Arithmetic & Algebraic Problems - A DREAMLAND CLOCK

Question: In a dream, I was travelling in a country where they had strange ways of doing things. One little incident was fresh in my memory when I awakened. I saw a clock and announced the time as it appeared to be indicated, but my guide corrected me.

He said, "You are apparently not aware that the minute hand always moves in the opposite direction to the hour hand. Except for this improvement, our clocks are precisely the same as. those you have been accustomed to."

Since the hands were exactly together between the hours of four and five o'clock, and they started together at noon, what was the real time?

Answer: The hour indicated would be exactly 23%3 minutes after four o'clock. But because the minute hand moved in the opposite direction, the real time would be 36 1'13 minutes after four. You must deduct the number of minutes indicated from 60 to get the real time.

Arithmetic & Algebraic Problems - ROBINSON'S AGE

Question: "How old are you, Robinson?" asked Colonel Crackham one morning.

"Well, I forget exactly," was the reply; "but my brother is two years older than I; my sister is four years older than he; my mother was twenty when I was born; and I was told yesterday that the average age of the four of us is thirty-nine years."

What was Robinson's age?

Answer: Robinson's age must have been 32, his brother's 34, his sister's 38, and his mother's 52.

Arithmetic & Algebraic Problems - THE BIRTH OF BOADICEA

Question: A correspondent (R. D.) proposes the following little puzzle:

Boadicea died one hundred and twenty-nine years after Cleopatra was born. Their united ages (that is, the combined years of their complete lives) were one hundred years. Cleopatra died 30 B.C. When was Boadicea born?

Answer: There were 129 years between the birth of Oeopatra and the death of Boadicea; but, as their united ages amounted to 100 years only, there must have been 29 years when neither existed-that is, between the death of Oeopatra and the birth of Boadicea. Therefore Boadicea must have been born 29 years after the death of Cleopatra in 30 B.C., which would be in the year 1 B.C.

Arithmetic & Algebraic Problems - FINDING A BIRTHDAY

Question: A correspondent informs us that on Armistice Day (November 11, 1928) he had lived as long in the twentieth century as he had lived in the nineteenth. This tempted us to work out the day of his birth. Perhaps the reader may like to do the same. We will assume he was born at midday.

Answer: The man must have been born at midday on February 19, 1873, and at midday on November 11, 1928, he had lived 10,17616 days in each century. Of course, the century ended at midnight on December 31, 1900, which was not a leap year, and his age on November 11, 1928, was 55 years and nearly 9 months.

Arithmetic & Algebraic Problems - IN THE YEAR 1900

Question: A correspondent, in 1930, proposed the following question. The reader may think, at first sight, that there is insufficient data for an answer, but he will be wrong:

A man's age at death was one twenty-ninth of the year of his birth. How old was he in the year 1900?

Answer: The man was born in 1856 and died in 1920, aged 64 years. Let x = age at death. Then 29x = date of birth. The date of birth + age = date of death, so that 29x + x = 30x, or date of death. Now, from the question he was clearly alive in 1900, and dead by 1930. So death occurred during or between those dates, and as the date is 30x, it is divisible by 30. The date can only be 1920,

which, divided by 30, gives 64. So in 1900 he was 44 years of age.

Arithmetic & Algebraic Problems - A SQUARE FAMILY

Question: A man had nine children, all born at regular intervals, and the sum of the squares of their ages was equal to the square of his own. What was the age of each? Every age was an exact number of years.

Answer: The ages of the nine children were respectively 2, 5, 8, 11, 14, 17,20,23,26, and the age of the father was 48.

Arithmetic & Algebraic Problems - BROTHER AND SISTER

Question: A boy on being asked the age of himself and of his sister replied:

"Three years ago I was seven times as old as my sister; two years ago I was four times as old; last year I was three times as old; and this year I am two

and one-half times as old."

What are their ages?

Answer:The boy's age was ten and his sister's four.

Arithmetic & Algebraic Problems - THEIR AGES

Question: Rackbrane said the other morning that a man on being asked the ages of his two sons  stated that eighteen more than the sum of their ages is double the age of the elder, and six less than the difference of their ages is the age of the younger. What are their ages?

Answer: Thirty years and twelve years, respectively.

