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# Reverse Syllogism Tricks & Tips

With the introduction of new pattern questions on syllogism in Bank exams it has become important to use short method to solve such lengthy syllogism questions. Most of the students use Venn Diagram method to solve syllogism. No doubt the method is easy to use, but it is time taking to solve the new Reverse Syllogism questions by Venn Diagram, in which you have to check each option to satisfy the condition given in the question. So switching to RULES for SOLVING SYLLOGISM is best way to solve such question in short time.

Suppose you take 3 minutes to solve a single question of new pattern syllogism, you can solve the same question in 1 minute using rules after proper practice. In this way you can easily save 6-8 minutes in 5 such questions. You can use this saved time for solving other questions. This time will fetch you extra marks. Extra marks enough to clear sectional cutoff and to score good.

## Day 1 Topic: Arrangement of Sentence (FIRST & LAST RULE)

Today we will discuss the basic methods that will help us in solving syllogisms in a fast and effective manner. The real rules will start from Day 3. But without the knowledge of the concepts that we will discuss today, you cannot move on to rules directly. So understand these concepts well.

Basic Terms used in syllogism that we will follow in our post

Sample example of sentence in Syllogism: Some A are B

• Entity/Term: The terms between which relation is given is referred as entity. Here A and B are entity.
• Relation: Some, All, No, Some Not are examples of relation.
• Proposition/Sentence: The statement from which we have to derive the conclusion. We can refer to it by statement also in general.
• Conclusion: The derived solution from the proposition.

The main concept of Syllogism: FIRST AND LAST RULE

Our main aim in syllogism is to arrange the given proposition in such a way that the entities which are common in the multiple propositions always come together. Let’s understand it by an example.

Example: Consider these three statements:

But this may not be the case always. Some time in question, we may get propositions which are not arranged. We have to manually arrange them and then start solving the question. So we will use the FIST AND LAST RULE as our starting step in solving syllogism.

FIRST & LAST RULE: The propositions should be arranged in such a manner that, the two entities between which the conclusion is to be found out should come at the beginning and at the last while the common term should be in middle. i.e One should be come at the beginning and the other should be at the last.

Example: In the above example 1: The entity at the beginning is A and at the last is D. So using these three statements together we can find the relation between A and D.
If we consider only statements b) and c) then the first entity is B and Last entity is D. So using these two propositions we can find the relation between B and D.
If we consider only statement a) and b) then the FIRST entity is A and the last entity is B, so using these two together we can find the relation between A and B.

We must note that in Example 1, the propositions are already arranged in the desired manner. But in reality it may not be arranged. We have to arrange them by using certain methods. We will discuss two methods to achieve this.

(i) Rearrangement of Proposition – Day 1
(ii) Reversal of Proposition – Day 2

(i) Rearrangement of Proposition

First identify the common terms. And arrange them accordingly. With non-common entity at FIRST and LAST and the common entity coming together as shown in example 1 (b) highlighted entity B,B and C,C comes together. Example:

Now if we want to find relation between G and C, we will use just proposition (a) and (b)
If we want to find relation between F and Z, we will use the proposition (c) and (a)
For relation between F and C we will use all (c), (a) and (b).

For concept of rearrangement to work, one of the common entity should be at the last of the proposition and the other should be starting.

Example: All A are B. Some B are C. Here the common entity is B. It comes at the end in proposition 1 while at the start in Proposition 2.
Example: All A are B. Some A are C. Here the common term is A. But in both the proposition A is at the beginning. So in such case we cannot use the concept of rearrangement. The method to solve such a case will be discussed later.

More Examples:

Exercise 1) Arrange these propositions in such a way that common entity comes together:

(a) No tiger is lion
(b) Some elephant are tiger
(c) All lion are monkey

Solution: The common terms are tiger and lion. Arrangement is (b),(a),(c) i.e
Some elephant are tiger. No tiger is lion. All lion are monkey.

Exercise 2) Arrange these propositions in such a way that common entity comes together:

(a) Some red are blue
(b) All black are pink
(c) some blue are black
(d) No pink is violet

Solution:  (a),(c),(b),(d)
so we have now
(a) Some red are blue
(c) some blue are black
(b) All black are pink
(d) No pink is violet

Ques) Which propositions are required to get a relation between blue and violet.
Answer: (c),(b),(d) [FIRST and LAST RULE] – The two entity between which relation is to be found out should come at beginning and at last.

Ques) Which propositions are required to get a relation between red and pink.
Answer: (a),(c),(b) [FIRST and LAST RULE

LIMITATION of REARRANGEMENT

Consider this example:

We will discuss the solution to this problem on Day 2.

