a drunkard walking ina narrow lane takes 5 steps forward and 3 steps backward followed again 5 steps forward and 3 steps backward and so on. if each step is 1m and requires 1 second. determine how long would the drunkard take to fall in to pit 13 m away from starting point?
each cycle advances the drunk 2 meters
So, after 4 cycles and 4(5+3)=32 seconds, the drunk has advanced 8 meters.
The next 5 steps take him to the end, so it takes 32+5 = 37 seconds.
To determine how long it would take for the drunkard to fall into the pit, we first need to understand the pattern in their movement.
The drunkard takes 5 steps forward and then 3 steps backward repeatedly. This creates a sequence of steps: +5, -3, +5, -3, and so on.
Now, the drunkard needs to cover a distance of 13 m to fall into the pit. Let's break down their movement into cycles:
In one cycle, the drunkard takes 5 steps forward, covering a distance of 5 m. Then, they take 3 steps backward, moving 3 m in the opposite direction. So, within one cycle, the drunkard covers a net distance of 5 m - 3 m = 2 m.
Thus, the drunkard covers 2 m in each cycle. To find out how many cycles are required to cover a distance of 13 m, we divide the distance to be covered by the distance covered in one cycle:
13 m / 2 m = 6.5 cycles.
Since we cannot have half of a cycle, we need to take the ceiling value of 6.5, which is 7. This means the drunkard would need 7 cycles to reach a distance of 13 m.
Now, we need to calculate the time it takes for each cycle:
One step takes 1 second, and within one cycle, the drunkard takes (5 + 3) = 8 steps. So, one cycle takes 8 seconds.
Finally, to find the total time required for the drunkard to fall into the pit, we multiply the number of cycles by the time per cycle:
Total time = 7 cycles × 8 seconds/cycle = 56 seconds.
Therefore, the drunkard would take 56 seconds to fall into the pit, which is 13 m away from the starting point.