i think 'D" is answer.

we can assume 1997 as 100, then 1998 becomes 100+x, and 1999 becomes 100 + (x-y) - xy/100. we need to find if term (x-y) - xy/100 is positive or not. both the statement are SUFF

sorry..its B

A is a trap!!

It took me 20 mins to solve this one.

Any other method
??
Stuart,Pls help!!

thanks

D is the answer. Moreover, both A and B lead to the same conclusion that X> Y, so if B is the answer, A should be the answer as well. statement B doesn't hold true if X< y since it will give a positive value. Atleast I assumed that we can only consider positive values for both x and y since they are percentages. I hope it's an OK assumption.

oops, my bad..I think I compared 1999 to 1998 instead of to 1997. In anycase, I want to understand if my approach is right or wrong.

Ian, can you please correct me if I'm wrong with the 1999> 1998 comparision. I chose some unknown value "a" for 1997(eventhough 100 would have made it easier , so for 1998, we have "a + 0.0xa" and in 1999, we have (a+0.0xa)(1-0.0y). Comparing 1999 to 1998 =>

is (a+0.0xa)(1-0.0y) > (a+0.0xa), we can divide a + 0.0xa since both x and a are postive values and the should be a positive value without worrying about reversing the inequalities.,

=> is 1-0.0y > 1 ; since y is a positive value as well, would this equation be always true and so 1999 > 1998.

Now if I have compared 1999> 1997, this is how I would have went about solving it,

is (a+0.0xa)(1-0.0y) > a ; simplifying furthur,

is a(1+ 0.0x)(1-0.0y) > a => is (1+0.0x)(1-0.0y)> 1; which I think would be insufficient if x> y. But I did spend some time trying to figure out what values of x and y that can result in the value less then 1.

oh yes... answer has to be 'B'. thankyou people.