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help put that puzzle together. - Melody Thorntonzz

Book 4

Table of Contents Puzzle #01: Arshveen and her new motorbike ......................................... 1

Puzzle #02: Christopher's Birthday Gift .................................................... 2

Puzzle #03: Silly Summer Games .............................................................. 3

Puzzle #04: An Eggy Situation ................................................................... 4

Puzzle #05: The Cricket-Girls ..................................................................... 5

Puzzle #06: Excuse me, I want some Toffees ............................................ 6

Puzzle #07: Hey Dude, What's the Time? ................................................. 7

Puzzle #08: Puzzling the Prisoners ............................................................ 8

Puzzle #09: A Plan to Wipe Out a Kingdom .............................................. 9

Puzzle #10: By God's Grace ..................................................................... 10

Puzzle #11: Breaking In Exact Fraction ................................................... 11

Puzzle #12: Trading in Animals ............................................................... 12

Puzzle #13: The Quirky Farmer ............................................................... 13

Puzzle #14: The Game with too many coincidences ............................... 14

Puzzle #15: Two Poles Apart ................................................................... 15

Puzzle #16: The Old Postman .................................................................. 16

Puzzle #17: Walking along the beach ..................................................... 17

Puzzle #18: Designing the Die ................................................................. 18

Puzzle #19 : Marketing in Rural MP ........................................................ 19

Puzzle #20 : Jamie and the See-saw ........................................................ 20

Puzzle #21 : The Squared Army .............................................................. 21

Puzzle #22 : It pays to Bargain ................................................................ 22

Puzzle #23 : Generous Mr. Luis ............................................................... 23

The LEADER in Online Test Prep

Puzzle #24 : The Tired Ant....................................................................... 24

Puzzle #25 : The Faulty Clocks................................................................. 25

Puzzle #26 : The Weird Scholastic Tradition ........................................... 26

Puzzle #27 : The Mixed Up Caps ............................................................. 27

Puzzle #28 : Curious Little Keith .............................................................. 28

Puzzle #29 : Mystery of the Missing Money ........................................... 29

Puzzle #30 : The L-Shaped Table at Mrs. Shah's Home .......................... 30

ANSWERS ................................................................................................ 31

Puzzle #01: Arshveen and her new motorbike ....................................... 32

Puzzle #02: Christopher's Birthday Gift .................................................. 33

Puzzle #03: Silly Summer Games ............................................................ 34

Puzzle #04: An Eggy Situation ................................................................. 34

Puzzle #05: The Cricket-Girls ................................................................... 35

Puzzle #06: Excuse me, I want some Toffees .......................................... 36

Puzzle #07: Hey Dude, What's the Time? ............................................... 36

Puzzle #08: Puzzling the Prisoners .......................................................... 37

Puzzle #09: A Plan to Wipe Out a Kingdom ............................................ 38

Puzzle #10: By God's Grace ..................................................................... 39

Puzzle #11: Breaking In Exact Fraction ................................................... 40

Puzzle #12: Trading in Animals ............................................................... 40

Puzzle #13: The Quirky Farmer ............................................................... 40

Puzzle #14: The Game with too many coincidences ............................... 41

Puzzle #15: Two Poles Apart ................................................................... 41

Puzzle #16: The Old Postman .................................................................. 41


Puzzle #17: Walking along the beach ..................................................... 42

Puzzle #18: Designing the Die ................................................................. 43

Puzzle #19 : Marketing in Rural MP ........................................................ 43

Puzzle #20 : Jamie and the See-saw ........................................................ 43

Puzzle #21 : The Squared Army .............................................................. 44

Puzzle #22 : It pays to Bargain ................................................................ 45

Puzzle #23 : Generous Mr. Luis ............................................................... 45

Puzzle #24 : The Tired Ant....................................................................... 46

Puzzle #25 : The Faulty Clocks................................................................. 47

Puzzle #26 : The Weird Scholastic Tradition ........................................... 48

Puzzle #27 : The Mixed Up Caps ............................................................. 48

Puzzle #28 : Curious Little Keith .............................................................. 49

Puzzle #29 : Mystery of the Missing Money ........................................... 49

Puzzle #30 : The L-Shaped Table at Mrs. Shah's Home .......................... 50

About TestFunda.com ............................................................................. 52

PREFACE

For the past couple of years, CAT and other MBA entrance exams have

shown a trend towards questions testing a student’s ability to apply

Mathematical Principles and Analytical Reasoning to solve problems.

The unpredictable nature of CAT has ensured that most students are

never fully prepared to ace the exam. This is because students limit their

preparation to just the learning and practice of core concepts of

Mathematics, Verbal Ability and Data Interpretation & Logical

Reasoning. However, to bell the CAT, divergent thinking is required

which is why experts also recommend solving an eclectic mix of Puzzles,

Crosswords, Riddles and Brain Teasers. These enhance the problem-

solving skills of CAT aspirants and encourage them to think out-of-the-

box.

We, at TestFunda.com, feel that solving puzzles not only helps sharpen

one’s logical acumen, but also gives immense pleasure and satisfaction.

The puzzles in this book will give students that extra edge and

confidence needed to be ready for any surprise that CAT might throw

their way.

We are sure that our readers will benefit greatly from these books. They

shall provide a much-needed break from long study hours as well as

high-quality cerebral recreation.

1 © http://www.TestFunda.com


Book 2 of TestFunda Puzzles

Puzzle #01: Arshveen and her new motorbike

On the occasion of Women's Day on March 8th, Rutuja bought herself a motorcycle for Rs. 60,000. The motorcycle gave her excellent value for money with its mileage. Last week, she decided to go to Pune for the weekend with her roommate, Purbasha. On their way, she noticed that while travelling downhill, she went at 72 kmph (kilometers per hour). While driving on level ground, she did so at 63 kmph. She travelled uphill on the motorcycle at only 56 kmph. It took her 4 hours to travel from Mumbai to Pune. The return trip took her 4 hours and 40 minutes. Using this information, you need to calculate the distance between the two cities. Provide a detailed explanation along with your answer.




Puzzle #02: Christopher's Birthday Gift

Christopher was a fan of Rock music. For his birthday, Priyanka gifted

Christopher the latest album of his favorite band. The album consisted of three

CD-ROMs. Each CD played for a different time. There were four songs on each

CD. Each CD had a total playing time of less than an hour. Each of the 4 songs

on all the three CDROMs lasted for time period that was a prime number of

minutes. Also, on a particular CDROM, each song was of a different length. If

you take a combination of three different songs on all the CDs, they play for a

prime number of minutes.

Using this information, determine the lengths of the songs on each of these

three CDs. Provide a detailed explanation along with your answer.




