Question

A drunkard walking in a narrow lane takes $5$ steps forward and $3$ step backward, followed again by $5$ steps forward and $3$ steps backward and so on. Each step is $1m$ long and requires $1s$. How long the drunkard takes to fall in a pit $13m$ away from the start?

Answer 1

Step 1: Given Data

1. The steps taken by the drunkard in the forward direction in one cycle are $5$ steps.
2. The steps taken by the drunkard in the backward direction in one cycle are $3$ steps.
3. The length of one step is $1m$.
4. The time taken for one step is $1s$.
5. The distance between the pit and the drunkard is $13m$.

Step 2: Find the distance covered and time taken by the drunkard in one cycle:

1. According to the given data, $1$ step means $1m$ distance covered in $1s$.
2. Hence, the total distance covered in one cycle is $d=(5-3)\times 1=2m$
3. The total time taken to cover one cycle is $t=(5+3)\times 1=8s$

Step 3: Find the number of cycles and steps required for him to fall in the pit:

1. After the second cycle, the drunkard has moved $4m$ forward.
2. After the third cycle, he has moved $6m$ forward.
3. After the fourth cycle, he has moved $8m$ forward.
4. Now, the drunkard has moved $8m$ forward in total. Hence, he is $13-8=5m$ away from the pit.
5. As $1$ step means $1m$ distance, he will fall into the pit when he takes $5$ more steps in the forward direction.
6. Thus, $4$ cycles were completed and $5$ steps were taken in the forward direction by the drunkard to fall into the pit.

Step 4: Calculate the total time taken:

1. The drunkard completed $4$ cycles and took $5$ steps in the forward direction in total before falling into the pit.
2. The drunkard took $1$ step in $1s$ and completed $1$ cycle in $8s$.
3. Therefore the total time taken by the drunkard will be:
   $$\begin{gathered} t=(4\times 8)+(5\times 1) \\ =32+5 \\ =37s \end{gathered}$$

Final Answer:

The drunkard will take $\mathbf{37}\mathbf{}\mathit{s}$ to fall into a pit which is $13m$ away from the start.