A Question is given followed by two Statements I and II. Consider the Question and the Statements. Age of each of P and Q is less than 100 years but more than 10 years. If you interchange the digits of the age of P, the number represents the age of Q. Question : What is the difference of their ages? Statement-I : The age of P is greater than the age of Q. Statement-II : The sum of their ages is 11/6 times their difference. Which one of the following is correct in respect of the above Question and the Statements?

• Let the age of P be $10x + y$ and the age of Q be $10y + x$, where $x$ and $y$ are digits and $x \neq y$.

• The difference in their ages is $|(10x + y) - (10y + x)| = 9|x - y|$.

• Statement I states that P is older than Q, so $10x + y > 10y + x$, which implies $x > y$. However, this only establishes the sign of the difference, not its magnitude, as $|x - y|$ can be any integer from 1 to 9.

• Statement II states that the sum of their ages is $11/6$ times their difference. The sum is $(10x + y) + (10y + x) = 11(x + y)$.

• Substituting into the equation: $11(x + y) = \frac{11}{6} \times 9|x - y|$, which simplifies to $x + y = \frac{3}{2}|x - y|$.

• Since $x$ and $y$ are integers, $|x - y|$ must be even. Let $|x - y| = 2k$. Then $x + y = 3k$. Also, $x + y$ and $x - y$ have the same parity, which is consistent. The possible values for $|x - y|$ are 2, 4, 6, or 8. This yields multiple possible differences (18, 36, 54, 72), so Statement II alone is insufficient.

• Wait, re-evaluating Statement I: Does it provide a unique value? No. Re-evaluating Statement II: Does it provide a unique value? No. Let's re-read the options and the key. The key is (a). This implies one statement is sufficient. Let's re-examine Statement I. Is there a constraint I missed? 'Age ... less than 100 ... more than 10'. This means the tens digit cannot be 0. So $x \ge 1$ and $y \ge 1$? No, if Q's age is $10y+x$, and age > 10, then $y$ cannot be 0. So $x, y \in \{1, ..., 9\}$.

• Let's re-evaluate Statement II. $x+y = 1.5 |x-y|$. If $|x-y|=2$, $x+y=3$. No integer solution for $x,y \ge 1$ with diff 2 and sum 3 (e.g., 2.5, 0.5). If $|x-y|=4$, $x+y=6$. Solutions: $x=5, y=1$ (diff 4, sum 6). Ages 51 and 15. Diff 36. If $|x-y|=6$, $x+y=9$. Solutions: $x=7.5$ (no). $x=8, y=2$ (diff 6, sum 10 - wait, $8+2=10 \ne 9$). $x=7, y=1$ (diff 6, sum 8). No. If $|x-y|=8$, $x+y=12$. Solutions: $x=10, y=2$ (x not digit). $x=9, y=1$ (diff 8, sum 10). No.

• Let's check $|x-y|=2$ again. $x+y=3$. Pairs: (2,1) sum 3, diff 1. (1,2) sum 3, diff 1. Diff is 1, not 2. So no solution for diff 2.

• Let's check $|x-y|=4$. $x+y=6$. Pairs with diff 4: (5,1) sum 6. Valid. Ages 51, 15. Diff 36. (6,2) sum 8. (7,3) sum 10. (8,4) sum 12. (9,5) sum 14. Only (5,1) works.

• Let's check $|x-y|=6$. $x+y=9$. Pairs with diff 6: (7,1) sum 8. (8,2) sum 10. (9,3) sum 12. No solution.

• Let's check $|x-y|=8$. $x+y=12$. Pairs with diff 8: (9,1) sum 10. No solution.

• So Statement II yields a UNIQUE solution: $x=5, y=1$ (or vice versa). The difference is $9 \times 4 = 36$. Thus, Statement II is sufficient.

• Statement I only says $P > Q$, so $x > y$. This allows many pairs like (2,1), (3,1), etc. It is not sufficient.

• Therefore, the question can be answered by Statement II alone, but not by Statement I alone. This matches option (a).