Puzzle #03: Silly Summer Games

While passing time post exams, Vimble decided to play a game with his

brother, Vible. The rules of the game were as follows:

1. One of the players would lay a bet on a number between 1 and 6 (both

numbers included.

2. The player would then roll, in succession, 3 dice.

3. If he won, he would get back his bet, of course. In addition to that, he would

get 1, 2 or 3 times the amount that he had originally bet. That additional

amount would be decided on how many times the player managed to roll out

his number using the dice. Of course, if it didn't come up at all, he got nothing.

4. The sum of the numbers on the dice wouldn't be considered. He would need

to roll out his number on each die separately.

Using this data, you need to deduce the expected gain or loss per game if

Vimble/Vible bet Re. 1. Provide a detailed explanation along with your answer.




Puzzle #04: An Eggy Situation

I made some chocolate Easter eggs to sell. I had made 10 baskets with eggs

ranging from 10 - 20 in each basket. Each egg weighed 50 grams. However, my

assistant mixed up a different batch of eggs that weighed 60 grams each along

with the eggs in these baskets. The 50 and 60 g eggs can't be distinguished

from each other by looking at them. I have a one empty basket and one set of

scales. And, I have 9 baskets with eggs that weigh 50 g each and one basket

with eggs that weight 60 g each. The client who had placed an order for 10 eggs

is at the door. I have enough time to weigh just once using the scales. How do I

identify the incorrect basket and have it replaced in just one weighing? Provide

a detailed explanation along with your answer.




Puzzle #05: The Cricket-Girls

Four friends, Anisha, Bella, Candice and Damyanti, lived in London a few years

ago. They once played four games of cricket. Each girl batted just once during a

round.

At the end of the day's play, all their cumulative scores were tied. However, the

individual scores of the 4 girls for the 16 innings were all different.

The data given is :

1. All the 16 scores were between 60 and 80.

2. All four of Anisha's scores were prime numbers.

3. All four of Bella's scores were semiprimes. (A semiprime is defined as the

product of two primes).

4. None of Candice's or Damyanti's scores was either prime or semiprime.

5. Candice's lowest round was better (lower)than Damyanti's lowest round.

Candice's worst round was better than Damyanti's worst round. (In these

games, the one to score lower would have played better.)

Using this data, find the scores of each of the girls for the four rounds. Provide

a detailed explanation along with your answer.




Puzzle #06: Excuse me, I want some Toffees

In 1960, Bhushan Singh went to London and started his own toffee shop. He had 3 varieties of toffees: Butterscotch (that cost a Penny each), Caramel (that cost 2 Pennies each) and Fudge (that cost 2 Pence- half Penny each). One day, a small girl named Diana came to his shop and gave him a Crown and said: "I want some Caramel toffees, six times as many Butterscotch toffees and for the remaining amount, give me Fudge toffees. Bhushan for a second was stumpted at the request. Then, he just smiled and handed her the toffees she wanted. How many of each variety of toffees did Bhushan give her? Provide a detailed explanation along with your answer.




Puzzle #07: Hey Dude, What's the Time?

Once upon a time, there was a little boy named Arjun. He was very lazy and

hated studying. Among all his subjects, he disliked mathematics the most. He

had an older brother named Rakesh, who kept trying to tell him to mend his

ways, but Arjun did not heed him. In fact, to teach Rakesh a lesson, he kept

annoying Rakesh by asking him silly questions all day long. One day, Rakesh

decided that it was payback time! As usual, Arjun asked him, "Rakesh, could

you tell me what time it is? I'm hungry and want to know how long will I have

to wait till dinner is served." Dinner used to be served at 10:00 pm in their

home. Rakesh answered, "If you add one fourth of the time that has elapsed

since noon until now to one half the time that it will take to be noon tomorrow,

your time starting now, you will arrive at the current time. Using that, figure

out how long is it until dinner." Arjun was shocked, perplexed, angry and

hungry, all at the same time. But there was nothing he could do about it. Help

him solve Rakesh's riddle by deducing how long it will take until dinner is

served. Provide a detailed explanation along with your answer.




Puzzle #08: Puzzling the Prisoners

On an island inhabited by tribals, 100 Englishmen were held captive. One day the leader of the tribe decided that the deserved one chance to save themselves. He ordered for 100 wooden boxes and placed into each of them a name of one of the 100 prisoners. The boxes were then lined up on a table in one of the huts.

The rules were simple:

1. The Englishmen would be led into the hut one by one and one at a time.

2. He was allowed to open at most 50 boxes. After that, he had to shut the boxes and leave the hut and wait in another hut for the remaining men to finish their turns.

3. Once he entered the hut with the boxes, he was not permitted to communicate with any of the others.

4. The Prisoner's can't change the positions of the boxes or disturb the arrangement in any way.

The prisoners are, however, given an opportunity to plan out their strategy a day prior to the day when they have to enter the hut. The stakes for them were very high as each prisoner would HAVE to find a box with his name in it. Failing to do so would mean that all of them would be hanged to death. Help them chalk out a plan that has at least a 30% probability of success. Provide a detailed explanation along with your answer




Puzzle #09: A Plan to Wipe Out a Kingdom

Edisonia was a kingdom right next to the kingdom of Newtonia. The residents

of Edisonia had something peculiar about them. Each individual was born with

a black or purple "beauty" spot on his (or her) forehead. However, as per their

societal laws, they were never supposed to know the colour of the spot on

their heads. In case he/she ever figured it out, the penalty was death and each

citizen was to end his/her own life. Edisonia was a very prosperous kingdom.

Every evening at twilight, all the Edisonians would meet up at a tavern to

discuss the day's happenings.

Jealous of Edisonia's prosperity, the king of Newtonia sent one of his spies to

Edisonia to disrupt life at Edisonia. The spy decided to attend the meeting at

the tavern (un-invited, of course). He, very casually, announced to the

Edisonians them something related to the number of purple spots.

Using this data prove that ultimately, every Edisonian kills himself.




Puzzle #10: By God's Grace

Mario, a local tennis player, prayed to God to help him win the French Open this year. God heard his prayers and with God's blessings, he reached the finals of the French Open. God, however, decided it was unfair on HIS part to help Mario win the final against Rafael Nadal. So, HE told Mario to choose a score where Mario would feel confident that he would be able to win th game from that point onwards, without God's help and God will ensure that it happens. Help Mario choose the score in a way as to maximise his chances of winning. Provide a detailed explanation along with your answer.




