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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.
Puzzle 6
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?
Hint: The number of boys and girls doesn't matter.
Answer: 3,672.
The actual number of girls and boys doesn't actually matter!
If all of the children were girls then half of them (306) would be given 12 sweets:
306 x 12 = 3672
If all of the children were boys then three quarters of them (459) would be given 8 sweets:
459 x 8 = 3672
If there were 512 girls (so 256 would get 12 sweets = 3072) and 100 boys (so 75 would get 8 sweets = 600):
256 x 12 + 75 x 8 = 3672
We can change the numbers of girls and boys, but it doesn't change the answer. The reason for this lies in the fact that 1/2 girls x 12 sweets = 3/4 boys x 8 sweets (both are 6).
Puzzle 7
A large fresh water reservoir has two types of drainage system: small pipes and large pipes.
6 large pipes, on their own, can drain the reservoir in 12 hours.
3 large pipes and 9 small pipes, at the same time, can drain the reservoir in 8 hours.
How long will 5 small pipes, on their own, take to drain the reservoir?
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. [1]
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 [1] 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.
Puzzle 8
Last weekend I was given my pocket money, which is meant to last me all week.
On Monday, I spent a quarter of my money on clothes.
On Tuesday, I spent one third of my remaining money on a CD.
On Wednesday I spent half of my remaining money on sweets.
Finally on Thursday I spent my last £1.25 on a comic.
On Wednesday I had £1.25 x (1 ÷ one half) = £2.50.
On Tuesday I had £2.50 x (1 ÷ two thirds) = £3.75.
On Monday I had £3.75 x (1 ÷ three quarters) = £5.00.
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