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Aug 13 2008, 07:48 AM
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Advanced Member ![]() ![]() ![]() Group: Admin Posts: 337 Joined: 17-November 03 From: China Member No.: 58 |
A snail creeps 5 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 2 ft while sleeping.
If this happens every day, how many days will the snail take to reach the top of a wall 14 ft in height? Answer 4 Solution: On the first day, the snail climbs up 5 ft and slips down 2 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 3 days, it has traveled a distance of 9 ft from the bottom. Here lies the catch to the problem! On the last day, the snail travels 5 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 14 ft in height. On the last day, the snail will reach the top by traveling 5 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 5 ft and slips down 2 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 5 + 3 (x | 1) = 14 On solving the above equation, we get 3 (x | 1) = 14 | 5 = 9; or x = (9 / 3) + 1 = 4. -------------------- |
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