Question

UV
14)
2
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 meters away from
the starting point. The drunkard will fall into the
pit after :
(1) 29 s
(2) 21 s
(3) 37s
(4) 31 s
A car runs at constant speed on a circular trac​

Answer 1

Explanation:

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

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Answer 2

He

takes 5 steps forward and 3 steps backward, followed

by 5 steps forward and 3 steps ( total steps walked by him is 5+3 = 8 and totak distance travelled by him in 8 seconds is 2m as Each step is one meter long and takes one second and also 5-3=2 he walks 2m in 8 seconds and 1 m in 4 seconds) backward so after walking 8 steps he moves 2 m forward .

Now he will cover 11 m distance in 4×11 =44 seconds.

6m in 6×4 =24 seconds and after walking another 5 steps in another 5 seconds

he will fall so the answer is 29. Seconds.

So the answer will be 29 s.

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