I get x>y, then how to go about it?

OA 24, why not 23?

MBA.Aspirant Posted Wed Aug 17, 2011 4:04 am

I get x>y, then how to go about it?

I think the answer should be (E).

I reckon this is how:
Let us assume following values of x and y:
1)x=-1, y=-2 2)x=2, y=1

x^5--- x^4--- x^3--- x^2--- x^1--- y^1--- y^2--- y^3--- y^4 (taken from the question itself)
-1 1 -1 1 -1 -2 4 -8 16 (for 1)x=-1, y=-2)
32 16 8 4 2 1 1 1 1 (for 2)x=2, y=1)


For the above table, one should be able to make out that only condition in option (E) holds true.

Hope this helps.

Thanks.

Hi,
Q1)x=0, x=3 both should satisfy.

Q3)y=x+5
So, x+y = 2x+5
If you are getting 23, you are assuming that x is an integer and hence max(x) = 9 which leads to max(2x+5) = 23.
Now, 5<x<10. So, 10<2x<20
10+5<2x+5<20+5 => 15<x+y<25
So, 24 is the greatest integer value of (x+y).