walker wrote:

Tough one...

I think the fastest way to solve the problem - try to construct examples.

1) 0--XXMAX-------------------------------------------------------------------YYYMAYY

Median of a new set is greater than average.

2)

0--XXMAX--

0---YYYMAYY

Median of a new set is smaller than average.

hope not to find questions like this in the actual test

anyway, this could be an explanation:

(1) y > x. We know that the average of the elements in X U Y is between x

and y. We can say that at least three of the elements of X are less than

this average. If one element in Y is very large, it could be that all other

elements of Y are less than this average, so However, if all elements of

Y are nearly same, all will be above this average.

For example, suppose x=3 , y=5 and X = { 2,2,2,2,7} .

If Y= {3,3,3, ... , 3 , 17} . The average of the combined set X U Y is

greater than 4, but the median is 4.

Conversely, if Y= {4.9, 4.9,4.9,...,4,9, 5.6}, the median, 4.9 is greater

than the mean of X U Y. NOT SUFF.

(2) Since the median of X is less than the median of Y, the median of X U Y

will be less than the median of Y, but greater than the median of X.

Without information about the averages of the elements in X and those

in Y, we cannot answer the question. In each of the two examples cited

in (1), the median of X is less than the median of Y. NOT SUFF

(T) NOT SUFF