Syvum Home Page

Home > Math Puzzles & Brain Teasers > Print Preview

Math Puzzles & Brain Teasers


The Snail on the Wall

View Puzzle Solve Puzzle

Hide all answers   View all answers   Print   Try the Quiz

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?
Answer: 4

Solution:

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.

  Try the Quiz :     Puzzles & Brain Teasers : The Snail on the Wall

puzzle : image for snail

Contact Info © 1999-2019 Syvum Technologies Inc. Privacy Policy Disclaimer and Copyright
Previous
-
Next
-