di imp questions

8/8/2019 DI Imp Questions

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-Pouring water twice from glass #3 filled glass #7.

-Pouring water twice from glass #4 filled glass #8.

How much water does each glass hold?

Here's the Answers...Question:2.

The glasses hold this much water:

-Glass #1 holds 4 ounces.








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Question No- 3 : Family Tree.I was going through some old family photos in the attic when I stumbled upon our family tree. Istudied it for a couple minutes then went back down stairs to tell my mom about the family tree.The problem is I didn't study it long enough to remember the whole thing. I only remembered acouple things about it, and recent memories. Can you help me figure out my family tree? Thereare two grandparents, who had two children, who both got married and had 2 more children each.Totalling 10 people in all (Alex, David, Jamie, Jessica, John, Justin, Lincoln, Martha, Mary andTina).

1. One of Jamie's ancestors was David.

2. John's sister gave birth to Tina.

3. Mary went bowling with her nephew last Saturday.

4. Alex is cousins with one of the girls.

5. Justin married Mary.

6. Jessica is not an ancestor, nor cousin of Tina.



7. Lincoln's brother showed Justin's son his baseball cards.

Answers:

Martha and David were the grandparents.

They gave birth to John and Mary.

John married Jessica, who gave birth to Lincoln and Alex.

Mary married Justin and gave birth to Tina and Jamie.

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Question No- 4 : Order Five swimmers (Adam, Brad, Carl, Doug, and Eric) have been preparing for the Olympics. It isnow time for the swimming time trials. The five swimmers each compete in the four differentstrokes (backstroke, breaststroke, butterfly, and freestyle). The top three finishers in each eventwill qualify for the Olympic swim team in that stroke. Using the following clues, determine the

order of finish in each of the four strokes.

1) Only one contestant qualified in all four strokes.2) No contestant finished last in more than one event.3) Adam finished better in the backstroke than he did in the butterfly.4) Brad finished better than Doug in the butterfly.5) Adam finished just behind Brad and just ahead of Eric in the breaststroke.6) Doug finished just ahead of Carl in the freestyle.7) Neither Brad nor Eric finished third in any event.

Eric's finish in the backstroke was the same as Doug's in the butterfly.9) Doug only finished in the same position in the backstroke and the freestyle.10) Carl finished in a different position in each event.11) Brad finished only two events in the same position.

12) The contestant who finished second in the butterfly beat Doug in the freestyle.13) The contestant who finished first in the freestyle did not qualify in the backstroke.14) The contestant who finished fifth in the backstroke did not finish third in the butterfly.15) No contestant finished in the same position in both the breaststroke and the butterfly.

Anyway this is Answer key for the Question.No: 4

Backstroke: Adam, Carl, Doug, Brad, EricBreaststroke: Doug, Brad, Adam, Eric, CarlButterfly: Eric, Adam, Carl, Brad, DougFreestyle: Eric, Adam, Doug, Carl, Brad

Hint: Here clues 5,6 & 14 are very imp please observe it carefully..you will get..clincher..

Clue 5th is the most imp 1, as it fixed the positions of Adam, Brad & Eric i.e. they all 3together...ok

Now look at clue: 7th it says, Brag & Eric shouldn’t be in 3rd place…so these 2guys never occupies...Isn’t it.

Another clue: 10th says, Carl always occupied different positions….ok...Always unique...



Steve --- FTFTFGeorge --FFTFTJohn ---- FTFTF

So Answer --- Tom

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Today's Question.No:7 - CAT family

On a beautiful spring day in the park, there were five female cats walking together and gossipingabout their humans. Up ahead, there were five males walking together bragging about their conquests. When the two groups met, there were five cupids flying through the air. Next thing yaknow, there were five couples walking away in separate directions. The couples began talkingand soon found that they had amazing things in common. Well, long story short, we all know whathappens to cats that fall in love in the springtime, oh my what a racket! .

Male: Punkin, Batman, Billy, Jake, DibiiFemale: Pepper, Sally, Ruby, Spot, Starbuck

Snack: Chicken, Mice, Tuna, Ice Cream, ChipsActivity: Laser, String, Sleep, Beat Up, Ball# in Litter: 1, 3, 5, 7, 9

1. The five males were: Jake, the one that liked tuna fish & had five kittens, the one that likedSpot, the one that liked to beat up other cats, and the one that loved ice cream.

2. The five females were: Starbuck, the one that had nine kittens, the one that liked to eat chicken& chase a ball, the one that liked to sleep, and the one that liked Punkin.

3. Dibii ( who had three kittens) liked to chase the laser light but not mice. On the other hand,Sally loved to chase mice but not string.

4. Batman didn't have a thing for Sally and he didn't have seven kittens. Billy, who thought of himself as very big & tough, had the most kittens & wasn't about to chase some stupid stringaround - what a waste of time!

5. Ruby loved to cuddle up to her male for a long afternoon nap in the sun but hated it when heate his chips in bed. That may be the reason they didn't have the most kittens (they didn't havethe least either).

6. Pepper & Punkin didn't have one kitten but they did like tuna fish

Can you figure out who the couples were, what treats they liked, what their favorite activitywas and how many kittens each had in their litter?

Answers for Question.No:7

Here's the CAT family...

#Male # Female #Snack # Activity # Litter Punkin #Pepper # Tunafish # String # 5Batman #Spot #Chicken # Ball # 1Billy #Sally # Mice # Beat Up # 9



Jake # Ruby # Chips # Sleep # 7Dibii # Starbuck #Ice Cream # Laser #3

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Question. No: 8 AGE of Brothers

Jordan, Justin and Jeff are brothers whose ages are 29, 30 and 31 (in any order). Your job is todetermine the brothers' ages from the clues below. There's one catch though....only ONEclue is true. The rest are false. Good luck!

1. Justin is 302. Jordan is the youngest3. Jordan is 314. Justin is one year older and one year younger than his brothers.5. Justin is not the youngest.6. Jordan is not two years older than Justin.7. Justin is the oldest8. Jeff is older than Jordan.

ANSWER:Justin, 29Jeff, 30Jordan, 31.Make 3rd clue TRUE.

I’m sure explanation is not much require for this prob but still…look at below..

Clue 1: If this is true, so is clue 4. Therefore, it is false and Justin is not 30.Clue 2: If this is true, so is clue 5. Therefore, it is false and Jordan is not 29.Clue 3: This clue is true

Clue 4: If this is true, so is clue 1. Therefore, it is false.Clue 5: If this is true, Justin would have to be 31, because we have already determined from clue# 1 that he is not 30, and clue 7 would also be true. Therefore, it is false.Clue 6: If this is true, Jordan would be 30 and clue # 8 would also be true. Therefore it is false.

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Another Question. No: 9

Question No: 9 FIVE Businessmen

These five businessmen represented different companies at a recent trade fair. Unfortunately, thehotel they were all due to stay in had accidentally double booked their rooms. They tried other hotels in the area, but all were fully booked. Consequently, they all agreed to share the only tworooms available in the hotel - one twin and one triple.

From the clues, can you work out each man's name, company and official title?

1. When the five men realized their dilemma, they drew straws to see who would share withwhom. The outcome for four of them was that the CEO shared with the businessman from ABMInc. and Edgar shared with the Developer.



2. Alan does not work for Reed Right and is not the Director. The Director does not work for Lantel or Blue Teeth.3. Neither Clarkson nor Grimaldi works for Reed Right. Grimaldi is either the IT Analyst or theCEO.4. The businessman from Reed Right did not share a room with Thomas.5. Clarkson, who does not work for Lantel, shared with either Thomas or Grimaldi but not both.6. Edgar's surname is either Casson or Graves. Edgar did not share with Carl.7. The businessman from Chiptech shared with the businessman from Lantel.8. The CEO shared with the IT Analyst.9. Casson works for either Blue Teeth or Chiptech.10. Neither Joshua nor Alan works for ABM Inc.11. Fielder shared with the Director.12. The Accountant from Blue Teeth shared with the Developer.

First names: Alan, Carl, Edgar, Joshua, Thomas.Last names: Casson, Clarkson, Fielder, Graves, Grimaldi.Company: ABM Inc., Blue Teeth, Chiptech, Lantel, Reed

Answers

Guy's atleast find my analysis for this Puzzle....please go through it..

#First Name, #Last Name, #Company, #Position:

# Carl , # Grimaldi , #ABM Inc. , #IT Analyst#Edgar , # Casson , #Chiptech , #Director #Joshua , # Graves , #Reed Right, #CEO#Alan , # Clarkson , # Blue Teeth, #Accountant#Thomas , # Fielder , # Lantel , #Developer

To solve this teaser,you first need to determine who shared which room with whom. Three delegates shared the tripleand two shared the twin. The CEO shared with the businessman from ABM in one room and

Edgar shared with the Developer in the other (1). Since Grimaldi is either the IT Analyst or theCEO (3) and the CEO shared with the IT Analyst, Grimaldi is in the same room as whoever is theCEO, IT Analyst and ABM delegate although Grimaldi could have been one of these people.Edgar did not share with Carl (6) so Carl is in the same bedroom as, and possibly is, the CEO, ITAnalyst, Grimaldi and/or the ABM delegate. Grimaldi and Thomas were in different rooms soThomas shared a room with Clarkson (5) and the Reed Right delegate was in a different room toThomas. At this stage we know:

Room 1CEOIT AnalystABM delegateGrimaldi

CarlReed Right delegate

Room 2Edgar Developer ThomasClarkson

Since the Chiptech delegate shared with the Lantel delegate (7), both of these must have been in



room 2 otherwise there would be four delegates in room 1 which is not possible.

Now this has been ascertained, the puzzle can be solved:

The CEO shared with the IT Analyst so the IT Analyst must work for ABM (Clues 1 and so theCEO must work for Reed Right. Since Grimaldi does not work for Reed Right, he must be the IT

Analyst. Clarkson, Fielder and Casson are in room 2 so the CEO must be Graves. Graves is notEdgar (1), Thomas (4), Carl (IT Analyst) or Alan (2) so must be Joshua. Clarkson is not Thomas(5), Edgar (6), Carl or Joshua (in room 1) so must be Alan. The Director does not work for Lantelor Blue Teeth (2), Reed Right or ABM (room 1) so must work for Chiptech. This is not Graves or Grimaldi (room 1), Clarkson (2) or Fielder (11) so must be Casson. Since Graves; first name isJoshua, Casson must be Edgar (6). Alan Clarkson does not work for Reed Right or ABM (room1); Lantel (5) or Chiptech so must work for Blue Teeth. Since he is not the Developer (12) hemust be the Accountant. This leaves Thomas Fielder who is the Developer for Lantel.

Well..i'm sure, you might have sensed it now....its a good teaser..isn't it

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Q.No: 10

Survival of the Sheep

There is an island filled with grass and trees and plants. The only inhabitants are 100 lions and 1sheep.

The lions are special:1) They are infinitely logical, smart, and completely aware of their surroundings.2) They can survive by just eating grass (and there is an infinite amount of grass on the island).3) They prefer of course to eat sheep.4) Their only food options are grass or sheep.

Now, here's the kicker:

5) If a lion eats a sheep he TURNS into a sheep (and could then be eaten by other lions).6) A lion would rather eat grass all his life than be eaten by another lion (after he turned into asheep).

Assumptions:1) Assume that one lion is closest to the sheep and will get to it before all others. Assume thatthere is never an issue with who gets to the sheep first. The issue is whether the first lion will geteaten by other lions afterwards or not.2) The sheep cannot get away from the lion if the lion decides to eat it.3) Do not assume anything that hasn't been stated above.

So now the question:Will that one sheep get eaten or not and why?

Answer for the Q.No: 10

The sheep would remain untouched….Of course it’s a very easy problem...

In fact, the sheep would remain untouched if there is an even number of lions on the island, andwould be eaten immediately if there is an odd number of lions on the island.



I was playing a game of five card draw poker with a bunch of logicians. By the time we hadfinished bidding and were just about to reveal our cards, I was pretty confident I would win of thefour of us remaining. I had three nines, some face card (I can't remember what suit or even

whether it was a jack, queen, or king) and a four. (Or was it a five? I can't remember.)

I was even more sure when two of my opponents laid down their cards. One had a pair of foursand a pair of sevens, the other had a pair of twos and a pair of eights. My third opponent,however, laid down his five cards face down in a row. He said, "I have a straight, and the cardsare, from lowest to highest: a ten, a jack, a queen, a king, and an ace. I have at least one card of each of the four suits: clubs, spades, hearts, and diamonds. I am fairly certain that this is the

winning hand, but I'm feeling generous today, and I will give a third of the pot to whoever candetermine which suit I have two cards of.

