Answer : Option C

Explanation :

Let the distance be x km ,
the speed in which he moved = v kmph
Time taken when moving at normal speed - time taken when moving 3 kmph faster = 40 minutes

$MF#%\begin{align}
&\Rightarrow \dfrac{x}{v} - \dfrac{x}{v + 3} = \dfrac{40}{60}\\
&\Rightarrow x \left[\dfrac{1}{v} - \dfrac{1}{v + 3}\right] = \dfrac{2}{3}\\
&\Rightarrow x \left[\dfrac{v + 3 - v}{v\left(v + 3\right)}\right] = \dfrac{2}{3}\\
&\Rightarrow 2v(v+3) = 9x\text{................(Equation1)}\\\\\\\\
\end{align} $MF#%

$MF#%\begin{align}
&\Rightarrow \dfrac{x}{v-2} - \dfrac{x}{v} = \dfrac{40}{60}\\
&\Rightarrow x \left[\dfrac{1}{v-2} - \dfrac{1}{v}\right] = \dfrac{2}{3}\\
&\Rightarrow x \left[\dfrac{v - v + 2}{v\left(v - 2\right)}\right] = \dfrac{2}{3}\\
&\Rightarrow x \left[\dfrac{2}{v\left(v - 2\right)}\right] = \dfrac{2}{3}\\
&\Rightarrow x \left[\dfrac{1}{v\left(v - 2\right)}\right] = \dfrac{1}{3}\\
&\Rightarrow v(v-2) = 3x\text{................(Equation2)}\\\\\\\\
&\dfrac{\text{Equation1}}{\text{Equation2}}\Rightarrow \dfrac{2(v+3)}{(v-2)}= 3\\
&\Rightarrow 2v + 6 = 3v -6 \\
&\Rightarrow v = 12\\\\\\\\
&\text{Substituting this value of v in Equation 1}\Rightarrow 2 \times 12 \times 15 = 9x\\
&=> x = \dfrac{2 \times 12 \times 15}{9} = \dfrac{2 \times 4 \times 15}{3} = 2 \times 4 \times 5 = 40\\
&\text{Hence distance = 40 km}\end{align} $MF#%