Wickedly Tough Reasoning Brainteasers

Seemingly Tough Yet Simple Puzzles on Logical Reasoning, a la Poirot!

Hercule Poirot was a wildly popular character created by Agatha Christie, who believed in solving problems by working systematically to eliminate possibilities and arrive at a final conclusion. Where Holmes excelled at solving problems by establishing abstract relationships based on limited or disjointed information, Poirot was in a league of his own in working methodically and step-by-step, taking his time to work out his conclusions.

In today’s world of instant gratification and instant reactions, it is becoming extremely difficult for us to hold a train of thought for any length of time. Logical thought is getting increasingly beyond the grasp of people, fully evidenced in the mindless exchanges we witness so often on social media platforms as well as the lack of thought behind instant forwards on various messaging platforms.

Is the current generation more intelligent, or is it merely -and perhaps dangerously- over-dependent on use of technology to find answers? Would you like to find out if you are left with sufficient capacity to reason things out in a logical and systematic manner?

Put on your thinking caps, then, and let’s go. Don’t forget to have pen and paper ready, as you may be required to jot down your thoughts in most such puzzles.

Be forewarned: Some of these will be the toughest puzzles you are likely to encounter anywhere, and will really test your fortitude!

We start with a few simple ones, just to keep morale high :))


5.1    Who is the tallest and who is the shortest?

Amar is taller than Rohan, but shorter than Krish. Amy is shorter than Rohan, but taller than Sheena. Krish is shorter than Ram.


5.2    Dudeney was a master puzzler, and we have already seen a few of his puzzles earlier on. Reproduced below is another of his classic mind-bogglers, published in Strand magazine in 1929.

Arrange all the 10 digits in three arithmetical sums, employing three of the four operations of addition, subtraction, multiplication, and division, and using no signs except the ordinary ones implying those operations


5.3    You are in a room which has three switches. You cannot see or hear anything outside, and there is only one closed door through which you may exit the room. Once past the door, you cannot touch it again.

Each switch controls one lightbulb in the corridor outside the room. You cannot see the bulbs or the light from inside.

You have to identify which switch controls which bulb. How do you do it?


5.4    There are Eighteen boys in a class, along with some girls. Thirteen of the kids wear a Yellow shirt, but three of the kids neither wear a yellow shirt nor are boys. If eight boys wear a white shirt, how many kids are there in total?


5.5    Four playing cards are placed face down on the table. Each card is of a different suit and each has a different face value: a Three, a Four, a Five and a Six.

Some information about them is given below, the task is to find out each card’s suit, face value and position basis this.

  1. The sum of the two cards in the middle is an even number.
  2. The Club is not in the middle.
  3. The Four has black suits on either side of it
  4. The Club lies to the right of the Three but is not placed next to it
  5. The Spade lies to the left of the Heart


5.6    Sometimes, the way a puzzle is presented makes it difficult for us to comprehend it. The key, of course, is to fix a starting point and then fit in the facts provided to find the solution. Try your hand at the one below:

What day would tomorrow be if Sunday was five days after the day before yesterday?


5.7    Seated in the Club lounge are, X, Y and Z. Their professions, in random order, are: Doctor, Engineer and Businessman. In the same lounge there are three Lawyers, who are Mr X, Mr. Y and Mr. Z. They each stay in one of three places: Delhi, Shimla or a town exactly midway between them.

Given the clues below, can you establish the identities of X,Y and Z?

  1. The lawyer with the same name as the Engineer lives in Delhi
  2. Mr. Z earns Rs. 16 lakh in a year
  3. Mr Y is a resident of Shimla
  4. The Engineer lives exactly midway between Delhi and Shimla
  5. The Engineer’s nearest neighbour is a lawyer who earns exactly three times as much as the Engineer
  6. The Doctor was beaten by X in a game of Chess


5.8    If all Bings are Bongs and some Bungs are Bings, which of the below statements is definitely true?

(i) All Bings are Bungs.

(ii) Some Bongs are Bungs.

