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. QED.