Question

A drunkard is walking along a straight road....

A drunkard is walking along a straight road. He takes 5 steps forward and 3 steps backward, followed by 5 steps forward and 3 steps backward and so on. Each step is one meter long and takes one second. There is a pit on the road 11 meter away from the starting point then after how much time drunkard will fall into the pit?

Answer 1

heya.....

For each wobbly "cycle" he makes 8 total steps and takes 8 seconds, but moves in total 2 steps forward (2 m) 
So let's see. 
3*2 m = 6 m 
So at the end of the 3rd cycle he's at 6 m (elapsed time 3*8 = 24 s) 

And the he starts cycle #4 and walks 5 m forward, taking him 5 s. He is now at 11 m. 
Time = 24 + 5 = 29 s 

Got it? 


2. 
I'm too baffled by the same thing. I don't know the purpose of the "15 km northeast". It's definitely not correct either. 
A is moving north, B is moving west. Last time I checked the compass, that would look like this: 
............................^ A 
............................| 
............................| 
............................| 
............................| 
<<------------------------| 
B 

Either B has a teleportation device or you should recheck your problem. Your approach is otherwise correct. Maybe that one sentence was a glitch in your book.

tysm.@kundan the GOZMIT

Answer 2

Hey mate!

Here's your answer!!

For 8 total steps it takes 8 seconds,
and moves in total 2 steps forward (2 m).

Therefore,
3 × 2 m = 6 m

At 3rd cycle,
Time Elapsed = 24 s

After walking 5 m forward, he takes 5 s more.
It becomes 11 m.

Time = 24 + 5 = 29 s.

$hope \: it \: helps \: you.$
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