Syllogisms or Statements and Conclusions

A syllogism question contains 2 or more Question statements, and 2 or more conclusions followed by 4 options. 

Statement 1: - - - - 
Statement 2: - - - - 
Conclusion 1: - - - -
Conclusion 2: - - - -

Answers options: 
Option 1: If only conclusion 1 can be drawn from the given statements.
Option 2: If only conclusion 2 can be drawn from the given statements.
Option 3: If both conclusions 1 and 2 can be drawn from the given statements.
Option 4: None of the conclusions can be drawn from the given statements.

Example Question:
Statements:
All MBAs are Graduates
All graduates are Students
Conclusions:
1: All MBAs are Students
2: Some students are MBAs

There are two types of Conclusions: 
1. Immediate Inferences
2. Logical Conclusions.

Immediate inferences are the conclusion which are drawn from only one statement. For example, From the statement All A's are B's, we can draw some A's are B's, Some B's are A's.   These are easy to draw.
For Logical conclusions you have to follow complete theory.

Understanding Syllogism question:

Questions on syllogisms contains only the following 4 types of statements:
1. The universal positive : Eg: All X√$\stackrel{\surd }{X}$ are Y×$\stackrel{\times }{Y}$
2. The universal negative  :  Eg: No X√$\stackrel{\surd }{X}$ is Y√$\stackrel{\surd }{Y}$
3. The particular positive:   Eg:  Some X×$\stackrel{\times }{X}$ are Y×$\stackrel{\times }{Y}$
4. The particular negative :  Eg: Some X×$\stackrel{\times }{X}$ are not Y√$\stackrel{\surd }{Y}$’s

Here Checkmark (√$\surd$) indicates "Distribution".  If a term is distributed means It covers each and every element of it. All X are Ys means X ∈$\in$ Y, But Y need not be a subset of X. So Y does not have check mark.

You should commit to memory, how to put √$\surd$ marks  and ×$\times$ marks and to distinguish positive and negative statements, universal and particular statements.

Here two statements are universal (1 and 2), and two statements are particular (3 and 4).
Two statements are positive (1 and 3) and two statements are negative (2 and 4).

I.  The Universal Positive:   All X√$\stackrel{\surd }{X}$ are Y×$\stackrel{\times }{Y}$
It states that every member of the first class is also a member of the second class.  Take a statement "All Tamilians are Indians".  It does not necessarily follows All Indians are Tamilians.  So Indians is not distributed on Tamilians.  

The general diagram for Universal  Affirmative ‘All X√$\stackrel{\surd }{X}$ are Y×$\stackrel{\times }{Y}$’ is

Immediate inference: 
1. Some X×$\stackrel{\times }{X}$ are Y×$\stackrel{\times }{Y}$
2. Some Y×$\stackrel{\times }{Y}$ are X×$\stackrel{\times }{X}$

II.  The universal Negative: No X√$\stackrel{\surd }{X}$ is Y√$\stackrel{\surd }{Y}$
It states that no member of the first class is a member of the second class.  This proposition takes the form - No X is Y.  
The general diagram for Universal Negative ‘No X is Y’ is

Immediate Inferences: 
1. No Y√$\stackrel{\surd }{Y}$ is X√$\stackrel{\surd }{X}$ 
2. Some X×$\stackrel{\times }{X}$ are not Y√$\stackrel{\surd }{Y}$’s, 
3. Some Y×$\stackrel{\times }{Y}$ are not X√$\stackrel{\surd }{X}$’s

III.  The Particular Affirmative:  Some X×$\stackrel{\times }{X}$ are Y×$\stackrel{\times }{Y}$
It states that at least one member, but never all, of the term designated by the class ‘X’ is also a member of the class designated by the term ‘Y’. This proposition takes the form Some Xs are Ys.  This possible diagrams as shown by the Euler’s circles for this proposition are:

Immediate Inferences: 
Some Y×$\stackrel{\times }{Y}$ are X×$\stackrel{\times }{X}$

IV.  The particular Negative: Some X×$\stackrel{\times }{X}$ are not Y√$\stackrel{\surd }{Y}$’s.
It states that at least one member of the class designated by the term ‘X’ is excluded from the whole of the class designated by the term ‘Y’.  This proposition takes the form Some Xs are not Ys.  The Euler’s circle diagrams for this proposition are as follows.

