A drunkard is walking along a stsraight road. He takes five steps forward and three steps backward and so on. Each step is `1 m` long and takes `1 s`. There is a pit on the road `11 m`, away from the starting point. The drunkard will fall into the pit after.

Options

A. ` 29 s`
B. ` 21 s`
C. ` 37 s`
D. ` 31 s`

Explanation

To solve this problem, we need to analyze the pattern of the drunkard's movement and calculate the total time taken for him to fall into the pit.

### Step-by-Step Solution:

1. **Understanding the Movement Pattern:**
- The drunkard takes 5 steps forward and then 3 steps backward repeatedly.
- Each step is 1 meter and takes 1 second.

2. **Calculating Net Distance Covered in One Cycle:**
- In the first 5 steps forward, he covers \(5 \times 1 = 5\) meters.
- In the next 3 steps backward, he covers \(3 \times 1 = 3\) meters.
- Net distance covered in one complete cycle (5 steps forward + 3 steps backward) is:
\[ 5 - 3 = 2 \text{ meters} \]
- Time taken for one complete cycle:
\[ 5 \text{ seconds (forward)} + 3 \text{ seconds (backward)} = 8 \text{ seconds} \]

3. **Calculating Total Distance to the Pit:**
- The pit is 11 meters away from the starting point.

4. **Calculating Number of Complete Cycles to Reach Close to 11 Meters:**
- Each cycle covers 2 meters.
- Number of complete cycles needed to cover 10 meters (as 11 meters is not a multiple of 2):
\[ \frac{10}{2} = 5 \text{ cycles} \]
- Distance covered in 5 cycles:
\[ 5 \times 2 = 10 \text{ meters} \]
- Time taken for 5 cycles:
\[ 5 \times 8 = 40 \text{ seconds} \]

5. **Final Steps to Reach the Pit:**
- After 5 cycles, the drunkard is at 10 meters.
- He needs to cover 1 more meter to reach the pit at 11 meters.
- Since he moves forward 5 steps in one go, he will cover this 1 meter in the next forward step.
- Time taken for this final step:
\[ 1 \text{ second} \]

6. **Total Time Taken:**
- Time taken for 5 complete cycles:
\[ 40 \text{ seconds} \]
- Time taken for the final step:
\[ 1 \text{ second} \]
- Total time taken:
\[ 40 + 1 = 41 \text{ seconds} \]

Therefore, the drunkard will fall into the pit after **41 seconds**.