Now I know you can't figure it out without some clues. Here they are:

1. The king is next to at least one diamond.2. The queen is next to exactly one heart.3. The jack is next to at least one spade, but is not next to any hearts.4. The ten is next to at least one club.

5. The ace does not border any black cards, nor does it border any diamonds.6. My two cards of the same suit are not next to each other.7. Of the ten possible pairings of cards, only one pair, when removed, leaves three cards inascending order from left to right.8. My ace is not the card on the far left."

There was a minute's silence. One of the other logicians said, "I give up! There's no way to figurethat out!"The other agreed. But I didn't. I had just figured out which suit he had two of.

Which suit is it?

HINT:

The first step is to determine where the ace goes. Also, consider this: How could I have figured it out when two expert logicians couldn't?

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Question.No: 14th - Reindeer

Santa always leaves plans for his elves to determine the order in which the reindeer will pull hissleigh. This year, for the European leg of his journey, his elves are working to the followingschedule, which will form a single line of nine reindeer:

Comet behind Rudolph, Prancer and Cupid. Blitzen behind Cupid and in front of Donder, Vixenand Dancer. Cupid in front of Comet, Blitzen and Vixen. Donder behind Vixen, Dasher andPrancer. Rudolph behind Prancer and in front of Donder, Dancer and Dasher. Vixen in front of Dancer and Comet. Dancer behind Donder, Rudolph and Blitzen. Prancer in front of Cupid,Donder and Blitzen. Dasher behind Prancer and in front of Vixen, Dancer and Blitzen. Donder behind Comet and Cupid. Cupid in front of Rudolph and Dancer. Vixen behind Rudolph, Prancer and Dasher.

Can you help the elves work out the order of the reindeer?



Answer for the Q.No: 14th

From front to back: Prancer, Cupid, Rudolph, Dasher, Blitzen, Vixen, Comet, Donder thenDancer.

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Question.No:15th

Henry and Gretchen are going to play a game. Henry explains, "You and I will take turns sayingnumbers. The first person will say a number between 1 and 10. Then the other person will say anumber that is at least 1 higher than that number and at most 10 higher. We will keep going backand forth in this way until one of us says the number 50. That person wins. I'll start."

"Not so fast!" says Gretchen. "I want to win, so I will start."

What number should Gretchen say to start?

Answer for the Q.No:15th

She i.e. Gretchen wants to say Number, "6."

The series of numbers she will say is 6, 17, 28, 39, 50.

Since she wants to say 50, she needs Henry to say a number between 40 and 49. Therefore, shewants to say 39.

Knowing she wants to say 39, she needs Henry to say a number between 29 and 38. So shewants to say 28. Following this same logic recursively, she will want to say 17, and she will wantto say 6 to start the game, and be assured victory.

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Question.No: 16th

Billiards Tournament

Seven amateur billiards players gather to play a billiards tournament. The players were dividedinto two sections: Section A has 4 players; and Section B has 3 players. Each player playedagainst every other player in his/her Section.

Players: Betty, Chris, Greg, Jill, John, Mary, and Tina.

1. Every player from Section B has the same win-loss record.

2. John lost all of his games.

3. Mary, who defeated Jill, won more games than exactly 5 of the players.

4. Jill lost to Greg, and Tina defeated Chris.

With the information below, can you determine each person's section, and to whom theywon and lost.



Here’s the answer for the Q.No:16th

Section A:

John: Lost to everyone.Jill: Beat John, Lost to Mary and Greg.Mary: Beat John, and Jill. Lost to Greg.Greg: Beat everyone.

Section B:

Betty: Beat Tina. Lost to Chris.Tina: Beat Chris. Lost to Betty.Chris: Beat Betty. Lost to Tina.----------------------------------------------------------------------------------------------------------------

Question.No: 17th -- Motor Bike Race Tournament

For Questions 1 to 5 read the following instructions:

Nine Bikes participated in a F1-MotorBike tournament. Five races were held in this tournament.The winner of a race gets 5 points, 2nd gets 3 and the 3rd gets 1. The table gives the point’s tally.

Bike: PointsA # 5B # 5C # 8D # 2E # 5F # 6G # 4H # 10

I # 0

One race is held between five Bikes. The first race is held among the first 5 Bikes from the left inthe above table. Out of these Bikes, A drops out of the race and a new Bike. I.e. F enters into therace. In the next race, B drops out and G enters and so on. It is also given that H is the only Bikethat scored in 2 consecutive races.

Q.1) What was the position of Bike F in race 4?a. 4th b. 2nd c. 3rd d. cannot be determined e.1st

Q.2)What is the ratio of the points scored by “G” and E in race 5?a. 1:3 b. 3:1 c. 1:5 d. cannot be determined e. 5:1

Q.3) The first 3 rankers of the race 3 in some order?a. C,E,G b. D,E,F c. C,D,F d. C,E,F e. C,G,F

Q.4) D could have come third in which of the following races?a. Race 1 b. Race 3 c. Race 4 d. Race 5 e. cannot be determined



Q.5) Suppose if C had 6 points and E had 7 points while the points of the rest of the Bikes remainunchanged then E could’ve come second ina. One race b. Two races c. At most one race d. At most 2 races e. cannot be determined

Answer key for the Q.No: 17th.

1-b, 2-d, 3-a, 4-c 5-dThe table of the winners is as follows:Race1A

C

E

Race2B

F

D

Race3C

E/G

G/E

Race4H

F

D

Race5H

G/E

E/G

So 1-b, 2-d, 3-a, 4-cFor 5 interchange the places of C and E in race 1. Hence the answer is d, 5-d

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Question.No: 18th : It's a Jungle Out There.

There is a plot of land that is 16 square acres, arranged in a 4 X 4 grid. There are 16 mammals,two each of the following: beaver, cat, dog, goat, horse, lion, tiger, and walrus. There is onemammal in each acre. The acres are labeled A through P, as shown below.

A B C D



E F G HI J K LM N O P

Clues:1) A and D do not contain the same mammal, but one of them has a dog.2) E and O do not contain the same mammal, but one of them has a horse.3) B and L contain the same mammal, but not a tiger.4) C and M contain the same mammal, but not a beaver.5) F, K, N, and P are four different mammals.6) The tigers are in different columns.7) The goats are located in two of the following locations: B, C, K, and O.

The lions are located in two of the following locations: A, E, F, and M.9) Cats don't get along with dogs, so neither cat is next to a dog.10) Tigers have goats on their menu, so each tiger must be next to at least one goat and eachgoat must be next to at least one tiger.11) The beavers and walruses live on the same body of water, so all four must be connectedtogether in some fashion.

Given the above clues, determine where each mammal is located. The terms "next to" and"connected together" mean vertically, horizontally, or diagonally

Answer for Q.NO: 18th: A = lion, B = cat, C = walrus, D = dogE = horse, F = lion, G = beaver, H = tiger I = dog, J = beaver, K = goat, L = catM = walrus, N = tiger, O = goat, P = horsePer #3, B & L are the same; per #4, C & M are the same; therefore, per #7, K & O are goats;therefore, per #2, E is a horse; therefore, per #8, A & F are lions; therefore, per #1, D is a dog.Per #4, C & M are not beavers; per #9, C is not a cat; per #10, M is not a tiger; therefore, C & Mare walruses; therefore, per #11, G & J are beavers.Per #3, B & L are not tigers; therefore, B & L are cats. Per #6, N is a tiger; per #9, H & P are not a

dog; therefore, I is a dog.Per #5, P is a horse; therefore, H is a tiger ----------------------------------------------------------------------------------------------------------------

One more..Question.No: 19th: The Seven SportsmenSeven sportsmen are standing in a row. Each sportsman has a different 1-digit number. Theproduct of the first three, the product of the middle three, and the product of the last three are allequal.

What's the middle sportsman's number?

Answer for Q.NO: 19th: 2 Let their numbers be a through g, we have abc=cde=efg.5 or 7 cannot occur, because they are prime and no other 1-digit number contains them.Therefore, the numbers are 1, 2, 3, 4, 6, 8, 9.Since abc=efg, abcdefg/d is a perfect square. 1*2*3*4*6*8*9=2^7*3^4, and d can be only 2 or 8.If d=2, the numbers 1, 3, 4, 6, 8, 9 can be made into 1*8*9=3*4*6=72, and we can have c=9, e=4:9*2*4=72.If d=8, the numbers 1, 2, 3, 4, 6, 9 can be made into 1*4*9=2*3*6=36, but 36 is not divisible by 8.Therefore, the middle sportsman's number is 2.



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There are a number of long fuses, of which you only know that they burn for exactly one hour after you light them at one end. However, you don't know whether they burn with constant speed.Fore example, the first half of the fuse is burned in only 10 minutes while the rest takes the other fifty minutes to burn completely. With what least number of fuses can you measure exactly seven

eighths of an hour in time? You can safely assume that you have a lighter.

Answer:number of candles required is 3.

lit all the three together, but the first one in bth the ends.so the first one will be burnt out in 1/2 hr. the other two are left over to be burnt for 1/2 hr more.now lit the second candle in the other end also. so this will burn it in 1/4 hrs. the last one is left tobe burnt for 1/4 hrs.lit @ the other end as well. it will take time of 1/8 hrs.so total time = 1/2+1/4+1/8 = 7/8

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Four coworkers are all going on different vacations during the month of February. Each

proclaimed that they wanted to go somewhere warm for awhile � they�d seen enough snow for

one winter. The rest of their departments watched them come and go with envy. Each coworker had a wonderful time and returned to work with a tan and lots of pictures of their trip. Determinethe full name of each vacationer, the city each is traveling to, the means of transportation beingused (one was an airplane), and how long each vacationer plans to stay (3 to 6 days).

1. Mr. Cook didn�t go to Las Vegas but he was gone longer than the person who used the bus.

The person whose last name was Wakefield was gone longer than the person who went toAtlanta.

2. The person who traveled by train spent five days on vacation.

3. Fred rented a car for his vacation. Melinda�s last name wasn�t Wakefield.

4. Larry went to Orlando. The person whose last name was Surrey didn�t travel to Atlanta but

was gone for three days.

5. Harriet West didn�t travel by bus. The person who visited Atlanta for four days wasn�t Fred.

6. Larry�s last name wasn�t Wakefield. The person who went to Houston was gone for six

days.

Answer:

First name # Last name #City # Transport # DurationFred , Wakefield , Houston, Car , 6 daysHarriet, west, ,Atlanta , Airplane ,4 daysLarry, Cook , Orlando, Train , 5 daysMelinda, Surrey, Las Vegas , Bus , 3 days



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There are ten gnomes. They are in the dungeon. Their captor has given the gnomes a chance of survival. Here is the offer:

He lines the gnomes up in a single-file row. This means that the tenth gnome sees the back of theperson in front of him, and there is no gnome behind the tenth gnome. The ninth gnome has thetenth gnome behind him and the eighth gnome directly in front of him, and so on. Finally, the firstgnome has the second gnome directly behind him, but there is no one in front of the first gnome.

The captor has a bag full of many black hats and many white hats. There is not necessarily thesame number of black hats as white hats. (important) He randomly reaches into his bag andplaces a hat on each of the gnomes. This means that the tenth gnome can see everyone's hatexcept his own, the ninth gnome can see everyone's hat except his own and the tenth gnome'shat, and so on. The first gnome can see no one's hat.

The captor then takes out his gun and puts it to the temple of the tenth gnome. He asks thegnome, "What color is your hat?" If the gnome answers correctly, he lives and gets freed from thedungeon. If he does not, he dies. He continues up the line in this progression.

However, before placing the hats on the gnomes, he allows the gnomes to meet as a group anddiscuss a strategy to save as many of the gnomes as possible. How many gnomes can youguarantee to save, and with what strategy?

REMEMBER: When it is your turn to say the color of your hat you must ONLY say "white" or "black." If you say anything else, the king will shoot you and all of the remaining gnomes.

Answer 9 gnomes can be saved for sure at any given time.

StrategyLets sayBlack = 0

White = 1

Now, all that the last person has to do is add the number of White caps in front of him, say x.The last person will call out the color of his cap as depending on x mod 2. That is, if x is eventhen the color he'll call out is Black else White.This person will have 50% probability of survival as whatever color he says has nothing to withhis cap's color.

Now this is where it begins.Hearing the color said by last person, the second last person would realize whether the number of white caps seen by last were even or odd. Now this person would calculate the number of White caps in front of him and would check if he has a White cap or a Black Cap. This he willmanage to save himself.

The third last person will now take into account the number of White caps as told by the lastperson and knwoing the color of second last person's cap and also the number of White caps infront of him would evaluate whether his cap is White or not.So on would the sysle continue till the first person is saved.

