CAT 2018 Question Paper Shift 1 Solved Paper

Let the other two numbers be $\mathit{y}$ and $\mathit{z}$.
As per the condition
$73\mathit{y}\mathit{z}-37\mathit{y}\mathit{z}=720$
Or $36\mathit{y}\mathit{z}=720$
Or $\mathit{y}\mathit{z}=20$
Minimum possible sum of the squares of the other two numbers would occur when $\mathit{y}=\mathit{z}$ i.e.
$\mathit{y}=\mathit{z}=\sqrt{20}$
Hence the required sum $=40$.