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first think that he ought to be told "which pair is which pair," or something of that sort, but it is quite unnecessary. Can you give the five correct weights? (2/2)

and 121. Now, how was Farmer Tompkins to find out from these figures how much every one of the five trusses weighed singly? The reader may at (1/2)

SOLUTION TO 99. THE THIRTY-THREE PEARLS.

The value of the large central pearl must have been £3,000. The pearl at one end (from which they increased in value by £100) was £1,400; the pearl at the other end, £600.

...

"I am sure you are all wrong," insisted Mr. Wilson, "for I consider that Eve 8 1 4 Adam, and Adam 8 1 2 4 Eve, so we get a total of 8,938."

"But, look here," broke in Herbert. "If Eve 8 1 4 Adam and Adam 8 1 2 4 2 oblige Eve, surely the total must have been 82,056!"

At this point Uncle Walter suggested that they might let the matter rest. He declared it to be clearly what mathematicians call an indeterminate problem. (4/4)

...

have figured it out differently, and hold that Eve 8 and Adam a total of 16. Yet the most recent investigators think the above figures entirely wrong, for if Eve 8 and Adam 8 2, the total must be 90."

"Well," said Harry, "it seems to me that if there were giants in those days, probably Eve 8 1 and Adam 8 2, which would give a total of 163."

"I am not at all satisfied," said Maud. "It seems to me that if Eve 8 1 and Adam 8 1 2, they together consumed 893."

(3/4)

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string is worth £65,000. What is the value of that large pearl?"

"Pearls and other articles of clothing," said Uncle Walter, when the price of the precious gem had been discovered, "remind me of Adam and Eve. Authorities, you may not know, differ as to the number of apples that were eaten by Adam and Eve. It is the opinion of some that Eve 8 (ate) and Adam 2 (too), a total of 10 only. But certain mathematicians (2/4)

"A man I know," said Teddy Nicholson at a certain family party,

"possesses a string of thirty-three pearls. The middle pearl is the largest and best of all, and the others are so selected and arranged that, starting from one end, each successive pearl is worth £100 more than the preceding one, right up to the big pearl. From the other end the pearls increase in value by £150 up to the large pearl. The whole (1/4)

SOLUTION TO 67. AVERAGE SPEED.

The average speed is twelve miles an hour, not twelve and a half, as most people will hastily declare. Take any distance you like, say sixty miles. This would have taken six hours going and four hours returning. The double journey of 120 miles would thus take ten hours, and the average speed is clearly twelve miles an hour.

SOLUTION TO 108. THE LEAP-YEAR LADIES. (2/2)

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difficult, by algebra, when once we have succeeded in correctly stating it.

SOLUTION TO 108. THE LEAP-YEAR LADIES. (1/2)

The correct and only answer is that 11,616 ladies made proposals of marriage. Here are all the details, which the reader can check for himself with the original statements. Of 10,164 spinsters, 8,085 married bachelors, 627 married widowers, 1,221 were declined by bachelors, and 231 declined by widowers. Of the 1,452 widows, 1,155 married bachelors,

and 297 married widowers. No widows were declined. The problem is not

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declined. All the widows were accepted. Thirty-five forty-fourths of the widows married bachelors. One thousand two hundred and twenty-one spinsters were declined by bachelors. The number of spinsters accepted by bachelors was seven times the number of widows accepted by bachelors. Those are all the particulars that I was able to obtain. Now, how many women proposed? (2/2)

108. THE LEAP-YEAR LADIES. Last leap-year ladies lost no time in exercising the privilege of making proposals of marriage. If the figures that reached me from an occult source are correct, the following represents the state of affairs in this country.

A number of women proposed once each, of whom one-eighth were widows. In consequence, a number of men were to be married of whom one-eleventh were widowers. Of the proposals made to widowers, one-fifth were (1/2)

SOLUTION TO 7. THE WIDOW'S LEGACY.

The widow's share of the legacy must be £205, 2s. 6d. and 10/13 of a penny.

SOLUTION TO 379. THE FIVE DOMINOES.

There are just ten different ways of arranging the dominoes. Here is one of them:--

(2--0) (0--0) (0--1) (1--4) (4--0).

I will leave my readers to find the remaining nine for themselves.

...

(1--0) (0--0) (0--2) (2--1) (1--3)

(4--0) (0--0) (0--2) (2--1) (1--0)

(2--0) (0--0) (0--1) (1--3) (3--0)

Now, how many similar arrangements are there of five dominoes that shall give six instead of five in the two additions? (2/2)

(1/2)

SOLUTION TO 30. TWO QUESTIONS IN PROBABILITIES. (4/4)

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one must draw at least a shilling--there being no blanks.

SOLUTION TO 30. TWO QUESTIONS IN PROBABILITIES. (3/4)

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only 12 to 20 in your favour, or (which is the same thing) 3 to 5. Put another way, you have only 3 chances out of 8.

The amount that should be paid for a draw from the bag that contains three sovereigns and one shilling is 15s. 3d. Many persons will say that, as one's chances of drawing a sovereign were 3 out of 4, one should pay three-fourths of a pound, or 15s., overlooking the fact that

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Puzzles from Henry Ernest Dudeney's "Amusements in Mathematics"

Source: https://www.gutenberg.org/ebooks/16713

Joined Sep 2017