median??

alivapriyada wrote:Given distinct positive integers 1, 11, 3, x, 2, and 9, which of the following could be the median?

A. 3

B. 5

C. 7

D. 8

E. 9

I think it should be B,

Nice work, goyalsau - here's why you're right:

The median value is the middle value of a set if it has an odd number of terms, or the average of the two middle values if it has an even number of terms.

Here, we have six terms, so the median will be the average of terms 3 and 4.

From our current set (we'll add x in a second) of 5 fixed terms, the middle term is 3:

1, 2, 3, 9, 11

Our possibilities for x are:

x < 2 -----> 2 and 3 become the middle values and the median is 2.5 (which isn't an answer choice)
3 < x < 9 ------> 3 and x are the middle values and the average of 3 and x is the median, so:

x = 4; median = 3.5 (note, if x is even we get a .5, and there are none in the solutions so we can skip those)
x = 5; median = 4 (not an answer choice)
x = 7; median = 5 (an answer choice - this is right!)

x > 9 -----> median is the average of 3 and 9, which is 6 (not an answer choice)

You can obviously do it less labor-intensively than I just did, but I did that for the sake of comprehensiveness. Basically,your steps are to recognize that you need to take the average of x and 3 if x is between 3 and 9, and then you can do those calculations until you match an answer choice.