(iii) Some Bings are not Bungs


5.9    This is a very interesting problem from the Central New York Mathletics in 2011: Mrs. Barnard has three sons, Bartholemew, Barton and Benjamin.  One is an avid bowler in Baltimore, a second is in Bainbridge, and the third is in Binghampton. One is boating and another is barbequeing. Bartholemew is not in Baltimore. Benjamin is not in Bainbridge, and the son who barbeques is not in Binghampton. The bowling enthusiast is not Benjamin.

Who is barbequeing and where is he?


5.10 Here’s another one from the Central New York Mathletics, 2010:  One rainy evening, five military men were murdered in the old mansion on Willow Lane (a General, a Captain, a Lieutenant, a Sergeant, and a Corporal). The murders took place in the bedroom, basement, pantry, den, and attic of the house. No two men were murdered in the same room or with the same weapon. The weapons used were poison, a poker, a gun, a knife, and a shovel. From the clues given, try to determine the room in which each man was killed and the weapon used to do him in.

1.       The murder with the shovel was not done in the den or the attic; neither the captain nor the lieutenant was killed with the shovel, nor was either killed in the den or the attic.

2.      The captain was not murdered in the bedroom.

3.      The poker was not the murder weapon used in the attic.

4.      Neither the general nor the corporal was murdered with poison, a gun, or a shovel.

5.      The man murdered in the basement had just had dinner with the corporal, the captain, the man done in with poison, and the victim of the poker.


5.11 Abhi, Bob, Jill, Sonu and Amy have birthdays between Jan to May, with only one person in each month. They each like a different type of dessert on their birthday: Sweets, Chocolates, Pastries, Sundaes and Dry fruits.

The one who likes Pastries is born in the middle of all months, while the favorite of the May born is Dry fruits. Jill doesn’t like Dry fruits or Sundaes. Amy loves Sweets and is born in the month immediately after Jill’s birthday month. Sonu doesn’t like Sundaes but gifts Chocolates to Abhi in February.

Can you find out whose birthday is in which month, and what do they each like for dessert?


5.12 Three countries A, B and C participate in a sporting event which has 10 events in total. A gold medal is worth 3 points, a silver medal is worth 2 points and a bronze medal is worth 1 point. C wins more gold medals than either A or B. The total number of medals won by C is also 1 more than B and 2 more than A. However, A finishes the event on top with overall 1 point more than B and 2 points more than C.

Can you find out which country won how many of each medal, and also the total points won by each?


5.13 In a three-storeyed store, A, B, C, D, E and F are six staff of which three are female. There are three departments – Accounts, Administration and Personnel, on the three floors. All the females work on different floors and in different departments. Persons working in the same department are not together on the same floor, and each floor has exactly two staff.  

B and E work in the same department but not in Personnel. D is a female who works in Administration, but is not on the 2nd Floor. E and A are on the 1st and 3rd floor respectively and work in the same department. C is a male who works on the 1st floor. A total of two people work in Admin.

Can you establish the gender of each person and find out who works where?


5.14 There are 70 people of which 30 are females.

a) 30 people are married.

b) 15 males are married of which 8 are below 30 years of age

c) 19 married people are above 30 years

d) 12 males are above 30 years of age.

e) 24 people are above 30 years of age.

Can you find out the count of males and females as per their age and marital status?          


5.15 This puzzle is one allegedly devised by Albert Einstein, and hence also known as Einstein’s puzzle or the Zebra riddle. It is alleged that only 2% of the population can solve it. There is no conclusive evidence as to the origin of this puzzle, but it remains a favorite of puzzlers worldwide. Have a go at it, and see for yourself.

There are 5 houses in 5 different colors. All the 5 owners are of different nationality. The 5 owners drink different beverages, smoke different brand of cigars, and own different pets. No owners have the same pet, smoke the same brand of cigar, or drink the same beverage.

• The Brit lives in the red house.
• The Spaniard owns the dog.
• Coffee is drunk in the green house.
• The Ukrainian drinks tea.
• The green house is immediately to the right of the ivory house.
• The Old Gold smoker owns snails.
• Kools are smoked in the yellow house.
• Milk is drunk in the middle house.
• The Norwegian lives in the first house.
• The man who smokes Chesterfields lives in the house next to the man with the fox.
• Kools are smoked in the house next to the house where the horse is kept.
• The Lucky Strike smoker drinks orange juice.
• The Japanese smokes Parliaments.
• The Norwegian lives next to the blue house.

