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Puzzle 1
Here we have a rectangular room, measuring 30 feet by 12 feet, and 12 feet high.
There is a spider in the middle of one of the end walls, 1 foot from the ceiling (A).
There is a fly in the middle of the opposite wall, 1 foot from the floor (B).
What is the shortest distance that the spider must crawl in order to reach the fly?
The Spider and the Fly, The Canterbury Puzzles, Henry Ernest Dudeney.
Hint
There are more than 37, look at the different sizes.
Answer
There are 64 hexagon-type shapes in total.
Reasoning
37 single hexagons
+ 19 hexagons that contain 7 smaller hexagons
+ 7 hexagons that contain 19 smaller hexagons
+ 1 large hexagon that contains all of the smaller hexagons.
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Puzzle 4
A large fresh water reservoir has two types of drainage system, small pipes and large pipes.
6 large pipes on their own, can drain the reservoir in 12 hours.
3 large pipes and 9 small pipes, at the same time, can drain the reservoir in 8 hours.
How long will 5 small pipes on their own, take to drain the reservoir?
Hint
How many large pipes are required to drain the reservoir in 24 hours?
Answer
21 hours and 36 minutes.
Reasoning
Looking at the first clue:
in 12 hours, 6 large pipes can drain 1 reservoir
in 24 hours, 6 large pipes can drain 2 reservoirs
in 24 hours, 3 large pipes can drain 1 reservoir (*)
Looking at the second clue:
in 8 hours, 3 large + 9 small pipes can drain 1 reservoir
in 24 hours, 3 large + 9 small pipes can drain 3 reservoirs
But, by (*), we know that in those 24 hours, 3 large pipes can drain 1 of those reservoirs.
Therefore, the other 2 reservoirs can be drained by the small pipes on their own:
in 24 hours, 9 small pipes can drain 2 reservoirs
in 24 hours, 1 small pipe can drain 2⁄9 reservoirs
multiply the hours by 9:
in 216 hours, 1 small pipe can drain 2 reservoirs
in 216 hours, 5 small pipes can drain 10 reservoirs
divide the hours by 10:
in 21.6 hours, 5 small pipes can drain 1 reservoir