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?
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 6
As the auditor for my local theme park, I noticed that on Saturday there were 4,296 children and 2,143 adults and the takings were £98,718.
However, on Sunday, there were 5,146 children and 2,807 adults and the takings were £122,570.
How much were the children's tickets and adult's tickets?
Hint
This is quite a tricky puzzle, and knowledge of algebra would certainly help.
Answer
The children tickets were £14, and the adult tickets were £18.
Reasoning
There are a number of methods for solving this problem, including:
Using a spreadsheet.Using a computer program.Using the intersection of lines on a graph.Using an online equation solver.Solving simultaneous equations using algebra.Solving simultaneous equations using inverse matrices.
Here is my solution using simultaneous equations and algebra.
First construct two algebraic equations, where 'c' is the number of children, and 'a' is the number of adults:
Workers with the surnames Butcher, Baker, Carpenter and Plumber are currently attending an annual convention.
No-one is currently, nor ever has been in the same profession as their name and no-one has had the same profession twice.
Alan is now a butcher, whereas Mr Brian Butcher used to be a baker. The person who is now a carpenter used to be a butcher. Charles has never been a baker, Mr Darren Carpenter has never been a butcher and Mr Baker is not now a carpenter.
Can you determine their full names, along with their current and previous professions?
