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A snail creeps 10 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 6 ft while sleeping.
If this happens every day, how many days will the snail take to reach the top of a wall 30 ft in height? Answer:Answer:6
On the first day, the snail climbs up 10 ft and slips down 6 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 5 days, it has traveled a distance of 20 ft from the bottom.
Here lies the catch to the problem! On the last day, the snail travels 10 ft upwards and hence reaches the top of the wall in a total of 6 days.
Alternative Solution through Equation:
Let x be the number of days the snail takes to reach the top of the wall 30 ft in height.
On the last day, the snail will reach the top by traveling 10 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 10 ft and slips down 6 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
10 + 4 (x − 1) = 30
On solving the above equation, we get
4 (x − 1) = 30 − 10 = 20; or
x = (20 / 4) + 1 = 6.