Data Sufficiency

Sets X and Y consist solely of positive integers. Each set contains at least two elements, and no element appears more than once within a set. Sets X and Y contain the same number of elements. Is the standard deviation of set X greater than the standard deviation of set Y?

(1) The positive difference between the range of set X and set Y is 12
(2) Each element of set Y is the square if an element of set X

OA is B

Please let me know if i am correct in s(1).

My working for S(1):-
X = {1,22,23,24,25} range of set X = 25 - 1 = 24
Y = {2,5,8,11,14} range of set Y = 14 - 2 = 12
The positive difference between the range of set X and set Y = 24 - 12
Here, SD(X) < SD(Y). So, NO

X = {1,25} range of set X = 25 - 1 = 24
Y = {0,12} range of set Y = 14 - 2 = 12
The positive difference between the range of set X and set Y = 24 - 12
But here SD(X) > SD(Y) . So, YES
Because after getting the average of both the sets the measure of dispersion will be {1,23,25} and {0,6,12}

Please let me if i am wrong.

Regards

Can anyone please let me know whether my approach is correct ? or i am missing something. I always lack in coming up with different set of numbers in SD, but these days i am trying with different set of numbers, thereby I am taking time in such questions but accuracy is going up.
Can you please tell me what is the correct approach or sets of numbers that i can use for this question?

Thanks & Regards

GMATGuruNY wrote:
Statement 1:
We don't know which set has the greater range.
INSUFFICIENT.

Mitch,

Thanks for your reply. From your explanation what i have understand is that "The positive difference between the range of set X and set Y" does not only mean X - Y = 12,
but it can also mean Y - X = 12.
X - Y = 12
Y - X = 12
& SD can change

Suppose
X = {1,22,23,24,25} range of set X = 25 - 1 = 24
Y = {2,5,8,11,14} range of set Y = 14 - 2 = 12
Here, SD(X) < SD(Y).

But, if i change
X = {2,5,8,11,14} range of set Y = 14 - 2 = 12
Y = {1,22,23,24,25} range of set X = 25 - 1 = 24
Here, SD(X) > SD(Y).

Am i correct.