Best Preparation Strategy - 30 Days to CMAT

The countdown has begun. With just 30 days to go for CMAT  2016, this is probably going to be the most important phase of your CMAT preparations. This is quite a tricky phase, because almost all of you are already done with solving all the modules at least twice and are pretty much confident about your concepts but, still there is a that stupid feeling of unpreparedness or fear of exam that keeps lurking around in some corner of your mind. 

Believe me when I say this, the only way to overcome your fear of exams is to face it as Dale Carnegie aptly said “Do the thing you fear, and continue to do so. This is the quickest and surest way of all victory over fear”.

Read Complete Strategy at - http://www.endeavorcareers.com/gurus-speak/761-30-days-to-cmat-preparation-strategy

Arithmetic & Algebraic Problems - A LEGACY PUZZLE

Question: A man left legacies to his three sons and to a hospital, amounting in all to $1,320.00. If he had left the hospital legacy also to his first son, that son would have received as much as the other two sons together. If he had left it to his second son, he would have received twice as much as the other two sons together. If he had left the hospital legacy to his third son, he would have received then thrice as much as the first son and second son together. Find the amount of each legacy.

Answer: The legacy to the first son was $55.00, to the second son $275.00, to the third son $385.00, and to the hospital $605.00, making $1,320.00 in all.

Arithmetic & Algebraic Problems - THE STAIRCASE RACE

Question: This is a rough sketch of the finish of a race up a staircase in which three men took part. Ackworth, who is leading, went up three steps at a time, as arranged; Barnden, the second man, went four steps at a time, and Croft, who is last, went five at a time. Undoubtedly Ackworth wins. But the point is, how many steps are there in the stairs, counting the top landing as a step?

I have only shown the top of the stairs. There may be scores, or hundreds, of steps below the line. It was not necessary to draw them, as I only wanted to show the finish. But it is possible to tell from the evidence the fewest possible steps in that staircase. Can you do it?

Answer: If the staircase were such that each man would reach the top in a certain number of full leaps, without taking a reduced number at his last leap, the smallest possible number of steps would, of course, be 60 (that is, 3 X 4 X 5). But the sketch showed us that A. taking three steps at a leap, has one odd step at the end; B. taking four at a leap, will have three only at the end and C. taking five at a leap, will have four only at the finish. Therefore, we have to find the smallest number that, when divided by 3, leaves a remainder I, when divided by 4 leaves 3, and when divided by 5 leaves a remainder 4. This number is 19. So there were 19 steps in all, only 4 being left out in the sketch.

Arithmetic & Algebraic Problems - TIMING THE CAR

Question: "I was walking along the road at three and a half miles an hour," said Mr. Pipkins, "when the car dashed past me and only missed me by a few inches." "Do you know at what speed it was going?" asked his friend. "Well, from the moment it passed me to its disappearance round a corner.

I took twenty-seven steps and walking on reached that corner with one hundred and thirty-five steps more." "Then, assuming that you walked, and the car ran, each at a uniform rate, we can easily work out the speed."

Answer: As the man can walk 27 steps while the car goes 162, the car is clearly going six times as fast as the man. The man walks 3 1/2 miles an hour: therefore the car was going at 21 miles an hour.

Arithmetic & Algebraic Problems - HILL CLIMBING

Question: Weary Willie went up a certain hill at the rate of one and a half miles per hour and came down at the rate of four and a half miles per hour, so that it took him just six hours to make the double journey. How far was it to the top of the hill?

Answer: It must have been 6 3/4 miles to the top of the hill. He would go up in 4 1/2 hours and descend in 1 1/2 hours.

Arithmetic & Algebraic Problems - WESTMINSTER CLOCK

Question: A man crossed over Westminster Bridge one morning between eight and nine o'clock by the tower clock (often mistakenly called Big Ben, which is the name of the large bell only, but this by the way). On his return between four and five o'clock he noticed that the hands were exactly reversed. What were the exact times that he made the two crossings?

Answer: The times were 8 hours 23 71/143 minutes, and 4 hours 41 137/143 minutes. We are always allowed to assume that these fractional times can be indicated in clock puzzles.

Arithmetic & Algebraic Problems - AT RIGHT ANGLES

Question: Rackbrane asked his young friends at the breakfast table one morning this little question:

"How soon between the hours of five and six will the hour and minute hands of a clock be exactly at right angles?"

