CAT 2022 Slot 1 Paper

Let the original number of students be ' $\mathit{n}$ ' whose average weight is $\mathit{x}$
Let the number of students added be ' $\mathit{m}$ ' and the average weight will be $\mathit{x}+3$
We need to find the value of $\mathit{n}:\mathit{m}$
It is given, average weight of students in a class increased by 0.6 after new students are added.
Therefore,
$\frac{\mathit{n}\mathit{x}+\mathit{m}\left(\mathit{x}+3\right)}{\mathit{n}+\mathit{m}}=\mathit{x}+0.6$
$\mathit{n}\mathit{x}+\mathit{m}\mathit{x}+3\mathit{m}=\mathit{m}\mathit{x}+\mathit{n}\mathit{x}+0.6\mathit{n}+0.6\mathit{m}$
$2.4\mathit{m}=0.6\mathit{n}$
$4\mathit{m}=\mathit{n}$
$\frac{\mathit{n}}{\mathit{m}}=\frac{4}{1}$
The answer is option C