Sir Edwyn De Tudor. From Amusements In Mathematics by Henry Ernest Dudeney (1917).
In the illustration we have a sketch of Sir Edwyn de Tudor going to rescue his lady-love, the fair Isabella, who was held a captive by a neighbouring wicked baron.
Sir Edwyn calculated that if he rode fifteen miles an hour he would arrive at the castle an hour too soon, while if he rode 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 lady would be taking her afternoon tea.
The puzzle is to discover exactly how far Sir Edwyn de Tudor had to ride.
Hint: The gap between the two options is 2 hours.
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.
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. QED.