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# M20 Q35

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M20 Q35 [#permalink]  18 Nov 2008, 03:49
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What is $$x$$ ?

1. The median of set $$(x, 1, -1, 3, -x)$$ is 0
2. The median of set $$(x, 1, -1, 3, -x)$$ is $$\frac{x}{2}$$

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Re: M20 Q35 [#permalink]  20 Nov 2008, 06:43
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In both the cases, x=0
hence D
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Re: M20 Q35 [#permalink]  20 Nov 2008, 19:34
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ventivish wrote:
What is $$x$$ ?

1. The median of set $$(x, 1, -1, 3, -x)$$ is 0
2. The median of set $$(x, 1, -1, 3, -x)$$ is $$\frac{x}{2}$$

A.

1: x must be 0.
2: x could be 2 or 0.

If x = 2, then the set is [-2, -1, 1, 2, 3]. hence the median, x/2, is 1.
If x = 0, then the set is [-1, 0, 0, 1, 3]. hence the median, x/2, is 0.
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Re: M20 Q35 [#permalink]  23 Nov 2008, 11:27
Hi thanks for answering this Q
The OA is A
And this is the OE
The median of a five-element set is necessarily a member of this set. No other member of the given set but $$x$$ can be 0. S2 is not sufficient. $$x$$ can be 2 or 0.

I could not get my head around to why the second statement was insufficient, I guess the median for an odd number of elements in a set must be a member ofthe set. And if x/2 is the median, then x/2 must be an integer, so x must either be 0 or 2. However the question stem does not state that x is an integer!

Thanks
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Re: M20 Q35 [#permalink]  29 Apr 2010, 06:08
I just picked some random numbers and hit S2 as invalid for x=2... But i will like a better approach.

Anyway, Answer is definitely option A (S1 suff.)

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Re: M20 Q35 [#permalink]  29 Apr 2010, 09:43
OA is A.

Median for set of odd numbers must be middle number in the sorted order.
Thus 0 means x or -x is middle number which implies x=0.

However St2 is never possible.

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Re: M20 Q35 [#permalink]  29 Apr 2010, 10:17
jvaidya wrote:
OA is A.

Median for set of odd numbers must be middle number in the sorted order.
Thus 0 means x or -x is middle number which implies x=0.

However St2 is never possible.

-Jvaidya

Why wouldn't st2 be possible? With st2, x could be 0, making the median 0, or x could be 2, making the median 1.

I ordered the values in each way possible to find the answer. This was easy to do since there were an odd number of values, and since the values were the same for each statement.

-x, -1, 1, x, 3
-1, -x, x, 1, 3

The median would either have to be x or 1.
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Re: M20 Q35 [#permalink]  29 Apr 2010, 11:59
mmphf wrote:
jvaidya wrote:
OA is A.

Median for set of odd numbers must be middle number in the sorted order.
Thus 0 means x or -x is middle number which implies x=0.

However St2 is never possible.

-Jvaidya

Why wouldn't st2 be possible? With st2, x could be 0, making the median 0, or x could be 2, making the median 1.

I ordered the values in each way possible to find the answer. This was easy to do since there were an odd number of values, and since the values were the same for each statement.

-x, -1, 1, x, 3
-1, -x, x, 1, 3

The median would either have to be x or 1.

St2 alone will never be able to decide the value for x as it can be both 2 or 0 which is not true. However, St1 can give you 1 value i.e. x=0 which is correct.
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Re: M20 Q35 [#permalink]  30 Apr 2010, 04:49
I am a little bit confused here. The definition of a set is "A collection of distinct objects". Now if we choose A, then we are no longer dealing with a set.
And B does not help in identifying the answer either.
I feel that E is the answer, neither of these statements are sufficient to answer this question.
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Re: M20 Q35 [#permalink]  30 Apr 2010, 12:38
aielman wrote:
I am a little bit confused here. The definition of a set is "A collection of distinct objects". Now if we choose A, then we are no longer dealing with a set.
And B does not help in identifying the answer either.
I feel that E is the answer, neither of these statements are sufficient to answer this question.