Who owns the zebra? And who drinks water?


5.16 This is what is known as a self-referential puzzle, wonderful for boosting the way of logical thought. We start with this practice problem, then you can attempt the really hard one which will follow.

Q 1) What is the answer to the second question?

A. A

B. B

C. C

D. D

Q.2) How many correct answers are option B?

A. 0

B. 1

C. 2

D. 3

Q.3) Is there a question with the correct answer A?

A. No

B. Yes,1

C. Yes, 2

D. Yes, all 3

Answers: A, A and C


5.17 This puzzle will be a truly challenging one. Do keep a pen and paper handy, and keep filling in the options as you go along. Read all the questions fully, and try to understand which questions are dependent on each other and what should be the logical answer to them. This will give you a starting point for solving this type of puzzle.

There is only one answer to each question, and a total of 20 questions. The key to solving this is to arrange your data in a solution grid, which will keep things organized and give you a handle on the inter-relationships between the various questions and answers.

Given that the answer to Q. 20 is E. Now, find the answers to Questions 1-19.

1.      The first question whose answer is B is question:

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

2.      The only two consecutive questions with identical answers are questions:

(A) 6 and 7

(B) 7 and 8

(C) 8 and 9

(D) 9 and 10

(E) 10 and 11

3.      The number of questions with the answer E is:

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

4.      The number of questions with the answer A is:

(A) 4

(B) 5

(C) 6

(D) 7

(E) 8

5.      The answer to this question is the same as the answer to question:

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

6.      The answer to question 17 is:

(A) C

(B) D

(C) E

(D) none of the above

(E) all of the above

7.      Alphabetically, the answer to this question and the answer to the following question are:

(A) 4 apart

(B) 3 apart

(C) 2 apart

(D) 1 apart

(E) the same

8.      The number of questions whose answers are vowels is:

(A) 4

(B) 5

(C) 6

(D) 7

(E) 8

9.      The next question with the same answer as this one is question:

(A) 10

(B) 11

(C) 12

(D) 13

(E) 14

10.    The answer to question 16 is:

(A) D

(B) A

(C) E

(D) B

(E) C

11.    The number of questions preceding this one with the answer B is:

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

12.    The number of questions whose answer is a consonant is:

(A) an even number

(B) an odd number

(C) a perfect square

(D) a prime

(E) divisible by 5

13.    The only odd-numbered problem with answer A is:

(A) 9

(B) 11

(C) 13

(D) 15

(E) 17

14.    The number of questions with answer D is

(A) 6

(B) 7

(C) 8

(D) 9

(E) 10

15.    The answer to question 12 is:

(A) A

(B) B

(C) C

(D) D

(E) E

16.    The answer to question 10 is:

(A) D

(B) C

(C) B

(D) A

(E) E

17.    The answer to question 6 is:

(A) C

(B) D

(C) E

(D) none of the above

(E) all of the above

18.    The number of questions with answer A equals the number of questions with answer:

(A) B

(B) C

(C) D

(D) E

(E) none of the above

19.    The answer to this question is:

(A) A

(B) B

(C) C

(D) D

(E) E

20.    Standardized test is to intelligence as barometer is to:

(A) temperature (only)

(B) wind-velocity (only)

(C) latitude (only)

(D) longitude (only)

(E) temperature, wind-velocity, latitude, and longitude




5.1      Basis the three height comparisons provided, the facts can be represented visually like this:

 So, Ram will be the tallest and Sheena will be the shortest

5.2      To form three logical equations using each digit just once, there must be two equations involving only three digits, and one equation will use 4 digits.

We see that 0 cannot be used with any operator, hence can only be used as the second digit of a two digit resultant of any operation. Thus, the equation involving 4 digits must necessarily have zero in the units place of a two digit resultant.

With zero in the units place of the resultant, the only possible candidates for the digit in the tens place of the resultant are 1,2,3 and 4 ( this is because the largest number possible with a zero in units place is 8*5=40)

Only possibilities for our first equation are:

8+2=10, 7+3=10, 6+4=10, 5*2=10, 5*4=20, 5*6=30 and 5*8=40

We can immediately see that 9 is missing here. So, 9 must be part of one of the other two equations. What are the options for an equation using 9?

