Puzzle 45
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. From Amusements In Mathematics by Henry Ernest Dudeney (1917).
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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 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
Time2 - Time1 = 2
D ÷ 10 - D ÷ 15 = 2
Multiply throughout by 30:
3D - 2D = 60
D = 60 miles.
Puzzle 46
A kind old person decided to give 12 sweets to each of the adults in the town and 8 sweets to each of the children.
Of the 612 people in the town, exactly half of the adults, and exactly three quarters of the children took the sweets.
How many sweets did the kind old person have to buy?
Puzzle Copyright © Kevin Stone
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Hint
The number of adults and children doesn't matter.
Answer
3,672.
Reasoning
The actual number of adults and children doesn't actually matter.
If all of the people were adults, then half of them (306) would be given 12 sweets:
306 x 12 = 3672
If all of the people were children, then three quarters of them (459) would be given 8 sweets:
459 x 8 = 3672
If there were 512 adults (so 256 would get 12 sweets = 3072) and 100 children (so 75 would get 8 sweets = 600):
256 x 12 + 75 x 8 = 3672
We can change the numbers of adults and children, but it doesn't change the answer.
The reason for this lies in the fact that 1/2 adults x 12 sweets = 3/4 children x 8 sweets (both are 6).
Puzzle 47
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?
Puzzle Copyright © Kevin Stone
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Hint
What would happen in 24 hours?
Answer
21 hours and 36 minutes.
Reasoning
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.
(fix the number of hours, and divide pipes and reservoirs by 9)
In 24 hours: 1 small pipe can drain 2/9 reservoirs.
(fix the numbers of pipes, and multiply hours and reservoirs by 9)
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 48
A very long time ago, at the beginning of the week, I was given some money for my birthday.
On Monday, I spent a quarter of the money on clothes.
On Tuesday, I spent one third of the remaining money on music.
On Wednesday, I spent half of the remaining money on food.
Finally, on Thursday, I spent the last £1.25 on a book.
How much birthday money did I receive?
Puzzle Copyright © Kevin Stone
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It might be easier to work backwards.
Answer
£5.00.
Reasoning
Working backwards:
On Thursday, I had £1.25.
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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