Why are we no longer dealing with a set if we pick "A" ? The value of x is the only thing that's in question. The numbers of the set do not change.
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Re: M20 Q35 [#permalink]  30 Apr 2010, 22:42
mmphf wrote:
aielman wrote:
I am a little bit confused here. The definition of a set is "A collection of distinct objects". Now if we choose A, then we are no longer dealing with a set.
And B does not help in identifying the answer either.
I feel that E is the answer, neither of these statements are sufficient to answer this question.

Why are we no longer dealing with a set if we pick "A" ? The value of x is the only thing that's in question. The numbers of the set do not change.

If the value of x is 0, then the set(lets call it A) now becomes
A={-1,-0,0,3}
There is no such thing as -0 and a +0
Therefore now the set A becomes
A={-1,0,0,3}
Now A can no longer be called a set, because the basic definition of a set is "A collection of distinct objects"
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Re: M20 Q35 [#permalink]  01 May 2010, 10:36
aielman wrote:
mmphf wrote:
aielman wrote:
I am a little bit confused here. The definition of a set is "A collection of distinct objects". Now if we choose A, then we are no longer dealing with a set.
And B does not help in identifying the answer either.
I feel that E is the answer, neither of these statements are sufficient to answer this question.

Why are we no longer dealing with a set if we pick "A" ? The value of x is the only thing that's in question. The numbers of the set do not change.

If the value of x is 0, then the set(lets call it A) now becomes
A={-1,-0,0,3}
There is no such thing as -0 and a +0
Therefore now the set A becomes
A={-1,0,0,3}
Now A can no longer be called a set, because the basic definition of a set is "A collection of distinct objects"

I would say, at least on GMAT-Land, until and unless it's explicitly mentioned that set contains distinct objects..DO not take it for granted. In many question, you fill find the concept of same elements in a given set..!!
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Re: M20 Q35 [#permalink]  02 May 2010, 01:54
Stmt 1 - x can only be zero

Stmt 2- here i need some explanation

Is 0/2 = 0 ????
If yes then 2/0 = ??
and 0/0 = ??

Can anyone please explain
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Re: M20 Q35 [#permalink]  20 May 2010, 08:13
Stmt 1. The median of set $$(x, 1, -1, 3, -x)$$ is 0

There are 5 no in the set and only x and -x are unknown. So x must be zero.

Stmt 2. The median of set $$(x, 1, -1, 3, -x)$$ is $$\frac{x}{2}$$
x is variable here.

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Re: M20 Q35 [#permalink]  03 May 2011, 04:29
(1)

-1,1,3 are remaning elements, so x = 0, for x to median (the 3rd element)

Sufficient.

(2)

-1,1,3 are remaning elements, but x can be 2, so that 1 = 2/2 is meian

or x = 6, so that 6/2 = 3 is a median.

Not Sufficient.

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Re: M20 Q35 [#permalink]  03 May 2011, 11:20
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timmyd wrote:
Not realy clear on the answer. Does anyone have another take on an explanation? Thanks!

My explanation is not going to be very different from others, but let me give it a shot. Please let me know if something is not clear.

Median of elements of any set arranged in ascending order will be the middle term if the number(count) of elements is odd.
Median of elements of any set arranged in ascending order will be the average of two middle terms if the number(count) of elements is even.

e.g.
A={-2,45,-56,0,23,-9,34}
Let's find the median of A:
Count= 7(Odd)
Arrange elements in ascending order:
A={-56,-9,-2,0,23,34,45}
Median is the middle term(because Count=7(odd)), i.e. {(Count+1)/2}th term.
Here, count=7; median will be (7+1)/2 i.e. 4th term.
4th term = 0 and is the median of the set.

One point worth noticing is that if the set has odd number of elements, Median must be one of its element. Like we saw, that "0" was one of the elements in the set.

case II:
B={-2,45,-56,0,23,-9,34,1}
Let's find the median of B:
Count= 8(Even)
Arrange elements in ascending order:
A={-56,-9,-2,0,1,23,34,45}
Median is the average of middle terms(because Count=8(Even)), i.e. Average of (Count/2)th term and ((Count/2)+1)th term.
Here, count=8; Count/2=4th term=0; And Count/2+1=4+1=5th term=1
Median = (0+1)/2 = 0.5
Thus, we saw that in case of even number of elements, median may be something other than any of the elements. Here 0.5 was not among the elements.