Immediate Inferences: None
The shaded portion in each is that part of X that is not Y.

Most of the students get some doubt why "Y" is distributed here.  Here is the explanation.  Suppose take an example statement "Some students are not hardworking".  This can be phrased as All the hardworking students are not some students".  Or let us say, Rama is one of the student who is not hardworking.  So not even a single hardworking student cannot be Rama. So All hardworking students cannot be Rama.
Read this article for more information: Click Here

How to answer Syllogisms:

There are two methods to answer syllogisms.  
1. Euler venn diagram method 
2. Aristotle's rules Method
If all the statements are Universal you can easily draw venn diagrams and solve the questions. But If there are more particular statements, then you better learn Aristotle method. But Aristotle's method initially seems to be a bit difficult to understand, as one practices good number of questions, one can easily crack these questions.  

Aristotle's Rules to solve syllogisms: 

1.   If both the statements are particular, no conclusion possible
 (Explanation: Statements starting with "Some" are particular)
2.   If both the statements are negative, no conclusion possible
3.   If both the statements are positive, conclusion must be positive
4.   If one statement is particular, conclusion must be particular
5.   If one statement is negative, conclusion must be negative
6.   Middle term must be distributed in atleast one of the premises
(Explanation: Middle term is the common term between two given premises, and A terms is distributed means it must have the "✓$✓$" mark above it)
7.   If a term is distributed in the conclusion, the term must be distributed in atleast one of the premises.
(Explanation: If any term is having star mark in the conclusion, it term must have star mark in the given premises)
8. If a term is distributed in both the statements only particular conclusion possible. 

Solved Example 1:
Statements:
All MBAs are Graduates
All graduates are Students
Conclusions:
1: All MBAs are Students
2: Some students are MBAs
Explanation:
Statement 1: All MBA√$\stackrel{\surd }{\mathrm{M}\mathrm{B}\mathrm{A}}$s are Graduates×$\stackrel{\times }{\mathrm{G}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{s}}$
Statement 2: All Graduates√$\stackrel{\surd }{\mathrm{G}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{s}}$ are Students×$\stackrel{\times }{\mathrm{S}\mathrm{t}\mathrm{u}\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s}}$
C1: All MBA√$\stackrel{\surd }{\mathrm{M}\mathrm{B}\mathrm{A}}$ are students×$\stackrel{\times }{\mathrm{s}\mathrm{t}\mathrm{u}\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s}}$
C2: Some Students×$\stackrel{\times }{\mathrm{S}\mathrm{t}\mathrm{u}\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s}}$ are MBA×$\stackrel{\times }{\mathrm{M}\mathrm{B}\mathrm{A}}$
Now Let us apply rules:
1. Both statements are positive, conclusion must be positive
2. Common term is Graduate and it has check mark in the second statement
Conclusion 1: MBA in the conclusion has got a check mark so it must have check mark in atleast one of the premises.  MBA in S1 has got star mark.  It satisfied all the rules. It is valid conclusion
Conclusion 2: No term in the conclusion has got a check mark so no need to check anything.  It followed all the rules.  This statement is a valid conclusion.

