Question

A DRUNKARD IS WALKING IN A NARROW LANE TAKES 5 STEPS FORWARD AND 3 STEPS BACKWARD EACH STEP IS 1m LONG per second . determine how long drunkard takes to fall in a pit 15m away from starting point.

Answer 1

In each forward motion of 5 steps and backward motion of 3 steps,
net distance covered = 5 - 3 = 2 m

=> Time taken = 5 + 3 = 8 s.

=> Time required to cover a distance of 8 m = 8/2 x 8 = 32s

Remaining distance of the pit  = 13 - 8 = 5 m

In next 5 s, as he moves 5 steps forward, he falls into the pit.

Hence, total time taken = 32 + 5 = 37 s  

Answer 2

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Distance covered with 1 step = 1 m

Time taken = 1 s

Time taken to move first 5 m forward = 5 s

Time taken to move 3 m backward = 3 s

Net distance covered = 5 – 3 = 2 m

Net time taken to cover 2 m = 8 s

Drunkard covers 2 m in 8 s.

Drunkard covered 4 m in 16 s.

Drunkard covered 6 m in 24 s.

Drunkard covered 8 m in 32 s.

In the next 5 s, the drunkard will cover a distance of 5 m and a total distance of 13 m and falls into the pit.

Net time taken by the drunkard to cover 13 m = 32 + 5 = 37 s

I hope, this will help you

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