An ant is at the bottom of a 12-foot deep well and is trying to get to the top. During the day he climbs 4 feet IP up, but at night he slides back 2feet. How long does it take for him to get out of the well?

To find out how long it takes for the ant to get out of the well, we need to calculate the number of steps the ant takes to reach the top.

1. Let's first calculate how many feet the ant climbs during the day. The ant climbs up 4 feet during the day.

2. Now, let's calculate how many feet the ant slides down at night. The ant slides back 2 feet at night.

3. To calculate the net distance covered by the ant in one day, we subtract the distance the ant slides down from the distance it climbs up: 4 feet - 2 feet = 2 feet.

4. Therefore, the ant moves 2 feet per day toward the top of the well.

5. Since the well is 12 feet deep, the ant needs to climb a total of 12 feet.

6. To calculate the number of days it takes for the ant to climb out of the well, we divide the total distance by the distance covered per day: 12 feet / 2 feet = 6 days.

Therefore, it will take the ant 6 days to climb out of the well.

To determine how long it takes for the ant to get out of the well, we can calculate the number of cycles (day and night) the ant needs to climb the distance of 12 feet.

During the day, the ant climbs 4 feet up. However, at night, it slides down 2 feet. So, each cycle (day and night) the ant makes a net progress of 4 feet - 2 feet = 2 feet.

To find out how many cycles are required to cover a distance of 12 feet, we can divide the total distance by the net progress made per cycle.

12 feet / 2 feet per cycle = 6 cycles

Thus, the ant needs 6 cycles to climb out of the well. Since each cycle consists of both a day and night, it will take the ant 6 x 2 = 12 time intervals to reach the top.

Therefore, it will take the ant 12 time intervals to get out of the well.

16

5 days

net 2 ft/day for 4 days, ends up at 8 ft.
On day 5, up to 12, and out