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 $\frac{11}{6}$ times their difference.

Which one of the following is correct in respect of the above Question and the Statements?

The correct option is: The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone

Let us analyze the problem step-by-step.

We are given that the age of each of P and Q is less than 100 years but more than 10 years. This means both ages are two-digit numbers.

Let the age of P be represented as a two-digit number:
$P=10x+y$
where $x$ and $y$ are digits from 1 to 9 (since the ages are greater than 10, the tens digit cannot be 0, and because interchanging the digits also gives a two-digit age for Q, the units digit $y$ also cannot be 0).

If we interchange the digits of the age of P, we get the age of Q:
$Q=10y+x$

Now, let us find the difference between their ages:
$\text{Difference}=|P-Q|=\left|\right(10x+y)-(10y+x\left)\right|=|9x-9y|=9|x-y|$
Thus, the difference in their ages is always a multiple of 9.

Let us evaluate Statement-I: The age of P is greater than the age of Q.
This statement implies $P>Q$, which means $x>y$. Therefore:
$\text{Difference}=9(x-y)$
However, Statement-I alone does not give us the specific values of $x$ and $y$ or the value of $x-y$. So, the difference cannot be uniquely determined using Statement-I alone.

Let us evaluate Statement-II: The sum of their ages is $\frac{11}{6}$ times their difference.
The sum of their ages is:
$\text{Sum}=P+Q=(10x+y)+(10y+x)=11(x+y)$
According to Statement-II:
$\text{Sum}=\frac{11}{6}$
Substituting the expressions for Sum and Difference:
$11(x+y)=\frac{11}{6}\times 9|x-y|$
Dividing both sides by 11:
$x+y=\frac{9}{6}|x-y|$
Simplifying the fraction:
$x+y=\frac{3}{2}|x-y|$
$2(x+y)=3|x-y|$

Since the left side is positive and symmetric, let us assume without loss of generality that $x>y$ (the difference remains the same whether $x>y$ or $y>x$). This gives:
$2(x+y)=3(x-y)$
$2x+2y=3x-3y$
$5y=x$

Since $x$ and $y$ must be single-digit integers from 1 to 9, the only possible integer values that satisfy $x=5y$ are:
$y=1\Rightarrow x=5$
(Any other positive integer for $y$ would make $x\ge 10$, which is not a single digit).

Thus, the digits must be 5 and 1.
The difference between their ages is:
$\text{Difference}=9|x-y|=9|5-1|=9\times 4=36$
We obtained a unique value of 36 for the difference using Statement-II alone, without needing Statement-I.

Hence, the Question can be answered by using Statement-II alone, but cannot be answered using Statement-I alone.