Puzzle #11: Breaking In Exact Fraction

Clinton went to London on a holiday last year. When he returned from his trip, he gave his brother Carsten three coins — One sovereign, one shilling, and one penny. On examining the coins, Carsten saw that each of the coins was chipped. Interestingly, exactly the same fraction of each coin had been chipped off. Using this data, you need to deduce what proportion of each coin had been lost if the value of the three remaining fragments is exactly one pound. Here we assume that the original intrinsic value of each coin is the same as its face value. For example, the shilling worth a shilling, and so on. Provide a detailed explanation along with your answer.




Puzzle #12: Trading in Animals

Three farmers - Darpan, Gautam and Jayanti - once assembled at the weekly village market. They started eyeing each other's animals that they had brought along with them to sell in the village market. Said Gautam to Jayanti, "If I give you six of my chickens for one of your cows, you'll have twice the number of animals that I've got." Hearing that Darpan got an idea and said to Gautam, "If that's how you people to trade animals, here's a deal: I'll give you fourteen of my goats for a cow. In this case, you'll have three times the number of animals that I have." Jayanti didn't want to remain behind and immediately butted in with a plan of his own. Said Jayanti to Darpan; "I'll give you four donkeys for a cow, and then you'll have six times as many animals as I've got here." Using this information, deduce the number of animals owned by the three of them together. Provide an explanation along with your answer.




Puzzle #13: The Quirky Farmer

Uncle Sam was a farmer. When he was young, he was really good at math and

loved solving puzzles, especially those related to arithmetic. Last week, when

he went to Church, he met his neighbour’s wife, Martha, on the way back. The

poor, unsuspecting woman had no idea that he loved math and asked him a

very simple question. "How many goats are left on your farm?" she said. He

decided to confound her. All he said as, "I can divide my goats into two parts

(not necessarily equal) such that the difference between the numbers of the

two parts is equal to the difference between their squares. Have a nice day!"

And, he walked of chuckling to himself. Help poor Martha figure this one out.

Provide an explanation along with your answer.




Puzzle #14: The Game with too many coincidences

Seven friends, Arjun, Benny, Carl, Danny, Edwin, Firoz, and Gopi, once played a little game. One of the rules of the game was that whenever a player won a game he should double the money of each of the other players; i.e. he would have to give the other players just as much money as they had already in their pockets. On that particular day, the seven friends played seven games, and, miraculously, each of them won a game in turn, that too in the order in which their names are given (What are the chances???). But, the biggest coincidence of them all was that at the end of the 7 games, each of them had exactly the same amount — two shillings and eight pence — in his pocket. From this data, you need to deduce how much money each man had with him before he sat down to play. Provide an explanation along with your answer.




Puzzle #15: Two Poles Apart

Last week, Radha, my neighbor's maid decided to dry out their clothes on the

building terrace. On the terrace there were 2 poles. The taller pole was 7 feet

above the ground, while the shorter one was five feet tall. She tied a clothes

line from the top of each of two poles to the base of the other. Using this data,

determine the height (in feet and inches) from the ground where the two cords

crossed one another. Provide an explanation along with your answer.




Puzzle #16: The Old Postman

Fr. William, during his sermon, often spoke of an old postman who travelled by

foot to deliver letters to people in the village. One of his friends liven atop a

hill. Ramlal (the postman) would travel uphill from the post office to the house

at a speed of a mile and a half an hour. On delivering the letter, he would

traverse downhill at a speed of four and a half miles an hour. Despite his old

age, he would take just six hours to make the double journey. Using this data,

determine the distance between the post office and the house on top of the

hill. Provide an explanation along with your answer.




Puzzle #17: Walking along the beach

Last week, Kareena and Saif went to Gorai beach. After walking for a long time,

Kareena observed that along the beach, there were a few poles equidistant

from each other. The poles had been numbered 1, 2, 3, 4, and so on. Kareena

walked from the first pole to the last one and back. At the same, Saif was doing

that in the opposite direction. Both Saif and Kareena started moving at the

same time. Both of them walk with a constant velocity. Also, both of them need

not have moved with the same velocity0. They first encountered each other at

pole number 10 and, later, their second encounter (when they were both on

the way back) took place at pole number 20. Using this data, you need to

determine how many poles are standing along the beach. Provide an

explanation along with your answer.




Puzzle #18: Designing the Die

Boney's school teacher was annoyed with her incessant questioning. So, to

keep her busy for a while, she gave her a small cube and told her to convert

that cube into a die in as many unique ways as possible. The only constraint she

put on that was that the numbers "1 & 6", "2 & 5" and "3 & 4" should always

be on opposite sides? Using this information, deduce the number of ways in

which Boney can do this. Provide a detailed explanation along with your

answer.




Puzzle #19 : Marketing in Rural MP

In a small village in rural Madhya Pradesh, there lived four married couples.

Last Monday, owing to the weekly sale, they decided to buy some stuff to take

home. However, together they has only possessed forty 10 Rupee notes. The

fact is, Anya spent Rs. 100., Maya spent Rs. 200, Jamuni spent Rs. 300., and

Karisma spent Rs. 400. Their husbands, however, tracked down their wives

were rather more extravagant than their wives, for Nikhil Sharma spent as

much as his wife, Tarun Bhatia twice as much as his wife, Billu Jhanwar three

times as much as his wife, and Jaspinder Birla four times as much as his wife.

Once they were done, their friend, Bhavik said that they should divide what

notes they had left equally among them. Using this data find out the surname

of each woman? Also, if possible, pair off the four couples. Provide a detailed

explanation along with your answer.




Puzzle #20 : Jamie and the See-saw

A few years ago, on a bright shiny day, I happened to see fat, little six year old Jamie attmept to play see-saw. Unfortunately, kids his age weren't as well-built as he was. So, he couldn't find someone to balance him on the other side. So, he hit upon a small, yet clever idea to play the game. He kept tying together and then attaching bricks to the other end of the see-saw, assuming he would sit at the other end. When he started tying the bricks to the longer end of the see saw, he noticed that he needed 11 bricks to balance himself. Whereas when he fixed the bricks to the see saw's shorter end, he needed as, any as 16 bricks to do the needful. The weight of a single brick is equal to the weight of three quarters of a brick and 750 grams. Using this, deduce little Jamie's weight. (Round off the weight to the nearest integer.) Provide a detailed explanation along with your answer.




Puzzle #21 : The Squared Army

Saurabh, the King of Pervenia, loved the concept of squares in mathematics.

Once, when defending his kingdom against the King of Thievenia, he sent into

battle a huge army. The peculiarity of the army was that it could be formed into

two perfect squares in twelve different ways. Using this data, deduce the

smallest possible number of members in that army. Provide a detailed

explanation along with your answer. (For example, if there were 130 members

in the king's army, the army could be split into into two squares in only two

different ways — 81 and 49, or 121 and 9. Needless to say, every member

should be used on every occasion.)




