Let the usual speed and the usual time taken to travel the distance be $s$ and $t,$ respectively.
We know that, $\boxed{\text{Speed} = \frac{\text{Distance}}{\text{Time}}}$
$ \Rightarrow \text{Speed} \propto \frac{1}{\text{Time}} \quad [\because \text{Distance constant}]$
We get, $\boxed{\frac{s_{1}}{s_{2}} = \frac{t_{2}}{t_{1}}}$
$\begin{array}{cc} \underline{\text{Speed}} & \underline{\text{Time}} \\ S & t \\ \frac{1}{3}S & t + 30 \\ \frac{1}{3}S & 3t \end{array}$
Here, $3t = t + 30$
$\Rightarrow 2t = 30$
$\Rightarrow \boxed{ t = 15 \; \text{minutes}}$
$\begin{array}{cc} \underline {\text{Speed}} & \underline{\text{Time}} \\ S & 10\;\text{minutes} \\ \frac{10}{6}S & 6\;\text{minutes} \\ \frac{5}{3}S & 6 \;\text{minutes}\end{array}$
Required percentage $ = \dfrac{\left(\frac{5}{3}s – s \right)} {s} \times 100 \%$
$ \qquad = \frac{2}{3} \times 100 \% = 66.66 \% \simeq 67 \%$
Correct Answer$: \text{B}$