Question

A drunkard walking in a narrow lane takes 5 step forward & 4 steps backward and then stay for 2 s and repeat the same process. Each step is 1 m long and require 1 s. The time taken by drunkard to fall in a pit 10 m away from start is
(1) 45 s
(2) 60 s
(3) 75 s
(4) 65 s ​

Answer 1

You got the initial answer as 110 s, right?

Answer: 60 s

Explanation:

We need to calculate the time taken by the drunkard to cover 5 m, i.e., AB. (B is taken mid point of AC = 10 m.)

Displacement in a process = 5 m - 4 m = 1 m

And time taken = 5 × 1 s + 4 × 1 s + 2 s = 11 s.

Therefore, he will take 11 × 5 s = 55 (A) s to go at point B.

Now, from B, he will move 5 m and fall at the pit which is at C. [He can't return as he has completed the rest 5 m.]

So, time taken to go at C = 5 × 1 s = 5 s. (B)

Hence, total time = 55 s + 5 s = 60 s [A and B].