Question

A drunkard walking in a narrow lane takes 8 steps forward and 6 steps backward, followed again by 8 steps forward and 6 steps backward and so on. Each step is 1 m long and requires 1 s. Determine how long the drunkard takes to fall in a pit 18 m away from the start

• A. 18 s
• B. 126 s
• C. 78 s
• D. 62 s

Answer 1

Answer: (C)

78 s

Answer 2

When the drunkard walks 8 steps forward and 6 steps backward, the displacement in the first 14 steps $= 8\, m - 6\, m = 2\, m$ Time taken for first 14 steps $= 14\, s$ Time taken by drunkard to cover first 10 m of journey $= \frac{14}{2}\times10 = 70\,s$ If the drunkard takes 8 steps more, he will fall into the pit, so the time taken by the last 8 steps $= 8 \,s$ Total time taken $= 70 \,s + 8\, s = 78\, s$