Puzzle #22: It pays to Bargain An old Indian aunty went grocery shopping once in the UK. On her way, she saw a man selling apples. He offered her some apples for a shilling. However, she argued that they were small and demanded more apples. After a heated argument, he threw in two more apples. After all the time and effort she put in while arguing, she realised (on going home) that at the end of the day. she made the apples cost merely a penny a dozen less than what the man had demanded in at the beginning. Using this data, deduce how many apples she got for her shilling. Provide an explanation along with your answer.




Puzzle #23: Generous Mr. Luis Old Mr. Joseph Arthur Luis was known for his generosity. One evening, when returning home from church, he came across three urchins on the street. Each of them, in succession, appealed to him for alms as they were hungry. To the first urchin, he gave one penny more than half the money he had in his pocket. The second urchin received two pence more than half the money he then had in his pocket. The last urchin got three pence more than half of what he had left. When he reached home, he had only one penny left in his pocket, which he gave to his grand-daughter. Using this data, determine how much money he had on him when he set out homeward from church. Provide an explanation along with your answer.




Puzzle #24 : The Tired Ant

Anu, the ant, decided to take a stroll in a square shaped field. He reached a point and decided that he had walked enough. This point is located 13 meters from one of the field's corners, 17 meters from the corner that is diagonally opposite that one, and 20 meters from a third corner post. Using this data, deduce the area of the field. (Assumption: The land is flat.) Provide a detailed explanation along with your answer.




Puzzle #25 : The Faulty Clocks On 31st March 1898, Kevin was on his way home as it was his birthday on the 1st of April. However, due to the infamous train delays, he got stranded at the railway station. At midnight, he marvelled at the fact that the 3 clocks at the station showed precisely the same time, (12:00). The following day, again he was stranded at the same place and at the same time. However, he noticed that the clock in the centre had kept the perfect time, while to clock to its left had lost exactly one minute and the clock to its right had gained exactly one minute. Using this data, answer Kevin's little query: on what date and at what time of day would all three pairs of hands again point at the same moment at twelve o'clock? The date needs to be exact - not a day before, not a day after. You can assume that the railway authorities are oblivious to the fact that the other clocks are going slow/fast and also assume that the clocks maintained the same rates of progress without stopping. Provide a detailed explanation along with your answer




Puzzle #26: The Weird Scholastic Tradition

After his father got transferred to Japan, Deepak was forced to enrol in a school there. In that school, they had a weird custom. Being a co-ed school, all the boys and girls had to assemble in the school grounds by 7 am along with the principal. Every boy had to bow to every other boy, to every girl and to the principal of the school. Post that, every girl would bow to every other girl, to every boy and to the principal. After all the bows, they'd go to class and the lectures would begin. In that school, there were half as many boys as girls. In all, there would be nine hundred bows made in every morning. Using this data, deduce the number of boys that were there in the school. Provide a detailed explanation along with your answer.




Puzzle #27: The Mixed Up Caps Last week, I had lunch with 7 of my friends from college. Due to the sweltering summer heat, each one of us arrived at the restaurant wearing a cap. However, post lunch, when we reached our respective homes, we realised that each one of us was wearing a cap different from what we had entered with. (In totality, the eight caps we left with were the same as the eight caps we entered with and there were no exchanges with others who had lunch in the same restaurant.) According to Atul, there were 40,320 different ways in which we could have picked the caps (randomly). This was easily obtained by multiplying the numbers 1 to 8. To this, a puzzle lever, Saurabh, responded saying that all we had to do was find out in how many of these cases did no one wear the same cap as he had entered with. People immediately decided that this math was too tedious to get involved in and, thus, gave up trying to answer Saurabh's question. Using the given data, try to answer Saurabh's question. Provide an explanation along with your answer.




Puzzle #28 : Curious Little Keith

Last week, little Keith wanted to know the age of his parents. So, he asked his mom her age. She very smartly (so as to not reveal her age) replied that all three of their ages (hers, her husband's and Keith's) added up to 70 years. Keith then decided to pester his father. So he walked up to his father and asked him his age. To which, his father replied, "I'm just 6 times as old as you." Keith still wasn't satisfied with the information he got. So he asked his father again,

"Dad, will I ever be half as old as you?" To which his dad replied in the affirmative, saying that when that happens the sum of their ages would be exactly twice they were on that day. Before Keith could ask anyone any more questions, his mom came along and packed him off to bed. Using the given data, deduce the ages of Keith and his parents on that day. Provide an explanation along with your answer.




Puzzle #29 : Mystery of the Missing Money

Ronen & his friend Anil opened shops next to each other. Both of them sold eggs in their respective shops. Ronen sold his eggs at the rate of 3 eggs for Rs. 10 whereas Anil sold eggs at 2 eggs for Rs. 10. One evening, they both had 30 unsold eggs each. They decided to go out for dinner and, hence, handed the 30 eggs each to a friend and asked her to sell the eggs at 5 eggs for Rs. 20. (As per Ronen's calculations, 3 for Rs. 10 and 2 for Rs. 10 was synonymous with 5 for Rs. 20).

The next morning, they expected to get Rs. 250 as they would have earned had they sold their eggs separately. However, when their friend handed them their money, they were shocked to know that they had got Rs. 240 instead. Using this data, determine where they had gone wrong. Provide a detailed explanation along with your answer.




Puzzle #30 : The L-Shaped Table at Mrs. Shah's Home

Mr. and Mrs. Shah invited four couples (all married) to their home for Diwali. In

their dining room, they had a long L-shaped table (as shown in the figure) and

the chairs were placed around the table (also depicted in the figure).

Mrs. Shah ensured that the seating arrangements were such that:

1. Every lady sat next to her husband.

2. Every man sat directly across a lady.

3. Mrs. Shah sat to her husband's right.

4. Mrs. Shah was the only woman seated between 2 men.

Using this data, you need to deduce in which chair exactly Mrs. Shah sat (Chairs

- a, b, c, d, e, f, g, h, i, or j). Provide a detailed explanation along with your

answer.




ANSWERS




Puzzle #01: Arshveen and her new motorbike

273 kmph.

Let the total distance travelled downhill, on the level, and uphill, on the

outbound journey, be x, y, and z, respectively.

The time taken to travel a distance s at speed v is s/v.

Hence, for the outbound journey

x/72 + y/63 + z/56 = 4

While for the return journey, which we assume to be along the same roads

x/56 + y/63 + z/72 = 14/3

We need to calculate the value of x + y + z.

