Medium Puzzles 

Workings

Hint
Look at the number of sides of each of the shapes.

Answer
Square.

Reasoning
If we add the shape's sides, the total increases by 8, then 10, then 12, etc (2 more each time).
   triangle (3) + pentagon (5) + square (4) = 12

   square   (4) + hexagon (6) + hexagon (6) + square (4) = 20

   pentagon (5) + hexagon (6) + hexagon (6) + hexagon (6) + square (4) + triangle (3) = 30

Therefore:
   hexagon  (6) + octagon (8) + octagon (8) + octagon (8) + octagon (8) + square (4) = 42

Alternative Reasoning
Square the first shape's sides, and the remaining shapes add up to this number.
   triangle (3 x 3) = pentagon (5) + square (4)

   square   (4 x 4) = hexagon (6) + hexagon (6) + square (4)

   pentagon (5 x 5) = hexagon (6) + hexagon (6) + hexagon (6) + square (4) + triangle (3)

Therefore:
   hexagon  (6 x 6) = octagon (8) + octagon (8) + octagon (8) + octagon (8) + square (4)

Workings

Hint
One of the words is gone.

Answers
bo + nd = bond
br + ew = brew
fo + am = foam
go + ne = gone
ho + pe = hope
jo + ke = joke
le + af = leaf
lo + af = loaf
lo + ud = loud
so + re = sore

Workings

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.

Workings

Hint
The first word begins with the letter B.

Answers
-e-rd = beard (BA) -u-et = quiet (QI) wa-t- = waltz (LZ) -um-y = jumpy (JP) -e-er = fever (FV) a-o-e = awoke (WK) mi--d = mixed (XE) -hew- = chewy (CY) hu-i- = humid (MD) -abi- = habit (HT) co--h = cough (UG) ho--e = horse (RS) arg-- = argon (ON)
This puzzle isn't guaranteed to have a unique answer, can you find another?