Let the original speed $ = S$ and the original time $ = T$ minutes.
$\text{Speed(S)} = \dfrac{\text{Distance(D)}}{\text{Time(T)}}$
$\implies S \propto \dfrac{1}{T}$
$\implies \dfrac{S_{1}}{S_{2}} = \dfrac{T_{2}}{T_{1}}$
$\implies \dfrac{S}{\left(\dfrac{5}{7}\right)S} = \dfrac{T + 16}{T}$
$\implies \dfrac{7}{5} = \dfrac{T + 16}{T}$
$\implies 7T = 5T + 80$
$\implies 2T = 80$
$\implies T = 40\,\text{minutes}.$
$$\textbf{(OR)}$$
Let usual speed $ = S$
and, new speed $ = \left(\dfrac{5}{7}\right)S$
If usual time $ = T$ minutes, then new time $ = \left(\dfrac{7}{5}\right)T$ minutes.
$\therefore \left(\dfrac{7}{5}\right)T - T = 16$
$ \implies T = 40$ minutes.
So, the correct answer is $(C).$