## Exercise for the Day

Q1) Arrange the given statement according to the FIRST & LAST rule to find a relation between CONE and RECTANGLE.
(a) All cones are cylinders.
(b) No square is a rectangle.
(c) All cylinders are squares.

Correct Order is: (a)(c)(b)
Explanation
: As we have to find relation between CONE and RECTANLGE, they should come at the beginning and at the end. So start with the statement that has CONE in the start. i.e (a).  Now the common term cylinder is in (a) and (c). So next proposition is (c) followed by (b).

Q2) Arrange the given statement according to the FIRST & LAST rule to find a relation between GRAPE and SWEET
(a) No orange is apple.
(b) All grapes are oranges.
(c) Some apples are sweet.

(b) (a) (c)

Q3) Arrange the given statement according to the FIRST & LAST rule to find a relation between PRETTY and BOYS

(a) All girls are pretty.
(b) Some smart are girls.
(c) Some boys are smart.

(c) (b) (a)
Explanation
: Start with the proposition which has either BOY or PRETTY in it. We have (c) and (a) . But choose (c) because it has boy in the starting. It will be easy. While (a) has PRETTY at the end. (c) follow by (b) [due to common term smart] followed by (a) [] due to common term girls

Q4) Arrange the given statement according to the FIRST & LAST rule to find a relation between BUSSES and PLANE

(a) No plane is metro
(b) All busses are rails.
(c) Some rail is plane.

(b) (c)
Explanation
: Start with (b) it has BUSSES in starting. Common term rails is in (b) (c) SO we now have BUSSES and PLANE in starting and end. So FIRST LAST RULE is satisfied.

Q5) Arrange the given statement according to the FIRST & LAST rule to find a relation between SILVER and PLATINUM

(a) No gold is silver.
(b) Some platinum is gold.
(c) Some silver are diamond.

(b) (a)

## Day 2 Topic: Reversal of Sentence (FIRST & LAST RULE)

(ii) Reversal of Proposition

How to solve such statement? Here the common term is in the starting in both the proposition. For this we must know these four rules.

 Original Statement Reversed Statement All A is B Some B is A Some A is B Some B is A No A is B No B is A Some A is not B No Conclusion (Reversal Not Possible)

Let’s again take example 3:
Statement (a) Some A are B
(b) All A are C
Reverse statement (a) we get => Some B are A.
Now take the proposition in the order (a), (b) we get
Some B are A. All A are B [Now common entity are together.  So FIRST & LAST RULE is applicable.]

Doubt??????
In the above example, we reversed Some A are B. Can we reverse All A is C=> Some C are A????
Then arrange the proposition in the order (b), (a) ????
Some C are A + Some A are B. [Note here also FIRSt & LAST RULE is applicable.] But which method is correct?? First , Second or Both????

To get the answer we will discuss the preference order for reversal rule

Preference Order for reversal of statement

The table below shows the order of preference for reversal.

 Order Relation Comments 1 Some, No Both Some and No have equal preference 2 All All has less preference than Some and No 3 Some Not Some Not cannot be reversed

Understand it by example.

Consider the following example:

(a) Some A are B
(b) No A are C
(c) All A are D
(d) All A are E

Example 1: Suppose we need to find relation between B and D. Then we need only proposition (a) and (c). But one of these must be reversed for FIRST & LAST RULE to be applicable.

According to above table Some has higher preference than All. So the statement with Some i.e proposition (a) must be reversed. We cannot reverse All here. So we get: Some B are A + All A are D

Example 2: Suppose we need to find relation between C and D. Then we need proposition (b) and (c) only. As per the table No has higher preference than All . So we have to reverse (b) we cannot reverse (c). So we will reverse proposition (b) and arrange in the order (b),(c) to get. No C is A + All A is C.

Example 3: Suppose we need to find relation between B and C. Then we need only proposition (a) and (b). Now Some and No both have equal preference. So we can reverse any one of them to get the desired result. i.e
Case 1: reverse (a) => Some B are A. and add it to (b)
Some B are A + No A is C
Case 2: reverse (b) => No C is A. and add it to (a)
No C is A + Some A are B

Example 4: Suppose we need to find relation between D and E. Then we need only proposition (c) and (d). here both the proposition contains ALL. So we can reverse either of them. Same as above.
Suppose we reverse (c) => Some D are A. + All A are E [FIRST & LAST RULE applicable now.]

So now our above discussed doubt has been cleared.

