Puzzle 5
Can you find common anagrams of the following words:
BINARY
ABROAD
RASCAL
ALTARS
BADGER
BARKED
MARBLE
UNABLE
TABLET
CALLER
Puzzle Copyright © Kevin Stone
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Hint
The first letters of the words are: B, A, S, A, B, B, R, N, B, C.
Answer
BINARY = BRAINY
ABROAD = ABOARD
RASCAL = SCALAR
ALTARS = ASTRAL
BADGER = BARGED
BARKED = BRAKED
MARBLE = RAMBLE
UNABLE = NEBULA
TABLET = BATTLE
CALLER = CELLAR or RECALL
There are some less common anagrams such as:
RASCAL = SACRAL
ALTARS = TARSAL
ALTARS = RATALS
ALTARS = TALARS
BADGER = GARBED
BARKED = DEBARK
MARBLE = BLAMER
MARBLE = AMBLER
MARBLE = LAMBER
UNABLE = UNBALE
TABLET = BATTEL
So you can have an extra half point if you found one of these! There are some even rarer anagrams, but I don't think I'll allow those!
Puzzle 6
I started a journey to my Aunt's yesterday with a full tank of gasoline, 10 gallons, but at that exact moment my gasoline tank sprang a leak.
I managed to drive 50 miles, in exactly one hour, before coming to a complete stop with an empty tank.
I know that the car manages 25 miles per gallon.
How quickly was I losing gasoline?
Puzzle Copyright © Kevin Stone
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Hint
How many gallons did the journey use?
Answer
8 gallons an hour.
Reasoning
I travelled 50 miles in one hour, at 25 miles per gallon this must have used 2 gallons.
So I must have lost the other 8 gallons through the hole.
Puzzle 7
What letter comes next in this sequence:
E O E R E ?
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Hint
This is a very common sequence.
Answer
X.
The last letters of the words one, two, three, four, five, six.
Puzzle 8
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.
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Hint
Going down and across the floor isn't the only route.
Answer
40 feet.
Explanation Diagram
If you imagine the room to be a cardboard box, you can 'unfold' the room in various ways, and each route gives a different answer.
We can use Pythagoras' theorem (a2 + b2 = c2) to calculate the distances:
distance2 = horizontal2 + vertical2
distance = √(horizontal2 + vertical2)
Route #1
distance = 1 + 30 + 11 = 42 feet.
Route #2
horizontal = 6 + 30 + 6 = 42 feet.
vertical = 10 feet.
distance = √(422 + 102) ≈ 43.174 feet.
Route #3
horizontal = 1 + 30 + 6 = 37 feet.
vertical = 6 + 11 = 17 feet.
distance = √(372 + 172) ≈ 43.178 feet.
Route #4
horizontal = 1 + 30 + 1 = 32 feet.
vertical = 6 + 12 + 6 = 24 feet.
distance = √(322 + 242) = 40 feet.
