: M20-35  [#permalink]

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New post 13 Oct 2017, 01:06
shweta.aarya@gmail.com wrote:
Hi,

what if x is negative eg ( -4,-1,1,3,4) median here is 0. isnt st1 insufficient?


The median of a set with odd number of elements is the middle term, when arranged in ascending/descending order. The media of {-4, -1, 1, 3, 4} is 1, not 0.
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Re: M20-35  [#permalink]

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New post 31 Oct 2017, 05:28
+1 for A
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Re: M20-35  [#permalink]

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New post 31 Oct 2017, 06:45
What is the value of \(x\)?


(1) The median of set \(\{x, -1, 1, 3, -x\}\) is 0

(2) The median of set \(\{x, -1, 1, 3, -x\}\) is \(\frac{x}{2}\)

we need a single value of X

condition 1 : 0 is the median then zero is a part of the set mentioned :so x and -x =0 hence an unique answer :sufficient
condition 2: x/2 is median :X can be any any number 0 or 2 itself : so not sufficient

A is correct option
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Re: M20-35  [#permalink]

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New post 08 Nov 2017, 00:13
Bunuel wrote:
Official Solution:


Note that the median of a set with odd number of elements is just the middle element, when arranged in ascending/descending order.

(1) The median of set \(\{x, -1, 1, 3, -x\}\) is 0. The median of this set of 5 (odd) elements must be the middle term, hence \(x=0\). Sufficient.

(2) The median of set \(\{x, -1, 1, 3, -x\}\) is \(\frac{x}{2}\). It could be that \(x=0\) (in this case the set is \(\{-1, 0, 0, 1, 3\}\)) as well as it could be that \(x=2\) (in this case the set is \(\{-2, -1, 1, 2, 3\}\)). Not sufficient.


Answer: A

Hi Bunuel,
Are we missing x=-2 as a probable option? Still statement A is correct but just for clarity..
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Re: M20-35  [#permalink]

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New post 08 Nov 2017, 00:26
ManishKM1 wrote:
Bunuel wrote:
Official Solution:


Note that the median of a set with odd number of elements is just the middle element, when arranged in ascending/descending order.

(1) The median of set \(\{x, -1, 1, 3, -x\}\) is 0. The median of this set of 5 (odd) elements must be the middle term, hence \(x=0\). Sufficient.

(2) The median of set \(\{x, -1, 1, 3, -x\}\) is \(\frac{x}{2}\). It could be that \(x=0\) (in this case the set is \(\{-1, 0, 0, 1, 3\}\)) as well as it could be that \(x=2\) (in this case the set is \(\{-2, -1, 1, 2, 3\}\)). Not sufficient.


Answer: A

Hi Bunuel,
Are we missing x=-2 as a probable option? Still statement A is correct but just for clarity..


Are you talking about (2)? If x = -2, then the set is \(\{-2, -1, 1, 2, 3\}\). The median = 1, but we are told that it should be x/2 = -1.
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Re: M20-35  [#permalink]

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New post 08 Nov 2017, 00:31
Bunuel wrote:
ManishKM1 wrote:
Bunuel wrote:
Official Solution:


Note that the median of a set with odd number of elements is just the middle element, when arranged in ascending/descending order.

(1) The median of set \(\{x, -1, 1, 3, -x\}\) is 0. The median of this set of 5 (odd) elements must be the middle term, hence \(x=0\). Sufficient.

(2) The median of set \(\{x, -1, 1, 3, -x\}\) is \(\frac{x}{2}\). It could be that \(x=0\) (in this case the set is \(\{-1, 0, 0, 1, 3\}\)) as well as it could be that \(x=2\) (in this case the set is \(\{-2, -1, 1, 2, 3\}\)). Not sufficient.


Answer: A

Hi Bunuel,
Are we missing x=-2 as a probable option? Still statement A is correct but just for clarity..


Are you talking about (2)? If x = -2, then the set is \(\{-2, -1, 1, 2, 3\}\). The median = 1, but we are told that it should be x/2 = -1.



Got it..Thanks a lot..:)
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Re: M20-35  [#permalink]

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New post 13 Nov 2017, 07:32
Good afternoon Bunuel,

I was wondering when a set is written, it is all the time written in an ascending order.

In our case: (X<-1<1<3<-X) and thus I understood that -X>3 which mean X=O for (2).

Shall I understood that in a set when a variable is present it is not always presented in ascending order?

Thanks in advance,

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Re: M20-35  [#permalink]

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New post 13 Nov 2017, 07:35
Tpral wrote:
Good afternoon Bunuel,

I was wondering when a set is written, it is all the time written in an ascending order.

In our case: (X<-1<1<3<-X) and thus I understood that -X>3 which mean X=O for (2).

Shall I understood that in a set when a variable is present it is not always presented in ascending order?

Thanks in advance,

Regards,


A set, by definition, is a collection of elements without any order. While, a sequence, by definition, is an ordered list of terms.
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Re: M20-35 &nbs [#permalink] 13 Nov 2017, 07:35
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