Practise sets for Reversal of Proposition

Example 1:
(a) Some A is B
(b) All C is B
(c) No D is A
Q1) Arrange the sentence to get a conclusion between D and B.
Solution 1) For B and C we have common term A in the proposition (a) and (c). Take them in order (c),(a) so that FIRST LAST RULE is applicable.
=> No D is A + Some A is B

Q2) Arrange the sentence to get relation between C and D.
Solution 2) Now we need to use both the rearrangement and reversal concept to achieve this goal. Follow this step:

(i) As per preference order we cannot Reverse statement (b) as it contains ALL. So fix statement as it is. (ii) Now the common entity between (b) and (a) is B. So reverse (a) so that FIRST LAST RULE is applicable.
We now have (b) + reverse(a) => All C is B + Some B is A.
(iii) Now the only proposition left is (c). The common entity between (a) and (c) is A. But for FIRST LAST RULE to be applicable (c) must be reverse.  So we have now
All C is B + Some B is A + No A is D.
(b) + reverse(a) + reverse(c)

Note:These are only for practice. When you get used to these concepts, you can simply solve the above questions in mind without using pen and paper. So practice these concepts regularly.

Example 2:
(a) All A is B
(b) No A is D
(c) All B is C

Q1) Arrange the above propositions to get a conclusion between D and C.

Steps:
(i) Note that here the highest preference is of proposition (b).On reversal it can be joined with (a) with the common entity A. So we have Reverse(a)+ (b)
(ii) Now the common entity between (a) and (c) i.e B, is already arranged. So we have: Reverse(a+(b)+ (c)

=> No D is A + All A is B + All B is C

## Exercise for the Day

Direction (1-3): Using the given statements answer the following questions:

(a) No C is B
(b) All A is B
(c) Some D is C

1. Arrange the propositions to get a conclusion between B and D

Explanation
: (c) + (a)=> Some D is C+ No C is B
2. Arrange the propositions to get a conclusion between A and C

Explanation
: (b) + reverse(a) =>All A is B + No B is C
3. Arrange the propositions to get a conclusion between A and D

Explanation
: (b) + rev(a)+rev(c) =>All A is B + No B is C + Some C is D

Direction (4-5): Using the given statements answer the following questions:

(a) All A is B
(b) No C is B
(c) All A is D

1. Arrange the propositions to get a conclusion between A and C

Explanation
: (a) + reverse(b)
2. Arrange the propositions to get a conclusion between D and C

Explanation
: reverse(c) + (a) + reverse(b)

## Day 3: Actual Rules Syllogism

Rule Table: Learn it so that next time you don’t have to refer this table

 S.No First Proposition Second Proposition Conclusion 1 ALL ALL ALL 2 ALL NO NO 3 ALL SOME NO CONLCUSION* 4 SOME ALL SOME 5 SOME NO SOME NOT 6 SOME SOME NO CONCLUSION* 7 NO ALL SOME NOT REVERSED 8 NO SOME SOME NOT REVERSED 9 NO NO NO CONCLUSION* 10 Some Not Anything(Some/No/All/Some Not) NO CONLCUSION* *No Conclusion= Any Possibility is True

Note that we will have to use the concept of Rearrangement and Reversal that we have learned on Day 1 and Day 2 to solve Syllogism by rules.

Examples

1) All+All=All

All Red are Green + All Green are Blue. => Conclusion=  All Red is Blue
Note: The reverse of the obtained conclusion i.e Reverse of All Red is Blue = Some Blue is Red will also be true.

2) All + No= No

All Red is green + No Green is Blue => Conclusion= No Red is blue.
Reverse conclusion= No Blue is Red

3) All+Some=No Conclusion

All Red is Green + Some Green is Blue => Conclusion= No conclusion can be drawn between Red And Blue. (But No conclusion means that any possibility between Red and Blue will be true)
Example: All Red being blue is a possibility.
Some Red are blue is a possibility
Some red are not blue is a possibility
No red is blue is a possibility.
All these possibility are true.

4) Some + All= Some
Some Red is green + All green is blue. => Conclusion= Some Red is blue.
Reverse conclusion=some blue is red.

5) Some + No = Some Not

Some Red is green + No green is blue=> Conclusion= Some Red is Not Blue

6) Some + Some= No Conclusion

Some Red is Green + Some Green is blue. => Conclusion= No Conclusion = Any possibility between Red and Blue is true (Same as in (3))

7) No + All = Some Not Reversed
No Red is Green + All Green is blue => (Apply Some Not on Reversed entity. i.e reverse Red and blue)
Means First Blue then Red at last and put Some Not
=> Conclusion= Some Blue are Not Red

8) No + Some = Some Not Reversed
No Red is Green + Some Green is blue => Conclusion= Some Blue is not Red

9) No + No = No
No Red is Green + No Green is blue => No Conclusion.

## Exercise for the Day

Solve only by rules. Don’t use venn diagram.

