from (2) we get xy/100<x-y
if we take the value of xy = 100 taking individual values will predetmine the ratio x/y but product xy leave us with possibilities
then 1<x-y or x-y > 1 will be possible only if x > y by an +ve integer which is Option (1) option C is out
let us take value of xy as (20,5) (25,4) (50,2) factor of 100 are 2²X5² choosing value as close to each other the best option is (20,5) we get
1997 let the value of the building be 100

1998 x=20 value = 120 x=50

1999 y=5 value = 114 option A , D and E are out

more than 2 mins are up option B or D.

This is what i would find a quick way.

Rent in 1997 be R
Rent in 1998 = R*(100+x)/100
Rent in 1999 = R(100+x)/100*(100-y)/100

Reverse approcah , if we can prove stmnt II holds for Rent in 1999 >Rent in 1997 , then its SUFFICIENT
So , assume R(100+x)/100*(100-y)/100 > R
==> 100^2 + 100(x-y) -xy >100^2
==>(x-y)>xy/100

This is what stmnt II is (x-y)>xy/100 , so we proved stmnt II holds if Rent in 1997 <Rent in 1999 , so you can take the converse of this that if stmnt II holds then Rent in 1999 is surely greater than Rent in 1997

IMO D.

i) x > y is true and always gives us that R(97)< R(99)

ii) Stmt II will be true only if x>y, as mentioned by KSTV. and give the same answer.

So I think answer should be D.

Can anyone provide a case where Stmt I Provides YES and NO condition to rule it out?

Should the ans be D instead of B?