Mamma: "Let me think, Tommy. Well, our three ages add up to exactly seventy years."
Tommy: "That's a lot, isn't it? And how old are you, papa?"
Papa: "Just six times as old as you, my son."
Tommy: "Shall I ever be half as old as you, papa?"
Papa: "Yes, Tommy; and when that happens our three ages will add up to exactly twice as much as to-day."
Tommy: "And supposing I was born before you, papa; and supposing mamma had forgot all about it, and hadn't been at home when I came; and supposing--"
Mamma: "Supposing, Tommy, we talk about bed. Come along, darling. You'll have a headache."
Now, if Tommy had been some years older he might have calculated the exact ages of his parents from the information they had given him. Can you find out the exact age of mamma?
Mamma's Age, Amusements In Mathematics, Henry Ernest Dudeney.
Hint
The answer isn't a whole number of years, and algebra might be required.
Answer
29 years 2 months.
Reasoning #1
This answer is taken directly from the original book.
The age of Mamma must have been 29 years 2 months; that of Papa, 35 years; and that of the child, Tommy, 5 years 10 months. Added together, these make seventy years. The father is six times the age of the son, and, after 23 years 4 months have elapsed, their united ages will amount to 140 years, and Tommy will be just half the age of his father.
Reasoning #2
Here's my answer, with a little algebra.
If we call Tommy T, Mamma M and Papa P we can see that:
"our three ages add up to exactly seventy years" gives us:
(1) T + M + P = 70
"Just six times as old as you" gives us:
(2) P = 6 x T
In an unknown number of years (Y) "Shall I ever be half as old as you" gives us:
(3) P + Y = 2 x (T + Y)
and "our three ages will add up to exactly twice as much as today" gives us:
(T + Y) + (M + Y) + (P + Y) = 140
which can be written as
(4) T + M + P + 3Y = 140
We can see from (4) and (1) that
3Y = 70
so
(5) Y = 70 ÷ 3
Using (2) and (5) in (3) we have
P + Y = 2 x (T + Y)
6 x T + 70 ÷ 3 = 2 x (T + 70 ÷ 3)
4 x T = 70 ÷ 3
(6) T = 70 ÷ 12
We can now use (6) in (2) to see that:
P = 6 x T
P = 6 x 70 ÷ 12
P = 70 ÷ 2
And using the values for T and P in (1) we have:
T + M + P = 70
70 ÷ 12 + M + 70 ÷ 2 = 70
Multiply throughout by 12 to give:
70 + 12 x M + 420 = 840
12 x M = 840 − 420 − 70
12 x M = 350
M = 350 ÷ 12
So:
Tommy = 70 ÷ 12 = 5.83333 = 5 years 10 months. Papa = 70 ÷ 2 = 35 = 35 years. Mamma = 350 ÷ 12 = 29.1666 = 29 years 2 months.
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Puzzle 50
The Miller next took the company aside and showed them nine sacks of flour that were standing as depicted in the sketch.
"Now, hearken, all and some," said he, "while that I do set ye the riddle of the nine sacks of flour.
And mark ye, my lords, that there be single sacks on the outside, pairs next unto them, and three together in the middle thereof.
By Saint Benedict, it doth so happen that if we do but multiply the pair, 28, by the single one, 7, the answer is 196, which is of a truth the number shown by the sacks in the middle.
Yet it be not true that the other pair, 34, when so multiplied by its neighbour, 5, will also make 196.
Wherefore I do beg you, gentle sirs, so to place anew the nine sacks with as little trouble as possible that each pair when thus multiplied by its single neighbour shall make the number in the middle."
As the Miller has stipulated in effect that as few bags as possible shall be moved, there is only one answer to this puzzle, which everybody should be able to solve.
The Miller's Puzzle, The Canterbury Puzzles, Henry Ernest Dudeney.
Hint
The two left numbers multiplied, or the right two numbers, should create the central number.
Answer
The way to arrange the sacks of flour is as follows: 2, 78, 156, 39, 4. Here each pair when multiplied by its single neighbour makes the number in the middle, and only five of the sacks need to be moved.
There are just three other ways in which they might have been arranged (4, 39, 156, 78, 2; or 3, 58, 174, 29, 6; or 6, 29, 174, 58, 3), but they all require the moving of seven sacks.
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Puzzle 51
In the local wood there are a number of trees. On these trees there are a number of birds.
A local bird spotter knows that there are as many birds on each tree as there are trees in the wood, and that there are between 2,000 and 2,100 birds in total.
Reasoning
We are after a number whose square lies between 2,000 and 2,100.
45 is the only possible number.
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Puzzle 52
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, Amusements In Mathematics, Henry Ernest Dudeney.
If Sir Edwyn left at noon and rode 15 miles an hour, he would arrive at four o'clock, which is an hour too soon. If he rode 10 miles an hour, he would arrive at six o'clock, which is 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, which is 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
