Question

A drunkard takes a step of $1\text{m}$ in 1 second. He takes 5 steps forward and 3 steps backward, and so on. The time taken by him to fall in a pit $13\text{m}$ away from the starts is

• A. $26$ seconds
• B. $31$ seconds
• C. $37$ seconds
• D. $41$ seconds

Answer 1

The correct option is C $37$ seconds

When he takes 8 steps the displacement is $(5-3)=2\text{m}$
Time taken for 8 steps $=8\text{s}$
Drunkard covers $2\text{m}$ in $8\text{s}$
Drunkard covers $4\text{m}$ in $16\text{s}$
Drunkard covers $6\text{m}$ in $24\text{s}$
Drunkard covers $8\text{m}$ in $32\text{s}$
In the next $5\text{s}$ , the drunkard will cover a distance of $5\text{m}$ and a total distance of $13\text{m}$ and falls into the pit.
Net time taken by the drunkard to cover $13\text{m}=32+5=37\text{s}$