Solved Example 2: 
Statements: 
All Cats are Dogs
No Dog is Fish
Conclusions:
1. No Cat is Fish
2. Some Cats are Fish
Explanation:
S1: All Cats√$\stackrel{\surd }{\mathrm{C}\mathrm{a}\mathrm{t}\mathrm{s}}$ are Dogsχ$\stackrel{\chi }{\mathrm{D}\mathrm{o}\mathrm{g}\mathrm{s}}$
S2: No Dog√$\stackrel{\surd }{\mathrm{D}\mathrm{o}\mathrm{g}}$ is Fish√$\stackrel{\surd }{\mathrm{F}\mathrm{i}\mathrm{s}\mathrm{h}}$
C1: No Cat√$\stackrel{\surd }{\mathrm{C}\mathrm{a}\mathrm{t}}$ is Fish√$\stackrel{\surd }{\mathrm{F}\mathrm{i}\mathrm{s}\mathrm{h}}$
C2: Some Catsχ$\stackrel{\chi }{\mathrm{C}\mathrm{a}\mathrm{t}\mathrm{s}}$ are Fishχ$\stackrel{\chi }{\mathrm{F}\mathrm{i}\mathrm{s}\mathrm{h}}$
Now Let us apply rules:
1. S2 is negative, so conclusion must be negative. So C2 is ruled out, as rule says that one statement is negative conclusion must be negative.
2. Common terms is Dog and it has check mark in both the premises
Conclusion 1: In the conclusion, both the terms Cat, Fish have check marks and These two terms have check marks in at least one of the premises.  So Conclusion 1 is valid
Conclusion 2: As one of the premises is negative, conclusion must be negative.  So this conclusion is not valid

Solved Example 3:
Statements:
Some books are toys.
No toy is red.
Conclusions:
1. Some toys are books.
2. Some books are not red.
Explanation:
S1: Some Booksχ$\stackrel{\chi }{\mathrm{B}\mathrm{o}\mathrm{o}\mathrm{k}\mathrm{s}}$ are Toysχ$\stackrel{\chi }{\mathrm{T}\mathrm{o}\mathrm{y}\mathrm{s}}$
S2: No Toy√$\stackrel{\surd }{\mathrm{T}\mathrm{o}\mathrm{y}}$ is Red√$\stackrel{\surd }{\mathrm{R}\mathrm{e}\mathrm{d}}$
C1: Some Toysχ$\stackrel{\chi }{\mathrm{T}\mathrm{o}\mathrm{y}\mathrm{s}}$ are Booksχ$\stackrel{\chi }{\mathrm{B}\mathrm{o}\mathrm{o}\mathrm{k}\mathrm{s}}$
C2: Some Booksχ$\stackrel{\chi }{\mathrm{B}\mathrm{o}\mathrm{o}\mathrm{k}\mathrm{s}}$ are not Redχ$\stackrel{\chi }{\mathrm{R}\mathrm{e}\mathrm{d}}$
We can easily draw conclusion one as it is a direct inference from Statement 1.
Now by applying the rules,
Statement 2 is negative so conclusion must be negative. and middle term "Toy" should have a √$\surd$ mark. It has √$\surd$mark in statement 2. Now If a term has a √$\surd$ mark in the conclusion it should have √$\surd$ mark in atleast one of the statements. Here "red" has √$\surd$ mark in the conclusion and also has √$\surd$ mark in statement 2.
So Statement 2 also following all the rules.
Both conclusions are valid.

Three Statement Types:
We can apply all the rules we learnt above while solving 3 statement questions too. Look at the options and see from which two statements those words are derived. If Once word is from first and another is from the third, we have to check that there are two middle terms with ✓$✓$.

Solved Example 4:
Statements:
All cats are dogs
some pigs are cats.
All dogs are tigers
Conclusions:
some tigers are cats
some pigs are tigers
all cats are tigers
some cats are not tigers
Answer options:
1. Only 1 and 2
2. Only 1, 2 and 3
3. All follow
4. None Follow
Explanation:
Let us rearrange for easy understanding.
S2: Some Pigsχ$\stackrel{\chi }{\mathrm{P}\mathrm{i}\mathrm{g}\mathrm{s}}$ are Catsχ$\stackrel{\chi }{\mathrm{C}\mathrm{a}\mathrm{t}\mathrm{s}}$
S1: All Cats√$\stackrel{\surd }{\mathrm{C}\mathrm{a}\mathrm{t}\mathrm{s}}$ are Dogsχ$\stackrel{\chi }{\mathrm{D}\mathrm{o}\mathrm{g}\mathrm{s}}$
S3: All Dogs√$\stackrel{\surd }{\mathrm{D}\mathrm{o}\mathrm{g}\mathrm{s}}$ are Tigersχ$\stackrel{\chi }{\mathrm{T}\mathrm{i}\mathrm{g}\mathrm{e}\mathrm{r}\mathrm{s}}$

