CAT 2020 Exam Shift 1 Solved Paper

Given $\mathit{a}\mathit{b}=432,\mathit{b}\mathit{c}=96$ and
To find the minimum value for $\mathit{a}+\mathit{b}+\mathit{c}$, the two larger numbers should be as close as possible.
The closest combination whose product is 432 is $18\times 24$. For $\mathit{b}=24$, we get $\mathit{c}=4$ and $\mathit{a}=18$
Hence the least value for $\mathit{a}+\mathit{b}+\mathit{c}=46$.