Hint: Think about the terms that are not explicitly written.
Since one of the terms is (x - x).
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).
Direct Link: www.brainbashers.com?ZBVT
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.
A kind old gentleman decided to give 12 sweets to each of the girls in his town and 8 sweets to each of the boys.
Of the 612 children in his town, only half the girls and three quarters of the boys took the sweets.
How many sweets did the kind old gentleman have to buy?
In 12 hours: 6 large pipes can drain 1 reservoir.
In 24 hours: 6 large pipes can drain 2 reservoirs.
In 24 hours: 3 large pipes can drain 1 reservoir. 
In 8 hours: 3 large + 9 small pipes can drain 1 reservoir.
In 24 hours: 3 large + 9 small pipes can drain 3 reservoirs.
But, by  we know that in those 24 hours 3 large pipes can drain 1 reservoir.
Therefore the other 2 reservoirs can be drained by the small pipes on their own:
In 24 hours: 9 small pipes can drain 2 reservoirs.
In 24 hours: 1 small pipe can drain 2/9 reservoirs.
In 216 hours: 1 small pipe can drain 2 reservoirs.
In 216 hours: 5 small pipes can drain 10 reservoirs.
Therefore 5 small pipes can drain 10 reservoirs in 216 hours.
216 hours ÷ 10 = 21.6 hours.
21.6 hours = 21 hours and 36 minutes.