aditi_bc wrote:pl solve
Here is the solution:
AM = (x+2+7+11+16)/5 = (x+36)/5
given AM(x,2,7,11,16) = Median
so lets think the possibilities of x from the (x+36)/5:
x (x+36)/5
-- ----------
4 8 --8 is not there in the set
9 9 --9 can be included in place of x which can also be the median
14 10 --10 is not there in the set and so on..
19 11 --19 can be included in the place of x as 11 can be median
Statement 1 --> 7<x<11
Only 9 is fits in x(as per above possibilities). Hence this statement is sufficient.
Statement 2 --> x is the median of 5 numbers.
From the above possibilities only 9 and 19 can be included in the set. But as x is also a median..19 cannot be median. Because if 19 is included in the set, then 2,7,11,16,19 --> 19 is AM and 11 is median which fails. If you consider 9, then 2,7,9,11,16 --> 9 is both AM and median
hence this statement is sufficient.
So OA is D