What a martyr the last Gnome would be..

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There are five men on a deserted island. The only things on the island are banana trees and amonkey. The men realize they need food to survive, so they start gathering bananas. At the endof the day, the men put all the bananas in a huge pile at the foot of a tree. Too exhausted todivide the bananas, the men determine that they'll evenly divide the bananas the next morning,and all go off to sleep.

At 1:00 AM, Man #1 wakes up. He determines that he is unable to trust the other men. Therefore,he divides the bananas evenly into five piles. After dividing the bananas, there is one left over, sohe gives it to the monkey. He then takes and hides his share of the bananas (one of the five piles)and aggregates the remaining four piles back under the tree.

At 2:00 AM, Man #2 wakes up. He also decides that he doesn’t trust the other men. Unaware thatMan #1 already took his share, Man #2 also divides the bananas into five piles. After splitting thebananas, there is one left over, so he gives it to the monkey. He takes and hides his share of thebananas (one of the five piles) and aggregates the remaining four piles back under the tree.

At 3:00 AM, 4:00 AM, and 5:00 AM, the three other men carry out the same action.

In the morning, all the men wake up and try to look innocent. No one remarks about thediminished pile of bananas and no one admits to already having taken a share. Instead, they

divide the remaining bananas into five piles (for the sixth time) and find that there is one bananaleft over, so they give it to the monkey.

Assuming no fractional divisions of bananas, what is the smallest amount of bananas thatcould’ve been in the original pile?

Answer Answer :Before all the iterations: 195311.39062.7813.156

4.315.6

6 bananas will be divided into 5 piles of 1 banana each and so the left out banana will be given tothe monkey.

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Question.No: 20th

Substitute the digits for the letters to make the following addition true.

G E O R G EZ G I N N Y+ Z Z R O N---------------------P O T T E Rwhere Z = zero

identify all the Alphapets in it..???*note: Though, these kind of Q's may not come in CAT but improves calculation skills for sure..



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DI Set # : 21

For Questions 1 to 5 read the following instructions:Six people A,B,C,D,E and F are standing in a row(from left to right) in that order. A is at place 1, Bis at place 2 and so on. They are rehearsing a dance sequence for a dance competition. Whenthe signal is given the following 3 steps are performed.Step 1: They break away and form a triangular formation with 4 and 5 in front of 1, 2 and 3 andbetween 1 and 2 and 2 and 3 respectively, while 6 is in front of 4 and 5.Step 2: Then 6 goes a step backward and joins between 4 and 5 who form the front row now.Step 3: After a few dance moves, both these rows merge with the person in front of 1 joiningbetween 1 and 2 and so on.These 3 steps form a round. These steps are repeatedly performed till the dancers attain their original positions. This is called one sequence.Q.1 How many times are step 1, step2 and step 3 are performed before one sequence?a. 8 b. 6 c. 4 d. 5 e. 10

Q.2 At which place is C after the completion of the penultimate round?a. 1 b. 3 c. 5 d. 2 e. 6Q.3 How many members come to their original positions at least once before the sequence ends?a. 1 b. 2 c. 3 d. 4 e. 5Q.4 Suppose their initial positions were in the order mentioned before but from right to left, andthe steps 1 and 2 remain the same while in step 3 the person in front of 3 joins between 3 and 2and so on to form a single row. How many rounds before this sequence ends?a. 4 b. 6 c.8 d. 10 e. They never come back to their original positionsQ.5 In the above mentioned sequence, how many members come back to their original positionat least once before the sequence ends?a. 2 b. 3 c. 4 d. 5 e. 6

the answers of the DI Set : 21

1-d, 2-d, 3-a ,4-b, 5-e just after the first 3 steps u will discover a pattern like which serial # will go to which place.simillarly for the last two question also. as u saw it is just one more type of arrangement, and ucan easily decipher the pattern with the first 3 steps itself.....

so for this question if u r able to get the pattern in the very first question then go for all, otherwiseits better to leave the last 2 and move on....

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Question.No: 22th - Ships

There are 5 ships in a port.1. The Greek ship leaves at six and carries coffee.2. The ship in the middle has a black chimney.3. The English ship leaves at nine.4. The French ship with a blue chimney is to the left of a ship that carries coffee.5. To the right of the ship carrying cocoa is a ship going to Marseille.6. The Brazilian ship is heading for Manila.7. Next to the ship carrying rice is a ship with a green chimney.



4. The first two numerals (as one number) as well as the two middle numerals (as one number)are a multiple of the last two numerals (as one number).

What is the code? (more than 1 solution)

Answer for Q.No: 23rd

Code - solution

The possible 2 last numerals are as follows: 03, 05, 07, 09, 14, 16, 18, 25, 27, 29 and 30. At leasttwo multiples less than 100 (this condition is already accomplished), which consist of even andodd numeral (respecting all other conditions) are for 03, 07, 09 and 18 as follows:03 – 27, 63, 69, 8107 – 49, 6309 – 27, 63, 8118 – 36, 72, 90

There are 5 numbers that can be made of these pairs of numerals to create the Code: 692703,816903, 496307, 816309 and 903618. (If we assume, that also in the number 903618 is

accomplished the requirement to alternate even and odd numbers, despite the opposite order.)

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Harish and Sachin are playing a game of matchsticks.There are N matchsticks on the table tostart with.Each player in his turn picks up atleast one matchstick and at most 8 matchsticks.Thtwo players takes turn alternately.The player who clears the table loses.Assume that each player plays intelligently with an objective of winning.The first move is made by Harish.No player isallowed to pass his turn without picking up any matchsticks,

Question 1:- If for some N,it is known that the number of matchsticks picked up by Harish in hisfirst four moves were 6,4,3 and 6 resp. then how many matchsticks wud Sachin have picked up inhis third move given that Harish wo the game?

1) 3 2) 5 3)6 4) Data Insufficient

Question 2:- If it is known that the game was completed in 8 moves (By each of the twoplayes),what is the maximum possible value for N?1)63 2)71 3)72 4)81

Addition information for the following two questions:Instead of minimum of one matchstick if each player has to pick up at least 2 matchsticks in histurn.The only instance that a player is allowed to pick up one matchstick is when there is only onematchstick left on the table.

Question 3:- It it is known that N is greater than 6 but less than 44,for how many values of N willHarish certainly lose the game,irrespective of how he plays?1)4 2)5 3)6 4)8

Question 4:- If it is known that N is greater than 128 but less than 138 and that Harish picked up 6matchsticks in his first move and eventually won the game then what is the value of N1)137 2)133 3)134 4)Can't be determined



Answers

These type of problem can be solved by finding the number which will ensure the total # of picksof matchstick with each round.The required # is : = [Min Pick + Max Pick]in this case it is = 1+8 = 9Q1) it is clear that Haris has to make sure, when he picks for the 1st time that, after the first pickthe remaining Matchstick become of form 9k. i.e the total number of matchstick is = 9k+6now evry round (ending with haris) shd be such that the total pick is = 9to determine Sachin's 3rd pick, we have to look for Haris' 4th pick,which is : 6. so Sachin will pick

9-6 = 3 to ensure that Haris' wins (which is very unlikely )

q2)this can be solved as explained above,we know that the total pick with each round is = 9so the max #of matchstick for 8 rounds are = 9*7+(8+1) = 72

Q3)here the ensure pick number shd be (2+ = 10now for haris to loose the total number of matchstick shd be in the form of 10k+1 and 10k+2so thr are total 8 such numbers.(11,12,21,22,31,32,41,42)so the answer is = 8

Q4)in order to win, haris must make sure that he shd pick in such manner that the number of matchstick remains in the form of 10k+1 or 10k+2he picks up 6, so the required # of matchsticks are 137.

Explanation:dude,the basic idea is that the first person who picks shd make sure that he wins irrespective of wat the

second non-puy picks.

now the idea behind [max pick + min pick] is that,suppose the second guy picks 3 in this case, then the first one can pick 6 and make the total sumas 9. the extreme case can be the second guy picks 8 or 1. in that case the first guy will pick 1 or 8 respectively.hope u r getting the point. in this manner the first guy makes sure that the picks remains in aparticular order,9 in this case.and to make himself winning the match the first guy will pick accordingly in the first pick itself, sothat the remaining no of matchstick is 9k+1 in this case. so that after all the iteration is over (withthe first guy), the nxt guy has nothin but 1 matchstick to pick, so he losses.thats the trick.hope this clears stuff for u.

suppose the number is 17 (greater than 16).so in the first pick the first guy will pick 7, so the remaining sticks is 10 which is 9k+1 form.so the nxt guy can pick ne thing between 1-8, whatever he picks the first guy will surely win...if hewants to

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Here is one of my favorite questions for the DI marathon for 2007:

The 5 greedy thieves

Once 5 greedy thieves stole 100 gold coins. They came up with the following idea for distribution



of the gold coins. They numbered themselves from 1 to 5. First thief will propose a distribution,and others would vote. If he gets 50% or more votes, what he said will be accepted; otherwise hewould be killed, in which case second will propose , and so on.

Assume that all the thieves were intelligent and blood thirsty. For each thief order of priorities indecreasing order is his own life, then gold coins and then other's life. And the proposing thief can'tparticipate in voting.

1. What is he maximum number of gold can the first thief can take away?a. 100 b. 98 c. 97 d. 50 e. 49

2. What will the maximum number of gold coins 5th thief can get?a. 0 b. 1 c. 2 d. 3 e. 4

3. Who all will be in favour of thief 1's proposal whatever be the case?a. 2 b. 3 c. 4 d. all e. none

4. In how many ways can the gold coins be distributed(considering that 1st thief's proposal isaccepted)?a. 1 b. 2 c. 3 d. 4 e. 5

The answer to the 5 Greedy Thieves

Lets go from the reverse ..

4-5i) Lets say only 4 and 5 are present , 5 will veto whatever 4 says so that he can get all the gold for himself. Therefore 4 will never want to come to this position.

3-4-5ii) So if only 3,4,5 are present - 4 will accept to anything 3 says to save his life and hence 3 canwalk away with all 100 coins (since with 4 accepting , he's got 50% of the votes)

2-3-4-5iii) Therefore 3 will refuse any distribution 2 says. Now 2 must make the other two accept andthey'll do that only if he makes it worthwhile for them. So he offers them 1 coin each to 4 and 5and walks away with 98 gold coins.

iv) So he'll refuse anything 1 proposes. 1 will offer 3 one gold coin for him to accept because in2's optimal distribution , he does not get anything. To swing one more vote , 1 needs to make itmore worthwhile to one of 4 or 5 to vote for this. So he has to offer them one of them 2 goldcoins.

So final distribution1: 97

2: 03: 14 (or) 5 : 2Other 0

The answers follow from this ..

1. What is he maximum number of gold can the first thief can take away?a. 100 b. 98 c. 97 d. 50 e. 49



2. What will the maximum number of gold coins 5th thief can get?a. 0 b. 1 c. 2 d. 3 e. 4

3. Who all will be in favour of thief 1's proposal whatever be the case?a. 2 b. 3 c. 4 d. all e. none

4. In how many ways can the gold coins be distributed(considering that 1st thief's proposal isaccepted)?a. 1 b. 2 c. 3 d. 4 e. 5

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Step I :Total Number of Eggs = 6 and 1/3rd of a Dozen = 19/3 * 12 = 76 eggs

Step II :A and D together carry a number of eggs which is a multiple of 4 and has 6 factors.

Now you could check each and every multiple of from 8 to 72 to see if it fits but here’s analternate method:

We know it is a multiple of 4 , so 2^2 is a factor.As you know the number of factors for any number P which can be factorized into prime factorsa^p * b^q * c^r …. Where a,b,c.. are prime is (p+1)(q+1)(r+1)…So it is quite clear 2^2*p where p is another prime number will have 6 factors. So the numbers weneed are12 = 2^2 *320 = 2^2 *528 = 2^2 *744 = 2^2 *11



52 = 2^2 *1368 = 2^2 *17(76 also fits but that would mean B and C carry nothing which does not fit)The only catch with this is that you will miss 32 = 2^5 which also has 6 factors.So A and D could have carried 12,20,28,32,44,52 or 68 eggs.

Step III:Tip :Before this , it might help to list out all prime numbers from 2 to 75 – The coaching institutes tellyou to memorize all primes from 2 to 100 but I found that difficult so if you fall in that category –by listing them down you can avoid silly mistakes and surprisingly save a bit of time too.2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61, 67,71,73

I shall solve the first question – the rest follow the same pattern.If B and C have to deliver the maximum number of eggs A and D have to deliver the minimumnumber of eggs.So A/D deliver 12 eggs. Possible prime pairings are (7,5) and (5,7) – meaning the number of Eggs A and D can carry can either be 7 and 5 respectively or 5 and 7 respectivelyB/C deliver 76-12=64 eggs. Possible prime pairings are (61,3) , (59,5), (53,11) , (47,17) , (41,23)and the reverse pairs.

