sairamGmat wrote:. If x and y are integers between 10 and 99, inclusive, is (x-y)/9 an integer?
(1) x and y have the same two digits, but in reverse order.
(2) The tens' digit of x is 2 more than the units digit, and the tens digit of y is 2 less than the units digit.

OA is A. But not sure why it is not D

(1) from 1, if K1 and K2 are two digits, then x = 10k1+k2 and y = 10K2+K1. (x-y) = 9(k1-k2) and (x-y)/9 is an integer. If k1<k2, then it is a negative integer and if k1>k2, then it is positive integer. - SUFFICIENT

(2) from 2, if K1 is tens' digit and K2 is units digit

K1 = 2+K2;
K2 = K1-2;

Now as x = 10K1+K2 and y = 10K2+K1; x-y = [10k1+(K1-2)] - [10 (k1-2) + k1] = [11k1-2] - [11k1 -20] = -2 + 20 = 18.

(x-y)/9 is 2 which is an integer. So, why not it is D?

sumanr84 wrote: Correcting a typo,
Let the unit digit of x is 'a' and unit digit of y is 'b'.
Then x = 10(a+2) + a and y = 10(b-2) + b
x - y = 11a +20 - 11b + 20
x - y = 11(a-b) +40

Let try getting a YES and a NO.
1. a = 4, b =3
11 * 1 + 40 = 51 not divisible by 9 - NO

2. a = 8 , b = 1
11 * 7 + 40 = 117 divisible by 9 - YES

So, B is not sufficient