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

Question

A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward followed again by 5 steps forward and 3 steps backward and so on. Each step is 1 m long and requires 1 s. The time the drunkard takes to fall in a pit 13 m away is (A) 13 s (B) 32 s (C) 37 s (D) 49 s

Tardigrade Admin · Accepted Answer

The drunkard advances (+5-3) m =2 m in 8 s. Hence 8 m in 32 s. Then in the next 5 s, he would move 5 m more or reach 13 m or the pit. Hence he reaches the pit in 37 s.

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

13 s

32 s

37 s

49 s

The drunkard advances $(+ 5 - 3) m = 2 m$ in $8 s$ . Hence $8 m$ in $32 s$ . Then in the next $5 s$ , he would move $5 m$ more or reach $13 m$ or the pit. Hence he reaches the pit in $37 s$ .

Q. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward followed again by 5 steps forward and 3 steps backward and so on. Each step is 1m long and requires 1s. The time the drunkard takes to fall in a pit 13m away is