A Strategic Decision

Let the distance between Nitin's house and school be $d$ .
Let the speed of Nitin when he walks be $v$ and his speed when he bicycles be $kv$ .
We have the time taken for him to walk to and fro from his home to school as $\frac{2d}{v}$ .
Note that $r$ percent of $d$ is $\frac{dr}{100}$ .
So in the case given in the question, he walks distance $\frac{2dr}{100}$ and cycles distance $2d$ .

Hence, this will take time $\frac{2dr}{100v}+\frac{2d}{kv}=\frac{2kdr+200d}{100kv}$ .

From the conditions of this problem, $\frac{2kdr+200d}{100kv} \le \frac{2d}{v}$ .

which reduces to $r \le \frac{100k-100}{k} = 100(\frac{k-1}{k})=a(\frac{k-1}{k})$ .

yielding $a=100$ .
So our answer is $\boxed{100}$