(Fastest way to do it is to start with 64 , look at the top of the list and go down to half of that.That’s not well phrased – Here’s an example.First prime less than 64 = 63 . Difference = 1 . Not a primeNext prime less than 64 = 61. Difference = 3. Prime . Pairing (61,3).Next prime less than 64 = 59. Difference = 5. Prime . Pairing (59,5).Next prime less than 64 = 53. Difference = 11. Prime . Pairing (53,11).Next prime less than 64 = 47. Difference = 17. Prime . Pairing (47,17).Next prime less than 64 = 43. Difference = 21. Not a primeNext prime less than 64 = 41. Difference = 23. Prime . Pairing (41,23).Next prime less than 64 = 37. Difference = 27. Not a primeAnd since we’ve reached half way mark 32 , we do not have to go any lower. The other pairs arethe reverse pairs of the above i.e (3,61),(5,59),(11,53),(17,47) and (23,41). )

So A/D in 2 ways , B/C in 10 ways. Total = 2*10 = 20 ways.Do try the other questions before proceeding below if you did not get them the first time …If you solve fully you’ll get a table like the one below ..



So the answers are

78.(3)79.(4)80.(1)81.(4)

Optimal :78 and 80 are easily doable and must be attempted irrespective of the paper. 79 and 81 are justnot worth it IMO for the time investment - and this is not just hindsight - after you solve 78 and80 , you get an approximate judge of the time for the other two. An easy 8 marks for 5-10 min.But a trap for 16 marks in a time of 30 min. Avoid the trap , get the easy ones right and go to theother sets.

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Five neighbors, Turtle, Pony, cat, Parrot and Pigeon,each have one animal as a pet. Five animalsare turtle, pony,cat, parrot,and pigeon. None of the neighbors has a pet with same name as his.One day every neighbor gave his pet to another neighbor and received another from a neighbor.Once again no one received a pet of his own name. the following informations are known...I. The pony was given away by a neighbor known by the name of the animal given away byParrot.

II. the turtle was given away by the neighbor known by the name of the animal received by Cat.



III. Pony gave away a cat and received a pigeon.THE ANIMALS GIVEN AND RECEIVED BY PIGEON WERE RESPECTIVELY...

a) the parrot and the cat.b) the parrot and the turtle.c) the pony and the cat.d) the turtle and the pony.

Answer:I feel the answer is b) Parrot and turtle;Person-->Animal given-->Animal received;Turtle----->PONY----->CATPony------>CAT------>PIGEONCat------->PIGEON--->PARROTParrot----->TURTLE--->PONYPigeon---->PARROT--->TURTLE

2:

According to 3rd condition:Pony --> Cat|Pigeon

Accoriding to 1st condidtion:

Assume X as one animal: Parrot --> X and X --> PonyLet X be Turtlewe get three conditions:Parrot --> Turtle ; Turtle --> Pony and Pony --> Cat

Remaining two conditions are: cat --> Pigeon and Pigeon --> Parrot [as pigeon cannot be with

pigeon]

Hence we get pairs as follows:

Parrot Turtle Pony Cat PigeonTurtle Pony Cat Pigeon Parrot

Hence the option is b.

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A certain number of towns are connectedthrough21 roads in all, one road connecting exactly twotowns. Six of the towns are connected through exactly 3 roads whereas rests of the towns are

connected through exactly 4 roads. The total number of towns is..1) 122) 203) 184) 15

Actually I didn't solve it systematically but with some tricks..I first took 6 towns up and triedconecting them with 3 roads exactly and after 2 or 3 permutations it was clear that in total 9 roadsare required to suffice the condition.



So I assumed a formula that No of roads(9) = [6(No of towns) * 3( No of roads)]/2

So if this logic holds true then for remaining 12 roads = [N * 4]/2 that gives N = 6 towns

So the total towns are 6+6=12

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Question.No: 24th

FIVE FRIENDS.

Five friends have their gardens next to one another, where they grow three kinds of crops: fruits(apple, pear, nut, cherry), vegetables (carrot, parsley, gourd, onion) and flowers (aster, rose, tulip,lily).

1. They grow 12 different varieties.2. Everybody grows exactly 4 different varieties

3. Each variety is at least in one garden.4. Only one variety is in 4 gardens.5. Only in one garden are all 3 kinds of crops.6. Only in one garden are all 4 varieties of one kind of crops.7. Pears are only in the two border gardens.8. Paul's garden is in the middle with no lily.9. Aster grower doesn't grow vegetables.10. Rose grower doesn't grow parsley.11. Nuts grower has also gourd and parsley.12. In the first garden are apples and cherries.13. Only in two gardens are cherries.14. Sam has onions and cherries.15. Luke grows exactly two kinds of fruit.

16. Tulips are only in two gardens.17. Apples are in a single garden.18. Only in one garden next to the Zick's is parsley.19. Sam's garden is not on the border.20. Hank grows neither vegetables nor asters.21. Paul has exactly three kinds of vegetable.

Who has which garden and what is grown where?

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Q: Ashish, Munish, Rakesh played a Dart game. Each one threw nine darts that lodged in theboard as shown. Each person’s total score was the same as any other man’s score. No number of points scored by a dart was scored by more than one person. Ashish scored all the 5’s andMunish scored all the 10’s.Who scored all the 100’s?



(1) Ashish (2) Munish (3) Rakesh (4) Can’t not bedetermined

answer to the dart board question is Munish

Ashish= 5*5 + 4*50 = 225 from 9 shots hits only fives and 50's

Rakesh= 9*25= 225 from 9 shots hits only 25

Munish= 100*2+ 10*2+ 1*5= 225 from 9 shots. hits all the tens, hundreds and ones.

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there are 81 similar horses out of which u have to select top 4 horses ...u can have N number of races WHERE in each race exactly 9 horses will run.....

SO WAT IS D MINIMUM POSSIBLE VALUE OF N BY WHICH U CAN SELECT TOP 4 HORSES

Answer

Questions like this are best solved by elimination rather than selection i.e Always eliminatehorses for which you know there are 4 faster horses ..

Let there be 9 races. Call these races A,B,C,D,E,F,G,H,I.

Let the finishers in the races be denoted by numbers followed by race i.e 1A,2A,3A,4A are the topfour finishers in race A.

The key to this is to try to eliminate horses. In the first 9 races , we can eliminate 45 horses.Because 5A,6A,7A,8A,9A and similar finishers for the other races cannot be the top four.

Let there be a 10th race with all the first position horses racing.

You can assume any order but i'm assuming they finish in this order -1A,1B,1C,1D,1E,1F,1G,1H,1I

Step I : You can eliminate all horses 1E,2E,3E,4E,1F.....3I,4I since we have four faster horsesthan them. this eliminates 20 horses.

Step II : Horses 2D,3D,4D can be eliminated since 1D is faster than them and 1A,1B,1C arefaster than 1D. This eliminates 3 horses.

Step III : Horses 3C,4C can be eliminated using same logic -> 2 horses eliminated



Step IV : Horses 4B can be eliminated using same logic -> 1 horse eliminated.

So we have eliminated 71 horses . we have 10 horses remaining -> we know 1A is the fastest of all the horses. Hold a race between the other horses to determine the second ,third and fourthfastest horses.

So Total number of races required = 9+1+1 = 11 races

Explanation 2

I take 9 horses at a time and keep a race between them. For the sake of notation i call the horses<position in race><followed by race letter> . So if a horse got third place in the second race , itwould be 3B.

The order in the First race is:1A > 2A > 3A > 4A > 5A > 6A > 7A > 8A > 9A

We are trying to find the four fastest horses. Take 5A - All the horses above it are faster i.e1A,2A,3A,4A are faster than 5A so it can never be one of the four fastest horses , so i ignore that

horse from now on.To rephrase , all horses finish the race but i am interested in only the horses which finishedfirst,second,third and fourth since the other horses can be eliminated.

So the horses left are1A > 2A > 3A > 4A (Race 1)1B > 2B > 3B > 4B (Race 2)...1I > 2I > 3I > 4I (Race 9)

Of course at this point i do not know the comparisons between the horses from different races i.e

3B could be faster than 1A for all we know.

So i hold a Race 10 between all the top finishers . I assume they finish in this order 1A > 1B> 1C > 1D > 1E > 1F > 1G > 1H > 1I(Note this does not matter , even if you assume 1E > 1F > 1G > 1H > 1I > 1A > 1B> 1C > 1D , thelogic remains the same)

Now why did i eliminate 5A in the first round of matches? Because i could find four faster horses1A,2A,3A and 4A than it. In this case 1A,1B,1C and 1D are faster than 1E.

So we can eliminate 1E for the same reason we eliminated 5A.Also we can eliminate all slower horses than 1E ie 2E,3E,4E. Twenty horses (All the 1E - 4I) canbe eliminated by this.

Now1A > 1B > 1C > 1D from Race 10 But we already have 1D > 2D > 3D > 4D from Race 4Combine1A > 1B > 1C > 1D > 2D > 3D > 4DSo 2D , 3D , 4D can be eliminated.In a similar way 3C,4C and 4B can be eliminated. (Try it out - Combine Race 10 info with Race 3and Race 2 info respectively)



So horses left are1A 2A 3A 4A 1B 2B 3B 1C 2C 1D -> 10 HorsesWe already know 1A is the fastest of all the horses ..

Keep a final race to determine the other fastest horses.

eg: If Race 11 finish order is2A > 1B > 1D > 2B > 1C > 3B > 2C > 3A > 4A

The four fastest horses areFirst : 1ASecond : 2AThird : 1BFourth : 1D

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There are 4 children in a family aged 6,8,11 and 14 years old.Their names are

Ali,Mohammed,Dipak and Nemisa.Use the clues below to find the age of each child:1)Dipak is three years older than Ali.2)Mohammed is older than Dipak.

Mohammad---14Deepak- ------11Ali------------- 8Nemisa--- ---- 6

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The following table gives the number of studens who secured more than 90% marks in each of the five subjects - English , Physics , Chemistry , Mathematics and Biology from Class 6 to Class10 at a school , in the year 2006 .

The number of students in the different classes in the year were class 6 -30 , class 7 - 35, class 8- 28 , class 9 - 36 and class 10 - 40

1. In the class 7, the number of students who scored more than 90 % in at least twosubjects is at least ?

a.12 b. 15 c.18 d. 20 e. None of these

2.The number of students in class 10 who scored more than 90 % , in exactly 3 subjects isat most

a.32 b.34 c.35 d.38 e. None Of these

3. If a scholarship is awarded to all the students from class 6 to class 10 who score morethan 90% in each of the five subjects , then the number of students who won thescholarship is at most



a. 120 b. 104 c.62 e.59 e.55

4. The number of students who scored more than 90% in exactly four subjects in allclasses together is at most

a.28 b.61 c. 96 d.101 e. None of these

5. The number os students in class 6 who scored more than 90 % in at most 2 subjects isatmost

a. 19 b. 21 c. 22 d.26 e.30

Answers

1) ans is (b) i.e. 15 students. We need the minimum no. of students who have scored > 90% in aminimum of 2 subjects.Total students in class 7=35.Taking any 2 subjects say: e with 12 students & p with 16 students. Since we are concerned withmin. we get 12 students (at least) get > 90% in atleast 2 subjects.--> 1On similar lines we see that for p+c we get 15 students.--->2for c+m we get 15--->3for c+b we get 15 --->4for others wither the sum goes above 35 or it falls in one of the above 2. So if we take 12 (it beingthe least) we see that 12 students got >90% in only 1 subject. To the least value should be 15,

which included students >90% in atleast 2 subjects.

3) ans.(d) The least number of students who scored > 90% in all the 5 subjects for a particular class is the value of the minimum students in any subject.i.e. add up the value of the column for a class which has the least number of students getting>90%.=> 12+15+7+10+15 = 59



5. The number os students in class 6 who scored more than 90 % in at most 2 subjects isatmost

c. 22

2.The number of students in class 10 who scored more than 90 % , in exactly 3 subjects isat most

c.35

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There are 80 balls which look identical. All the balls are of same weight except one which isheavier. In how many minimum weighing with a bean balance can you find the heavier ball?

a) 4 b) 6 c)7 d)5

Answer

New reasoning-Breakup the 80 balls into 3 sets of (30,30,20).(a) Weigh (30,30). If they are equal in weight. Then the heavier ball must be in the third set of 20balls --->1.this is again div. 10-10 and weighed --->2the heavier 10balls are again broken down into 5-5 and weighed ---->3.

