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Hide all answers View all answers Print Try the Quiz A snail creeps 7 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 11 ft in height? Answer: 3
Solution:
On the first day, the snail climbs up 7 ft and slips down 5 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 2 days, it has traveled a distance of 4 ft from the bottom. Here lies the catch to the problem! On the last day, the snail travels 7 ft upwards and hence reaches the top of the wall in a total of 3 days. Alternative Solution through Equation:
Let x be the number of days the snail takes to reach the top of the wall 11 ft in height.
On the last day, the snail will reach the top by traveling 7 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 7 ft and slips down 5 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 7 + 2 ( x − 1) = 11
On solving the above equation, we get 2 ( x − 1) = 11 − 7 = 4; or
x = (4 / 2) + 1 = 3. Try the Quiz : |