Management Aptitude Test Dec 2013 Paper

(a) Radius of inner track = 100 m
Radius of outer track = 102 m
∴ Distance covered by
A = 2 x π x 100 = 200 π m
Distance covered by
B = 2 x π x102 = 204 π m
Time taken by A = 1 min 30 s = 90 s
Time taken by B = 1 min 32 s = 92 s
∴ ${\mathit{S}}_{\mathit{A}}=\frac{200}{90}\mathit{\pi }$
${\mathit{S}}_{\mathit{B}}=\frac{204}{92}\mathit{\pi }$
${\mathit{S}}_{\mathit{A}}:{\mathit{S}}_{\mathit{B}}$ = $\frac{200}{90}\mathit{\pi }$ : $\frac{204}{92}\mathit{\pi }$
= 18400 : 18360
So, ${\mathit{S}}_{\mathit{A}}>{\mathit{S}}_{\mathit{B}}$ i.e., speed of A is greater than B