Answer: To be at right angles the minute hand must always be exactly fifteen minutes either behind or ahead of the hour hand. Each case would happen eleven times in the twelve hours-i.e., every 1 hour 5 5/11 minutes. Starting from nine o'clock, the eighth addition will give the case 5 hours 43 7/11 minutes. In the other case, starting from three o'clock, the second addition gives 5 hours 10 10/11 minutes.

These are the two cases between five and six, and the latter will, of course, be the sooner.

Arithmetic & Algebraic Problems - EQUAL DISTANCES

Question: A few mornings ago the following clock puzzle was sprung on his pupils by Professor Rackbrane. At what time between three and four o'clock is the minute hand the same distance from VIII as the hour hand is from XII?

Answer: At 23 1/13 minutes past three o'clock.

Arithmetic & Algebraic Problems - MISTAKING THE HANDS

Question: "Between two and three o'clock yesterday," said Colonel Crackham, "I looked at the clock and mistook the minute hand for the hour hand, and consequently the time appeared to be fifty-five minutes earlier than it actually was. What was the correct time?"

Answer: The time must have been 5 5/11 minutes past two o'clock.

Arithmetic & Algebraic Problems - WHEN DID THE DANCING BEGIN?

Question: "The guests at that ball the other night," said Dora at the breakfast table, "thought that the clock had stopped, because the hands appeared in exactly the same position as when the dancing began. But it was found that they had really only changed places. As you know, the dancing commenced between ten and eleven oclock. What was the exact time of the start?"

Answer: The dancing must have begun at 59 83/143 minutes past ten, and the hands were noticed to have changed places at 54 l38/l43 minutes past eleven.

Arithmetic & Algebraic Problems - THE AMBIGUOUS CLOCK

Question: A man had a clock with an hour hand and minute hand of the same length and indistinguishable. If it was set going at noon, what would be the first time that it would be impossible, by reason of the similarity of the hands, to be sure of the correct time?

Readers will remember that with these clock puzzles there is the convention that we may assume it possible to indicate fractions of seconds. On this assumption an exact answer can be given.

Answer: The first time would be 5 5/143 minutes past twelve, which might also (the hands being similar) indicate 60/143 minutes past one o'clock.

Arithmetic & Algebraic Problems - WHAT IS THE TIME?

Question: At what time are the two hands of a clock so situated that, reckoning as minute points past XII, one is exactly the square of the distance of the other?

Answer: The time is 6 3/4 minutes past IX, when the hour hand is 45 9/16 (the square of 6 3/4) minutes past XII. If we allow fractions less than a minute point, there is also the solution, five seconds (one-twelfth of a minute) past XII o'clock.

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.

Arithmetic & Algebraic Problems - DIVIDING THE LEGACY

Question: A man left $100.00 to be divided between his two sons Alfred and Benjamin. If one-third of Alfred's legacy be taken from one-fourth of Benjamin's, the remainder would be $11.00. What was the amount of each legacy?

Answer: The two legacies were $24.00 and $76.00, for if 8 (one-third of 24) be taken from 19 (one-fourth of 76) the remainder will be II.

Arithmetic & Algebraic Problems - PUZZLING LEGACIES

Question: A man bequeathed a sum of money, a little less than $1,500.00, to be divided as follows: The five children and the lawyer received such sums that the square root of the eldest son's share, the second son's share divided by two, the third son's share minus $2.00, the fourth son's share plus $2.00, the daughter'S share multiplied by two, and the square of the lawyer's fee all worked out at exactly the same sum of money. No dollars were divided, and no money was left over after the division. What was the total amount bequeathed?

Answer: The answer is $1,464.oo-a little less than $1,500.00. The legacies, in order, were $1,296.00, $72.00, $38.00, $34.00, and $18.00. The lawyer's fee would be $6.00.

CAT 2015 Preparation - How to Adopt New Test Format

With just 1 month left for CAT 2015, we bring you video of several format changes in CAT history in last 5-6 years and how one should see the latest change in normalization as a opportunity to score well in all the sections.

