Alcoholic

Suppose that it takes $n~seconds$ for that alcoholic person to go by $13-5=8~metres$ .So, he'll just go forward for $5$ secs and fall. Hence, he'll need to go forward for $8~m$ .But, in that $8~m$ , he has made some of action which is 5 m forward and 3 m backward without spare walk. Then, for each action , he has walk forward for $5-3=2~metres$ and have used $5+3=8~secs$ .So, the amount of time that we need is: $\frac{\color{#3D99F6}{8}(metres)}{\color{#3D99F6}{2}(metres)}*\color{#D61F06}{8}(secs)+\color{#D61F06}{5}(secs)$ $=\color{#20A900}{4}(times)*\color{#D61F06}{8}(secs)+\color{#D61F06}{5}(secs)$ $\color{#D61F06}{32}(secs)+\color{#D61F06}{5}(secs)$ $=\boxed{\color{#D61F06}{37}~secs}$

This man is walking 5m forward and 3m backward which means total distance that he covered is 2m in 8 seconds,repeating it 4 times he will cover 8mm in 32 seconds(8 seconds *4 times),now he has to cover 5m,now if we will read the question again it is written that the person is walking 5m forward and after that 3m backward, so when he will move 5m forward he will reach that manhole and fall into it,and the last step is to calculate and add the time taken by him to cover that last 5m that will be 32s+5s=37s.