We know the resultant of each equation will be > zero and in single digits. The only possibilities are given below:

9-1=8. 9-2=7, 9-3=6, 9-4=5            Or, alternatively:       8+1=9, 7+2=9, 6+3=9, 5+4=9.

Whether we take subtraction or addition, will not really matter in finalizing the options for the third equation as what concerns us is mainly the digits remaining to be used for the final equation.

So let’s see the possibilities we have after accounting for the first two equations:

The first column gives the possibilities we have for the first equation. The second column contains the available options for the second equation corresponding to the first equation, while the third column gives the possibilities of forming a third equation using the remaining digits and operators available for use in the third equation:   

Hence we can see that the only available option for Equation 1 is 5*4=20

Equation 2 may be either 9-3=6 , or 6+3=9

Accordingly, equation 3 will be either 1+7=8, or 8-1=7, depending on which operator we have selected for the second equation.

5.3      Any bulb when left on for a period of time will turn hot, even if just a little. So, all you need to do is to keep one switch in one position for 5 minutes, then switch it off. Then you switch on any one of the remaining switches, and go out of the door and check the bulbs outside. One bulb will be off and cool, second will be off but hot, and the third bulb will be on. Thus, you can immediately relate the bulb to the switch.

5.4       Think in terms of set theory and venn diagrams for such puzzles. The facts provided can be represented like this:

We have to calculate the value of ‘N’, which is the total number of kids in class. x,y,and z are also unknown.

From the figure, we get:

x= no. of boys in yellow shirt = 18-8 = 10

x+y=13, on substituting the value of x we get: y=13-10=3

Therefore, z=y+3=6

Hence, we get:  N =Total ( boys + Girls) =18+z =  24

Thus, there are 24 kids in class.

5.5      The four cards are 3,4,5 and 6. Let’s examine our conditions:

For the sum of two numbers to be even, either both numbers will be even or both numbers will be odd. Since it is mentioned that the 4 has black suits on either side, hence 4 has to be one of the cards in the middle. Thus, the second card in the middle has to be even, which must be 6.

Therefore 3 and 5 will be the outer cards.

Since the Club lies to the right of the 3, therefore 3 has to be the outer card on the left-hand side. Also, since the Club is not in the middle, it has to be the outer card on the right which must therefore be the 5 of Clubs

 And since the 4 lies between the two black suits, hence the card adjacent to the 5 on its left will be the 4.

Therefore the 6 will be the second card from the left and will lie in between the 3 and the 4. Also, since on both sides of the 4 are black suits, hence it has to be the 6 of Spades.

Finally, since the Spade lies to the left of the Heart, therefore we conclude that the 4 is the Heart and therefore the 3 is the Diamond.

The final solution will look like this:

5.6      We are not used to juggling to-and-fro with data, and tend to throw us our hands in exasperation when confronted with something like this. Patience is not a major virtue of these times.

However, the moment we start putting down the data presented in a systematic manner, things immediately become clear.

Representing the facts given pictorially, the answer is: Tomorrow will be Friday.:

5.7      The plethora of facts presented can be bewildering to most, but putting down the facts and the conclusions in a logical order will immediately begin to untangle the puzzle:

  • Since both Shimla and Delhi are equidistant from Midway, hence the nearest neighbor of the Engineer will live in Midway town only.
  • Mr Y resides in Shimla so he cannot be the Engineer’s nearest neighbor
  • Mr. Z earns exactly 16lakh. Since 16 lakh is not exactly divisible by 3, hence the Engineer’s nearest neigbor will not be Mr Z
  • Thus, Mr X must be the nearest neighbor of the Engineer.
  • Therefore, the lawyer with the same name as the Engineer will be Mr Z. and hence, Z is the Engineer.
  • Since the Doctor was beaten by X, hence the only remaining person Y must be the Doctor.
  • Therefore, X is the Businessman

Not really as difficult as it looked at first, wasn’t it? We only had to follow a logical chain of thought and the solution just appeared!

5.8      Once again, let’s represent the facts given as a venn diagram:

One thing that is immediately clear is that if all Bings are Bongs, then it is impossible for some Bungs to be Bings without being Bongs. So the first statement is false.