Now make it into 3 sets of (2,2,1) balls.Weighing 2-2. ---->4Two more cases arise here---->If both these sets of balls are of equal weight than we know the fifth ball in the set is theheavier ball.

---->If both these sets of balls are not equal. Then take the heavier side which has 2 balls andweigh again. Now we get 1-1. Surely we could figure out the heavier ball from this..Hence 4 weighings.

(b) The other case being 30-30 are not equal. Meaning the heavier ball is in one of these sets. --->1Divide 30 balls into 3 sets of (10,10,10) and weigh 10-10. If they are equal... The whole thingfollows the same reasoning as case (a) and at the end we could do it in 4 weighings. Hope it'sclear.

Alternate soln

Weighings Set 1 Set 2 Set 3

First 27 27 26

Second 9 9 9

Third 3 3 3

Fourth 1 1 1



Divide the ball into diff sets as shown in the attachment

First measure Set 1 and Set 2

Case:1:======If one of the scale tilts,we know which set has the heavy ball.

Case2:====If the scale balances equally we know the heavy ball is in Set 3.So Set 1 and Set2 balls are of normal size.So add a ball frm set1/set2 to set3 (27balls) and continue the weighings.

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A dealer Mr. Bond purchases Jaggery from two villages A and B depending on availability. Invillage A cost price is Rs. 100 per 10 kg & in village B the cost price is Rs. 110 per 10 kg. The

jaggery is traded in the blocks of 10, 20 and 30 Kg only. He sells the jaggery in 2 different

markets X and Y depending on demand. He selles jaggery at Rs 110 per 10 kg at market X andRs 121 per 10 Kg at market Y, such that he gets at least 10% profit on each block.He invested i.e. bought jaggery with an amount of Rs. 500 on the first day and furthur traded withsale proceeds only. The purchase/sale of jaggery is done only in cash.The jaggery which hepurchased on a particular day cannot be sold on that day itself. On any day revenue obtained byselling jaggery cannot be used for purchase of jaggery on that day itself.On 2nd and 6th day he sold 10 Kg and 60 Kg of jaggery respectively.On 3rd, 4th & 8th days he bought 10kg, 20kg, 50kg of jaggery respectively.On 9th day he did not buy any jaggery but got Rs 440 from the sales of jaggery.On 10th day he sold all the stock left with him at minimum profit. He did not buy any jaggery on10th day.

Q1) What is the maximum possible number of 30 kg blocks he traded in?

1) 4 2)3 3)2 4)1 5) None of these

Q2) What is the maximum possible quantity of jaggery that he sold in market Y?1)120kg 2)130kg 3)150kg 4)140kg 5)none of these

Q3) What is the maximum possible profit (in Rs.)?1) 296 2)324 3) 253 4)334 5)none of these

Q4) What can be the minimum profit (in Rs.)?1) 296 2)334 3) 231 4)303 5)none of these

Q5)Which of the following statements is false?1) Mr Bond sold 20 kg of jaggery on 3rd day

2) Mr Bond sold 30 kg of jaggery on 4th day3) Mr Bond sold no jaggery on 5th day4) Mr Bond bought 20 kg of jaggery on 5th day5) Mr Bond sold no jaggery on 8th day

Answers:



Hence forth i will refer to each 10kgs as an "unit"

Cost price at A = 100 per unitCost price at B = 110 per unit

Selling price at X = 110 per unitSelling price at Y = 121 per unit

First Key point : "such that he gets at least 10% profit on each block"Therefore he can never buy at B and sell at X since he will not make the 10% profit.

So he has sold 1 on Day 2 . So on Days 3 and 4 he has to sell at least 4. He can sell more for more profit.

Also he cannot sell anything on Day 5 as he needs everything for Day 6. He also has to buy 4 onDay 5.

So we've already solved the 5th question since we have to buy 4 on Day 5.

Questions 1,2,3 all refer to the same thing asa) You only get maximum profit when you buy everything at A and sell at Y for a profit of 21b) You can maximize 3 unit packets only when you buy lots.On the 7th day you can buy 2 packets and sell on the 8th. (remember for maximum profits alwaysbuy as much as you can cheaply and sell it for maximum profit)

I've maximized the number of units your sell or buy which are 3+ .. These are the blocks which hetrades inBuys on First day - Sells on ThirdBuys on Fifth day - Sells on sixthBuys on Eight Day - Sells on NinthSo Totally 3 blocks

Remember for maximum profit you sell everything at Y , but its given for Day 9 that you sell 4units at 440 rs which means Village X. (this has been cell marked with yellow) . Everything else issold at Y (marked in green) - So totally 15 units = 150 Kgs

And the cash profit comes to 834 - 500 = 334 Rs

I'll leave the minimum profit for you to do (Look at below points for hints - see only if you're stuckor not getting the answer)



a) Remember even though buying at B and Selling at X would give you 0 profit - it cannot bedone as the question explicitly states 10% profitb) Buying at A and selling at X better than Buying at B and Selling at Y because for the former its10Rs profit while for the latter it is 11 Rs profit.c) You need to make an extra 100 by day 5 even when calculating minimum profit !!d) Leave Day 7 alone - when you do nothing , you do not make profits ..

I'm getting the answer as 214 for minimum profit.

Q1) What is the maximum possible number of 30 kg blocks he traded in?1) 4 2)3 3)2 4)1 5) None of these

Q2) What is the maximum possible quantity of jaggery that he sold in market Y?1)120kg 2)130kg 3)150kg 4)140kg 5)none of these

Q3) What is the maximum possible profit (in Rs.)?1) 296 2)324 3) 253 4)334 5)none of these

Q4) What can be the minimum profit (in Rs.)?1) 296 2)334 3) 231 4)303 5)none of these

Q5)Which of the following statements is false?1) Mr Bond sold 20 kg of jaggery on 3rd day2) Mr Bond sold 30 kg of jaggery on 4th day3) Mr Bond sold no jaggery on 5th day4) Mr Bond bought 20 kg of jaggery on 5th day5) Mr Bond sold no jaggery on 8th day

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"Half the average is a game played with 8 players.

In this game, everyone starts with 100 points. Each round, each person chooses a integer

between 1 and 100 inclusive and writes it on a slip of paper. After each round, half the average of all the choices is calculated. Anyone below half the average will be eliminated. Anyone above or equal to the half the average will have half the average subtracted from their guess and theresulting total subtracted from their points. If someone runs out of points that person is out.

Note : Half the Average will always be rounded to the nearest integer (.5 rounds up) "

Q1) What is the minimum possible rounds that the game can take?a)1 b)2 c)3 d)4 e)7

Q2) What is the maximum possible rounds that the game can extend upto?a)4 b)7 c)8 d)16 e)None of these

A person who is "paranoid" assumes all other 7 players are colluding against him. He alwayschooses a number such that he cannot get eliminated by being lower than half the average.

Q3) What will he choose on the first round of the game?a)100 b)99 c)50 d)49 e)47

Q4) Suppose you know all the 7 other players in the game are "paranoid" . What is your bestchoice in the first round of the game?a)100 b)75 c)25 d)13 e)None of these



• Amit lives in the city, which has the shortest name amongst the above cities.

Q1]If Akansh lives in Vastrapur, then what is the average salary of the persons living in Banerghattaand Kunnamangalam?

1. Rs 9 lakh2. Rs 10 Lakh3. Rs 12 Lakh4. Rs 10.5 Lakhs5. Data Insufficient

Q2]Who says in Prabandhnagar?

1. Ashok2. Amit3. abhishek4. Ajay5. Akansh

Q3]If Amit and Ajay lives in cities with names starting with consecutive alphabets, then who lives inVastrapur?

1. Ashok2. Amit3. Abhishek4. Ajay5. Akansh

Q4]If person from Banerghatta does not earn the maximum or the minimum salary, then what is the

average salary of persons living in Kunnamanglam and Vastrapur?

1. Rs 10.5 lakhs2. Rs 10 lakh3. Rs 12 lkh4. Rs 9 lakh5. Cannot be determined.

Answers

1] i think we cant deduce from the given data that abhishek lives at prabhandnagar...we candeduce that person living in prabhandnagar earns 9 lacs but we cant say its abhishek...

by condition "The difference between slaries of Akansh and Ajay is the same as thedifference between salaries of Ashok and Abhishek" we can have two sets1. (13,11) & (9,7)2. (13,9) & (11,7)

now if we put together the conditionsajay-100K = (akansh+ashok)/2 &akansh must be one of 7,11,13



we will find that

ashok,7ajay,11amit,8, jokaakansh,13abhishekh,9

and then we can say that abhishek lives in prabhandnagar

so if Akansh lives in Vastrapur, Ajay and Ashok live in Banerghatta and Kunnamangalam... inany order...so their average salary is (1100000+700000)/2=900000

Hence (1)

2] Since Abhishek gets 900K salary. From given info. He has to stay in Prabandh...Hence (3)

3] Amit lives in Joka. So Ajay has to live in Kunnama… Remaining places are bannerghatta and

vastrapur. But Akansh cannot live in bannerghatta. So He lives in vastrapur. Hence (5)

4] Currently we have 2 open slots. For Ashok and Ajay, Akansh. Ashok gets (1300K or 700K). Butthe one living in bannergh… does not have min or max salary. so Ajay lives in bannerghatta and(Akansh, ashok) live in either of (vastrapur, kunnama…).Although they have salaries in pairs of (1300K or 700K) and (700K, 1300K) in both cases theaverage comes to 10K. Hence (2)

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For Questions 1 to 5 read the following instructions:Nine horses participated in a horse tournament. Five races were held in this tournament. The

winner of a race gets 5 points, 2nd gets 3 and the 3rd gets 1. The table gives the point’s tally.

Horse | A | B | C | D | E | F | G | H | IPoints | 5 | 5 | 8 | 2 | 5 | 6 | 4 | 10| 0

One race is held between five horses. The first race is held among the first 5 horses from the leftin the above table. Out of these horses, A drops out of the race and a new horse. I.e. F entersinto the race. In the next race, B drops out and G enters and so on. It is also given that H is theonly horse that scored in 2 consecutive races.

Q.1 What was the position of horse F in race 4?a. 4th b. 2nd c. 3rd d. cannot be determined e.1st

Q.2 What is the ratio of the points scored by “G” and E in race 5?a. 1:3 b. 3:1 c. 1:5 d. cannot be determined e. 5:1

Q.3 The first 3 rankers of the race 3 in some order?a. C,E,G b. D,E,F c. C,D,F d. C,E,F e. C,G,F

Q.4 D could have come third in which of the following races?a. Race 1 b. Race 3 c. Race 4 d. Race 5 e. cannot be determined



Q.5 Suppose if C had 6 points and E had 7 points while the points of the rest of the horsesremain unchanged then E could’ve come second ina. One race b. Two races c. At most one race d. At most 2 races e. cannot be determined

1)B2)D3)A4)C5)D

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You are one of 20 prisoners on death row with the execution date set for tomorrow. Your king is aruthless man who likes to toy with his people's miseries. He comes to your cell today and tellsyou:“I’m gonna give you prisoners a chance to go free tomorrow. You will all stand in a row (queue)before the executioner and we will put a hat on your head, either a red or a black one. Of courseyou will not be able to see the color of your own hat; you will only be able to see the prisoners infront of you with their hats on; you will not be allowed to look back or communicate together in

any way (talking, touching.....).

The prisoner in the back will be able to see the 19 prisoners in front of him. The one in front of him will be able to see 18…

Starting with the last person in the row, the one who can see everybody in front of him, he will beasked a simple question: WHAT IS THE COLOR OF YOUR HAT?

He will be only allowed to answer “BLACK” or “RED”. If he says anything else you will ALL beexecuted immediately.

If he guesses the right color of the hat on his head he is set free, otherwise he is put to death.And we move on to the one in front of him and ask him the same question and so on…

Well, good luck tomorrow, HA HA HA HA HA HA!”

Now since you all can communicate freely during the night, can you find a way to guarantee thefreedom of some prisoners tomorrow? How many?

Is this the answer - 19 prisoners will be able to save their lives and the 20th prisoner (thelast one) will have a chance of 50% to live.

I think 19 lives can be saved and last person will be having 50% chance.

My approach was

Let Red cap= +1Black Cap =-1

now 20th person will take product of corresponding value of caps for all the pepole standing infront of him and tell the color of cap corresponding to the product. so he will be having 50%chance

simillarly 19th person will take the product of all 18 cap values of pepole standing in front of himhe knows the product of 19 caps so he can tell the color of his cap



for 18th-2nd person, they can tell the color of cap using the equation

Product of corresponding cap values of pepole standing in front of him* his cap value* preoduct of corresponding values of color of cap told by pepole standing behind him starting from 19th=Corresponding value of cap color told by 20th person

for 1st person equation will behis cap color*preoduct of corresponding values of color of cap told by pepole standing behind himstarting from 19th= Corresponding value of cap color told by 20th person----------------------------------------------------------------------------------------------------------------

Three men, including Gianni and three woman, including Sachi are in line at the BrentWoodpost office. Each has two different pieces of business to conduct.

