Management Aptitude Test Sep 2020 Paper

Radius of inner track $=100\mathit{m}$
Radius of outer track $=102\mathit{m}$
$\therefore$ Distance covered by $\mathit{A}=2\times \mathit{\pi }\times 100=200\mathit{\pi }\mathit{m}$
Distance covered by $\mathit{B}=2\times \mathit{\pi }\times 102=204\mathit{\pi }\mathit{m}$
Time taken by $\mathit{A}=1\min 30\mathit{s}=90\mathit{s}$
Time taken by $\mathit{B}=1\min 32\mathit{s}=92\mathit{s}$
$\therefore {\mathit{S}}_{\mathit{A}}=\frac{200\mathit{\pi }}{90},{\mathit{S}}_{\mathit{B}}=\frac{204\mathit{\pi }}{92}$
${\mathit{S}}_{\mathit{A}}:{\mathit{S}}_{\mathit{B}}=\frac{200\mathit{\pi }}{90}:\frac{204\mathit{\pi }}{92}=18400:18360$
So, ${\mathit{S}}_{\mathit{A}}>{\mathit{S}}_{\mathit{B}}$ i.e., speed of $\mathit{A}$ is greater than $\mathit{B}$.