A snail fell on a dry well 20ft. high. This snail has the behavior of climbing up 3 ft high half of the day and going down 2 ft. half of the day . How many days this snail reach the top of the well?
the snail gaios 1 ft each day.
After 17 days, he has made it 17 ft.
On the 18th day he climbs the last 3 feet and is out.
To solve this problem, we need to analyze the snail's behavior and determine how many cycles it will take to reach the top of the well.
Let's break down the snail's behavior:
- It climbs up 3 ft. during half of the day.
- It goes down 2 ft. during the other half of the day.
This means that in a full day, the snail covers a net distance of 1 ft. (3 ft. - 2 ft.).
Now, let's calculate the number of days it will take for the snail to reach the top of the well.
Since the snail covers 1 ft. per day, we can calculate the number of days it will take to reach the top by dividing the height of the well (20 ft.) by the distance covered per day (1 ft.):
Days = Height of the well / Distance covered per day
Days = 20 ft. / 1 ft.
Days = 20
Therefore, it will take the snail 20 days to reach the top of the well.