1. The first person is a woman.2. Carlos wants to send an overnight package.3. Lau is just ahead of Pimentelli who is the same sex as Lau.4. Gianni is two places ahead of the person who wants to buy stamps.5. Knutson - who is the opposite sex than Rendler - isn't the person who wanted to

complain about a mail carrier.6. The six people, not necessarily in the same order are - Anthony, Donna, the personwho wants to fill out a change-of-address form, the one who wants to buy a money order, theone who wants to send Airmail to Tibet and the second person in the line.7. The four tasks of the last two people in line, not necessarily in the same order are- sending books fourth class, buying a money order, picking up a package and complainingabout a mail carrier.8. The person who wants to send books fourth class is just behind a person of the same sex.9. Mary is just behind a person who wants to send an insured package.10. The person who wants to send Airmail to Tibet is either two places ahead of or twoplaces behind the one who wants to add postage to his or her meter.11. Anthony isn't two places behind the who wants to pickup a registered letter.12. Toriseza is two places ahead of the person who wants to pick up a package.

13. Knutson isn't just ahead of the person who wants to send an item parcel post.Can you figure out where each customer is in the line, his or her full name (one surname isLoti) and the two things he or she wants to accomplish?

Provide your answer is POSITION - FIRST NAME - LAST NAME - BUSINESS format.

A very TOUGH puzzle !

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In a society, there are five buildings, A, B, C, D and E, in a straight line in the same order. Eachone of them is having different number of floors. The numbers of floors in two consecutive

buildings are not consecutive numbers. One day Mr. Sharma planned to assign a number to eachof these floors. He picked up one of the five buildings randomly and gave an arbitrary number toits ground floor. He then proceeded from the ground floor to the top floor, giving consecutivenumbers to the floors. For example, he could give the number 41 to the ground floor, 42 to thefirst floor, 43, to the second floor and so on. After reaching the top floor of the building, he chosethe next building randomly and numbered its ground floor as the next number after that of the topfloor of the previous building and proceeded again in the previous manner. After numbering allthe floors of all the buildings, he observed the following:

1. Sum of the numbers of all the top floors is 119.2. Sum of the numbers of all the ground floors is 109.



3. The sum of all the floors of building B is 81.4. The number of Mr. Sharma’s flat, which is at the top floor of building A, is 21.

1. Find out the sum of all the floors of building E.61 60 55 49 312. Find out the number of floors in building C1 2 3 4 53. What is the highest number assigned to a floor by Mr. Sharma?28 29 30 31 324. If sums of all the top floors and ground floors are 134 and 114, respectively, what is the sum of all the floors of all the building?

20 24 25 30 35

Answers

option5option1option5option2----------------------------------------------------------------------------------------------------------------

For Independence Day celebration, Deepak brought some candies to distribute among hisfavourite students Upasana, Pooja, Tanisha and Ramneet. But they were busy in flag hoistingceremony; Deepak sir put the candies box on a table and told them to share equally. Immediatelyafter the flag hoisting, Pooja, who is very fond of cadies took her share and 3 candies extra. Butnone of the other three observed her. After some time Ramnet, an honest girl took exactly one-fourth of the candies, from the box. Then, Upasana took one-third of the remaining candies.Finally, innocent Tanisha, who did not know that all the other three had taken their shares, tookexactly one-fourth of the remaining candies as her share, leaving the rest in the box.Answer the following questions.71. Which of the students got equal number of candies?(1) Upasana and Pooja (2) Pooja and Tanisha(3) Ramneet and Upasana \(4) Poooja and Ramneet(5) Upasana and Tanisha

72. What was the least possible number of candies that Deepak could bring?(1) 32 (2) 68 (3) 36(4) 56 (5) 7273. If the number of candies that Pooja got was equal to the number of candies that bothUpasana and Tanisha got together, how many candies did "Deepak bring?(1) 100 (2) 132 (3) 164 (4) 200 (5) Cannot be determined74. If Upasana got 18 candies, how many candies did Deepak bring?(1) 68 (2) 100 (3) 132 (4) 170 (5) 20075. If the number of candies left on the table, ',after/all the four girls took their shares, was equalto thenumber of candies that Pooja got; how many candies did Ramneet get?(1) 6 (2) 18 (3") 36 (4) 30 (5) 24

answers are

71. 372. 373. 274. 275. 5

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410 students of 5 different colleges have registered in TathaGat for a classroom program. Thesecolleges are NSIT, IIT, Jamia, JSS and North Campus (DU). TathaGat team has divided the cityin a grid network as shown in the figure below. The grid network follows the co–ordinate rule toidentify the points on it. One can travel along this grid only. The five colleges are shown by dots( .</span>)</span> and the number written in bracket below a college’s name represent thenumber of students from the respective college.

67. If TathaGat team decides to open only one center, find out the co-ordinate of this only center such that the sum of the total distance traveled by all the students is minimum.(0, 4) (8, 7) (6, 15) (6, 4) none of these68. After six months team proposed to start another center without shifting the first center of theabove problem. Find out the co-ordinate of the second center assuming that this proposal is putforward just to further minimize the sum of the distance traveled by all the students.(0, 4) (8, 7) (6, 15) (6, 4) none of these69. The ratio of the number of student in two centers of the first two problems is given by23:18 17:24 29:12 30:11 none of these70. If the team decided to open two centers in a way such that the sum of the total distancestraveled by all the students is minimum. Find out the coordinates of these two centers along thegrid line. (do not use data from the previous question)

(0, 4) & (8, 7) (0, 4) & (6, 15) (8, 7) & (6, 15) (6, 15) & (6, 4) none of these

Answers1) Option(D)2) Option(C)3) Option(C)4) Option(B)



1) Tried various distances. Main point being, to min. the cumulative distance for all the students, itmakes sense to have ti near a place(s) with max. students. Tried at (0,4) etc. But at (6,4) its min.i.e. almost the centroid. Hence (D)

2) This one is simple! It has to be a place most far off and has many students. (6, 15) satisfiesthis. Hence (C)

3) All other students (except North Campus) to to centre1 i.e. 290 students. In centre 2: 120students. Ratio: 29:12. Hence (3)

4) Obviously these 2 centers have to be opened at places which has max. students. i.e. Northcampus and NSIT. Hence (0, 4) and (6,15). Hence (B)

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A botanical experiment is being conducted on three varieties of sweet pea- Healthy, infected andresistant. In the three stages of the experiment, equal number of plants from two of the threevarieties are chosen and crossed with each other in pairs. Whenever, a plant of healthy variety iscrossed with that of infected variety, it results in the healthy plant also becoming infected.

Whenever a plant of infected variety is crossed with that of resistant variety, it results in theinfected plant dying out. Whenever a plant of resistant variety is crossed with that of healthyvariety, it results in the healthy plant also becoming resistant. The bar chart below shows theinitial and stage-wise percentage of plants of the three varieties out of the total number of plants.

The crossing between the healthy variety and the resistant variety is happening inChoose one answer.

a. stage 1

b. cannot be determined

c. stage 3

d. stage 2

76Marks: --/1What is the percentage change in the number of plants of the healthy variety in stage 1 comparedwith the initial number of plants?Choose one answer.

a. 16.6%

b. no change

c. 12.5%



1) a2) a3) d

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You are locked inside a room with 6 doors - A, B, C, D, E, F. Out of which 3 are Entrances onlyand 3 are Exits only. One person came in through door F and two minutes later second personcame in through door A. He said, "You will be set free, if you pass through all 6 doors, each door once only and in correct order. Also, door A must be followed by door B or E, door B by C or E,door C by D or F, door D by A or F, door E by B or D and door F by C or D."

After saying that they both left through you pass through the doors?

Answer

The correct order is CFDABEIt is given that one person came in through door F and second person came in through door A. It means that door A and door F are Entrances. Also, they both left through door B.

Hence, door B is Exit.As Exit and Entrance should alter each other and we know two Entrances, let's assume thatthe third Entrance is W. Thus, there are 6 possibilities with "_" indicating Exit.

(1) _W_A_F (2) _W_F_A (3) _F_W_A (4) _F_A_W (5) _A_W_F (6) _A_F_W

As door A must be followed by door B or E and none of them lead to the door F, (1) and (6)are not possible.Also, door D must be the Exit as only door D leads to the door A and door A is theEntrance.(2) _W_FDA (3) _F_WDA (4) _FDA_W (5) DA_W_FOnly door D and door C lead to the door F. But door D is used. Hence, door C must be theExit and precede door F. Also, the third Exit is B and the W must be door E.

(2) BECFDA (3) CFBEDA (4) CFDABE (5) DACEBFBut only door B leads to the door C and both are Exits. Hence, (2) and (5) are notpossible. Also, door F does not lead to door B - dis(4) i.e. CFDABE.

All who posted answer as CFDABE. It's the right answer !

----------------------------------------------------------------------------------------------------------------In the middle of the confounded desert, there is the lost city of "Ash". To reach it, Iwill have to travel overland by foot from the coast. On a trek like this, each person canonly carry enough rations for five days and the farthest we can travel in one day is 30miles. Also, the city is 120 miles from the starting point.What I am trying to figure out is the fewest number of persons, including myself, that Iwill need in our Group so that I can reach the city, stay overnight, and then return to the

coast without running out of supplies.

How many persons (including myself) will I need to accomplish this mission?

Answer

Yes the answer is 4; as many have you got! Here's my approach-

Total 4 persons (including you) required.



It is given that each person can only carry enough rations for five days. And there are 4 persons.Hence, total of 20 days rations is available.

1. First Day : 4 days of rations are used up. One person goes back using one day of rations for the return trip. The rations remaining for the further trek is for 15 days.

2. Second Day : The remaining three people use up 3 days of rations. One person goes backusing 2 days of rations for the return trip. The rations remaining for the further trek is for 10 days.

3. Third Day : The remaining two people use up 2 days of rations. One person goes back using 3days of rations for the return trip. The rations remaining for the further trek is for 5 days.

4. Fourth Day : The remaining person uses up one day of rations. He stays overnight. The nextday he returns to the coast using 4 days of rations.Thus, total 4 persons, including you arerequired.

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Q. a party is held at the house of the mehtas. there are 5 other couples present(besides mr & mrs

mehta), and many,but not all, pairs of people shook hands. nobody shook hands with anyonetwice, and nobody shook hands with his or her spouse. both the host and hostess shook somehands.at the end of the party, mr mehta poll each person present to see how many hands eachperson(other than himself) shook. each person gives a different answer.determine how many hands mrs mehta must have shaken.

Answer: 5

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There are 10 statements written on a piece of paper:

1. At least one of statements 9 and 10 is true.2. This either is the first true or the first false statement.3. There are three consecutive statements, which are false.4. The difference between the numbers of the last true and the first true statementdivides the number, that is to be found.5. The sum of the numbers of the true statements is the number, that is to be found.6. This is not the last true statement.7. The number of each true statement divides the number, that is to be found.8. The number that is to be found is the percentage of true statements.9. The number of divisors of the number, that is to be found, (apart from 1 and itself)is greater than the sum of the numbers of the true statements.10. There are no three consecutive true statements.

Find the minimal possible number?