1. Statement:
(a) Some A are B
(b) All A are C
Conclusion:
(i) Some B are C
(ii) Some C are A
(iii) Some B are A
A) Only (i)
B) Only (ii)
C) Both (i) and (ii)
D) Both (ii) and (iii)
E) All

Option E
Explanation
: For (i) => reverse(a) + (b) => Some B are A+ All A are C => Some + All= Some => Some B are C is true.
For (ii) true by reverse of (b) . For (iii) True by reverse of (a)
2. Statement:
(a) No pink is white
(b) All white is red
Conclusion:
(i) Some red are not pink
(ii) Some pink are not red
A) Only (i)
B) Only (ii)
C) Either
D) Both
E) Neither

Option A
Explanation
: No+All = Some not reversed=> Some red are not pink.
Please Note: Some A are not B DOESNOT MEANS THAT Some B are Not A.
3. Statement:
(a) All A are B.
(b) Some C are B
Conclusion:
(i) All A are C
(ii) Some A are C
(iii) All A are C is a possibility
A) Only (i)
B) Only (ii)
C) Only (iii)
D) None follows

Option C
Explanation
: For FIRST & LAST RULE to be applicable. We have to reverse (b). So we get
All A are B + Some B are C => All+ Some = No conclusion between A and C => any possibility between A and C is true. So only (iii) is true.
4. Statement:
(a)Some Samsung are Phone
(b) No Samsung is Vivo.
Conclusion:
(i) Some Vivo are not phone
(ii) Some Phone are not Vivo
(iii) No Vivo is samsung
A) (i) and (iii)
B) (iii)
C) (ii) and (iii)
D) (ii)
E) All

Option C
Explanation
: For (ii): Reverse any of the two statement, as the preference of both are equal. Here I am reversing (a). So we get=> Some Phone are Samsung + No samsung is vivo => Some + No = Some Not => Some phone are not vivo. So (ii) is true. You can try by reversing (b). The conclusion will be same.
For (iii) It is the reverse of (b). So true
5. Statement:
(a) Some A are B
(b) Some B are C
Conclusion:
(i) Some A are C
(ii) No A is C is a possiblity
(iii) Some C are B
A) Only (ii)
B) Only (iii)
C) Only (i)
D) None
E) Both (ii) and (iii)

Option E
Explanation
: For (ii) we have Some + Some = No conclusion between A and C. So any possibility is true between A and C. So (ii) is true.
(iii) is true by reverse of (b)

## Day 4: Solving 3 or more statements in one go by rules

Example: Some A are B + All B are C + All C are D.
Method 1: Take first 2 proposition: We have Some+ All = Some
Then this Some with the last proposition i.e Some +All= Some.
Hence We have Some A are D is the conclusion.

Method 2: Take 2nd and 3rd proposition i.e All+ All = All
Now solve it with first proposition Some+All=Some => Some A are D

In both case we get the same answer. But there are some cases where you have to follow a sequence in solving the proposition to get the conclusion. Let’s have a look on the next example.

We will discuss again about this irregularity and a special case related to it on the last day.

Solved Examples:

1) Statements: (a) Some A are B.  (b) All C are B.  (c)Some C are D.  (d)No E is C.
Conclusions:
I. Some B are not E.
II. No B is D
A) only I follows
B) only II follows
C) either I or II follows
D) neither I nor II follow
E) both I and II follow

Option A
Explanation
: In (I) we have to check conclusion between E and B. So see which propositions do we need as per the FIRST & LAST RULE. We have B in two statement, (a) and (b) while we have E in proposition (d). Note that the common entity between (b) and (d) is C. So we need just (b) and (d)
Solve in the order (d)+ (b) so that FIRST & LAST RULE is followed. No + ALL= Some Not reversed. Some B are not E. So (I) is true.
(II) we need to find relation between B and D. We need only proposition (b) and (c) for this
Take Reverse (c) + (b) = Some D are C + All C are B= Some D are B. Hence (II) is false

2) Statements: All A are B. Some C are B. Some C are D. Some E are C.
Conclusions:
I. Some A are C
II. No B is D
A) only I follows
B) only II follows
C) either I or II follows
D) neither I nor II follow
E) both I and II follow

Option D
Explanation
: For (I): (a) + reverse(b) = All + Some  No Conclusion between A and C. So I is false
(II): reverse(b)+(c)= Some+ Some= No conclusion between B and D. So II is false

3) Statements: Some A are B. All B are C. Some B are D. Some E is D.
Conclusions:
I. Some A are C
II. No E is A
A) only I follows
B) only II follows
C) either I or II follows
D) neither I nor II follow
E) both I and II follow

Option A
Explanation
: For (I): (a)+(b)= Some+All= Some = Some A are C is true. So I follows
(II): (a)+(b)+(c)+reverse(d)= Some+All+Some+Some  => solve one by one in mind.. Some+All= Some then add it to next Some, Some +Some No conclusion. Stop here. So there is no definite conclusion between A and E. So II doesnot follows.

