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A snail creeps 3 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 1 ft while sleeping.
If this happens every day, how many days will the snail take to reach the top of a wall 37 ft in height? Answer:Answer:18
On the first day, the snail climbs up 3 ft and slips down 1 ft while sleeping. So, next morning, it is 2 ft from where it started. The snail thus travels 2 ft upwards every day. Therefore, in 17 days, it has traveled a distance of 34 ft from the bottom.
Here lies the catch to the problem! On the last day, the snail travels 3 ft upwards and hence reaches the top of the wall in a total of 18 days.
Alternative Solution through Equation:
Let x be the number of days the snail takes to reach the top of the wall 37 ft in height.
On the last day, the snail will reach the top by traveling 3 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 3 ft and slips down 1 ft while sleeping, it travels 2 ft upwards on each of these remaining days. Thus,
Distance traveled on last day + Distance traveled on remaining days = Wall height; or
3 + 2 (x − 1) = 37
On solving the above equation, we get
2 (x − 1) = 37 − 3 = 34; or
x = (34 / 2) + 1 = 18.