Madhya Pradesh Police Exam Model Paper 6

Given:
${\left(101\times 72\right)}^2$
Concept:
If $\left(101\times \mathit{a}\right)$ , then
$\Rightarrow \left[\left(100+1\right)\times \mathit{a}\right]$
$\Rightarrow \left(100\mathit{a}+\mathit{a}\right)$
Calculation:
${\left(101\times 72\right)}^2$
$\Rightarrow {\left[\left(100+1\right)\times 72\right)}^2$
$\Rightarrow {\left(7200+72\right)}^2$
$\Rightarrow {\left(7272\right)}^2$
⇒ 52881984
$\therefore {\left(101\times 72\right)}^2=52881984$ .
Smart Method:
${\left(101\times 72\right)}^2$
Taking unit digits
${\left(1\times 2\right)}^2$
$\Rightarrow 2^2$
⇒ 4
Hence, answer must have 4 at its unit digit.
∴ 52881984 is the correct option.