Driving force for this year CAT will be -

-> Time Management
-> Selection of Questions
-> Sectional Consistency

Read More at - http://www.endeavorcareers.com/gurus-speak/747-cat-2015-preparation-how-to-adopt-new-test-format

Arithmetic & Algebraic Problems - THE SEVEN APPLEWOMEN

Question: Here is an old puzzle that people are frequently writing to me about. Seven applewomen, possessing respectively 20, 40, 60, 80, 100, 120, and 140 apples, went to market and sold all the apples at the same price, and each received the same sum of money. What was the price?

Answer: Each woman sold her apples at seven for I¢, and 3¢ each for the odd ones over. Thus, each received the same amount, 20¢. Without questioning the ingenuity of the thing, I have always thought the solution unsatisfactory, because really indeterminate, even if we admit that such an eccentric way of selling may be fairly termed a "price." It would seem just as fair if they sold them at

different rates and afterwards divided the money; or sold at a single rate with different discounts allowed; or sold different kinds of apples at different values; or sold the same rate per basketful; or sold by weight, the apples being of different sizes; or sold by rates diminishing with the age of the apples; and so on. That is why I have never held a high opinion of this old puzzle.

In a general way, we can say that n women, possessing an + (n - 1), n(a + b) + (n - 2), n(a + 2b) + (n - 3), ... ,n[a + b(n - 1) 1 apples respectively, can sell at n for a penny and b pennies for each odd one over and each receive a + b(n - I) pennies. In the case of our puzzle a = 2, b = 3, and n = 7.

Arithmetic & Algebraic Problems - MARKET TRANSACTIONS

Question: A farmer goes to market and buys a hundred animals at a total cost of $1,000.00. The price of cows being $50.00 each, sheep $10.00 each, and rabbits 50¢ each, how many of each kind does he buy? Most people will solve this, if they succeed at all, by more or less laborious trial, but there are several direct ways of getting the solution.

Answer: The man bought 19 cows for $950.00, 1 sheep for $10.00, and 80 rabbits for $40.00, making together 100 animals at a cost of $1,000.00.

A purely arithmetical solution is not difficult by a method of averages, the average cost per animal being the same as the cost of a sheep.

By algebra we proceed as follows, working in dollars: Since x + y + z = 100 

by subtraction, or 99x + 19y = 1900. We have therefore to solve this indeterminate equation. The only answer is x = 19, Y = l. Then, to make up the 100 animals, z must equal 80.

Arithmetic & Algebraic Problems - NAME THEIR WIVES

Question: A man left a legacy of $1 ,000.00 to three relatives and their wives. The wives received together $396.00. Jane received $10.00 more than Catherine, and Mary received $10.00 more than Jane. John Smith was given just as much as his wife, Henry Snooks got half as much again as his wife, and Tom Crowe received twice as much as his wife. What was the Christian name of each man's wife?

Answer: As it is evident that Catherine, Jane, and Mary received respectively $122.00, $132.00, and $142.00, making together the $396.00 left to the three wives, if John Smith receives as much as his wife Catherine, $122.00; Henry Snooks half as much again as his wife Jane, $198.00; and Tom Crowe twice as much as his wife Mary, $284.00, we have correctly paired these married couples and exactly accounted for the $1,000.00.

Arithmetic & Algebraic Problems - DIGGING A DITCH

Question: Here is a curious question that is more perplexing than it looks at first sight. Abraham, an infirm old man, undertook to dig a ditch for two dollars. He engaged Benjamin, an able-bodied fellow, to assist him and share the money fairly according to their capacities. Abraham could dig as fast as Benjamin could shovel out the dirt, and Benjamin could dig four times as fast as Abraham could do the shoveling.

How should they divide the money? Of course, we must assume their relative abilities for work to be the same in digging or shoveling.

Answer: A. should receive one-third of two dollars, and B. two-thirds. Say B. can dig all in 2 hours and shovel all in 4 hours; then A. can dig all in 4 hours and shovel all in 8 hours. That is, their ratio of digging is as 2 to 4 and their ratio of shovelling as 4 to 8 (the same ratio), and A. can dig in the same time that B. can shovel (4 hours), while B. can dig in a quarter of the time that A. can shovel. Any other figures will do that fill these conditions and give two similar ratios for their working ability. Therefore, A. takes one-third and B. twice as much-two-thirds.

