CAT 2023 Slot 2 Solved Paper

It is given that $\mathit{k}$ divides $\mathit{m}+2\mathit{n}$ and $3\mathit{m}+4\mathit{n}$.
Since $\mathit{k}$ divides $\left(\mathit{m}+2\mathit{n}\right)$, it implies $\mathit{k}$ will also divide $3\left(\mathit{m}+2\mathit{n}\right)$. Therefore, $\mathit{k}$ divides $3\mathit{m}+6\mathit{n}$.
Similarly, we know that $\mathit{k}$ divides $3\mathit{m}+4\mathit{n}$.
We know that if two numbers $\mathrm{a}$, and $\mathrm{b}$ both are divisible by $\mathrm{c}$, then their difference (a-b) is also divisible by $\mathrm{c}$.
By the same logic, we can say that $\left\{\left(3\mathit{m}+6\mathit{n}\right)-\left(3\mathit{m}+4\mathit{n}\right)\right\}$ is divisible by $\mathit{k}$. Hence, $2\mathit{n}$ is also divisible by $\mathit{k}$.
Now, $\left(\mathit{m}+2\mathit{n}\right)$ is divisible by $\mathit{k}$, it implies $2\left(\mathit{m}+2\mathit{n}\right)=2\mathit{m}+4\mathit{n}$ is also divisible by $\mathit{k}$.
Hence, $\left\{\left(3\mathit{m}+4\mathit{n}\right)-\left(2\mathit{m}+4\mathit{n}\right)\right\}=\mathit{m}$ is also divisible by $\mathit{k}$.
Therefore, $\mathrm{m}$, and $2\mathrm{n}$ are also divisible by $\mathrm{k}$.
The correct option is C