The number of pairs of natural numbers the difference of whose squares is 45 will be ?

Answer: Option B

Let the numbers be x and y
According to question,
$$\begin{array}{rl}& \left(x>y\right)\\ & x^2-y^2=45\\ & \left(x+y\right)\left(x-y\right)=45\\ & \text{Make factor of 45}\\ & \text{15}\times 3\\ & 9\times 5(3\text{pairs})\\ & 45\times 1\end{array}$$
These pairs gives the value of x and y which satisfy the given condition.