A fidget day of a business man

Let the distance between K and the man's home be $x$ km. Then that between A and K is $\dfrac x3$ km. Let the speed of the train by $v$ . Then that of the taxi is $\dfrac v2$ .

The time taken for the man to reach home from work is $t_{\text{usual}} = \underbrace{\dfrac {40}v}_{\text{train}} + \underbrace{10}_{\text{walk}}$ minutes. Since today, he was 5 minutes late, $t_{\text{today}} = \dfrac {40}v + 15$ . This time is the same as $t_{\text{today}} = \underbrace{\dfrac {x/3}v}_{\text{A-K}} + \underbrace{\dfrac x{v/2}}_{\text{taxi}}$ . Therefore,

$\begin{aligned} \frac {40}v + 15 & = \frac x{3v} + \color{#3D99F6} \frac {2x}v & \small \color{#3D99F6} \text{Since taxi ride is }\frac {2x}v = 30 \text{ minutes} \\ \frac {40}v + 15 & = 5 + \color{#3D99F6} 30 \\ \implies v & = 2 \\ \frac {2x}v & = 30 \\ \implies x & = \boxed {30} \end{aligned}$