Answer: $\mathbf A$
Explanation:
$x ≡ 2 \;\text{mod}\; 15$ means $ x-2=15\times t_1 \Rightarrow x=2+15\times t_1$
$x ≡ 4 \;\text{mod}\;21$ means $x-4 = 21 \times t_2 \Rightarrow x=4+ 21\times t_2$
where $t_1$ and $t_2$ are two constants.
For finding the common value of $x$ in both cases, both these equations can be equated with each other.
$2+15\times t_1=4+ 21\times t_2 \Rightarrow 15\times t_1-21\times t_2 = 2$
Hence, no combination of $t_1$ and $t_2$ can satisfy this equation.
So, the number of values of $x$ is 0.
$\therefore \mathbf A$ is the correct option.