Multiplying both equations by the least common multiple of denominators 56,

63, and 72, we obtain

7x + 8y + 9z = 4 * 7 * 8 * 9

9x + 8y + 7z = (14/3) * 7 * 8 * 9

Now it is clear that we should add the equations, yielding

16(x + y + z) = (26/3) * 7 * 8 * 9

Therefore x + y + z = 273; the distance between the two towns is 273

kilometers.

Discuss the solution with TestFunda users.

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Puzzle #02: Christopher's Birthday Gift

The lengths of the tracks are (5, 7, 17, 19), (7, 11, 13, 23) and (7, 13, 17, 23).

None of the track lengths can be 2 because any sum of the three different

primes, including 2, would not be prime. Therefore, the track lengths must be

chosen from the 13 primes between 3 and 43.

To get the answer, find all three prime combinations whose sum is a small

enough prime that another prime (larger than any of the three) can be added

to result in a sum of 60 or less.

There are 14 such combinations: (3, 5, 11), (3, 5, 23), (3, 7, 13), (3, 7, 19), (3, 11,

17), (5, 7, 11), (5, 7, 19), (5, 11, 13), (5, 13, 19), (7, 11, 13), (7, 11, 19), (7, 13,

19), (7, 13, 17) and (11, 13, 17).

If you try all possible fourth primes with those combinations, you'll find that

the only possibilities which meet the requirements (sum of 60 or less and sum

of any three is a prime) are the answers given above.


http://testfunda.com/examprep/learningresources/pod/puzzle-152--christophers-birthday-gift.htm?assetid=ed25ad12-0c6b-484c-9ca2-3ddd1ed76737




Puzzle #03: Silly Summer Games

The player would have an expected loss of 7.87 paise per game.

There are 216 (6³) possible dice throw permutations. If the player always bets

on the same number, let's say 1 and then there are four possible distributions

of 1 and "not 1." Their probabilities are like this, where x stands for "not 1."

x x x 125/216

1 x x 75/216

1 1 x 15/216

1 1 1 1/216

That leads to:

E = ((0)125 + (1 + 1)75 + (1 + 2)15 + (1 + 3)1) / 216 - 1 = -17/216 = -Re. 0.0787.


Puzzle #04: An Eggy Situation

Number Each Basket from 1 to 10.

Select 1 egg from basket 1

Select 2 eggs from basket 2

Select 3 eggs from basket 3

etc...

You will now have (1+2+...+10) = 55 eggs

If they all weighed 50 g, then the scales would show 55x50 = 2750 g

But if the weight was, say, 2800 g, then you would know that there 5 eggs that

weigh 10 g extra. So the basket with the heavier eggs must be basket 5.


http://testfunda.com/examprep/learningresources/pod/puzzle-153--silly-summer-games.htm?assetid=ba2d0e8b-64cc-4c84-a15d-670da08250d3

http://testfunda.com/examprep/learningresources/pod/puzzle-154--an-eggy-situation.htm?assetid=3e1fc7d3-e88b-4225-a60f-62961f3225a0




Puzzle #05: The Cricket-Girls

Anisha's scores for the four rounds were 61, 67, 71 and 70.

Bella's scores were 62, 65, 74 and 77.

Candice's scores were 63, 68, 72 and 75.

Damyanti's scores were 64, 66, 70 and 78.

Here is one solution:

The 20 possible scores are the integers 60 through 79. Of those, 61, 67, 71, 73

and 79 are prime, and 62, 65, 69, 74 and 77 are semiprime.

Anisha scores must be four of the five primes. Bella's scores must be four of the

five semiprimes. Because the primes add up to 351, and the semiprimes add up

to 347,

the unused prime must be four more than the unused semiprime. Therefore,

the numbers not used are 73 and 69 and each golfer's score for the four rounds

was 278.

Continuing, the sum of the ten possible scores that are not prime and not

semiprime is 692. The sum of the values must be 692 - 2(278) = 136. The only

ways that 136

can be eliminated are to not use (60, 76), (64, 72), or (66, 70). Since there are

only two odd numbers among the possible scores for Candice and Damyanti,

they must

have been scored by the same golfer.

From this point, trial and error will show that the only possible solution is the

one stated above.


http://testfunda.com/examprep/learningresources/pod/puzzle-155--the-cricketgirls.htm?assetid=0eea8cb5-11c6-475c-929f-153f6740ebf2




Puzzle #06: Excuse me, I want some Toffees

Bhushan supplied 5 Caramel, 30 Butterscotch, and 8 Fudge toffees. This distribution exactly fulfils the conditions and represents a cost of five shillings (one crown).


Puzzle #07: Hey Dude, What's the Time? 24 minutes until dinner. The time must have been 9.36 p.m. A quarter of the

time since noon is 2 hrs and 24 mins., and a half of the time till noon next day is

7 hrs. and 12 mins. These added together make 9 hrs. and 36 mins. From here,

it's 24 minutes to 10:00 pm.


http://testfunda.com/examprep/learningresources/pod/puzzle-156--excuse-me-i-want-some-toffees.htm?assetid=976ca531-c882-4042-aca8-7f8b89f044c5

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Puzzle #08: Puzzling the Prisoners

In order to solve the puzzle, all the prisoners must first agree on a random

labeling of the boxes by their own names.

(The point of making it random is that it makes it impossible for the warden to

place names in boxes in such a way as to foil the protocol described next.)

When admitted to the room, each prisoner inspects his own box (that is, the

box with which his own name has been associated). He then looks into the box

belonging to the name he just found, and then into the box belonging to the

name he found in the second box, etc. until he either finds his own name, or

has opened 50 boxes. That’s the strategy.

Explanation: The process which assigns to a box’s owner the name found in his

box is a permutation of the 100 names, chosen uniformly at random from the

set of all such permutations. Each prisoner is following a cycle of the

permutation, beginning with his box and (if he doesn’t run over the 50-box

limit) ending with his name on a piece of paper. If it happens that the

permutation has no cycles of length greater than 50, this process will work

every time and the prisoners will be spared.

In fact, the probability that a uniformly random permutation of the numbers

from 1 to 2n contains no cycle of length greater than n is at least 1 minus the

natural logarithm of 2—about 30.6853%.

To see this, let k > n and count the permutations having a cycle C of length

exactly k. There are (2n k) ways to pick the entries in C, (k-1)! ways to order

them, and (2n-k)! ways to permute the rest; the product of these numbers is

(2n)!/k. Since at most one k-cycle can exist in a given permutation, the

probability that there is one is exactly 1/k.

