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 2
Starting in the bottom left corner and moving either up or right, adding up the numbers along the way, what is the largest sum that can be made?
2
3
5
4
1
3
2
4
3
4
5
1
3
5
2
3
3
2
3
1
1
4
2
4
4
Note: this puzzle is not interactive, and the numbers cannot be clicked.
Reasoning
Q1 can't have A as its answer (otherwise it would contradict itself).
If Q1's answer was C (meaning Q4's answer was A), then Q3's answer would be C. However, Q2's answer should now be A, but this isn't allowed by Q1 (as Q4 is the first answer with A). This is a contradiction.
Therefore, Q1's answer is B (meaning Q3's answer is A).
For Q2's answer to be C, Q4's answer would have to be A, which would contradict Q4. Therefore, Q2's answer is B making Q4's answer B.