The second statement will hold true in all cases, as can be seen in the diagram.

The third statement won’t hold if the above diagram looks like this:

Hence option (ii) is the only correct answer.

5.9      Let’s take it step-by-step:

  • Avid bowler is in Baltimore
  • Son who barbeques is not in Binghampton, hence barbequeing is happening in the third town which is Banbridge
  • Thus, Boating will happen in Binghampton

Now the situation looks like this:

Now, let’s place the brothers:

  • Bartholemew is not in Baltimore, and Benjamin is not the bowling enthusiast . Hence, the only option left for Baltimore is Barton.
  •  Since Benjamin is not in Bainbridge, he can only be in Binghampton.

Hence the third brother Bartholemew is the person who is barbequeing in Bainbridge.

5.10    Such problems require a methodical approach, meticulously organizing the facts in a fashion that helps to eliminate possibilities one by one.

Plotting the facts provided in a tabular fashion often helps. Let’s see how the situation looks after we have put in all given facts here:

The ‘x’ in black is an option eliminated due to a fact provided to us in the statements given, and the number following the ‘x’ gives the statement number containing the fact.

The ‘x’ in red are eliminated logically due to the statement whose number follows immediately after the x.

  • We see that only the Sergeant could be killed by the Shovel – Conclusion 1
  • We also see that the Pantry is the only place where the Captain could have been murdered – Conclusion 2
  • The Captain could only have been killed with the Gun – Conclusion 3
  • Only the knife could have been used to kill the Corporal- Conclusion 4

Now the situation looks like this:

  • Since the Knife was used on the Corporal, hence deriving from statement 4, the General could only have been killed with the Poker – Conclusion 5
  • And thus, the only weapon remaining is Poison, which was used on the Lieutenant – Conclusion 6
  • The murder weapons have now been finalized, and our chart now looks like this:
  • Now we know from statement 5 that the man murdered in the basement had dinner with the man done in with poison (General) and the victim of the poker (Lieutenant). Hence in the chart above, the only candidate remaining for the basement is the Sergeant – Conclusion 7
  • This also tells us that the only location available for the Lieutenant is the Bedroom – Conclusion 8
  • The final two locations available for the General and the Corporal are the den and the attic. Since Statement 3 tells us that the poker was not used in the attic, hence the General could only have been killed in the den – Conclusion 9
  • Thus, the Corporal was killed in the attic with the knife – Conclusion 10

5.11    We begin by arranging the already given facts in a table:

  • Jill doesn’t like Sundaes or Dry fruits, so they are crossed out
  • Sonu doesn’t like Sundaes
  • Amy likes Sweets – Conclusion 1
  • Abhi likes Chocolates and his birthday is in Feb – Conclusions 2 and 3.

From given facts, we can also eliminate a few possibilities. Since Amy’s birthday has to fall in the month immediately after Jill’s, hence Jill cannot be born in Jan as Feb is occupied by Abhi.

Hence Amy’s birthday cannot fall in Jan, Feb or March. Also, since the March born is the person who likes Pastries, hence Amy does not like Pastries. The initial table looks like this:

(figures in red show the eliminated options, cells shaded green are the finalized options)

  • The above table immediately tells us that the only option left for Jill’s favorite dessert is Pastries – Conclusion 4
  • Since we know that the person who loves Pastries was born in the middle month, hence Jill was born in March – Conclusion 5
  • Since Amy’s birthday follows that of Jil, hence it is in April – Conclusion 6

The situation now looks like this:

  • Since Sonu does not like Sundaes, hence the only available option for him is Dry fruits – Conclusion 7
  • Hence Bob is the person who likes Sundaes – Conclusion 8
  • The May-born likes Dry fruits, hence Sonu was born in May – Conclusion 9
  • Thus, Bob was the person born in Jan – Conclusion 10

5.12    The total number of medals is equal to 30, and the total sum of points available equals 60 (10gold*2+10silver*2+10bronze*1= 60)

  • Assuming sA, sB and sC are the sum of points won by A, B and C respectively, we are given: sA = sB+1 = sC+2