It follows that the probability that there is no long cycle is

1 - 1/n+1 - 1/n+2 - ... - 1/2n = 1 - H2n + Hn where Hm is the sum of the

reciprocals of the first m positive integers, approximately ln m. Thus our

probability is about 1 - ln 2n + ln n = 1 - ln 2, and in fact is always a bit larger.

For n = 50, we get that the prisoners survive with probability 31.1827821%.


http://testfunda.com/examprep/learningresources/pod/puzzle-158--puzzling-the-prisoners.htm?assetid=5ee2bc77-5e4f-424c-8b75-b2f962e16b51




Puzzle #09: A Plan to Wipe Out a Kingdom

The proof uses backwards induction on the number of disallowed numbers of purple spots. In the base case, there are n−1 such numbers; where n is the population of Edisonia. In that case, everyone can immediately deduce his own spot-color, so the kingdom is wiped out immediately.

Say there are n residents and the stranger announces that the number of purple spots does not belong to the set K, where K is some non-empty subset of the numbers from 0 to n.

Suppose the actual number of purple spots is j, so that the black-spotted residents see j purple spots and the others see only j-1. If j−1 is in K then the purple-spotted residents will all kill themselves the first night. The remaining residents (if any) will deduce that their spots are black, and dispense with themselves the following night.

If j+1 is in K then the black-spotted residents will all kill themselves the first night, and the remaining residents will follow suit the next night after deducing that their spots are purple. If no one kills himself the first night, everyone can deduce that neither j−1 nor j+1 is in K, i.e. the actual number of purple spots is not within one of any number in K. This increases the number of forbidden numbers of purple spots, and the induction hypothesis can now be applied.

Note that the proof shows that n nights suffice; further, that the full n nights are needed only when K = {0} or {n} and either all spots are the same color or n <= 2.


http://testfunda.com/examprep/learningresources/pod/puzzle-159--a-plan-to-wipe-out-a-kingdom.htm?assetid=45a84201-0816-431f-9104-52920d20ffe0




Puzzle #10: By God's Grace

It sounds obvious that Mario should ask to be ahead two sets to love (it takes 3

out of 5 sets to win the men’s), and in the third set, ahead 5-0 in games and 40-

love in the sixth game. (Probably Mario may want to serve, but he might prefer

Roger to be serving the sixth game down 0-40 so that he can pray for a double

fault.)

These solutions, however, give him essentially 3 chances to get lucky and win,

but he can get six chances—with three services by Mario and three by Rafael.

He still may want to be up two sets to none, but let the game score be 6-6 in

the third set and 6-0—in his favor, of course—in the tiebreaker.

Based on the idea that traditionally the complete score of a tennis match

includes the game scores of all sets and, if the game score was 6-6, the

tiebreaker score as well. Then Mario could ask for example that the score be 6-

0, 6-6 (9999-9997), 6-6 (6-0). The theory here (dubious, granted) is that while

you God was helping Mario,

Rafael was wearing himself out in the second-set tiebreaker and is now more

likely to blow one of the six upcoming match points.


http://testfunda.com/examprep/learningresources/pod/puzzle-160--by-gods-grace.htm?assetid=1df2114e-7ba7-4d4d-a5c9-71072f42fcf2




Puzzle #11: Breaking In Exact Fraction

If the three broken coins when perfect were worth 253 pence, and are now in

their broken condition worth 240 pence, it should be obvious that 13/253 of

the original value has been lost. And as the same fraction of each coin has been

broken away, each coin has lost 13/253 of its original bulk.


Puzzle #12: Trading in Animals

Jayanti must have taken 7 animals to market, Gautam must have taken 11, and

Darpan must have taken 21. There were thus 39 animals altogether.


Puzzle #13: The Quirky Farmer

Sam had one goat only! If he divided this goat (which is best done by weight)

into two parts, making one part two-thirds and the other part one-third, then

the difference between these two numbers is the same as the difference

between their squares—that is, one-third. Any two fractions will do if the

denominator equals the sum of the two numerators.


http://testfunda.com/examprep/learningresources/pod/puzzle-161--breaking-in-exact-fraction.htm?assetid=54f688b9-a99c-4b0c-a6af-4a0e178336f9

http://testfunda.com/examprep/learningresources/pod/puzzle-162--trading-in-animals.htm?assetid=23d618f0-c3c8-4e14-bb5e-b4b28b247276

http://testfunda.com/examprep/learningresources/pod/puzzle-163--the-quirky-farmer.htm?assetid=255e4af7-40a2-404f-b9da-ed6c83f1cdfc




Puzzle #14: The Game with too many coincidences

Puzzles of this class are generally solved in the old books by the tedious process

of "working backwards." But a simple general solution is as follows: If there are

n players, the amount held by every player at the end will be m(2n), the last

winner must have held m(n+1) at the start, the next m(2n+1), the next

m(4n+1), the next m(8n+1), and so on to the first player, who must have held

m(2n-1n+1).

Thus, in this case, n = 7, and the amount held by every player at the end was 27

farthings. Therefore m = 1, and G started with 8 farthings, F with 15, E with 29,

D with 57, C with 113, B with 225, and A with 449 farthings.


Puzzle #15: Two Poles Apart

Multiply together, and also add together, the heights of the two poles and

divide one result by the other. That is, if the two heights are a and b

respectively, then ab/(a + b) will give the height of the intersection. In the

particular case of our puzzle, the intersection was therefore 2 ft. 11 in. from the

ground. The distance that the poles are apart does not affect the answer.


Puzzle #16: The Old Postman

The distance must be 6.75 miles.


http://testfunda.com/examprep/learningresources/pod/puzzle-164--the-game-with-too-many-coincidences.htm?assetid=d9e26a56-67ca-4f35-a8ef-426bef5613a6

http://testfunda.com/examprep/learningresources/pod/puzzle-165--two-poles-apart.htm?assetid=880eb042-29c6-40e2-b11c-b02a4999146b

http://testfunda.com/examprep/learningresources/pod/puzzle-166--the-old-postman.htm?assetid=92576963-ebc4-415a-8e26-99684772747f




Puzzle #17: Walking along the beach The distance between the poles is not important; take it for example 100 m. Let

us call the distance between the first and the last pole D.

At the first encounter, Kareena and Saif have together walked a distance D. At

the second encounter (as one can see by making a sketch) they have together

walked 3 times as much, so 3D. At the first encounter Kareena has walked 900

m, so at the second encounter she has walked 2700 m. Then she still has to go

back 1900 m to her start at pole 1. So in total Kareena walked 4600 m. From

this it follows that D = 2300 m, so there are 24 poles along the beach.

