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.
[Ref: ZBVT] Sir Edwyn De Tudor. 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.
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.
Below, 10 nine letter words have been broken into chunks of three letters. These chunks have been mixed up, no chunk is used twice and all chunks are used. Can you determine what the 10 words are?
som one dbr pen rtm ber
nes ent ive eph eti ens
mes off eak eid sep win
erd tel day ter wed tem
aft apa own car ern oon
Which square, which circle, and which triangle has the closest area to the doughnut shape on the left? The drawings are to scale, so you might be able to judge it, as well as working the actual areas out.
The larger circle has diameter = 4, therefore the radius is 2, and the area is π x 22 = 4π.
The smaller circle has diameter = 2, therefore the radius is 1, and the area is π x 12 = π.
Therefore the shaded area is 4π - π = 3π ≈ 9.42.
The area of a circle is π x Radius2.
The circle with diameter 3.1 has a radius of 1.55 and an area of π x 1.552 = 7.55.
The circle with diameter 3.5 has a radius of 1.75 and an area of π x 1.752 = 9.62 (** closest match).
The circle with diameter 3.9 has a radius of 1.95 and an area of π x 1.952 = 11.95.
The area of a square is Side x Side.
The square with side 3.1 has an area of 3.1 x 3.1 = 9.61 (** closest match).
The square with side 3.5 has an area of 3.5 x 3.5 = 12.25.
The square with side 3.9 has an area of 3.9 x 3.9 = 15.21.
The area of a triangle is ½ x Base x Height. Using Pythagoras' theorem it can be shown that the area of an equilateral triangle is Sqrt(3) x Base2 ÷ 4.
The triangle with side 3.9 has an area of Sqrt(3) x 3.9 x 3.9 ÷ 4 = 6.59.
The triangle with side 4.3 has an area of Sqrt(3) x 4.3 x 4.3 ÷ 4 = 8.01.
The triangle with side 4.7 has an area of Sqrt(3) x 4.7 x 4.7 ÷ 4 = 9.57 (** closest match).
Using the BrainTracker grid below, how many words can you find? Each word must contain the central M and no letter can be used twice. The letters do not have to be connected. Proper nouns are not allowed, however, plurals are. Can you find the nine letter word?