At last month's rehearsal, four top athletes competed in two qualifying 400 metre races.
As the results were expected to be mislaid, various notes were taken to ensure the accuracy of the overall placings:
No-one finished both races in the same position.
In both races Sam beat the runner whose surname was Donald.
Blake Curtail came third in the second race and Pat came last in the first race.
In the second race the athlete whose surname was Arnold won, and the athlete whose surname was Bowler came last.
In the first race, Blake beat Jamie, but Jamie beat Sam.
Can you determine who finished where in each of the races?
The local habitat around a railway track can be very interesting. For example, supporting the track is a sleeper, under which you can find the lesser spotted great weevil.
In the illustration we have a sketch of Sir Edwyn de Tudor going to rescue his love, who was held a 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.
[Ref: ZBVT] Sir Edwyn De Tudor. From Amusements In Mathematics by Henry Ernest Dudeney (1917).
Hint: The gap between the two options is 2 hours.
Answer: The distance must have been sixty miles.
If Sir Edwyn left at noon and rode 15 miles an hour, he would arrive at four o'clock - an hour too soon. If he rode 10 miles an hour, he would arrive at six o'clock - 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 - 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 using 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
Time2 - Time1 = 2
D ÷ 10 - D ÷ 15 = 2
Multiply throughout by 30:
3D - 2D = 60
D = 60 miles.