This puzzle can also be solved in a quite different less elegant way. Let us call

the velocity of Kareena v_a and that of Saif v_b and

the number of poles x. At the first encounter at time t_1 it holds that:

t_1 = 9 / v_a = (x - 10) / v_b

At the second encounter at time t_2 it holds that:

t_2 = (x - 1) / v_a + (x - 20) / v_a = (x - 1) / v_b + 19 /v_b

These equations can be rewritten into

9 v_b = (x - 10) v_a

(2x - 21) v_b = (x + 18) v_a

If we divide the upper equations by each other and multiply by the

denominators we obtain the following quadratic equation

x^2 - 25 x + 24 = 0.

This equation has as solution x = 1 and x = 24. x = 1 is not a correct solution,

since there are at least 20 poles along the beach. So we

can conclude that there are 24 poles along the beach.


http://testfunda.com/examprep/learningresources/pod/puzzle-167--walking-along-the-beach.htm?assetid=8d17a4f5-b898-4f9b-820c-8ad28da57ad2




Puzzle #18: Designing the Die 48 ways.

The 1 can be marked on any one of six different sides. For every side occupied

by 1 we have a selection of four sides for the 2. For every situation of the 2 we

have two places for the 3. (The 6, 5, and 4 need not be considered, as their

positions are determined by the 1, 2, and 3.) Therefore 6, 4, and 2 multiplied

together make 48 different ways—the correct answer.


Puzzle #19: Marketing in Rural MP As every person's purchase was of the value of an exact number of Rupees, and

as the party possessed when they started out forty Rupee coins altogether,

there was no necessity for any lady to have any smaller change, or any

evidence that they actually had such change. This being so, the only answer

possible is that the women were named respectively Anya Jhanwar, Maya Birla,

Jamuni Sharma, and Karisma Bhatia. It will now be found that there would be

exactly eight Rupees left, which may be divided equally among the eight

persons in coin without any change being required.

So, the names of the couples are as follows:

1. Anya & Billu Jhanwar

2. Maya & Jaspinder Birla

3. Karisma & Tarun Bhatia

4. jamuni & Nikhil Sharma


Puzzle #20 : Jamie and the See-saw

The boy's weight must have been about 40 kgs (39.80). A brick weighed 3 kgs.

Therefore 16 bricks weighed 48 kgs. and 11 bricks 33 kgs. Multiply 48 by 33 and

take the square root. That gives the boy's weight.


http://testfunda.com/examprep/learningresources/pod/puzzle-168--designing-the-die.htm?assetid=2feb07dd-0fd3-4079-8037-607440601028

http://testfunda.com/examprep/learningresources/pod/puzzle-169--marketing-in-rural-mp.htm?assetid=cd6bfd2b-f035-4dcc-9a4c-aaef42c55643

http://testfunda.com/examprep/learningresources/pod/puzzle-170--jamie-and-the-seesaw.htm?assetid=447e1dc3-6095-45b5-9ecd-f9b7a988f714




Puzzle #21 : The Squared Army

The smallest primes of the form 4n + 1 are 5, 13, 17, 29, and 37, and the

smallest of the form 4n - 1 are 3, 7, 11, 19, and 23. Now, primes of the first

form can always be expressed as the sum of two squares, and in only one way.

Thus, 5 = 4 + 1; 13 = 9 + 4; 17 = 16 + 1; 29 = 25 + 4; 37 = 36 + 1. But primes of

the second form can never be expressed as the sum of two squares in any way

whatever.

In order that a number may be expressed as the sum of two squares in several

different ways, it is necessary that it shall be a composite number containing a

certain number of primes of our first form. Thus, 5 or 13 alone can only be so

expressed in one way; but 65, (5 × 13), can be expressed in two ways, 1,105, (5

× 13 × 17), in four ways, 32,045,(5 × 13 × 17 × 29), in eight ways. We thus get

double as many ways for every new factor of this form that we introduce. Note,

however, that I say new factor, for the repetition of factors is subject to

another law. We cannot express 25, (5 × 5), in two ways, but only in one; yet

125, (5 × 5 × 5), can be given in two ways, and so can 625, (5 × 5 × 5 × 5); while

if we take in yet another 5 we can express the number as the sum of two

squares in three different ways.

If a prime of the second form gets into your composite number, then that

number cannot be the sum of two squares. Thus 15, (3 × 5), will not work, nor

will 135, (3 × 3 × 3 × 5); but if we take in an even number of 3's it will work,

because these 3's will themselves form a square number, but you will only get

one solution. Thus, 45, (3 × 3 × 5, or 9 × 5) = 36 + 9. Similarly, the factor 2 may

always occur, or any power of 2, such as 4, 8, 16, 32; but its introduction or

omission will never affect the number of your solutions, except in such a case

as 50, where it doubles a square and therefore gives you the two answers, 49 +

1 and 25 + 25.

Now, directly a number is decomposed into its prime factors, it is possible to

tell at a glance whether or not it can be split into two squares; and if it can be,

the process of discovery in how many ways is so simple that it can be done in

the head without any effort. The number I gave was 130. I at once saw that this




was 2 × 5 × 13, and consequently that, as 65 can be expressed in two ways (64

+ 1 and 49 + 16), 130 can also be expressed in two ways, the factor 2 not

affecting the question.

The smallest number that can be expressed as the sum of two squares in

twelve different ways is 160,225, and this is therefore the smallest army that

would answer the Sultan's purpose. The number is composed of the factors 5 ×

5 × 13 × 17 × 29, each of which is of the required form. If they were all different

factors, there would be sixteen ways; but as one of the factors is repeated,

there are just twelve ways. Here are the sides of the twelve pairs of squares:

(400 and 15), (399 and 32), (393 and 76), (392 and 81), (384 and 113), (375 and

140), (360 and 175), (356 and 183), (337 and 216), (329 and 228), (311 and

252), (265 and 300). Square the two numbers in each pair, add them together,

and their sum will in every case be 160,225.


Puzzle #22 : It pays to Bargain

She was first offered sixteen apples for a shilling, which would be at the rate of

nine pence a dozen. The two extra apples gave her eighteen for a shilling,

which is at the rate of eight pence a dozen, or one penny a dozen less than the

first price asked.


Puzzle #23 : Generous Mr. Luis Mr. Luis must have had 3s. 6d. in his pocket when he set out for home.


http://testfunda.com/examprep/learningresources/pod/puzzle-171--the-squared-army.htm?assetid=d793e614-5bf7-451e-8f26-6201758b8c24

http://testfunda.com/examprep/learningresources/pod/puzzle-172--it-pays-to-bargain.htm?assetid=43c1aa8e-3888-482d-bfcb-a87109f1cca4

http://testfunda.com/examprep/learningresources/pod/puzzle-173--generous-mr-luis.htm?assetid=d6b1ce46-9298-4bef-b6e5-3181eb60db48




Puzzle #24 : The Tired Ant

369 m^2.

