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.