Arithmetic & Algebraic Problems - MIKE'S AGE

Question: "Pat O'Connor," said Colonel Crackham, "is now just one and one-third times as old as he was when he built the pig sty under his drawing-room window. Little Mike, who was forty months old when Pat built the sty, is now two years more than half as old as Pat's wife, Biddy, was when Pat

built the sty, so that when little Mike is as old as Pat was when he built the sty, their three ages combined will amount to just one hundred years. How old is little Mike?"

Answer: Mike's present age is 101~1 years, Pat's is 291~1, and Biddy's is 242~1 years. When the sty was built (79121 years ago), Mike was 3~1, Pat was 22~1, and Biddy was 171\6 I. In 1 P \61 years Mike will be 22~ 1 (as old as Pat was when he built the sty), Pat will be 41~1, and Biddy will be 361%1, making 100 years together.

Arithmetic & Algebraic Problems - FAMILY AGES

Question: A man and his wife had three children, John, Ben, and Mary, and the difference between their parents' ages was the same as between John and Ben and between Ben and Mary. The ages of John and Ben, multiplied together, equalled the age of the father, and the ages of Ben and Mary multiplied together equalled the age of the mother. The combined ages of the family amounted to ninety years. What was the age of each person?

Answer: The father and mother were both of the same age, thirty-six years old, and the three children were triplets of six years of age. Thus the sum of all their ages is ninety years, and all the other conditions are correctly fulfilled.

Arithmetic & Algebraic Problems - ANCIENT PROBLEM

Question: Here is an example of the sort of "Breakfast Problem" propounded by Metrodorus in 310 A.D.

Demochares has lived one-fourth of his life as a boy, one-fifth as a youth, one-third as a man, and has spent thirteen years in his dotage. How old is the gentleman?

Answer: Demochares must be just sixty years of age.

Arithmetic & Algebraic Problems - "SIMPLE" ARITHMETIC

Question: When visiting an insane asylum, I asked two inmates to give me their ages. They did so, and then, to test their arithmetical powers, I asked them to add the two ages together. One gave me 44 as the answer, and the other gave 1,280. I immediately saw that the first had subtracted one age from the other, while the second person had multiplied them together. What were their ages?

Answer: Their ages were, respectively, sixty-four and twenty.

Arithmetic & Algebraic Problems - DE MORGAN AND ANOTHER

Question: Augustus De Morgan, the mathematician, who died in 1871, used to boast that he was x years old in the year x2• Jasper Jenkins, wishing to improve on this, told me in 1925 that he was a2 + b2 in a4 + b4 ; that he was 2m in the year 2m2; and that he was 3n years old in the year 3n4• Can you give the years in which De Morgan and Jenkins were respectively born?

Answer: De Morgan was born in 1806. When he was 43, the year was the square of his age-1849. Jenkins was born in 1860. He was 52 + 62 (61) in the year 54 + 64 (1921). Also he was 2 X 31 (62) in the year 2 X 312 (1922). Again, he was 3 X 5 (15) in the year 3 X 54 (1875).

Arithmetic & Algebraic Problems - THEIR AGES

Question: If you add the square of Tom's age to the age of Mary, the sum is 62; but if you add the square of Mary's age to the age of Tom, the result is 176. Can you say what are the ages of Tom and Mary?

Answer: Tom's age was seven years and Mary's thirteen years.

Arithmetic & Algebraic Problems - MRS. WILSON'S FAMILY

Question: Mrs. Wilson had three children: Edgar, James, and John. Their combined ages were half of hers. Five years later, during which time Ethel was born, Mrs. Wilson's age equalled the total of all her children's ages. Ten years more have now passed, Daisy appearing during that interval. At the latter event Edgar was as old as John and Ethel together. The combined ages of all the children are now double Mrs. Wilson's age, which is, in fact, only equal to that of Edgar and James together. Edgar's age also equals that of the two daughters.

Can you find all their ages?

Answer: The ages must be as follows: Mrs. Wilson, 39; Edgar, 21; James, 18; John, 18; Ethel, 12; Daisy, 9. It is clear that James and John were twins.

Arithmetic & Algebraic Problems - THE BANKER AND THE COUNTERFEIT BILL

Question: A banker in a country town was walking down the street when he saw a five-dollar bill on the curb. He picked it up, noted the number, and went to his home for luncheon. His wife said that the butcher had sent in his bill for five dollars, and, as the only money he had was the bill he had found, he gave it to her, and she paid the butcher. The butcher paid it to a farmer in buying a calf, the farmer paid it to a merchant who in turn paid it to a laundry woman, and she, remembering that she owed the bank five dollars, went there and paid the debt.

