Explanation

(c) Let $v_{1}$, $v_{2}$ be the initial speeds of first and second runners. Let $t$ be time by them when the first runner has completed 50m. During this time, the second runner has covered a distance $= 50 - 1 = 49$m.
So, $t = \frac{50}{v_{1}} = \frac{49}{v_{2}}$ ...(i)
Suppose, the second runner increases his speed to $v_{3}$ so that he covers the remaining distance $( = 51$m) in times $t$. So
$t = \frac{51}{v_{3}} = \frac{49}{v_{2}}$
or $v_{3} = \frac{51}{49}v_{2}$
or $v_{3} = \left( 1 + \frac{2}{49} \right)v_{2}$ or $\frac{v_{3}}{v_{2}} - 1 = \frac{2}{49}$
or $\frac{v_{3} - v_{2}}{v_{2}} = \frac{2}{49}$
or % increase $= \frac{2}{49} \times 100 = 4$.1%