Puzzle 17
At the local sweet shop, three particularly nice sweets are on special offer.
A Nobbler is over three times the price of a Sparkle.
Six Sparkles are worth more than a Wibbler.
A Nobbler, plus two Sparkles costs less than a Wibbler.
A Sparkle, a Wibbler and a Nobbler together cost 40p.
Can you determine the price of each type of sweet?
Puzzle Copyright © Kevin Stone
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Hint
Which sweet is the least expensive?
Answer
Sparkle = 4p
Wibbler = 23p
Nobbler = 13p
Reasoning
By (3) a Nobbler, plus two Sparkles costs less than a Wibbler, therefore a Wibbler must be the most expensive sweet.
By (1) a Nobbler is over three times the price of a Sparkle, therefore a Sparkle must be the cheapest sweet.
So the order of sweets, from the least to most expensive, is Sparkle, Nobbler, Wibbler.
If a Sparkle was 1p, by (2) a Wibbler could only be up to 5p, by (4) a Nobbler would cost at least 34p, which is more than a Wibbler and isn't allowed as the Wibbler is the most expensive sweet.
If a Sparkle was 2p, by (2) a Wibbler could only be up to 11p, by (4) a Nobbler would cost at least 27p, which is more than a Wibbler and isn't allowed as the Wibbler is the most expensive sweet.
If a Sparkle was 3p, by (2) a Wibbler could only be up to 17p, by (4) a Nobbler would cost at least 20p, which is more than a Wibbler and isn't allowed as the Wibbler is the most expensive sweet.
So a Sparkle must be at least 4p.
If a Sparkle was 4p, by (1) a Nobbler must be at least 13p, by (4) a Wibbler would cost 23p – this combination matches all of the clues and is a possible solution.
If a Sparkle was 4p and a Nobbler 14p, by (4) a Wibbler would cost 22p. This would not satisfy (3). And if we increase the price of a Nobbler, (3) is never satisfied.
If a Sparkle was 5p, by (1) a Nobbler must be at least 16p, by (4) making a Wibbler at most 19p. This would not satisfy (3).
If we increase the price of a Sparkle or Nobbler further, (3) is will never be satisfied.
Therefore, the only solution we came across must be the correct one.
Puzzle 18
A coin collector decides to divide his coin collection between his children.
The eldest gets 1/2 of the collection, the next gets 1/4 of the collection, the next gets 1/5 of the collection, and the youngest gets the remaining 49 coins.
How many coins are in the collection?
Puzzle Copyright © Kevin Stone
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Hint
What do the fractions add up to?
Answer
There are 980 coins in the collection.
Reasoning
Using fractions:
1/2 + 1/4 + 1/5 = 19/20
The remaining 1/20 is 49 coins.
Therefore, the 20/20 must equal 20 lots of 49 = 980.
Alternative Reasoning
Using percentages:
50% + 25% + 20% = 95%
The remaining 5% is 49 coins.
If 5% is 49 coins, 10% is 98 coins, 100% is 980 coins.
Double-Checking
1/2 of 980 is 490
1/4 of 980 is 245
1/5 of 980 is 196
and the remaining 49 coins.
And 490 + 245 + 196 + 49 = 980.
Puzzle 19
Can you find the sum of the second column?
Puzzle Copyright © Kevin Stone
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Answer
14.
Reasoning
The values are: @ = 7, & = 4, $ = 8, # = 3, % = 5.
But…
…you don't have to find the value of every symbol!
The rows add up to 80, which means that the columns must also add up to 80.
24 + ? + 17 + 25 = 80.
? = 14.
Puzzle 20
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.
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