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The Snail on the Wall
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A snail creeps 9 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 5 ft while sleeping.
If this happens every day, how many days will the snail take to reach the top of a wall 21 ft in height?
On the first day, the snail climbs up 9 ft and slips down 5 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 3 days, it has traveled a distance of 12 ft from the bottom.
Here lies the catch to the problem! On the last day, the snail travels 9 ft upwards and hence reaches the top of the wall in a total of 4 days.
Alternative Solution through Equation:
Let x be the number of days the snail takes to reach the top of the wall 21 ft in height.
On the last day, the snail will reach the top by traveling 9 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 9 ft and slips down 5 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
9 + 4 (x − 1) = 21
On solving the above equation, we get
4 (x − 1) = 21 − 9 = 12; or
x = (12 / 4) + 1 = 4.
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