Answer

The number is 420.If statement 6 is false, it creates a paradox. Hence, Statement 6 must be true.Consider Statement 2:* If it is true, it must be the first true statement. Otherwise, it creates a paradox.* If it is false, it must be the second false statement. Otherwise, it creates a



paradox.In both the cases, Statement 1 is false.As Statement 1 is false, Statement 9 and Statement 10 both are false i.e. there are threeconsecutive true statements.1 2 3 4 5 6 7 8 9 10False - - - - True - - False FalseLet\'s assume that Statement 3 is false i.e. there are no three consecutive falsestatements. It means that Statement 2 and Statement 8 must be true, else there will bethree consecutive false statements.1 2 3 4 5 6 7 8 9 10False True False - - True - True False FalseAlso, atleast two of Statements 4, 5 and 7 must be true as there are three consecutive truestatements.According to Statement 8, the number that is to be found is the percentage of truestatements. Hence, number is either 50 or 60. Now if Statement 7 is true, then the number of each true statement divides the number, that is to be found. But 7 and 8 do not divideeither 50 or 60. Hence, Statement 7 is false which means that Statement 4 and 5 are true.But Statement 5 contradicts the Statement 8. Hence, our assumption that Statement 3 isfalse is wrong and Statement 3 is true i.e. there are 3 consecutive false statements whichmeans that Statement 8 is false as there is no other possibilities of 3 consecutive false

statements.Also, Statement 7 is true as Statement 6 is not the last true statement.1 2 3 4 5 6 7 8 9 10False - True - - True True False False FalseAccording to Statement 7, the number of each true statement divides the number, that is tobe found. And according to Statement 5, the sum of the numbers of the true statements isthe number, that is to be found. For all possible combinations Statement 5 is false.There 3 consecutive true statements. Hence, Statement 2 and Statement 4 are true.1 2 3 4 5 6 7 8 9 10False True True True False True True False False FalseNow, the conditions for the number to be found are:1. The numebr is divisible by 5 (Statement 4)2. The numebr is divisible by 2, 3, 4, 6, 7 (Statement 7)

3. The number of divisors of the number, that is to be found, (apart from 1 and itself)is not greater than the sum of the numbers of the true statements. (Statement 9)The minimum possible number is 420.The divisors of 420, apart from 1 and itself are 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21,28, 30, 35, 42, 60, 70, 84, 105, 140, 210. There are total of 22 divisors. Also, the sum of the numbers of the true statements is 22 (2+3+4+6+7=22), which satisfies the thirdcondition.----------------------------------------------------------------------------------------------------------------

500 men are arranged in an array of 10 rows and 50 columns according to their heights. Tallestamong each row of all are asked to come out. And the shortest among them is A. Similarly after resuming them to their original positions, the shortest among each column are asked to come out.And the tallest among them is B.

Now who is taller A or B ?

Answer

As it is mentioned that 500 men are arranged in an array of 10 rows and 50 columns according totheir heights. Let's assume that position numbers represent their heights. Hence, the shortestamong the 50, 100, 150, ... 450, 500 is person with height 50 i.e. A.Similarly the tallest among 1, 2, 3, 4, 5, ..... 48, 48, 50 is person with height 50 i.e. BNow, both A and B are the person with height 50. Hence both are same.



No one is taller, both are same as A and B are the same person.

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A cricket team of 11 players lined up in a straight line to have their photograph. The captain wasasked to stand in the center of the line-up.

1) Bharat and Bhavin stood to the right of the captain2) Two players stood between Bhagat and Bhairav3) Seven players stood between Bhadrik and Bhanu4) Bhavesh stood to the right of Bhuvan5) Bhola and Bhumit stood either side of Bhagat6) Bhavik and Bhumit stood to the left of the captain7) Six players stood between Bhavin and Bhagat

Two players stood between Bhagat and Bhavik

Who is the captain? Can you tell the positions of all the palyers?

Answer

Players from left to right : Bhavik, (Bhadrik/Bhanu), (Bhola/Bhumit), Bhagat,(Bhola/Bhumit), BHUVAN, Bhairav, (Bharat/Bhavesh), (Bharat/Bhavesh), (Bhadrik/Bhanu),Bhavin

Let's number the positions 1 to 11 from left to right. Hence, the captain is at position 6.Now, looking at the clues 7, 5, 2 and 8 together:Poistion 1 - Bhavik or BhairavPosition 3 - Bhumit or BholaPosition 4 - BhagatPosition 5 - Bhumit or BholaPoistion 7 - Bhavik or Bhairav

Position 11 - Bhavin

From clue (3), the only possible positions for Bhadrik and Bhanu are Position 2 andPosition 10.Now there are 3 positions remaining - 6, 8 and 9 and remaining 3 players are Bhuvan, Bharatand Bhavesh. But from clue (1), Bharat stood to the right of the captain i.e. Bharat mustbe on position 8 or 9 as position 6 is for the captain. So either Bhuvan or Bhavesh is thecaptain.From (4), Bhavesh stood to the right of Bhuvan. Hence, Bhuvan is the captain.Players from left to right are : Bhavik, (Bhadrik/Bhanu), (Bhola/Bhumit), Bhagat,(Bhola/Bhumit), BHUVAN, Bhairav, (Bharat/Bhavesh), (Bharat/Bhavesh), (Bhadrik/Bhanu),Bhavin.Thus,

* Bhavik(1), Bhagat(4), Bhuvan(6), Bhairav(7) and Bhavin(11) are the players whosepositions are fixed.* Bhadrik and Bhanu are at position 2 or 10.* Bhola and Bhumit are at position 3 or 5.* Bharat and Bhavesh are at position 8 or 9.

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A woman took a certain number of eggs to the market and sold some of them.The next day, through the industry of her hens, the number left over had been doubled, andshe sold the same number as the previous day.On the third day the new remainder was tripled, and she sold the same number as before.On the fourth day the remainder was quadrupled, and her sales the same as before.On the fifth day what had been left over were quintupled, yet she sold exactly the same ason all the previous occasions and so disposed of her entire stock.What is the smallest number of eggs she could have taken to mmany did she sell daily? Note thatthe answer is not zero.

Answer

She took 103 eggs to market on the first day and sold 60 eggs everyday.Let's assume that she had N eggs on the first day and she sold X eggs everyday. Puttingdown the given information in the table as follow.Days Eggs at the start of the day Eggs Sold Eggs RemainingDay 1 N X N-XDay 2 2N-2X X 2N-3XDay 3 6N-9X X 6N-10XDay 4 24N-40X X 24N-41X

Day 5 120N-205X X 120N-206XIt is given that she disposed of her entire stock on the fifth day. But from the tableabove, the number of eggs remaining are (120N-206X). Hence,120N - 206X = 0120N = 206X60N = 103XThe smallest value of N and X must be 103 and 60 respectively. Hence, she took 103 eggs tomarket on the first day and sold 60 eggs everyday.

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Four couples are going to the movie. Each row holds eight seats. Betty and Jim don't wantto sit next to Alice and Tom. Alice and Tom don't want to sit next to Gertrude and Bill. On

the otherhand, Sally and Bob don't want to sit next to Betty and Jim.How can the couples arrange themselves in a row so that they all sit where they would like?

AnswersFrom the given data, it can be inferred that:(Sally & Bob) NOT (Betty & Jim) NOT (Alice & Tom) NOT (Gertrude & Bill)(A) NOT (B) means A and B can not seat next to each other.Now, it is obvious that (Betty & Jim) and (Alice & Tom) will occupy the corner seats asboth of them can have only one neighbour. Therefore,(Gertrude & Bill) will seat next to (Betty & Jim)(Sally & Bob) will seat next to (Gertrude & Bill)(Alice & Tom) will seat next to (Sally & Bob)Thus, there are two possible arrangements - a mirror images of each other.

1. (Betty & Jim) - (Gertrude & Bill) - (Sally & Bob) - (Alice & Tom)2. (Alice & Tom) - (Sally & Bob) - (Gertrude & Bill) - (Betty & Jim)

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Substitute digits for the letters to make the following relation true.

N E V E RL E A V E+ M E



-----------------A L O N E

Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-onemapping between digits and letters. e.g. if you substitute 3 for the letter can be 3 and all other Min the puzzle must be 3.

Ps: There seems to be some indentation prob. everything in the addition is right aligned.

Answer Since R + E + E = 10 + E, it is clear that R + E = 10 and neither R nor E is equal to 0 or 5. This is the only entry point tosolve it. Now use trial-n-error method.

2 1 4 1 93 1 5 4 1+ 6 1-----------------5 3 0 2 1

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Shahrukh speaks truth only in the morning and lies in the afternoon, whereas Salman speakstruth only in the afternoon and lies in the morning.A says that B is Shahrukh.Is it morning or afternoon and who is A - Shahrukh or Salman?

Answer It is Afternoon and A can be Salman or Shahrukh. If A is Salman, he is speaking truth. If Ais Shahrukh, he is lying.Want to confirm it? Consider following 4 possible answers and check for its truthnessindividually.

1. It is Morning and A is Shahrukh2. It is Morning and A is Salman3. It is Afternoon and A is Shahrukh4. It is Afternoon and A is Salman

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Three men, including Gianni and three woman, including Sachi are in line at the BrentWoodpost office. Each has two different pieces of business to conduct.

1. The first person is a woman.2. Carlos wants to send an overnight package.

3. Lau is just ahead of Pimentelli who is the same sex as Lau.4. Gianni is two places ahead of the person who wants to buy stamps.5. Knutson - who is the opposite sex than Rendler - isn't the person who wanted tocomplain about a mail carrier.6. The six people, not necessarily in the same order are - Anthony, Donna, the personwho wants to fill out a change-of-address form, the one who wants to buy a money order, theone who wants to send Airmail to Tibet and the second person in the line.7. The four tasks of the last two people in line, not necessarily in the same order are- sending books fourth class, buying a money order, picking up a package and complainingabout a mail carrier.



8. The person who wants to send books fourth class is just behind a person of the same sex.9. Mary is just behind a person who wants to send an insured package.10. The person who wants to send Airmail to Tibet is either two places ahead of or twoplaces behind the one who wants to add postage to his or her meter.11. Anthony isn't two places behind the who wants to pickup a registered letter.12. Toriseza is two places ahead of the person who wants to pick up a package.13. Knutson isn't just ahead of the person who wants to send an item parcel post.

Can you figure out where each customer is in the line, his or her full name (one surname isLoti) and the two things he or she wants to accomplish? ProvFIRST NAME - LAST NAME -BUSINESS format.

Answers:

Anyway let me post the answer to this atleast (if not the explanation

POS - FIRST NAME & LAST NAME - BUSINESS1. Sachi Loti - Fill Out a Change-of-Address Form - Add Postage to Meter 2. Gianni Lau - Pick Up a Registered Letter - Send an Item Parcel Post3 Carlos Pimentelli - Overnight Package - Send Airmail to Tibet4 Donna Toriseza - Buy Stamps - Send an Insured Package5 Mary Knutson - Buy a Money Order - Send Books fourth Class6 Anthony Rendler - Complain About a Mail Carrier - Pick Up a Package

In this puzzle clues (6) and (7) are the foundation clues. They get your initial blue-print ready.After that I got to clue (10).

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Aarav told Vineet " Today i found two interesting numbers, each consisting of three differentdigits, each is divisible by the product of its digits and the results of the division gives themonth( january=1,etc ) and the day of the month of my birthday".

Vineet already knew the birthday of Aarav and soon found two integers which satisfied theconditions, but he found that they were both different from the numbers Aarav found. What isAarav's birthday?

Answer August 18

There are ten three-digit numbers whose digits are distinct which are divisible by the product of their digits. These are:

128 / 16 = 8132 / 6 = 22216 / 12 = 18

312 / 6 = 52384 / 96 = 4432 / 24 = 18612 / 12 = 51624 / 48 = 13672 / 84 = 8816 / 48 = 17

Only two quotients are repeated; these are 8 and 18.



The only remaining digit set is {7,3,1}. We check all 6 orderings andfind none divisible by 21

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four persons ... meeta, smita, geeta and rita .. gave one statement each. Among the four there isatleast one person who always speaks the truth , atleast one person who always tell lies butexactly one person whose statement cannot be classified as either true or false. Following are thestatements made by them

smita .... I am a liar rita........ I am not a liar geeta... I neither speak the truth nor do i liemeeta.... Rita is a liar

1. whose statement can be classified as true nor as false ??

smitha ... geetha.... rita... cannot be determined

2. who is/are truth teller(s) ??

smitha .... geetha... rita.... either rita or meeta

3. who among the following always tells lies ??

meeta.... geeta....rita.....smita....

Answers

Smita is the person whose statement can be classified neither as true nor as false as the truthspeakers are Either Rita aur Meetaand as exactly only one person can classified neither as true nor as false

so left is Geeta with liar 1)Smita2)Either Rita aur Meeta3)Geeta

My answers :1) smitha... try like this - smitha says that I m a liar, now if this statement is true then smithashuld be truth teller but actually she alleges that she is a liar...so not possible; again if thestatement is lie then it implies that she is a truth teller but that again contradicts our assumption,thus her statement can not be classified as either true or lie bcoz of the inherent paradox..

2) either rita or meeta... Here Rita's statement can either be true or false. Now according to

condition only one's statement cannot be determined as truth or lie hence Githa's statement "Ineither speak the truth nor do i lie " is a lie... Now Meeta can either lie or speak truth,if she saystruth then rita becomes liar and her statement becomes a lie which is feasible, again i she speakslie then rita becomes truth teller and her statement becomes truth which again is feasible... AsMeeta has to be one of them thus if Meeta lies then Rita speaks truth and other way around

3) geetha : as proven above geetha always speaks lie...