Label the vertices of the square A, B, C, D. The ant is at point P, with PB = 13, PC

= 20, PD = 17.

Rotate triangle CDP 90 degrees counter-clockwise about C, so that D goes to B,

and P goes to Q.

By definition, angle QCP = 90. Hence, by Pythagoras' Theorem, PQ = 20(sqrroot

2).

Also, since triangle PQC is isosceles, angle CPQ = 45.

Applying the law of cosines (also known as the cosine rule) to triangle BQP

17^2 = 13^2 + (20(sqrroot 2))^2 − 2 * 13 * 20 * cos QPB.

Simplifying, we find cos QPB = 17(sqrroot 2)/ 26.

Then sin^2 QPB = 1 − cos^2 QPB = 49/338.

Hence sin QPB = 7(sqrroot 2)/26. (We will need this result below.)




We have CPB = QPB + CPQ = QPB + 45.

Hence cos CPB = cos (QPB + 45)

= cos QPB * cos 45 − sin QPB * sin 45, by trigonometric identity cos(a + b) = cos

a * cos b − sin a * sin b

= (cos QPB − sin QPB) / (sqrroot 2)

= (10 (sqrroot 2)/ 26) /(sqrroot 2)

= 5/13

Applying the law of cosines to traingle CPB

BC^S2 = 20^2 + 13^2 − 2 * 20 * 13 * (5/13) = 369.

Therefore the area of the field is 369 m^2


Puzzle #25 : The Faulty Clocks

In order that the hands shall all point to twelve o'clock at the same time, it is

necessary that B shall gain at least twelve hours and that C shall lose twelve

hours. As B gains a minute in a day of twenty-four hours, and C loses a minute

in precisely the same time, it is evident that one will have gained 720 minutes

(just twelve hours) in 720 days, and the other will have lost 720 minutes in 720

days. Clock A keeping perfect time, all three clocks must indicate twelve o'clock

simultaneously at noon on the 720th day from April 1, 1898. What day of the

month will that be? We therefore find that 720 days from noon of April 1,

1898, brings us to noon of March 22, 1900. (1900 is not a leap year!)


http://testfunda.com/examprep/learningresources/pod/puzzle-174--the-tired-ant.htm?assetid=0194fbe5-c24f-4791-a518-1870a4b6e9a7

http://testfunda.com/examprep/learningresources/pod/puzzle-175--the-faulty-clocks.htm?assetid=a6a987e6-f6fd-4899-b53f-acb4ceb3336a




Puzzle #26 : The Weird Scholastic Tradition

There must have been ten boys and twenty girls. The number of bows girl to girl was therefore 380, of boy to boy 90, of girl with boy 400, and of boys and girls to the principal 30, making together 900, as stated. P.S. It was not said that the principal himself returned the bows of any child.


Puzzle #27 : The Mixed Up Caps

14,833. The number of different ways in which eight persons, with eight caps, can each take the wrong cap, is 14,833. Here are the successive solutions for any number of persons from one to eight:— 1 = 0 2 = 1 3 = 2 4 = 9 5 = 44 6 = 265 7 = 1,854 8 = 14,833 To get these numbers, multiply successively by 2, 3, 4, 5, etc. When the multiplier is even, add 1; when odd, deduct 1. Thus, 3 × 1 - 1 = 2, 4 × 2 + 1 = 9; 5 × 9 - 1 = 44; and so on. Or you can multiply the sum of the number of ways for n-1 and n-2 persons by n-1, and so get the solution for n persons. Thus, 4(2 + 9) = 44; 5(9 + 44) = 265; and so on.


http://testfunda.com/examprep/learningresources/pod/puzzle-176---the-weird-scholastic-tradition.htm?assetid=406de7ce-f05f-48ce-aa4b-565eb9865b2b

http://testfunda.com/examprep/learningresources/pod/puzzle-177---the-mixed-up-caps.htm?assetid=155175b5-ad30-460e-856e-ac57eea916c2




Puzzle #28 : Curious Little Keith

The age of his mom must have been 29 years 2 months; that of his father, 35 years; and that of the Keith, 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 Keith will be just half the age of his father.


Puzzle #29 : Mystery of the Missing Money

The explanation is simple. The method of selling the eggs would have been identical had the number of eggs sold at 3 for Rs. 10 and 2 for Rs. 10 been in the proportion of 3:2. Therefore, if Ronen had handed over 36 eggs and Anil 24 eggs, the eggs would have fetched Rs. 240 irrespective of whether sold separately or together. Selling equal eggs has led to a loss of Rs. 10 when sold together, in every 60 eggs.

So, if they had sold 60 each, they would have incurred a loss of Rs. 20 together or if they had sold 90 each they would have incurred a loss of Rs. 30 together. In case of 30 eggs each (60 together), Ronen gains Rs. 20 and Anil loses Rs. 10. Ronen receives Rs. 95 in the transaction and Anil gets Rs. 145 in the transaction so that each loses Rs. 5.


http://testfunda.com/examprep/learningresources/pod/puzzle-178---curious-little-keith.htm?assetid=af366d87-7aae-4005-adb3-33f4f050a57f

http://testfunda.com/examprep/learningresources/pod/puzzle-179---mystery-of-the-missing-money.htm?assetid=bc1e5711-e023-49ff-85f3-31f0f0bd2f9d




Puzzle #30 : The L-Shaped Table at Mrs. Shah's Home

Chair j.

Let X represent one sex and Y represent the other sex, and place an X in chair a.

Then, from [1], the spouse of the person in chair a is either in chair b - case (1) or in chair j - case (2).




Conditions [1] and [2] can be used alternately to determine the couples (X-1 and Y-1, X-2 and Y-2, etc.) in each case, by beginning at chair a and following the arrows:

In case 1, thee members of each sex did not sit next to a member of the same sex; so, from [4], case 1 is eliminated.

Then case 2 is the correct case.

Then, from 4, Mrs. Shah sat in either chair i or chair j.

Then, from [3], Mrs. Shah sat in chair j.


http://testfunda.com/examprep/learningresources/pod/puzzle-180---the-lshaped-table-at-mrs-shahs-home.htm?assetid=b8a787ac-750f-4ef4-98b2-4e89ce1ea467




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files.testfunda.comfiles.testfunda.com/Content/ZeusToolsAssets/Content/...Crosswords, Riddles and Brain Teasers. These enhance the problem-solving skills of CAT aspirants and encourage