C1: Some Tigersχ$\stackrel{\chi }{\mathrm{T}\mathrm{i}\mathrm{g}\mathrm{e}\mathrm{r}\mathrm{s}}$ are Catsχ$\stackrel{\chi }{\mathrm{C}\mathrm{a}\mathrm{t}\mathrm{s}}$
C2: Some Pigsχ$\stackrel{\chi }{\mathrm{P}\mathrm{i}\mathrm{g}\mathrm{s}}$ are Tigersχ$\stackrel{\chi }{\mathrm{T}\mathrm{i}\mathrm{g}\mathrm{e}\mathrm{r}\mathrm{s}}$
C3: All Cats√$\stackrel{\surd }{\mathrm{C}\mathrm{a}\mathrm{t}\mathrm{s}}$ are Tigersχ$\stackrel{\chi }{\mathrm{T}\mathrm{i}\mathrm{g}\mathrm{e}\mathrm{r}\mathrm{s}}$
C4: Some Catsχ$\stackrel{\chi }{\mathrm{C}\mathrm{a}\mathrm{t}\mathrm{s}}$ are not Tigersχ$\stackrel{\chi }{\mathrm{T}\mathrm{i}\mathrm{g}\mathrm{e}\mathrm{r}\mathrm{s}}$

From statements 1 and 3, All Cats are Tigers is correct as dogs has √$\surd$ mark.  So conclusion 3 is correct.  Also from the above,  Some tigers are cats is also true. So conclusion 1 is correct.
Pigs is from 2nd statement and tigers is from third statements. If you observed our rearraged statements, In the second and first statements Cats has √$\surd$ markand first and third statements has dogs has √$\surd$ mark. So we can conclude that Some Pigs are tigers.
All the given statements are positive so conclusion should be positive. So option 4 is not correct.
Correct Answer option is 2.

Seating and Complex Arrangements

Seating and Complex Arrangements

Arrangement problems are quite common in all entrance examinations. If you follow the instructions given below, you can easily solve any problem given.  First we learn different types of arrangements and tips to crack.

There are three types of arrangements:

1. Linear Arrangements : There are people sit in a row

2. Circular Arrangements : People sit in a row facing center.  In most of the cases, problems with 6, 8 people given. 

3. Complex Arrangements : In these questions, There are some persons or things which eat different foods, wear different colored shirts, use different bikes, have some first and last names etc. We have to match these persons and their interests according to the conditions given. 

Tips in solving linear and circular arrangement problems:

1. Always start filling in the details with Specific Statements. 

There are two types of statements given: 1. Specific 2. Non specific

Specific statements always give only one type of arrangement.  For example: D sits immediate left of F. So we have to put the combination DF in the diagram. 

C sits opposite to A.  This also a specific arrangement in a circular diagram as there is no other way to represent this. 

Example for non-specific statements is B and E sits next to each other.  This gives two types of arrangement. BE and EB. 

A sits in between C and D. This gives CAD and DAC. 

So never solving question with statements like this.  

2. Search for some continuation statement:

If the first statement starts with D and F, search for another statement which has either D or F in that.  So that it will give you some continuation.  

3. Draw the diagrams in the circular arrangement according to the shown below

Remember: Left side in a circular arrangement is always clock wise and right side means anti - clock wise. 

Set 1:  
Six persons A, B, C, D, E and F are sitting around a circular table facing the center.

I.  C is sitting in-between A and F.

II.  B is sitting two places to the left of E.

III.  D is sitting two places to the right of F.

1. Between which two persons does D is sitting?
a. F-B
b. E-b
c. C-B
d. A-B

2. Who is sitting diagonally opposite to A ?
a. F
b. C
c. E
d. None of these

From statement 3, F _ DFrom statement 2, B _ ESol: From statement 1, ACF / FCA 

We can start with either statement 2 or 3, but starting with statement 3 gives us continuation with statement 1. 

D sits to the right of F so When we fix F, we have to write D two places after Anti clock wise direction. Now ACF or FCA possible. ACF is not possible as D occupied so FCA possible.  B sits to the left of E. So Fix E and after D fix B. 

So Option B and Option B are correct. 

