j's initital = x. final = (1+ p/100)x
similarly K's: (1 + p/100)y
A) y = 2000 + x
insufficient.
B) (1+ p/100)y = 2440 + (1+ p/100)x insufficient
A&B)
(1+ p/100)y = 2440 + (1+ p/100)x -> (1 + p/100) (y-x) = 2440
(1 + p/100)2000=2440. solve for p
IMO C

Ashetty wrote:J's salary and P's salary were each P% greater in 1998 than in 1995, what is value of P?
1)In 1995 K's salary was 2000 greater than J
2)In 1998 K's salary was 2440 greater than J

Ans:C

CAN ANYONE PLEASE EXPLAIN ME WHY?

Cans' solution, in a more qualitative analysis: between 1995 and 1998, not only the salaries increase, but also the DIFFERENCE between the salaries increased: from a differnce of $2000 to a difference of $2440. The $440 extra that K's salary earned in those 3 years is the p% interest earned just on the extra $2000.

So P% of 2000 is $440 - which should allow you to calculate P. But you need both the initial difference and the final difference.

Is that a typo in your statements?. Because you've mentioned another entity 'K'

sl750 wrote:Is that a typo in your statements?. Because you've mentioned another entity 'K'

Yes - Edited. Thanks.