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A drunkard walking along a road takes 5 steps forward and 3 steps backward, and so on. Each step is 1m long and takes 1s. There is a pit on the road 11m away from the starting point.
The drunkard fall into the pit after
(A) 29 steps
(B) 21 steps
(C) 37 steps
(D) 31 steps

Last updated date: 13th Jun 2024
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Hint: We will calculate the above question on the number line to make it easy to understand. Distance between each point on line is 1km.
On the 11th point there is a pit.

Complete step by step solution: According to question : He walks 5 steps forward and 3 backward each step is of 1m and takes 1 sec. So, let's make a number line.
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                    5 steps forward
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                        3 steps backwards

(3) In the above two steps he reached at point 2. But travelled $5 + 3 = 8$ steps
      So, total distance travelled by him $ = 5 - 3 = 2$
      Since 1 step takes 1 sec, therefore by taking 8 steps he took 8 sec.
      So we can say that he is moving 2 km in 8 sec.
(4) Now he will move from point 2, in the same manner. i.e., 5 forward and 3 backward.
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Now he is on point 4.
So, covered 4 km in total $8 + 8 = 16$ seconds.
(5) In another 8 sec he will cover another 2 km. Now he is at point 6.
i.e., he covered 6 km in total $16 + 8 = 24$ sec
(6) But from point 6 when he takes 5 steps forward he falls in a pit so can’t take backward steps.
So, in the next 5 steps he took 5 sec.
(7) Or total time taken by him is $24 + 5 = 29$ sec
Therefore, option (A) is i.e., 29 sec is the correct option

Note: We can solve the above question in this manner.
He moves 5 steps forward, 3 steps backward in 8 sec.
In this way he is 2m away from an initial position $x = 5 - 3 = 2m$ in 8 sec.
After 24 sec he will 6m away from an initial position.
He moves further 5m forward then he falls in a pit which is 11m away from the initial position.
Total time $T = 24 + 5 = 29\sec $