Set 2:  
Eleven students A, B, C, D, E, F, G, H, I, J and K are sitting in a row of the class facing the teacher. D, who is to the immediate left of F, is second to the right of C.  A is second to the right of E, who is at one of the ends. J is the immediate neighbor of A and B and third to the left of G. H is to the immediate left of D and third to the right of I. 

3. Who is sitting in the middle of the row?

a. C          
b. I          
c. B            
d. G   
e. None

4. Which of the following groups of friends is sitting to the right of G?

a. IBJA 
b. ICHDF            
c. CHDF 
d. CHDE

Sol: _ _ _ _ _ _ _ _ _ _ _ _

Let us code all the given statements into some notation format so that it saves lot of time in going back and forth to the question. 

1. D, who is to the immediate left of F, is second to the right of C.  

This implies, D is sitting immediate left of F and D is setting second to the right of C. 

DF, C _ D

2. A is second to the right of E, who is at one of the ends. 

If E sits at one of the end he must sits at left end. Then only the following arrangement possible. 

E _ A

3. J is the immediate neighbor of A and B and third to the left of G.

AJB / BJA possible and J _ _ G

Therefore, A J B _ G /  B J A _ G 

4. H is to the immediate left of D and third to the right of I. 

HD and I _ _ H

From 1, C _ D F 

From 4, I _ _ H D

From 1 and 4, I _ C H D F ----(1)

From 3, A J B _ G or B J A _ G possible

If we consider 2 also, above statement becomes, E _ A J B _ G  - - - - - (2)

Now from 1 and 2, we have three possibilities. 1. F sits to the left of E 2. I sits to the right of F.

These two are not possible as total places are becoming more than 11. Now I should occupy the position between B and G.

So E _ A J B I G C H D F is the right arrangement. 

The remaining person K occupy the position between E and A. 

Now answers for the above questions are Option B and C.

Solving Complex Arrangement Questions:

1. Amit, Bharati, Cheryl, Deepak and Eric are five friends sitting in a restaurant. They are wearing caps of five different colours — yellow, blue, green, white and red. Also they are eating five different snacks — burgers, sandwiches, ice cream, pastries and pizza.

I. The person wearing a red cap is eating pastries.

II. Amit does not eat ice cream and Cheryl is eating sandwiches.

III. Bharati is wearing a yellow cap and Amit wearing a blue cap.

IV. Eric is eating pizza and is not wearing a green cap.

8. What is Amit eating?

a. Burgers 
b. Sandwiches 
c. Ice cream 
d. Pastries

9. Who among the following friends is wearing the green cap?

a.  Amit 
b. Bharati  
c. Cheryl 
d. Deepak

10. Who among the following friends is having ice cream?

a. Amit 
b. Bharati 
c. Cheryl 
d. Deepak

Sol: In this question, there are 3 variables.  Name of the person, Color of the caps, and Snacks they take. 

Never try to write all the names and try to match them.  This is a bad habit.  Try this method. 

1.  Identify one variable and write all the names belong to it below it.  Only write the variable names on both sides of this column

Now try to fill in the details in the table according the conditions given.

After Filling in all the available details, table looks like above.  Now we have to fit Pastries and Red some where.  Only one place left.  It must be at D. Once you fit that one, C's color becomes Green and E's become white.  Similarly B takes Ice cream and A takes Burger.  So final table looks like this

So answers for the above questions are A, C and B respectively.  

Complex arrangement problems are not this much straight forward. But the procedure to solve any question is like this.  

If there are more variables, the complexity increases.  But with adequate practice you can solve the questions easily.  

Non - verbal reasoning (Series)

Non-Verbal reasoning appears in Bank exams, Infosys, MAT exams constantly. There are 5 Problem Figures (PF) will be given with 5 Answer Figures (AF).  We need to determine the next figure in the series.  There are certain rules which make solving these problems easy.  So study the rules and solved examples.

How to answer these questions: 

Step 1: 

For all the series problems the following rules apply.  If problem figures A and E are equal our answer is problem figure B. Similarly, the other rules as follows. 