Arithmetic & Algebraic Problems - A WEIRD GAME

Question: Seven men engaged in play. Whenever a player won a game he doubled the money of each of the other players. That is, he gave each player just as much money as each had in his pocket. They played seven games and, strange to say, each won a game in turn in the order of their names, which

began with the letters A, B, C, D, E, F, and G.

When they had finished it was found that each man had exactly $1.28 in his pocket. How much had each man in his pocket before play?

Answer: The seven men, A, B, C, D, E, F, and G, had respectively in their pockets before play the following sums: $4.49, $2.25, $1.13, 57¢, 29¢, 15¢, and 8¢. The answer may be found by laboriously working backwards, but a simpler method is as follows: 7 + 1 = 8; 2 X 7 + 1 = 15; 4 X 7 + 1 = 29; and so on, where the multiplier increases in powers of 2, that is, 2, 4, 8, 16, 32, and 64.

Arithmetic & Algebraic Problems - THE PERPLEXED BANKER

Question: A man went into a bank with a thousand dollars, all in dollar bills, and ten bags. He said, "Place this money, please, in the bags in such a way that if I call and ask for a certain number of dollars you can hand me over one or more bags, giving me the exact amount called for without opening any of the bags." How was it to be done? We are, of course, only concerned with a single application, but he may ask for any exact number of dollars from one to one thousand.

Answer: The contents of the ten bags (in dollar bills) should be as follows: $1,2,4, 8, 16,32,64, 128,256, 4S9. The first nine numbers are in geometrical progression, and their sum, deducted from 1,000, gives the contents of the tenth bag.

Scoring Pattern and Parameters for GRE Exam

The GRE is a Section Adaptive Test such that, a good performance on the first section will lead to a second section with questions of higher difficulty level and therefore higher score per question, thereby significantly improving the chances of scoring very well on the test.

The total scaled score for the Quantitative Reasoning Section varies from 130 to 170. There is a similar scaled score range for Verbal Reasoning Section. The score is determined by a test taker’s performance over two sections each of Quantitative Reasoning and Verbal Reasoning, totalling to 40 questions each (2 sections of 20 questions each). 

But each test taker gets five sections instead of four, which means, either the Quantitative Reasoning or the Verbal Reasoning Section appears three times instead of two. The additional section resembles the other sections, but is an undisclosed, un-scored section, colloquially called a dummy section.

Read More at - http://www.endeavorcareers.com/gurus-speak/739-scoring-pattern-and-parameters-for-gre-exam

Arithmetic & Algebraic Problems - UNREWARDED LABOR

Question: A man persuaded Weary Willie, with some difficulty, to try to work on a job for thirty days at eight dollars a day, on the condition that he would forfeit ten dollars a day for every day that he idled. At the end of the month neither owed the other anything, which entirely convinced Willie of the folly of labor. Can you tell just how many days' work he put in and on how many

days he idled?

Answer: Weary Willie must have worked 1673 days and idled l3Y.! days. Thus the former time, at $8.00 a day, amounts to exactly the same as the latter at $10.00 a day.

Arithmetic & Algebraic Problems - BUYING BUNS

Question: Buns were being sold at three prices: one for a penny, two for a penny, and three for a penny. Some children (there were as many boys as girls) were given seven pennies to spend on these buns, each child to receive exactly the same value in buns. Assuming that all buns remained whole, how many buns, and of what types, did each child receive?

Answer: There must have been three boys and three girls, each of whom received two buns at three for a penny and one bun at two for a penny, the cost of which would be exactly 7¢.

Arithmetic & Algebraic Problems - GENEROUS GIFTS

Question: A generous man set aside a certain sum of money for equal distribution weekly to the needy of his acquaintance. One day he remarked, "If there are five fewer applicants next week, you will each receive two dollars more." Unfortunately, instead of there being fewer there were actually four more persons applying for the gift. "This means," he pointed out, "that you will each receive one dollar less." How much did each person receive at that last distribution?

Answer: At first there were twenty persons, and each received $6.00. Then fifteen persons (five fewer) would have received $8.00 each. But twenty-four (four more) appeared and only received $5.00 each. The amount distributed weekly was thus $120.00.

Arithmetic & Algebraic Problems - LOOSE CASH

Question: What is the largest sum of money-all in current coins and no silver dollars-that I could have in my pocket without being able to give change for a dollar, half dollar, quarter, dime, or nickel?

