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The Snail on the Wall
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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 4 ft while sleeping.
If this happens every day, how many days will the snail take to reach the top of a wall 22 ft in height?
On the first day, the snail climbs up 6 ft and slips down 4 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 8 days, it has traveled a distance of 16 ft from the bottom.
Here lies the catch to the problem! On the last day, the snail travels 6 ft upwards and hence reaches the top of the wall in a total of 9 days.
Alternative Solution through Equation:
Let x be the number of days the snail takes to reach the top of the wall 22 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 4 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
6 + 2 (x − 1) = 22
On solving the above equation, we get
2 (x − 1) = 22 − 6 = 16; or
x = (16 / 2) + 1 = 9.
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