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
Each term alternates between a square and a cube, 8 3 = 512.
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Puzzle 3
In the illustration we have a sketch of Sir Edwyn de Tudor going to rescue his love, who was held captive by a neighbouring wicked baron.
Sir Edwyn calculated that if he rode at fifteen miles an hour he would arrive at the castle an hour too soon, while if he rode at ten miles an hour he would get there just an hour too late.
Now, it was of the first importance that he should arrive at the exact time appointed, in order that the rescue that he had planned should be a success, and the time of the tryst was five o'clock, when the captive would be taking afternoon tea.
The puzzle is to discover exactly how far Sir Edwyn de Tudor had to ride.
Sir Edwyn De Tudor, Amusements In Mathematics, Henry Ernest Dudeney.
If Sir Edwyn left at noon and rode 15 miles an hour, he would arrive at four o'clock, which is an hour too soon. If he rode 10 miles an hour, he would arrive at six o'clock, which is an hour too late. But if he went at 12 miles an hour, he would reach the castle of the wicked baron exactly at five o'clock, which is the time appointed.
The text above is the answer given in the book, and below is a method of finding the answer.
If we call the distance to the castle, D and use the fact that Time = Distance ÷ Speed, we have:
Travelling at 15 mph:
Time1 = D ÷ 15 (an hour too soon)
Travelling at 10 mph:
Time2 = D ÷ 10 (an hour too late)
The time gap between these two times is 2 hours, therefore
