Set S = {4, x, 1, 4, 1} Is x the median of Set S?

Question

Set S = {4, x, 1, 4, 1}

Is x the median of Set S?

(1) x^2 = 16
(2) x is the mode of Set S.

saswata4s · Answer

(1) x^2 = 16; So, x = 4, or x = -4;
 Case 1: x = 4 :
Set S = {4, 4, 1, 4, 1} = {1, 1, 4, 4, 4}; Median is 4 (x);

Case 2: x = -4 :
Set S = {4, -4, 1, 4, 1} = {-4, 1, 1, 4, 4}; Median is 1 (not x);

Hence (1) is insufficient.

(2) x is the mode of Set S.
x can be either 1 or 4;
 Case 1: x = 1 :
Set S = {4, 1, 1, 4, 1} = {1, 1, 1, 4, 4}; 
Therefore Median = mode = 1 = x;

Case 2: x = 4 :
Set S = {4, 4, 1, 4, 1} = {1, 1, 1, 4, 4}; 
Therefore Median = mode = 4 = x;

Hence, when x is the mode, irrespective of the value of x, mode = median = x. Sufficient.
Ans B.

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nks2611 · Answer

hi

can you clarify that whether i am right to what i thought about this problem, i think the question is asking is x can be median or not so in statement 2 ,we have two values for median 1 and 4 but we do need to think about it ,right, we need to answer that is x cane be median or not ?
please explain 
thanks

Alexey1989x · Answer

So, to be the median of set x must be 1<=x<=4
(1)|x|=4, thus x may be or may be not the median.  Insufficient. 
(2) to be the mode of subject set x should be equal either to 1 or 4, thus it will median of subject set for sure.  Sufficient. 
I do believe that it should be classified as an easy one, not (700+) for sure at least...

roadrunner · Answer

the question asks if x IS the median..the difference is that according to the question regardless of the value of x, it must  always  be the median and not only sometimes in which case answer would be D.

adsabz · Answer

In an exercise I came across in an official gmat mock, they assumed that square root of x^2 = absolute x. They only consider the positive root, which is not consistent with the example here and everyone's answers. Could you clarify?

Bunuel · Answer

1. When the GMAT (and generally in math) provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

\sqrt{9} = 3 , NOT +3 or -3;
 \sqrt {16} = 2 , NOT +2 or -2;

Notice that in contrast, the equation  x^2 = 9  has TWO solutions, +3 and -3. Because  x^2 = 9  means that  x =-\sqrt{9}=-3  or  x=\sqrt{9}=3 .

2.  \sqrt{x^2} = |x| . Note that |x| is always non-negative, so this does not contradict the above. For example, if x = -5, then   \sqrt{(-5)^2} = |-5|=5 .

Hope it helps.

Set S = {4, x, 1, 4, 1} Is x the median of Set S?

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Set S = {4, x, 1, 4, 1} Is x the median of Set S?

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