Answer: Sparkle = 4p
Wibbler = 23p
Nobbler = 13p
If we label the clues.
A Nobbler is over three times the price of a Sparkle. [1]
Six Sparkles are worth more than a Wibbler. [2]
A Nobbler, plus two Sparkles costs less than a Wibbler. [3]
A Sparkle, a Wibbler and a Nobbler together cost 40p. [4]
By [3] a Nobbler, plus two Sparkles costs less than a Wibbler, a Wibbler must be the most expensive sweet.
By [1] a Nobbler is over three times the price of a Sparkle, a Sparkle is cheaper than a Nobbler and hence 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.
At the local sweet shop, three particularly nice sweets are on special offer.
