Set X has 5 numbers, whose average is greater than their

Question

Set X has 5 numbers, whose average is greater than their median. Set Y has 7
numbers, whose average is also greater than their median. If the 2 sets have
no number in common and are combined to a new set, is the average of the
new set greater than its median?

(1) The average of Y is greater than the average of X.
(2) The median of Y is greater than the median of X.

abhijit_sen · Answer

What is OA? I think since it is not given numbers are +ve or -ve, so that might lead me to go for E.

sreehari · Answer

I am guessing it's C because question includes reference to both median and average, but couldn't comeup with a logical way to tackle it. will wait for someone to elighten me!

marcodonzelli · Answer

OA is E. anyway, how to tackle this type of ds statistics problem in just 2 minutes?

walker · Answer

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.

marcodonzelli · Answer

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

Set X has 5 numbers, whose average is greater than their

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