The banker recognized the bill as the one he had found, and by that time it had paid twenty-five dollars worth of debts. On careful examination he discovered that the bill was counterfeit. What was lost in the whole transaction, and by whom?

Answer: Since the identical counterfeit bill can be traced through all the transactions, these are all invalid. Therefore everybody stands in relation to his debtor just where he was before the banker picked up the note, except that the butcher owes, in addition, $5.00 to the farmer for the calf received.

Arithmetic & Algebraic Problems - PROSPEROUS BUSINESS

Question: A man started business with a capital of $2,000.00, and increased his wealth by 50 per cent every three years. How much did he possess at the expiration of eighteen years?

Answer: He possessed $22,781.25.

Arithmetic & Algebraic Problems - APPLE TRANSACTIONS

Question: A man was asked what price per 100 he paid for some apples, and his reply was as follows: "If they had been 4¢ more per 100 I should have got five less for $1.20." Can you say what was the price per lOO?

Answer: The apples cost 96¢ per 100.

Arithmetic & Algebraic Problems - A QUEER SETTLING UP

Question: Professor Rackbrane told his family at the breakfast table that he had heard the following conversation in a railway carriage the night before.

One passenger said to another, "Here is my purse: give me just as much money, Richard, as you find in it."

Richard counted the money, added an equal value from his own pocket, and replied, "Now, John, if you give me as much as I have left of my own we shall be square."

John did so, and then stated that his own purse contained $3.50, while Richard said that he now had $3.00. How much did each man possess at first?

Answer: Richard had $4.00, and John had $2.50.

Arithmetic & Algebraic Problems - THE COST OF A SUIT

Question: "Hello, old chap," cried Russell as Henry Melville came into the club arrayed in a startling new tweed suit, "have you been successful in the cardroom lately? No? Then why these fine feathers?" 

"Oh, I just dropped into my tailor's the other day," he explained, "and this cloth took my fancy. Here is a little puzzle for you. The coat cost as much as the trousers and vest. The coat and two pairs of trousers would cost $175.00. The trousers and two vests would cost $100.00. Can you tell me the cost of the suit?"

Answer: The cost of Melville's suit was $150.00, the coat costing $75.00, the trousers $50.00, and the vest $25.00.

Arithmetic & Algebraic Problems - BOYS AND GIRLS

Question: Nine boys and three girls agreed to share equally their pocket money. Every boy gave an equal sum to every girl, and every girl gave another equal sum to every boy. Every child then possessed exactly the same amount. What was the smallest possible amount that each then possessed?

Answer: Every boy at the start possessed 12¢, and he gave 1¢ to every girl. Every girl held 36¢, of which she gave 3¢ to every boy. Then every child would have 18¢.

Arithmetic & Algebraic Problems - A POULTRY POSER

Question: Three chickens and one duck sold for as much as two geese; one chicken, two ducks, and three geese were sold together for $25.00. What was the price of each bird in an exact number of dollars?

Answer: The price of a chicken was $2.00, for a duck $4.00, and for a goose $5.00.

Arithmetic & Algebraic Problems - THE MISSING PENNY

Question: Here is an ancient puzzle that has always perplexed some people. Two market women were selling their apples, one at three for a penny and the other at two for a penny. One day they were both called away when each had thirty apples unsold: these they handed to a friend to sell at five for 2¢. It will be seen that if they had sold their apples separately they would have

fetched 25¢, but when they were sold together they fetched only 24¢.

"Now," people ask, "what in the world has become of that missing penny?" because, it is said, three for l¢ and two for l¢ is surely exactly the same as five for 2¢.

Can you explain the little mystery?

Answer: He must have had 168 each of dollar bills, half dollars, and quarters, making a total of $294.00. In each of the six bags there would be 28 of each kind; in each of the seven bags 24 of each kind; and in each of the eight bags, 21 of each kind.

Arithmetic & Algebraic Problems - THE THRIFTY GROCER

Question: A grocer in a small business had managed to put aside (apart from his legitimate profits) a little sum in dollar bills, half dollars, and quarters, which he kept in eight bags, there being the same number of dollar bi1ls and of each kind of coin in every bag. One night he decided to put the money into only seven bags, again with the same number of each kind of currency in every bag. And the following night he further reduced the number of bags to six, again putting the same number of each kind of currency in every bag. 

