The HCF and LCM of $2$ numbers are $2$ and $60$ respectively. If one of the numbers is $14$ more than the other, find the smaller number.

Let the smaller number be $x$.
Then, the larger number $=x+14$.

We know the relation:

$$\text{Product of two numbers}=\text{HCF}\times \text{LCM}$$

$$x(x+14)=2\times 60$$

$$x^2+14x=120$$

$$x^2+14x-120=0$$

Factorize the quadratic equation:

$$x^2+20x-6x-120=0$$

$$x(x+20)-6(x+20)=0$$

$$(x-6)(x+20)=0$$

Since numbers must be positive matching the options:

$$x=6$$

Hence, the smaller number is $6$.