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five persons a,b,c,d,e went to a shop to buy five different goods. below are some additional factsabout how much they spend during their shopping

1... the person who spent Rs.3214 arrived before the person who spent Rs.2143, and he was notd.

2... one person spent Rs.4321 and he was not c.

3... One person spent Rs.189 less than b, who spent the maximum amount.

4... e spent Rs.2898 more than a.

5... each person spent at least Rs.1000, but not more than 5000

Q1)... what is the amount spent by c

2143... 3214... 1234... cannot be determined

Q2)... which of the following could be the amount spent by e

4321.... 4510.... either 1 or 2 ..... none of these

Q3).... which of the following statements may be true

amount spent by a is Rs.1423......

amount spent by b is Rs.4132......

amount spent by d is Rs.4321......

more than one of the above........

Answers1)Amt spent by C:-32142)4321 i.e. option 13)option 1

Person who spent 3214 arrived before person who spent 2143 and he's not d.so person whospent 3214 can be a,b or c (e is not possible as no one came after 'e') .... it cant be 'b' as weknow b is highest and we know someone spent 4321. Now e spent 2898 more then a so if a is3214 amount spent by e will exceed limit of 5000 so it cant be 'a' So only c is left...So c spentRs3214

Person who spent 2143 came after 'c' so it can be either d or e. but we know amount spent by e>2898 (e=a+2898 )....so d spent Rs 2143

Now b>e>a...Now we know someone spent 4321 and it cant be a (as e=a+2898 ).......but we cantbe sure out of b or e who spent 4321.... so 1st assume 'e' spent 4321 then b=4510 anda=1423....2nd assume b spent 4321 then e=4132 and a=1234

So now we haveCase 1(e=4321) Case 2(b=4321)a=1423.........................1234b=4510.........................4321c=3214..........................3214



d=2143..........................2143e=4321..........................4132

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There is a safe with a 5 digit number as the key. The 4th digit is 4 greater than thesecond digit, while the 3rd digit is 3 less than the 2nd digit. The 1st digit is thrice thelast digit. There are 3 pairs whose sum is 11.Find the number.

My Solution:

65292As per given conditions, there are three possible combinations for 2nd, 3rd and 4th digits.

They are (3, 0, 7) or (4, 1, or (5, 2, 9)

It is given that there are 3 pairs whose sum is 11. All possible pairs are (2, 9), (3, ,(4, 7), (5, 6). Now required number is 5 digit number and it contains 3 pairs of 11. So itmust not be having 0 and 1 in it. Hence, the only possible combination for 2nd, 3rd and 4thdigits is (5, 2, 9)Also, 1st digit is thrice the last digit. The possible combinations are (3, 1), (6, 2) and(9, 3), out of which only (6, 2) with (5, 2, 9) gives 3 pairs of 11. Hence, the answer is 65292.

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Ankit and Tejas divided a bag of Apples between them.Tejas said, "It's not fair! You have 3 times as many Apples I have." Ankit said, "OK, Iwill give you one Apple for each year of your age." Tejas replied, "Still not fair. Now,you have twice as many Apples as I have." "Dear, that's fair enough as I am twice older than you.", said Ankit.Ankit went to Kitchen to drink water. While Ankit was in Kitchen, Tejas took apples fromAnkit's pile equal to Ankit's age. Who have more apples now?

Answer:

At the end, Ankit and Tejas, both have the same number of apples.Let's assume that initially Tejas got N apples and his age is T years. Hence, initiallyAnkit got 3N apples and his age is 2T years.

Operation Ankit's Apples Tejas's ApplesInitially 3N NAnkit gave T apples to Tejas(equals age of Tejas) 3N - T N + T

Tejas took 2T apples from Ankit's pile(equals age of Ankit) 3N - 3T N + 3T

It is given that after Ankit gave T apples to Tejas, Ankit had twice as many apples asTejas had.3N - T = 2*(N + T)3N - T = 2N + 2TN = 3T

From the table, at the end Ankit have (3N - 3T) apples and Tejas have (N + 3T) apples.



Substituting N = 3T, we getAnkit's apples = 3N - 3T = 9T - 3T = 6TTejas's apples = N + 3T = 3T + 3T = 6T

Thus, at the end Ankit and Tejas, both have the same number of apples.

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Question

Five women decided to do shopping at M.G. Road, Banglore. They arrived at the designatedmeeting place in the following order. 1. Archana, 2. Chellamma, 3. Dhenuka , 4. Helen, and 5.Shahnaz. Each women spent at least Rs. 1000. below are some additional facts about how muchthey spent during their shopping spree.i. The woman who spent Rs. 2234 arrived before the lady who spent Rs.1193.ii. One woman spent Rs. 1340 & she was not Dhenuka.iii. One woman spent Rs. 1378 more than Chellamma.iv. One woman spent Rs. 2517 & she was not Archana.v. Helen spent more than Dhenuka.

vi. Shahnaz spent the largest amount & Chellamma the smallest.

1. What was the amount spent by Helen?(1) Rs. 1193 (2) Rs. 1340 (3) Rs. 2234 (4) Rs. 2517.

2. Which of the following amount was spent by one of them?(1) Rs. 1139 (2) Rs. 1378 (3) Rs. 2571 (4) Rs. 2718.

3. The woman who spent Rs. 1193 was(1) Archana (2) Chellamma (3) Dhenuka (4) Helen

Answers:

silly mistake on my part....

the correct answers are:

1.(2)2.(1)3.(3)

The correct order is:Archana Chellamma Dhenuka Helen Shahnaz

2239 1139 1193 1340 2517

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Q. 1 to 4 are based on the following information:

Ghosh Babu's new interest is psychology. He has identified various personalitypatterns and given them names. These personality patterns are inter-related asfollows:- All Alessandras, Belissimas, Cassandras, Desdemonas, Elissimas and Firdauses



for Q4, If peter is a Queen, he should be in either E or B or C. Given that he is not B, so definitelyhe is either C or E. So the answer is option 2. Option 3 is also correct if it is given that peter is aQueen in the question.

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for the figure refer to the attachment in this post

data given...............

The inner most regions represents revenue from Adsense and Total value = Rs. 20,0000

The second inner most regions represent revenue from Adbrite and Total value = Rs. 10,0000

The third inner most regions represent revenue from YPN and Total value = Rs. 12,0000

The outer most regions represent revenue from other sources and Total value = Rs. 14,0000

questions........Q1. What is the percentage increase/decrease in revenue from Q1 to Q4?

Q2. What is the difference in revenue earned from adsense and adbrite in Q2?

Q3. What is the ratio of revenue generated from YPN in Q1 to others in Q4?

Consider the following information for answering Q4 and Q5

The total employee strength of inoideas at the begging of the year was 600.

Q4. If inoideas recruits 70 employees in first 3 quarters then what is the RPE for Q4?RPE = Revenue per Employee= Total revenue in a period / Employees at the beginning of the period

Q5. If no new employee joins inoideas in Q1 and 30 employees joins it during Q2 then what isratio of RPE in Q1 to Q3 (Consider information given in earlier questions if required)

AnswersQ1) 92.61%

Q2) 16,600Q3) 75:91Q4) cannot be determined ( since we dont know what percent of others revenues is of inoideas,we cannot determine the total revenue of inoideas for Quarter 4 )Q5) cannot be determined ( since the percentage contribution of inoideas to others revenue maynot be same in Q1 and Q3 )

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The information in the table below about TCIL as on 31st March:(All figures in Rs. Crores)

Cash loss as a %age of turnover was minimum in2001-022004-052003-04



....Let's examine them one by one.

If the sum of two numbers is 11, Sachin will think that the numbers would be (2,9), (3, ,(4,7) or (5,6).Sachin : "As 11 is not expressible as sum of two primes, Prashant can't know the numbers."

Here, the product would be 18(2*9), 24(3* , 28(4*7) or 30(5*6). In all the cases except

for product 30, Prashant would know the numbers.- if product of two numbers is 18:Prashant : "Since the product is 18, the sum could be either 11(2,9) or 9(3,6). But if thesum was 9, Sachin would have deduced that I might know the numbers as (2,7) is the possibleprime numbers pair. Hence, the numbers must be 2 and 9." (OR in otherwords, 9 is not in thePossible Sum List)- if product of two numbers is 24:

Prashant : "Since the product is 24, the sum could be either 14(2,12), 11(3, or 10(4,6).But 14 and 10 are not in the Possible Sum List. Hence, the numbers must be 3 and 8."- if product of two numbers is 28:Prashant : "Since the product is 28, the sum could be either 16(2,14) or 11(4,7). But 16 isnot in the Possible Sum List. Hence, the numbers must be 4 and 7."- if product of two numbers is 30:

Prashant : "Since the product is 30, the sum could be either 17(2,15), 13(3,10) or 11(5,6).But 13 is not in the Possible Sum List. Hence, the numbers must be either (2,15) or (5,6)."Here, Prashant won't be sure of the numbers.Hence, Prashant will be sure of the numbers if product is either 18, 24 or 28.

Sachin : "Since Prashant knows the numbers, they must be either (3, , (4,7) or (5,6)." But hewon't be sure. Hence, the sum is not 11.

Summerising data for sum 11:Possible Sum PRODUCT Possible Sum2+9 18 2+9=11 (possible)3+6=93+8 24 2+12=143+8=11 (possible)

4+6=104+7 28 2+12=143+8=11 (possible)4+6=105+6 30 2+15=17 (possible)3+10=135+6=11 (possible)

Following the same procedure for 17:Possible Sum PRODUCT Possible Sum2+15 30 2+15=17 (possible)3+10= 135+6=11 (possible)

3+14 42 2+21=23 (possible)3+14=17 (possible)6+7=134+13 52 2+26=284+13=17 (possible)5+12 60 2+30=323+20=23 (possible)4+15=195+12=17 (possible)



6+10=166+11 66 2+33=35 (possible)3+22=256+11=17 (possible)7+10 70 2+35=37 (possible)5+14=197+10=17 (possible)8+9 72 2+36=383+24=27 (possible)4+18=226+12=188+9=17 (possible)

Here, Prashant will be sure of the numbers if the product is 52.Sachin : "Since Prashant knows the numbers, they must be (4,13)."For all other numbers in the Possible Sum List, Prashant might be sure of the numbers butSachin won't.

Here is the step by step explaination:Sachin : "As the sum is 17, two numbers can be either (2,15), (3,14), (4,13), (5,12), (6,11), (7,10)

or (8,9). Also, as none of them is a prime numbers pair, Prashant won't be knowing numberseither."

Prashant : "Since Sachin is sure that both of us don't know the numbers, the sum must be one of the Possible Sum List. Further, as the product is 52, two numbers can be either (2,26) or (4,13).But if they were (2,26), Sachin would not have been sure in advance that I don't know thenumbers as 28 (2+26) is not in the Possible Sum List. Hence, two numbersare 4 and 13."

Sachin : "As Prashant now knows both the numbers, out of all possible products - 30(2,15),42(3,14), 52(4,13), 60(5,12), 66(6,11), 70(7,10), 72(8,9) - there is one product for which list of allpossible sum contains ONLY ONE sum from the Possible Sum List. And also, no such two listsexist. Hence, two numbers are 4 and 13." Nakul figured out both the numbers just as we did by

observing the conversation betweenSachin and Prashant.

P.S: It is interesting to note that there are no other such two numbers. We checked all thepossible sums till 500 !!!

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If today is Wednesday, what is one day before the day, after the day, three days after the daybefore yesterday?

Answer

ThursdayStart backwards.Today is Wednesday.The day before yesterday is Monday.Three days after Monday is Thursday.The day after Thursday is FridayThe day before Friday is Thursday.

Also, note that the first two conditions cancel each other out as one day before the day,



one day after the day is the same day. Hence, it can be reduced to "three days after theday before yesterday".

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A king decides to give 100 of his prisoners a test. If they pass, they can go free. Otherwise, theking will execute all of them. The test goes as follows: the prisoners stand in a line, all facingforward. The king puts either a black or a white hat on each prisoner. The prisoners can only seethe colors of the hats in front of them. Then, in any order they want, each one guesses the color of the hat on their head. Other than that, the prisoners can not speak. To pass, no more than 1 of them may guess incorrectly. If they can make their strategy before hand, the method they chooseshould save maximum prisoners. How many get saved? Explain the method.

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Five pirates have 100 gold coins.They have to divide up the loot in order of seniority (supposepirate 5 is most senior, pirate 1 is least senior), the most senior pirate proposes a distribution of the loot. they vote and if at least 50% accept the proposal, the loot is divided as proposed.

otherwise the most senior pirate is executed, and they start over again with the next senior pirate.what solution does the most senior pirate propose? assume they are very intelligent andextremely greedy (and that they would prefer not to die).

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di imp questions

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