Let x be the number of days the snail takes to reach the top of the wall 30 ft in height.

On the last day, the snail will reach the top by traveling 6 ft upwards and there will not be any question of slipping down.

The number of remaining days excluding the last day are (x − 1). Since the snail climbs up 6 ft and slips down 2 ft while sleeping, it travels 4 ft upwards on each of these remaining days. Thus,

Distance traveled on last day + Distance traveled on remaining days = Wall height; or

6 + 4 (x − 1) = 30

On solving the above equation, we get

4 (x − 1) = 30 − 6 = 24; or

x = (24 / 4) + 1 = 7.

A snail creeps 6 ft up a wall during the daytime. After all the labor it does throughout the day, it stops to rest a while... but falls asleep!! The next morning it wakes up and discovers that it has slipped down 2 ft while sleeping.

If this happens every day, how many days will the snail take to reach the top of a wall 30 ft in height?

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A snail creeps 6 ft up a wall during the daytime. After all the labor it does throughout the day, it stops to rest a while... but falls asleep!! The next morning it wakes up and discovers that it has slipped down 2 ft while sleeping.

If this happens every day, how many days will the snail take to reach the top of a wall 30 ft in height?

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