## Day 5: All the cases of possibility and Condition of Complementary Pair (EITHER-OR)

We have already discussed the main concept of Possibility i.e The case of NO CONCLUSION. Now we will some more cases where Possibility will be true. We have presented the proposition with all possible outcomes. Learn them.

Before starting learn this as a RULE OF THUMB:
1) Restatement Follows
2) Possibility of Restatement or Definite Conclusion Doesn’t Follows

Example: Statement: Some A are B
Conclusion: Some A are B(Follows)
Some A are B is a possibility(Doesn’t Follows)

 Statement: All A are B Conclusions Some B are A Some A are B All B being A is a possibility Some B are not A is a possibility

 Statement: Some A are B Conclusions Some B are A All B being A is a possibility All A being B is a possibility Some A are not B is a possibility Some B are not A is a possibility

 Statement: No A are B Conclusions No B is A Some A are not B Some B are not A

 Statement: Some A are Not B Conclusions All B being A is a possibility  (Very Important) No A is B is a possibility No B is A is a possibility Some B are not A is a possibility Some A are B is a possibility Some B are A is a possibility

Case of Complementary Pair (EITHER-OR)

There are only three conditions for two propositions to be Complementary pair. But remember for two propositions to be complementary pair, both of them must be False.

 Case Example Condition 1 Some + Some Not Some A are B + Some A are not B OR Some B are A + Some A are not B Condition 2 Some + No Some A is B + No A is B OR Some B is A + No A is B Condition 3 All + Some Not All A is B + Some A are not BNote: All A is B + Some B are not A (is not complementary pair)[Very Important]

Questions for Practise

1. Statement: All radios are pencils
Some pencils are files
Conclusion: I. No radio is file
A) Only I follows
B) Only II follows
C) Either I or II follows
D) Neither I nor II follows
E) Both I and II follows

Option C
Explanation
: All + Some = No conclusion ; with means both I and Ii are false.
But they satisfy the condition of Complimentary pair (Some + No) : So Either Or is the answer
2. Statements: Some diggers are jokers
All jokers are cute.
Conclusion: Some diggers are cute.
II. No diggers are cure.
III. Some diggers are not cute
IV. All diggers are cute.
A) I and III follows
B) Either II or IV follows
C) II and either III or IV follows
D) Either III or IV and I follows
E) Either I or II and either III and IV follows

Option D
Explanation
: Using the statements we get the conclusion: Some +All= Some=> Some diggers are cute. So I follows. While II, III and IV donot follow. But III and IV make complimentary pair.
3. Statements: All cones are cylinders. Some cones are squares. No square is a rectangle.
Conclusions:
I. Atleast some cones being rectangle is a possibility.
II. Some cylinders are cones.
A) only I follows
B) only II follows
C) either I or II follows
D) neither I nor II follow
E) both I and II follow

Option E
Explanation
:
For I. Using second and third proposition : Some+ No= Some not=> Some Cone is not Rectangle. So SOme COne being rectanlge is a possibility
For II. It is the reverse of All cone are cylinder. So true

This is the end of our main course. On the last day i.e tomorrow we will discuss two exceptional case of syllogism and certain controversy and practice a lot of questions including the new pattern reverse syllogism. Revise all the content of these 5 days and be in regular touch with Syllogism by rules. Once you get habituated, you can solve any difficult question easily that too in less time.

## Day 6: Certain Controversy and Exceptional Case

Two Exceptional Case of SOME NOT

On Day 3 we learned a rule that SOME NOT + ANYTHING = NO CONCLUSION
Here we will see two exceptional cases in which SOME NOT and ALL are involved. Rest everything can be solved by using the table discussed on Day 3.

### Exception 2:

In Rest All case of SOME NOT + ALL or Any other proposition we will get NO CONCLUSION

Example 1: Some A are not B + All B is C. Here the common term B is neither in the starting in both the proposition neither at the end in both proposition. So exceptional case is not valid here.

Example 2: Tell which of the following conclusion follows
Statement:
All red is pink
Some red is not green
Conclusion:
I. Some green are not pink
II. Some Pink are not green

Explanation
: Only II. Follows
This is the case 1 of exception. Common term is in the starting. So arrange them in the order Some Not + All . Then take the term from down to up, as explained. We get Some Pink are Not Green

Example 3: Tell which of the following conclusion follows
Statement:
Some blue are not green
All red is green
Conclusion:
I. Some blue are not red
II. Some red are not blue.

