Puzzle 17
Which circle, square, and triangle, have the closest area to the blue doughnut shape?
The drawings are to scale, so you might be able to judge by eye, or you can work out the actual areas.
Puzzle Copyright © Kevin Stone
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Hint
The area of a circle is π x Radius2.
Answers
Circle: 35.
Square: 31.
Triangle: 47.
Doughnut
The area of a circle is π x Radius2.
The larger circle has diameter = 40, therefore the radius is 20, and the area is π x 202 = 400π.
The smaller circle has diameter = 20, therefore the radius is 10, and the area is π x 102 = 100π.
Therefore the shaded area is 400π – 100π = 300π ≈ 942.
Circle
The area of a circle is π x Radius2.
The circle with diameter 31 has a radius of 15.5 and an area of π x 15.52 ≈ 755.
The circle with diameter 35 has a radius of 17.5 and an area of π x 17.52 ≈ 962 (closest match).
The circle with diameter 39 has a radius of 19.5 and an area of π x 19.52 ≈ 1195.
Square
The area of a square is Side x Side.
The square with side 31 has an area of 31 x 31 = 961 (closest match).
The square with side 35 has an area of 35 x 35 = 1225.
The square with side 39 has an area of 39 x 39 = 1521.
Triangle
The area of a triangle is 1/2 x Base x Height. Using Pythagoras' theorem it can be shown that the area of an equilateral triangle is √3 x Base2 ÷ 4.
The triangle with side 39 has an area of √3 x 39 x 39 ÷ 4 ≈ 659.
The triangle with side 43 has an area of √3 x 43 x 43 ÷ 4 ≈ 801.
The triangle with side 47 has an area of √3 x 47 x 47 ÷ 4 ≈ 957 (closest match).
Puzzle 18
For being well-behaved at the garden fayre, four children were each given two sweets.
Jesse had an orange sweet.
One child who had a red sweet also had a blue one.
No child had two sweets of the same colour.
A child who had a green sweet also had a red one.
Jamie didn't have a red sweet, and Jo had a green one.
Jordan didn't have an orange one, and Jesse had no blue sweets.
Knowing that there were two sweets of each colour, can you tell the colours of the sweets each child had?
Puzzle Copyright © Kevin Stone
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Hint
Start by looking at Jesse's sweets.
Answer
Jamie Orange Blue
Jesse Orange Green
Jo Green Red
Jordan Red Blue
Reasoning
The children were: Jamie, Jesse, Jo, Jordan.
The colours were: Blue, Green, Orange, Red.
There were two sweets of each colour, and by (3) no child had two sweets of the same colour.
By (1), Jesse had an Orange sweet...
...by (6), didn't have Blue...
...by (2) and (4), Red was paired with Blue and Green...
...therefore Jesse's other sweet can't have been Red, so was Green.
Jamie
Jesse Orange Green
Jo
Jordan
By (4) the other Green sweet was paired with Red, by (5) this must have been Jo.
Jamie
Jesse Orange Green
Jo Green Red
Jordan
By (2) the other Red sweet was paired with Blue, but by (5) this wasn't Jamie, so must have been Jordan.
Jamie
Jesse Orange Green
Jo Green Red
Jordan Red Blue
Leaving Jamie with Orange and Blue.
Jamie Orange Blue
Jesse Orange Green
Jo Green Red
Jordan Red Blue
Puzzle 19
Can you scramble each of these 6-letter words, and then split them into two common 3-letter words?
For example: notify = fin toy.
helium
latest
almond
eldest
placed
radium
desert
wealth
tricky
funnel
Puzzle Copyright © Kevin Stone
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Answers
helium = hum + lie
latest = let + sat
almond = man + old
eldest = led + set
placed = cap + led
radium = air + mud
desert = red + set
wealth = law + the
tricky = cry + kit
funnel = elf + nun
Alternative Answers
The following answers use less common words.
helium = him + leu
helium = hue + mil
helium = hum + lei
latest = els + tat
almond = don + lam
almond = lam + nod
eldest = eld + set
eldest = els + ted
placed = pec + lad
radium = mud + ria
radium = amu + rid
radium = aid + rum
wealth = awl + the
wealth = haw + let
There are more answers with even rarer words, but I've chosen not to include those.
Puzzle 20
What letter comes next in this sequence:
M V E M J {?}
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Hint
Look to the skies for the answer.
Answer
S.
From Saturn.
Reasoning
These are the initial letters of the planets.
M = Mercury
V = Venus
E = Earth
M = Mars
J = Jupiter
S = Saturn
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