The next night the poor demented miser tried to do the same with five bags, but after hours of trial he utterly failed, had a fit, and died, greatly respected by his neighbors. What is the smallest possible amount of money he had put aside?

Answer: He must have had 168 each of dollar bills, half dollars, and quarters, making a total of $294.00. In each of the six bags there would be 28 of each kind; in each of the seven bags 24 of each kind; and in each of the eight bags, 21 of each kind.

Arithmetic & Algebraic Problems - BUYING TURKEYS

Question: A man bought a number of turkeys at a cost of $60.00, and after reserving fifteen of the birds he sold the remainder for $54.00, thus gaining 1O¢ a head by these. How many turkeys did he buy?

Answer: The man bought 75 turkeys at 80¢ each, making $60.00. After retaining 15 he sold the remaining 60 at 90¢ each, making $54.00, as stated. He thus made a profit of 1O¢ each on the 60 birds he resold.

Arithmetic & Algebraic Problems - HORSES AND BULLOCKS

Question: A dealer bought a number of horses at $344.00 each, and a number of bullocks at $265.00 each. He then discovered that the horses had cost him in all $33.00 more than the bullocks. Now, what is the smallest number of each that he must have bought?

Answer: We have to solve the indeterminate equation 344x = 265y + 33. This is easy enough if you know how, but we cannot go into the matter here. Thus x is 252, and y is 327, so that if he buys 252 horses for $344.00 apiece, and 327 bullocks for $265.00 apiece, the horses will cost him in all $33.00 more than the bullocks.

Arithmetic & Algebraic Problems - REDUCTIONS IN PRICE

Question: "I have often been mystified," said Colonel Crackham, "at the startling reductions some people make in their prices, and wondered on what principle they went to work. For example, a man offered me a motorcycle two years ago for $1,024.00; a year later his price was $640.00; a little while after he asked a level $400.00; and last week he was willing to sell for $250.00. The next time he reduces I shall buy. At what price shall I purchase if he makes a consistent reduction?"

Answer: It is evident that the salesman's rule was to take off three-eighths of the price at every reduction. Therefore, to be consistent, the motorcycle should be offered at $156.25 after the next reduction.

Arithmetic & Algebraic Problems - DISTRIBUTION

Question: Nine persons in a party, A, B, C, D, E, F, G, H, K, did as follows: First A gave each of the others as much money as he (the receiver) already held; then B did the same; then C; and so on to the last, K giving to each of the other eight persons the amount the receiver then held. Then it was found that each of the nine persons held the same amount.

Can you find the smallest amount in cents that each person could have originally held?

Answer: The smallest number originally held by one person will be (in cents) one more than the number of persons. The others can be obtained by continually doubling and deducting one. So we get their holdings as 10, 19,37, 73, 145, 289,577, 1,153, and 2,305. Let the largest holder start the payment and work backwards, when the number of cents in the end held by each person will be 29 or 512-that is, $5.12.

Arithmetic & Algebraic Problems - POCKET MONEY

Question: "When I got to the station this morning," said Harold Tompkins, at his club, "I found I was short of cash. I spent just one-half of what I had on my railway ticket, and then bought a nickel's worth of candy. When I got to the terminus I spent half of what I had left and ten cents for a newspaper. Then I spent half of the remainder on a bus and gave fifteen cents to that old beggar outside the club. Consequently I arrive here with this single nickel. How much did I start out with?"

Answer: When he left home Tompkins must have had $2.10 in his pocket.

Arithmetic & Algebraic Problems - A NEW PARTNER

Question: Two partners named Smugg and Williamson have decided to take a Mr. Rogers into partnership. Smugg has 116 times as much capital invested in the business as Williamson, and Rogers has to pay down $2,500.00, which sum shall be divided between Smugg and Williamson, so that the three partners shall have an equal interest in the business. How shall the sum be divided?

Answer: We must take it for granted that the sum Rogers paid, $2,500.00, was one third of the value of the business, which was consequently worth $7,500.00 before he entered. Smugg's interest in the business had therefore been $4,500.00 (l\-2 times as much as Williamson), and Williamson's $3,000.00. As each is now to have an equal interest, Smugg will receive $2,000.00 of Rogers's contribution, and Williamson $500.00.