Also, sA+sB+sC = 60

Substituting the values of sB and sC in terms of sA, we get

sA+ (sA-1) + (sA-2) = 60

which implies, sA = 21. Therefore, sB=20 and sC=19

  • Assuming a,b and c is the total count of medals won by A, B and C respectively, we are given c = b+1 = a+2

Also, a+b+c = 30

Substituting in terms of c we get

(c-2)+(c-1)+c = 30

Which implies, c = 11. Therefore, b=10 and a = 9

  • This was the easy part. Now, we need to determine how many of which type of medal was won by each country. Let us take a look at the medal tally as it stands after our initial calculations:
  • We know that C won more gold than either A or B. Thus, neither A nor B could have won more than 4 gold medals, since if A or B was 5 then C would be 6 or more, which would take the total tally of gold medals over 10.
  • Also, the number of golds won by C cannot be less than 4, otherwise any one of A or B would have equal or more number of gold medals than C.
  • The only possible combination of medals with 3 or less gold and a total of 9 medals adding up to 21 points is 3gold+6silver = 9+12=21 points. Hence, the tally of A must be 3 gold and 6 silver.
  • We are left with 7 gold, 4 silver and 10 bronze for B and C
  • As C won more gold than A or B, hence B also cannot have more than 3 gold. The only possible option for B to win 20 points from 10 medals with maximum 3 gold and maximum 4 silver is 3gold+4silver+3bronze= 9+8+3=20
  • We are now left with 4 gold and 7 bronze = 12+7=19, which is the points won by C from a total of 11 medals.

Hence the final tally looks like this:

5.13    As usual, we start with plotting the provided facts in tabular form. Let ‘y’ represent a true fact, and ‘x’ represent an eliminated choice.

The preliminary table will look like this:

  • As there is only one female on every floor, and every floor has only two people, hence E, who works on the same floor as C – who is a male- must be the female – Conclusion 1
  • Since D is not on the 2nd floor, therefore the only floor available for her will be the 3rd floor – Conclusion 2
  • Thus, B and F will have to be on the 2nd floor – Conclusion 3
  • Since A,B and E are not in Personnel, and Admin has two people only, hence all three will be in Accounts only.- Conclusion 4
  • Since D – a female – is on the 3rd  floor, hence A – who is also on 3rd flr – will be the male – Conclusion 5

Now the interim table looks like this:

  • Since all females work in different departments, hence B must be a male: Conclusion 6
  • Therefore, F will be a female: Conclusion 7
  • Since all females work in different departments, hence F must work in Personnel: Conclusion 8
  • Finally, since Admin must have 2 people, hence C must work in Admin only: Conclusion 9

Thus, the final solution looks like this:

5.14    Depicting given facts as a venn diagram, we get:

Let x = total no. of males

A= married males who are under 30 =8 (given)

B = married males over 30

C = married females over 30

D= married females under 30

E=unmarried males over 30

F=unmarried females over 30

G =unmarried males under 30

H =unmarried females under 30

We are given: Total persons =70, Total married = 30, Total females = 30, Total above 30 = 24, total married males under 30= 8

Now we can derive the numbers from above diagram using the other facts provided:

  • Since 15 males are married, it means A+B = 15. Substituting the value of A =8, we get B = 7
  • 19 Married people are above 30 years. Thus, B+C = 19. Substituting the value for B, we get C = 19-7 = 12
  • 12 males are above 30 years. Thus, E+B = 12.  Substituting the value for B, we get E = 12-7 = 5
  • Since 30 out of 70 are females, hence x = total number of males =  40. This means, A+B+E+G = 40. Thus, 8+7+5+G = 40, which gives G = 20
  • Now, total people above 30 = 24 = B+C+E+F = 7+12+5+F. Thus, F= 0. This means, there is no unmarried female above 30.
  • 30 people are married. This means A+B+C+D = 30. Thus, 8+7+12+D=30, which gives D=3
  • There are 30 females, this means C+D+F+H = 30. Therefore, 12+3+0+H = 30, which gives H = 15

Thus, the final breakup will look like this:

5.15    This puzzle requires painstaking arrangement of all given facts and then elimination of the options one by one. It requires the patience of a saint to solve such puzzles, but is guaranteed to give a huge dose of satisfaction once you succeed.

