Four friends were discussing the colours of the flowers they had for the local flower show, from this year, and last.
There were four different colours: red, green, blue and yellow.
No-one entered the same colour flower twice. The entrant who currently has the green flowers, used to have yellow. Betty, who is not Mrs Long, currently has the blue flowers. Mrs Kipper, who now has yellow flowers, used to have blue. Neither Mrs Annie Ford, nor Mrs Jester, have ever owned yellow flowers. Cybil is not Mrs Kipper. Diane and Annie have entered four different colours in total.
Can you determine the full names, and colours of the flowers the ladies currently have, and previously had?
Hint
Start with Betty, Mrs Kipper, and Ada, as these are three different people.
Answer
Now Last
Annie Ford Red Green
Betty Jester Blue Red
Cybil Long Green Yellow
Diane Kipper Yellow Blue
Reasoning
By (3), Betty currently has Blue flowers.
Now Last
Betty Blue
By (4), Mrs Kipper currently has Yellow flowers, and used to have Blue.
Now Last
Betty Blue
Kipper Yellow Blue
By (5), we know Annie Ford's name.
Now Last
Betty Blue
Kipper Yellow Blue
Annie Ford
By (6), Mrs Kipper must be Diane, and the remaining person's name is Cybil.
Now Last
Betty Blue
Diane Kipper Yellow Blue
Annie Ford
Cybil
By (2), whoever currently has the Green flowers used to have Yellow, but by (5) this can't be Annie Ford, so this must be Cybil. Which means that Annie must currently have the Red flowers.
Now Last
Betty Blue
Diane Kipper Yellow Blue
Annie Ford Red
Cybil Green Yellow
By (3), Betty isn't Mrs Long, so must be Mrs Jester. Therefore Cybil must be Mrs Long.
Now Last
Betty Jester Blue
Diane Kipper Yellow Blue
Annie Ford Red
Cybil Long Green Yellow
By (1), Annie Ford must have previously had Green, and Betty Jester had Red.
Now Last
Betty Jester Blue Red
Diane Kipper Yellow Blue
Annie Ford Red Green
Cybil Long Green Yellow
??
Puzzle 18
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.
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).
