[spoiler][A][/spoiler] is the right answer. I don't know what you are looking for but I can share my solution.

St 1) a=(9/4)c, b=(3/2)c & c=c after simplifying the equations
Since a,b & c are positive single digit number, therefore c must be equal to 4 else given conditions for a doesn't satisfy. finally abc = 964 -----sufficient

St 2) a = 1.5x + c, b = x+c & c=c
a = 2.5x + c, b = x+c & c=c
for a to be positive single digit number x must be even. c can take any value
Let's put in some values
c=1, x=2 => abc=631
c=2, x=2 => abc=742
-----therefore insufficient

ANS A

This question comes down to finding three single digits that fit the parameters of the statements:

Statement 1: a = 1.5b and b = 1.5c

In order to get a digit when another digit is multiplied by 1.5, the digit being multiplied has to be even.

So we are down to 2, 4, 6 and 8 for digits b and c.

If a = 1.5b, and b = 1.5c, then a = 1.5²c = 2.25 c.

Of 2, 4, 6 and 8, the only one that can be used to generate a digit when multiplied by 2.25 is 4.

So a = 9.

b = 9/1.5 = 6

c = 6/1.5 = 4

So abc = 964

Sufficient.

Statement 2: a = 1.5x + b and b = x + c, where x represents a positive single digit.

Once again, in order for 1.5x to be an integer value, x has to be an even number, 2, 4, 6, or 8.

If a = 1.5x + b, and b = x + c, then a = 2.5x + c.

Since 2.5(4), 2.5(6) and 2.5(8) are all > 9, x can only be 2, and 2.5x = 5.

So we can start with any digit c and add 2 and 5 to it to get b and a.

Because 9 - 5 = 4, c ≤ 4.

So, for instance, abc could be 964 or 853.

Multiple values are possible.

Insufficient.

The correct answer is A.

S1:

c can't be odd, or b won't be an integer. Likewise, b can't be odd, or a won't be an integer. So b and c are even.

We know that a > b > c, so let's try possible c's.

c = 0
then b = 0 and a = 0, which doesn't give us a three digit integer;

c = 2
then b = 3 and a = 4.5, nope;

c = 4
then b = 6 and a = 9, looks good;

c = 6
then b = 9 and a > 9, too big;

and we're out of options! So only c = 4, b = 6, a = 9 works; SUFFICIENT.

S2

If x = 2, then 1.5x = 3, and we have a = b + 3 and b = c + 2. This gives us lots of solutions, such as

c = 0, b = 2, a = 5
c = 1, b = 3, a = 6,
etc.

Since there's more than one possibility, this is NOT sufficient.