Question

A drunkard is walking along a straight road. He takes 5 steps forward and 3 steps backward, followed by 5 steps forward and 3 teps backward and so on. Each step is 1 m long and takes 1 second. There is pit on the road 11 m away from the starting point. The drunkar will fall into the pit after how much time?
(answer : 29s)
But how? Please help me out!

Answer 1

Hello


Intresting Question :)

→ According to the question ,the drunkard walks 5 m in forward direction and then 3 m in backward direction

=> Displacement = 5 m - 3 m = 2 m

=> total distance moved in the process is 8 m as 5m + 3m = 8 m

Each steps take 1 sec .

Now the pit is 11 m from the starting point


→ So , it takes 8 sec to cover 2 m

=> 2 m covered in 8 sec

again the drunkard will continue to walk like this ( 5 m forward and 3 m backward ) So again 2 m will be covered in 8 sec .


→ 2 m covered in 8 sec Again 2 m covered in 8 sec . That is = 4 m in 16 sec again 2 m in 8 sec . That is 6 m in 24 sec Now he just have to move 5 m more to get fall in the Pit . That is 6m + 5 m = 24 sec + 5 sec ( 5 m covers in 5 sec )


→ 11 m will be covered in $\bold{29 \:\: sec}$


★ Quick Recap –» The drunkard covers 2 m in 8 sec i.e 2 m + 2 m + 2 m = 6 m ..... and 8 sec + 8sec + 8 sec = 24 sec

The drunkard is at 6 m And Pit is at 11 m So 5 m more and time will be 5 sec

So , 6 m + 5 m = 11 m and Time Taken = 24 sec + 5 sec = 29 Seconds.





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Answer 2

Answer:

Explanation:

Here's the solution to your question.

Hope it helps you...!!!

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