Mathematical Puzzles 

Workings

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).

Workings

Hint
The answer is over 35.

Answer
37.

Workings

Hint
Which sweet is the least expensive?

Answer
Sparkle =  4p
Wibbler = 23p
Nobbler = 13p

If we label the clues.

[1]   A Nobbler is over three times the price of a Sparkle.
[2]   Six Sparkles are worth more than a Wibbler.
[3]   A Nobbler, plus two Sparkles costs less than a Wibbler.
[4]   A Sparkle, a Wibbler and a Nobbler together cost 40p.

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.

Workings

Hint
What do the fractions add to?

Answer
There are 980 coins in the collection.

To check this we add:

 1÷2 + 1÷4 + 1÷5 + 49

this works out as

 490 + 245 + 196 + 49 = 980, as required.

To work the answer out from the question, we observe that:

1÷2 + 1÷4 + 1÷5 = 19÷20

and we know that there were 49 coins left, so 1÷20 of the collection is 49 coins.

This means that the entire collection is 20 x 49 = 980 coins.