Answer: The largest sum is $1.19, composed of a half dollar, quarter, four dimes, and four pennies.

Best Tips for Preparing Quantitative Ability Section

With the pattern changes that CAT 2015 has come up with, the quantitative ability section would have 34 questions with a section time limit of 60 minutes.

The section would have questions from Geometry, arithmetic, algebra and modern mathematics. The reliance on mental calculation will decrease substantially owing to the on-screen calculator.

Any QA selection in CAT will be a mix of easy, moderate and difficult questions.

Even the difficult questions are no match for the difficulty level of the paper based CATs of 2006 or 2007. An easy question typically takes 1.5 to 2 minutes to solve whereas a tough question would take 6-8 minute for a normal test taker.

Read More at - http://www.endeavorcareers.com/gurus-speak/738-tips-for-preparing-quantitative-ability-section

Arithmetic & Algebraic Problems - DOLLARS AND CENTS

Question: A man entered a store and spent one-half ofthe money that was in his pocket. When he came out he found that he had just as many cents as he had dollars when he went in and half as many dollars as he had cents when he went in. How much money did he have on him when he entered?

Answer: The man must have entered the store with $99.98 in his pocket.

Arithmetic & Algebraic Problems - CONCERNING A CHECK

Question: A man went into a bank to cash a check. In handing over the money the cashier, by mistake, gave him dollars for cents and cents for dollars. He pocketed the money without examining it, and spent a nickel on his way home. He then found that he possessed exactly twice the amount of the check. He had no money in his pocket before going to the bank. What was the exact amount of that check?

Answer: The amount must have been $31.63. He received $63.31. After he had spent a nickel there would remain the sum of $63.26, which is twice the amount of the check.

What is GRE Exam? – Format and Test Structure

The GRE (Graduate Record Examination) is a standardized aptitude test conducted online (paper based only in some countries) on a daily basis across the world. The test is conducted in two variants:

1.    The GRE General Test: An approximately 4 hour test conducted online round the year that tests quantitative reasoning, verbal reasoning and analytical writing skills of the test taker.

2.    The GRE Subject Test: Required in some instances, GRE Subject Tests evaluate subject-specific knowledge of a test taker. GRE Subject Tests are paper based tests conducted on fixed dates in a year, in most countries of the world. Currently, GRE offers Subject Tests in seven subjects viz. Physics, Biology, Mathematics, Cell and Molecular Biology, Chemistry, Literature in English and Psychology.

Read in Detail at - http://endeavorcareers.com/gurus-speak/734-what-is-gre-exam-format-and-test-structure

1 Month to NMAT - Preparation Strategy

NMAT 2016 is conducted by NMIMS for MBA and PGDM/PGDBM programs in Mumbai, Bangalore and Hyderabad.

The test will be conducted in five windows between 6th October, 2015 and 19th December, 2015. A test taker can take the NMAT a maximum of three times. The best of the three will be considered for the merit list made to shortlist the candidates for further admission process.

The test taker will be to choose the order in which the sections would appear before the test starts.

As we can clearly see, that NMAT will require the test taker to answer with greater speed and accuracy. There will be sectional and overall cut-offs for the selection for the second stage of the admission process.

Read More at - http://endeavorcareers.com/gurus-speak/692-1-month-to-nmat-preparation-strategy

90 Days to CAT : What Should be Your Preparation Plan

With the form filling of CAT and other exams nearing an end and the college forms pouring in, 90 days to prepare for CAT 2015 may seem less. But with the right dream and the right direction to work for the dream, 90 days might just be enough.


First things first: Make sure that you have filled the CAT 2015 form. The last date for online registration is 20th September, 2015.

For the test-takers who have started their preparation in the recent past, the first thing we would like to suggest is to take a Mock CAT test. This will help you analyze your strengths and weakness and help you take a call if you are prepared enough for this year’s CAT or the next.

Read More at -http://www.endeavorcareers.com/gurus-speak/687-90-days-to-cat-what-should-be-your-preparation-plan

CAT 2015 - Preparation Strategy for New Pattern

The one thing predictable about CAT is its unpredictability. CAT is characterized with changing patterns and giving the test taker no common ground to work with. This year is no different. The CAT 2015 changes patterns again, but some might say that the sly little CAT is back to its old bag of tricks.