1. PF(A) = PF(E) then answer is PF(B)

2. PF(D) = PF(E) then answer is PF(C)

3. PF(A) = PF(C) = PF(E) then answer is PF(B)

4. PF(A) = PF(D), PF(B) = PF(E) then answer is PF(C)

5. PF(D) = inverse of PF(A) and PF(E) = inverse of PF(2) then answer is inverse of PF(C)

Step 2: 

In general, the items in the box takes different positions in the subsequent figures.  They may rotate certain degrees either clock wise or anti-clockwise.  Look at the following diagram. In some problems new items add to the existing figures and some existing figures vanish.  

Solved Examples

1. 

In this problem If PF(A) = PF(E) then answer is PF(B).  In the answer options AF(4) is same as PF(B) so option 4

2. 

 Here PF(C) and PF(E) are equal.  So Answer figure should be PF(B).  So correct option is c.

3. 

The arrow is changing its positions clock wise 90^o, 45^o, 135^o, 45^o, ....next should be 180^o. So option 3.

4. 

 Simple one.  A new arrow and a new line are adding alternatively.  In PF(E) a new line has added.  So in the next figure a new arrow must be added.  And total lines should be 6.  Option 5

5. 

 Small hand is moving anticlock wise 90^o, 45^o, 90^o, 45^o,... and Big hand is moving clock wise 135^o constantly.  So in the next figure, small hand must move 90^o anti clockwise, and big hand must move 135^o.  So option 4

 6.

Here the symbol is changing positions anti clockwise by  45^o and every time a new symbol is adding. The "C"s in the middle are rotating clock wise by 90^o. So the next figure must be option 4

7. 

 This is a simple analogy.  There is a relationship between 1 and 2, 3 and 4.  the small figures in the first diagram are getting bigger and vice-versa.  So Option 3

8. 

 All the three symbols in the dice are rotating clockwise.  So option 3

Alternative method:

We know that if  PF(A) = PF(D), PF(B) = PF(E) then answer is PF(C).  So option 3

9. 

 A new symbols is appearing in the middle of the previous figure and the previous figure is getting bigger. So option 4 is the right option.  3 and 5 options are ruled out as the figures in the middle are appeared already.

10. 

 A dot and line are adding constantly to the figures in left and right sides alternatively.  So option 3

11.

There appears to be no pattern on immediate look, but his problem can be solved by simple observation.  Have a look at the diagram below..

The positions of two symbols are not changing in 2 consecutive figures.  So option 5

12. 

 the arrow and small line inside the small square are rotating constantly anti clockwise and clockwise respectively by 90^o, 45^o, 90^o, 45^o,... and 45^o, 90^o, 45^o, 90^o .  So next figure would be option 3.

13.

 The line is rotating anti clock wise by 90^o, 180^o, 270^o, 360^o  so next figure should be 90^o from figure E and a new symbol must appear.  So option 1 is the correct.

14. 

 Symbols X is rotating clockwise by 45^o, 90^o, 45^o, 90^o.  So our options will be either 1 or 3 as in the next figure symbol X must move 45^o.  A new symbols is being added to X each time one at front and next time at back.  So option 3 is right one.

15.

 Simple.  Observe PF(A) and PF(E) are equal.  So next figure will be PF(B).  So option 5

16. 

 the symbols are changing constantly in clockwise direction and a new symbol is being added.

The red rounded circle is a place whenever a symbol appear in that position must not appear in the next.  And remaining positions are moving clockwise by 90^o.  A new symbol must come at the place shown by green arrow.  So our option will be 1.  Option 2 is ruled out as + symbol appeared earlier.

17.

 Circle is moving diagonally and triangle is moving clockwise by 90^o. So option 1 is correct one.


18. 

Here you can easily observe that the lines are rotating 90^o clockwise. also in PF(B) and PF(D), half line has added at the right most side and in figures PF(C) and PF(E) a new line has added. So in our answer half line has to be added and lines should rotate 90^o.  So answer option 2.

19. 

 Simple one.  Figures A and B changed their symbols opposite them.  C and D also did So.  So option 1

20. 

Symbols in  A, B are same except Symbols at bottom.  A new symbol is coming there.  Similarly in C, D.  So option 3.  Option 2 is ruled out as C appeared earlier.