MAT Exam 2020 Paper

Concept:Speed is the distance covered per unit time. The distance for one round on a circular track is its circumference, $2\pi r$.Explanation:For runner A on the inner track:Radius $= 105\ \mathrm{m}$, so distance covered $= 2\pi(105) = 210\pi\ \mathrm{m}$.Time taken by A $= 1\ \mathrm{min}\ 30\ \mathrm{s} = 90\ \mathrm{s}$.Speed of A, $S_A = \frac{210\pi}{90} = \frac{7\pi}{3}\ \mathrm{m/s}$.For runner B on the outer track:Radius $= 119\ \mathrm{m}$, so distance covered $= 2\pi(119) = 238\pi\ \mathrm{m}$.Time taken by B $= 1\ \mathrm{min}\ 32\ \mathrm{s} = 92\ \mathrm{s}$.Speed of B, $S_B = \frac{238\pi}{92} = \frac{119\pi}{46}\ \mathrm{m/s}$.Now compare the speeds:$S_A = \frac{7\pi}{3} \approx 2.333\pi\ \mathrm{m/s}$$S_B = \frac{119\pi}{46} \approx 2.587\pi\ \mathrm{m/s}$Since $2.587\pi > 2.333\pi$, speed of B is greater than speed of A.Answer:B. B