The changes introduced this year along with the changes in the strategy for the test takers are as follows:

1) One day, Two slots:

CAT 2013 was conducted in a span of twenty days with two slots each day. CAT 2014 was conducted on two days (16th November and 22nd November, 2014) two slots each day. CAT 2015 will be conducted on one day, 29th November, 2015, in two slots. This reduces the anxiety and the dependence on rumors. Psychologically, it is best when a test taker goes into the paper with a blank slate. This gives them an opportunity to make their own strategy without any biased opinions. Moreover this brings down the ill-effects of normalization.

Read More at - http://endeavorcareers.com/gurus-speak/683-cat-2015-preparation-strategy-for-new-pattern

CAT 2015 - Important Dates And Major Changes In Exam Pattern

Indian Institute of Management, Ahmedabad (IIM-A) announced the details for Common Admission Test (2015) on 26th July 2015.

Important Dates 

• Registration period:  06th August to 20th September 2015.
• Exam Date: 29th November, 2015 (one day), in two slots (Forenoon & Afternoon Sessions).
• Download of Admit Card: 15th October 2015, 1:00 PM to Exam Day
• Results Date: 2nd Week January

Important Features of CAT 2015

•  Exam will be conducted in 136 cities across India, in 650 test sites.
• Duration of the examination has been increased to 180 minutes instead of 170 minutes. 

There will be 3 sections:

1. ‘Quantitative Aptitude (QA) : 34 questions (60 minutes)
2. ‘Data Interpretation & Logical Reasoning (DILR): 32 questions (60 minutes)
3. ‘Verbal and Reading Comprehension (VRC) : 34 questions (60 minutes)

• Some questions may not carry multiple choices and will have to be typed on the screen.             Also, use of basic on-screen calculator for computation will be allowed. 
• One will not be able to shift among sections.
• Format of the examination will be available on the CAT website from 15th October 2015.
• The Normalization process to be implemented shall adjust for location and scale differences of score distributions across different forms. After normalization across different forms the scores shall be further normalized across different sections and the scaled scores obtained by this process shall be converted into percentiles for purposes of short listing.

Eligibility Criteria

• The candidate must hold a Bachelor’s Degree, with at least 50% marks or equivalent CGPA
• For the candidates  appearing in the final year of bachelor’s degree/equivalent qualification      examination and those who have completed degree, the last date of submission for programme certificate is June 30, 2016.
• IIMs may verify eligibility at various stages of the selection process, the details of which are     provided at the website www.iimcat.ac.in

Contact Information

Website: www.iimcat.ac.in
Email ID: cat2015@iimahd.ernet.in
Help Desk Number: 18002660207 (Monday to Saturday 09:00AM to 06:00PM)

NMAT by GMAC 2016 Notification and Important Dates

NMAT would be conducted by Graduate Management Admission Council from the academic year 2016-18.

Various institutes like NMIMS University, Alliance University, SRM University, BML Munjal University, VIT University, ICFAI University, Ansal University, Mody University, NIIT University, Thapar University, Shoolini University and Woxsen School of Business will be accepting NMAT scores from now onwards.

A candidate can take 3 attempts in the given admission cycle. The test window would be open for 75 days starting from October 06 to December 19, 2015.

The exam will be administered at 27 test centers across 18 cities within India.

The admission process includes 3 stages:


Stage 1: NMAT by GMAC exam

Registration date: 2nd July – 9th October 2015

Test window: 6th October – 19th December 2015

Result date: 15th January 2016

Registration link: www.nmatbygmac.formistry.com

Exam format:

Note: There will not be any negative marking.

Exam Features:

• Test takers can alter the sections.
• Each sections is individually timed and the test takers need to adhere to the timeline of respective sections.
• Incase if the test takes complete the section before the allotted time, he/she may revise the section or go to the next section.

Stage 2 – The exam will be followed by Case Discussion and Personal Interview for the shortlisted students.

Stage 3 – Candidates’ academic performance, relevant work experience and other similar inputs  would play a role in the final selection criteria.

NMAT Support Details:

    Website:  www.nmat.org.in

    Phone Support: 09213027791/09213027792 – Monday to Friday (except public holidays - 9:00 AM to 6:00 PM IST)

    Email Support: nmatsupport@gmac.com

Also Read:

 NMAT 2016 Important Dates, Pattern and 2015 Paper Analysis