Let’s start by first creating a table that captures the puzzle and the basic given facts:

  • Since the Norwegian lives next to the blue house, and his house is the first, hence the second house is blue.
  • Also, since the Brit lives in the red house, and the green house has to be immediately to the right of the ivory house, hence the Norwegian cannot live in the red, green or ivory houses too.
  • Hence, the Norwegian must live in the yellow house.
  • Since Kools are smoked in the yellow house, hence the Norwegian smokes Kools living next to the  blue house which has the horse
  • Since Milk is drunk in the middle house, the Ukrainian drinks Tea, the Lucky Strike smoker drinks orange juice, and the green house has the coffee drinker, so the Norwegian must be the water drinker.
  • The Japanese cannot live in the second house, because then he cannot drink coffee (green house), milk (middle house), orange juice (lucky strike smoker), water (Norwegian) or tea (Ukrainian).
  • The Spaniard has the dog,so cannot live in the blue house (horse). The Brit lives in the red house. This leaves only the Ukrainian for the blue house.
  • So, the Ukrainian cannot smoke Kools, Old Golds, Lucky Strikes or Parliaments. Hence he must smoke Chesterfields.

At this point we have reached a position where we must eliminate choices by testing out a couple of hypothesis.

  • If either the Japanese or Spaniard live in the middle house, the red house will have to be last house as the red house belongs to the Brit and the green house must be immediately to the right of the ivory house. Hence the ivory house will be the middle one in such a case.
  • If the Japanese were to live in the middle house, it will mean he drinks milk.
  • Since coffee is drunk in the green house, that will leave only orange juice for the red house (Brit)
  • That will mean the Brit smokes Lucky Strikes, leaving Old Golds for the Spaniard in the green house.
  • But the Old Golds smoker owns snails, whereas we know that the Spaniard is the owner of the dog, so this option is contradictory and is ruled out.
  • In case the Spaniard lives in the middle house, he owns the dog and drinks milk. Then the Japanese must be in the green house smoking Parliaments.
  • But the Spaniard cannot smoke Old Golds as he does not own snails, and he cannot smoke Lucky Strikes as he drinks milk. So he has no smoking option left, which is a contradiction and hence is ruled out.
  • Hence, we conclude that the Brit is the only possibility for the middle house. Then the Green house has to be the last as it has to be on the right of the ivory house.
  • The only drink available for the ivory house is thus orange juice. This means Lucky Strikes are smoked in the ivory hosue
  • Thus the Japanese will occupy the fith house (green), smoking Parliaments
  • Hece, the Old Golds are being smoked by the Brit, who is also the owner of the snails
  • The Spaniard, therefore, lives in the ivory house (4th) with his dog.
  • Thus the Norwegian owns the fox, living next to the person smoking Chesterfields, while the Japanese owns the zebra.

5.16    We start by establishing the dependencies of questions on each other. Once we know that, then we get the starting point of the reasoning process.

In this case, the first question is dependent on the second question. Also, the third question can be answered if we know the answers to both the preceding questions.

Hence, we start with the second question. We can see that the options for the first question mirror the answers for the second question, i.e. if we answer A to the second question then the answer to question 1 also becomes A, If the answer is B, then the answer to question 1 will also be B, and so on.  

Thus, we can easily see that option A for the second question is the only one that holds true for both question 1 and question 3. All other options are either contradictory or not possible for questions 1 and 3. Hence the answers to the three questions will be: A, A and C.

5.17     The questions given are:

  • Given Answer of Q 20 is E. Other options are therefore eliminated.
  • Then we can immediately rule out Option A of Q3, as number of questions with answer E will be definitely greater than zero.
  • Since, as per Q.2, the only two consecutive questions with identical answers range from Q.6 to Q11, hence option E for Q19 gets eliminated.
  • The answer to Q1 cannot be option A, so this is also ruled out.
  • The answer to Q5 must be E, which effectively states that the answer to Q5 is the same as the answer to Q5. Hence other options for Q5 are ruled out.
  • As per the negated options of Q5, the corresponding answers of Q2 (B), Q3 (C) and Q4 (D) also get ruled out.
  • Since Q2 (B) is eliminated, hence Q1(B) also is ruled out.
  • Since Q5 has the answer E, hence option Q1 (E) is also ruled out.
  • By ruling out Q2 (B), we also rule out the corresponding option of Q1 (B).

