Solution:

On the first day, the snail climbs up 5 ft and slips down 1 ft while sleeping. So, next morning, it is 4 ft from where it started. The snail thus travels 4 ft upwards every day. Therefore, in 10 days, it has traveled a distance of 40 ft from the bottom.

Here lies the catch to the problem! On the last day, the snail travels 5 ft upwards and hence reaches the top of the wall in a total of 11 days.

Alternative Solution through Equation:

Let x be the number of days the snail takes to reach the top of the wall 45 ft in height.

On the last day, the snail will reach the top by traveling 5 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 5 ft and slips down 1 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
5 + 4 (x − 1) = 45

On solving the above equation, we get

4 (x − 1) = 45 − 5 = 40; or
x = (40 / 4) + 1 = 11.