Coming to the question:

1.
Median of the set {x,1,-1,3,-x} is 0.
Count=5(Odd). Thus, we know that median must be one of the elements.
1 != 0
-1 != 0
3 != 0
Only, x or -x can be 0
x=0
Or
-x=0 i.e. x=0
We know for sure that x=0
Sufficient.

2.
Median of {x,1,-1,3,-x} is x/2.
Count=5(Odd)

Means; x/2 must be one among 1,-1,3,x or -x for we know that median of odd number of elements must be one of the elements itself.

Let's check one by one;
Say; Median = x
Means; x/2=x; It is possible only for x=0. If x=0
{x,1,-1,3,-x} becomes {-1,0,0,1,3}. Median=0, which is true.
So; x could be "0". But, we must try the same thing for other numbers as well.

Say; Median=1
i.e. x/2=1; x=2;
{x,1,-1,3,-x} becomes {-2,-1,1,3,2}. Median=1, which is true.
What we saw here that; x could be 2 as well.
If we made x=2; the median becomes 1, which is x/2; supporting the statement.

We already have two possible values of x; 0 and 2. We can stop here.
Not Sufficient.

Ans: "A"

Note: We could have also checked for other numbers:
Say; Median = 3
x/2=3
x=6
{x,1,-1,3,-x} becomes {-6,-1,1,3,6}. Median=1, which is false.
x/2=6/2=3; According to the statement, Median should be 3, but we got our median as 1. Thus, x can't be 6.

Say; Median = -1
x/2=-1
x=-2
{x,1,-1,3,-x} becomes {-2,-1,1,2,3}. Median=1, which is false.
x/2=-2/2=-1; According to the statement, Median should be -1, but we got our median as 1. Thus, x can't be -2.

Say; Median = -x
-x/2=-x, which is x/2=x. Same as first
x can be 0.
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Re: M20 Q35 [#permalink]  03 May 2011, 21:49
ventivish wrote:
What is $$x$$ ?

1. The median of set $$(x, 1, -1, 3, -x)$$ is 0
2. The median of set $$(x, 1, -1, 3, -x)$$ is $$\frac{x}{2}$$

[Reveal] Spoiler: OA
A

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We have five numbers: -1,-x,x,1,3 and median has to be the middle number that is the 3rd number in the list.

Statement One:

These five numbers can be arranged in the following fashion:

1) -x, -1, 1, x, 3 (x can be 2 say)
2) -x, -1, 1, 3, x (x can be 4 say)
3) -1, -x, x, 1, 3 (x is 0<= x <= 1)

Now for these three possibilities median will be 1, 1 and x but we know it is zero so it implies x = 0.

So Statement One alone is enough.

Statement two:

We have same set of numbers and same set of possibilities. So if median is $$\frac{x}{2}$$ it means

1) x is 2 and numbers are: -2, -1, 1, 2, 3. x can not be 1 right????
2) x cannot be 2 as 3 is not less than 2 so leave this here.
3) x can only be zero and 0 divided be zero is again zero.

So x can be 0 and 2. Statement 2 alone is not enough.

So Answer is A.
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Re: M20 Q35 [#permalink]  07 May 2012, 06:14
from 1. x has to be zero to satisfy the given condition
from 2. x may be 0 or 2.
hence 1 alone is sufficient to answer the question.
my take option A
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Re: M20 Q35 [#permalink]  08 May 2013, 05:10
Question: x = ?

S1: Set = {x, -1, 1, 3, -x} and Median = 0
There are five terms and the median will be zero only if x = 0
S1 is sufficient. Eliminate BCE.

S2: Set = {x, -1, 1, 3, -x} and Median = x/2
If x = 2 => Set = {-2, -1, 1, 2, 3} => Median = 1
If x = 0 => Set = {-1, 0, 0, 1, 3} => Median = 0
So, S2 is true for at least two values of x.
S2 is not sufficient. Eliminate D.

Correct answer is A.
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Re: M20 Q35 [#permalink]  08 May 2013, 05:38
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What is the value of $$x$$?

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.

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Re: M20 Q35   [#permalink] 08 May 2013, 05:38

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# M20 Q35

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