At this stage our solution grid will look like this:

  • Q10 and Q16 reference each other’s answers. We see that options Q10 (A) and A16 (D) are hold true, while the cross-references of the remaining options do not hold true. Hence the answers to these two questions will be Q10 (A) and Q16 (D). The other options stand negated.
  • After this, referencing the statement of Q2, we rule out options Q4 (E) , Q15 (D) and Q17 (D) as answers to consecutive questions after Q11 or before Q6 cannot be the same.
  • Q6 and Q17 also reference each other. Option E for both questions immediately gets ruled out as there are no multiple true answers to any question. Also, since we have already ruled out option  Q17 (D), hence the only cross-reference which logically holds is of Q6 (D) and Q17 (B). The other options for these two questions are thus eliminated.

At this stage out solution grid will look like this:

  • By Q1, there has to at least one question among the first five with an answer (B), Hence option Q11 (A) is ruled out. After this, we will also rule out option .Q2 (E), since the answers of Q10 and 11 are not the same.
  • By Q13, the only odd numbered questions that can have option A as answer are Q9,11,13,15 or 17. Therefore we can now rule out Q7 (A) and Q19 (A).
  • Now for Q13, option A is ruled out since if Q9 had A as an answer, then no other odd number question will have option A as answer, and Q13 itself is odd. Thus, we rule out A9 (A) and Q13 (A).
  • Q13 (C) is also ruled out, since Q13 cannot have both A and C as answer. We have also already ruled out Q11 (A) and Q17 (A) previously, hence the only option remaining for Q13 is D.
  • Referencing Q13 (D), we now know that the answer to Q15 will be A.
  • Q15 (A) references the answer to Q12, which must therefore be Q12 (A). Other options of Q15 and Q12 are eliminated.

Now the solution grid will look like this:

  • We are getting close to the final solution. Since we have already eliminated Q8 (A), therefore Q2 (D) also gets eliminated.
  • By Q12, since the number of questions with a consonant as the answer is an even number, hence the number of questions with a vowel as the answer must also be even as the total number of questions is even. Hence we eliminate options B and D of Q8.
  • Since we already have a total of 5 answers which are vowels, hence option Q8 (A) also gets eliminated.
  • Also, since we already have two answers with E as the answer, so Q3 (B) also gets eliminated.
  • Since we know that Q5 has E as an answer, hence Q1 (E) is eliminated.
  • With this, the only remaining answer for Q1 is D, which means that Q4 is the first question which will have B as the answer.

The grid will now look like this:

  • Now Q4 (B) tells us that there are 5 questions with A as the answer which is an odd number quantity. Since we have previously concluded that the number of vowels as an answer has to be even, it follows that the numbers of questions with E as an answer must be odd (since odd + odd = even).
  • Thus the answer to Q3 must be D, since option Q3 (E) has the value of 4 which is even and hence must be ruled out.
  • Now from Q3 and Q4, we know that the total number of questions with a vowel as a solution is 3+5 = 8. Thus the answer to Q8 will be option E, and option C stands eliminated.
  • Since we now already have 3 answers for option E, which is the maximum as per Q3, hence all other E options of remaining questions stand eliminated.
  • Now, for Q9, option C gets ruled out, since the answer to Q12 is option A.
  • Also, we already have an option B as the answer to Q4, hence option B for Q9 gets ruled out, because in this case there will be two option B answers before Q11 which will then become contradictory.
  • Thus the only option remaining for Q9 is option D.
  • Since the answers to Q8 and Q9 are not similar, hence option C for Q2 now also gets eliminated, leaving only option A as the answer.
  • This means that Q6 and Q7 must have similar answers. Hence the answer to Q7 will also be option D, same as Q6.
  • Thus the only possible answer to Q11 is option B, and other options are eliminated.
  • Now, since we already have 7 answers as option D, hence Q14 (A) gets ruled out.
  • Since we know that there have to be 5 answers as option A, hence the answer to Q18 must be option A.
  • From Q18, since the number of option B answers must be equal to option A answers, hence both the remaining questions must have option B as the answer.

Thus, we reach our final solution grid, as shown below: