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The Snail on the Wall
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A snail creeps 4 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 40 ft in height?
On the first day, the snail climbs up 4 ft and slips down 1 ft while sleeping. So, next morning, it is 3 ft from where it started. The snail thus travels 3 ft upwards every day. Therefore, in 12 days, it has traveled a distance of 36 ft from the bottom.
Here lies the catch to the problem! On the last day, the snail travels 4 ft upwards and hence reaches the top of the wall in a total of 13 days.
Alternative Solution through Equation:
Let x be the number of days the snail takes to reach the top of the wall 40 ft in height.
On the last day, the snail will reach the top by traveling 4 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 4 ft and slips down 1 ft while sleeping, it travels 3 ft upwards on each of these remaining days. Thus,
Distance traveled on last day + Distance traveled on remaining days = Wall height; or
4 + 3 (x − 1) = 40
On solving the above equation, we get
3 (x − 1) = 40 − 4 = 36; or
x = (36 / 3) + 1 = 13.
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