Explanation
: Only I. Follows
This is the case 2 of exception as the common term is at the end. So take the sentence in order All+ Some Not. Start from down move to Up as in the image of example 8. Some blue are not red

Certain Controversy

 Case Solution 1 Does restatement follows? As explained on Day 5 The Rule is 1) Restatement Follows 2) Possibility of Restatement Doesn’t Follows Example: Statement: Some A are B Conclusion: Some A are B(Follows) Some A are B is a possibility(Doesn’t Follows) 2. Does Some A is B means Some A are Not B or vice versa? NO. It does not means. 3. Does All + No makes complimentary pair? NO. They don’t. On Day 5 we have covered this topic 4. The statement All A are Not B ??? What does it means This is a controversial statement which can have two meaning. So you will never get this statement in exam.

Certain Special words used in Syllogism and their meaning

 Terms Meaning Example Atleast, Most, Many, few SOME Atleast some A are B* OR Most A are B OR Many A are B OR Few A are B => Some A are B *In exams you will mostly find the case of Atleast. Only Statement: Only A is B Conclusion: All B is A

Trick to solve the new pattern syllogism in short time using rules

By using rules we can solve the new pattern reverse syllogism questions in very less time. We can further reduce this time by having a correct thought process while solving the question. In the below example we will show the right approach to solve such question.

Example 4Conclusions: Some cars are not trucks. Some buses are trucks.
STATEMENTS:
A) All cars are scooters. No bus is scooter. Some busses are trucks.
B) All cars are scooters. No truck is scooter. All trucks are busses.
C) All cars are scooters. All bus are scooters. All trucks are busses.
D) Some cars are scooters. No truck is scooter. No bus is car.
E) None

Option B

Now how to tackle such questions? You can either check all the option and get the correct answer easily. Or you can simultaneously use elimination and reduce the number of options that you have to check. To eliminate wrong options follow this approach:

One of the Conclusion is SOME NOT : When do we get SOME NOT? We get one SOME NOT when one of the proposition is SOME and another is NO. OR we can also get SOME NOT when tthere is case of NO :example No A is B also means that Some A is Not B. So even If All + No is there we can get SOME NOT.
So check in each option at a glance which option does not contains Some + No OR All + No together And eliminate that option as that cannot be the answer.
So we get Option C) cannot be the ans as it has Only ALL.

Now go to the second conclusion: We have the case of SOME. When do we get SOME? We get Some when the first proposition is Some and the second is All. Or both is ALL. Now scan the remaining options which does not have either Some+ All or All  + All together. So see that Option D is also eliminated.

Now we are left with only Option A) and B) now use the rules and check both the case. And get the final answer. You can practice this though process now and gain expertise over it  and use it in real exam.

Questions for Practise

1. Conclusions:
(i) All red is blue is a possibility.
(ii) All green being yellow is a possibility.
STATEMENTS:
A) All red is yellow. All yellow is blue. No blue is green.
B) Some red is yellow. No yellow is blue. Some blue is green.
C) Some red is yellow. Some yellow is blue. No blue is green.
D) All red is yellow. No yellow is blue. All blue is green.
E) None

Option C

From option C, i) Some+Some = NO CONCLUSION and NO Conclusion means Any Possibility is True. => ) Some red is yellow. Some yellow is blue. = All red is blue is a possibility is true. AND ii) Some + No = Some Not => Some yellow is blue. No blue is green.= Some Yellow are Not Green => All green being yellow is a possibility. [Remember When SOME A is NOT B is true then All B being A is a possibility is also true.]
2. Conclusions:
(ii) Some express are times
STATEMENTS:
A) All business are standard. All standard is times. Some times are express.
B) All business is standard. Some standard are times. All times are express.
C) No business is standard. All standard are times. Some times are express.
D) Some business is standard. All standard is times. No times is express
E) None

Option E

Statement i) doesnot follow in any of the Option
3. Conclusions:
(i) Some heater are lamp.
(ii) Some fan are lamp.
STATEMENTS:
A) All lamp are bulb. All bulb are fan. Some fan are heater.
B) All lamp are bulb. Some bulb are fan. All fan are heater.
C) Some lamp are bulb. All bulb are fan. All fan are heater
D) All lamp are bulb. All bulb are fan. Some fan are heater.
E) None

Option C

From option C, i) Some+All+All= Some => Some lamp are bulb. All bulb are fan. All fan are heater= Some lamp are heater= some heater are lamp(reverse) AND ii) Some +All=Some => Some lamp are bulb. All bulb are fan. = Some lamp are fan= some fan are lamp(reverse)
4. Conclusions:
(i) Some tulip are not lily.
(ii) Some flower are rose.
STATEMENTS:
A) Some rose are not lily. All rose are tulip. All tulip are flower.
B) All tulip are rose. Some lily are not rose. Some tulip are flower.
C) All rose are lily. No lily is tulip. No tulip are flower.
D) Some rose are lily. No lily is tulip. Some tulip are flower
E) None

