Hint: The area of a circle is π x Radius².

Answer:
Circle: 3.5.

Square: 3.1.

Triangle: 4.7.

Doughnut

The area of a circle is π x Radius².

The larger circle has diameter = 4, therefore the radius is 2, and the area is π x 2² = 4π.

The smaller circle has diameter = 2, therefore the radius is 1, and the area is π x 1² = π.

Therefore the shaded area is 4π - π = 3π ≈ 9.42.

Circle

The area of a circle is π x Radius².

The circle with diameter 3.1 has a radius of 1.55 and an area of π x 1.55² = 7.55.

The circle with diameter 3.5 has a radius of 1.75 and an area of π x 1.75² = 9.62 (** closest match).

The circle with diameter 3.9 has a radius of 1.95 and an area of π x 1.95² = 11.95.

Square

The area of a square is Side x Side.

The square with side 3.1 has an area of 3.1 x 3.1 = 9.61 (** closest match).

The square with side 3.5 has an area of 3.5 x 3.5 = 12.25.

The square with side 3.9 has an area of 3.9 x 3.9 = 15.21.

Triangle

The area of a triangle is ½ x Base x Height. Using Pythagoras' theorem it can be shown that the area of an equilateral triangle is Sqrt(3) x Base² ÷ 4.

The triangle with side 3.9 has an area of Sqrt(3) x 3.9 x 3.9 ÷ 4 = 6.59.

The triangle with side 4.3 has an area of Sqrt(3) x 4.3 x 4.3 ÷ 4 = 8.01.

The triangle with side 4.7 has an area of Sqrt(3) x 4.7 x 4.7 ÷ 4 = 9.57 (** closest match).

Hint: The answer is over 35.

Answer: 37.

Hint: The top-left square contains the 6.

Answer:

Hint: Which sweet is the least expensive?

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.