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# A drunkard takes a step of 1m in 1s. He takes 5 steps forward and 3 steps backward and so on. The time taken by him to fall in a pit of 13m away from the start is:A. 26sB. 31sC. 37sD. 41s

Last updated date: 22nd Jun 2024
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Hint: As it is mentioned, the drunkard requires 1s to take each step of 1m. He takes 5 steps forward and 3 steps forward, thus we can keep aside the last 5 steps of the drunkard and calculate the rest of the distance covered by him and the time taken by him to cover that distance. Then, we can add the time taken by the drunkard for the initial journey and last 5 steps. This will give us the time taken by him to fall in pit 1m away.

Drunkard takes a step of 1m in 1s.
He takes 5 steps forward and 3 steps forward. Thus, he takes 2 steps forward i.e. 2m
He takes overall 8 steps. Time taken for 8 steps is 8 sec. It means that he moves 2m in 8s.
So, to fall in the pit, the man has to walk 8m and then take 5 steps.
Time taken by the man to reach 8m is $8 \times 4= 32s$
Thus, the time taken by the man to walk 8m and 5 steps will be,
$T= 32 + 5$
$\Rightarrow T= 37s$
Thus, the time by the drunkard to fall in the pit is 37s.

So, the correct answer is “Option C”.

Note:
Students should not get confused between the distance and displacement covered by the drunkard. He needs to travel 13m to fall into the pit. This 13m is the drunkard’s displacement and not the distance covered by him. Students should be careful while considering the total steps in reaching the final destination i.e. pit. Don’t consider the backward movement of the drunkard for the final cycle as he has already fallen in the pit.