Option A

From Option A, i) ) Some rose are not lily. All rose are tulip.=> Some tulip are not lily AND ii) All+All = All => All rose are tulip. All tulip are flower.= All rose are flower= Some flower are rose.
5. Conclusions:
(i) All water being river is a possibility
(ii) Some ocean are not lake.
STATEMENTS:
A) Some lakes are water. All water us ocean. No ocean is river.
B) All lakes are water. All water is ocean. Some ocean is river.
C) All lakes are water. No water is ocean. Some ocean is river.
D) Some lakes are water. Some water is ocean. All ocean is river.
E) None

Option C

From option C, i) No+Some = Some not reversed=> No water is ocean. Some ocean is river.= Some river are not water. = All water being river is a possibility. AND ii) All+ No = No => All lakes are water. No water is ocean. No Lake is ocean and hence Some Ocean are not lake is also true.

## Practice Problems On Syllogism

Direction (1-5): Each option contains some statements. Determine from which statements the given two conclusions follow.

1. Conclusions:
I. All dog being tiger is a possibility
II. Some tiger are not lion
Statements:
A) Some tiger is cat. All cat is lion. No lion is dog
B) No tiger is cat. Some cat is lion. Some cat is dog
C) All dog is cat. No cat is tiger. Some cat is lion
D) No lion is cat. Some cat is tiger. No cat is dog
E) None is correct

Option D
2. Conclusions:
I. Some Blue are Green
II. No red is White
Statements:
A) All blue are red. All red are green. Some green is white
B) All red is green. No green is white. All white is blue
C) All blue is red. Some red are green. All green is white
D) Some blue are red. All red is green. No green is white
E) None is correct

Option D
Solution:
In D) For I. Some blue are green. All red is green => Some+ All= Some=> Some blue are green
For II. All red is green. No green is white => All+ No = No=> No red is white
3. Conclusions:
I. All dollar are euro is a possibility
II. Some pound are dollar.
Statements:
A) Some dollar are rupee. No rupee is pound. Some pound are euro.
B) Some dollar are rupee. All rupee are pound. No rupee is euro.
C) Some dollar are rupee. All rupee are pound. Some pound are euro.
D) All dollar are rupee. Some rupee are pound. Some pound are euro.
E) None of these

Option C
Solution
:
In C)
I. Some + All = Some
Some + Some = No Conclusion but possibility is true so I follow.
II. Some + All = Some. => Some dollar are pound. Reverse Some pound are dollar is true
4. Conclusions:
I. Some pencil are eraser
II. No scale is pen
Statements:
A) No scale is pencil. Some pencil are pen. All pen is pencil
B) All pen is pencil. No pencil is scale. All scale is eraser
C) Some pencil are scale. All scale is eraser. No eraser is pen
D) Some scale are pencil. No pencil is pen. All pen is eraser
E) None is correct

Option C
Solution:
In C) For I. Some pencil are scale. All scale is eraser=> Some + All = Some=> Some pencil are eraser
For II. All scale is eraser. No eraser is pen => All+No= No=> No scale is pen
5. Conclusions:
I. Some white are red is a possibility
II. No red is blue
Statements:
A) Some white are green. No green is red. All blue is green
B) Some red is green. No green is white. All white is blue
C) No blue is green. All green is red. Some green are white
D) Some white are blue. Some blue are red. All red are green.
E) None is correct

Option A
Solution
:
In A) For I. Some white are green. No green is red.=> Some+ No= Some Not=> Some white are not red => Some white are red is a possibility
For II. All blue is green. No green is red.=> All + No => No blue is red => No red is blue

Direction (6-10):  In the following question, a set of four statements is followed by five conclusions, one of which definitely does not follow (or is not a possibility of occurrence). Choose that conclusion as the answer.

1. Statements: Some red are green. All green are blue. No blue is yellow. All yellow is black.
Conclusions:
I.  Some black are not blue
II. Some yellow is green
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option B
2. Statements: All green are white. No white is red. Some red is yellow. No yellow is orange
Conclusions:
I. All red being orange is a possibility
II.  Some yellow are red
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option A
3. Statements: No cat is dog. All dog is lion. No lion is tiger. Some tiger are rabbit
Conclusions:
I. All lion being dog is a possibility
II. All rabbit is lion
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option B
4. Statements: Some rabbit is lion. All lion is monkey. All monkey is tiger. No tiger is dog
Conclusions:
I. Some monkey is dog
II. All rabbit are dog
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option E
5. Statements: All dog is cat. All cat is rat. No rat is lion. All lion is tiger
Conclusions:
I. Some cat are not lion
II. All rat being tiger is a possibility
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option D

Directions: In the following question, a set of four statements is followed by five conclusions, one of which definitely does not follow (or is not a possibility of occurrence). Choose that conclusion as the answer.

1. Statements: Some lions are tigers. Some monkeys are dogs. All dogs are cats. No monkey is lion
Conclusions:
I. All cats are tigers
II. All dogs are lions
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option B
Solution:
If all dogs are lions, then some lions will be monkeys but no lion is monkey so II is definitely false.
2. Statements: All yellow are red. All red are brown. Some green are red. Some white are yellow.
Conclusions:
I. All dogs  are tigers
II. No green is yellow
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option D
3. Statements: Some copy are paste. Some clip are paste. No paste is cut. All add are cut.
Conclusions:
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option A
4. Statements: Some add are subtract. Some divide are subtract. All divide are multiply. No bracket is subtract.
Conclusions:
I. No divide is bracket
II. All bracket are add is a possibility
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option D
5. Statements: All yellow are red. Some brown are yellow. No brown is white. All green are white.
Conclusions:
I. Some green are yellow
II. No red is white
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option D
6. Statements: All lions are tigers. Some dogs are tigers. No cat is dog. All monkey are cat.
Conclusions:
I. All tigers are cats
II. All monkey are tigers
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option A
7. Statements: All A are B. Some A are C. No C is E. All D are E.
Conclusions:
I. Some D are C
II. Some A are D
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option A
Solution:
No C is E, so some D are C is definitely false
8. Statements: Some A are B. All B are C. No C is D. All D are E.
Conclusions:
I. All A are E
II. Some B are D
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option B
9. Statements: Some A are B. All A are C. Some B are D. No B is E.
Conclusions:
I.No A is E
II. All E are C.
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option D
10. Statements: Some A are B. All C are D. No B is D. All D are E.
Conclusions:
I. All A are D
II. Some C are B
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option E
Solution:
I. If all A are D, then some B will be D which is false
II. Similarly II definitely not follows
1. Statements: Some red are green. All red are blue. No red is white.
Conclusions:
I. All white are blue is a possibility.
II. All green are white is a possibility.
A) only I follows
B) only II follows
C) either I or II follows
D) neither I nor II follow
E) both I and II follow

Option A
2. Statements: Some pets are good. Some good are fine. All tricks are fine. No good is bad.
Conclusions:
I. All pets are bad is a possibility.
II. Some tricks are good.
A) only I follows
B) only II follows
C) either I or II follows
D) neither I nor II follow
E) both I and II follow

Option D
3. Statements:Some waters are plants. All plants are pots. Some pots are roots. All roots are stems.
Conclusions:
I. Some roots are not waters
II. All roots being plants is a possibility.
A) neither I nor II follow
B) only I follows
C) either I or II follows
D) only II follows
E) both I and II follow

Option D
4. Statements: Some sites are roads. No site is door. All roads are narrow.
Conclusions:
I. At least some roads are doors.
II. Some narrow are doors is a possibility.
A) only I follows
B) only II follows
C) either I or II follows
D) neither I nor II follow
E) both I and II follow

Option B
5. Statements: All foods are trees. No tree is a stem. Some stems are roots.
Conclusions:
I. Some trees are foods
II. No stem is food
I. Some trees are foods
II. No stem is food
A) only I follows
B) only II follows
C) either I or II follows
D) neither I nor II follow
E) both I and II follow

Option E
6. Statements: All blue are white. No blue is green. Some green are black.
Conclusions:
I. No blue is black
II. Some green are white is a possibility
A) Both I and II follow
B) Neither I nor II follow
C) Only I follows
D) Either I or II follows
E) Only II follows

Option E
7. Statements:All fonts are styles. All styles are paras.Someparas are boards. Some boards are files.
Conclusions:
I. Some paras are fonts.
II. Some files are paras
A) only I follows
B) only II follows
C) either I or II follows
D) neither I nor II follow
E) both I and II follow

Option A
8. Statements:Some tigers are lions. All tigers are rats. No lion is dog. All cats are dogs.
Conclusions:
I. All tigers are cats is a possibility.
II. No cat is lion.
A) only I follows
B) only II follows
C) either I or II follows
D) neither I nor II follow
E) both I and II follow

Option B
9. Statements: Some sites are roads. No site is door. All roads are narrow.
Conclusions:
I. At least some roads are doors.
II. Some narrow are doors is a possibility.
A) only I follows
B) only II follows
C) either I or II follows
D) neither I nor II follow
E) both I and II follow

Option B
10. Statements: Some pets are good. Some good are fine. All tricks are fine. No good is bad.
Conclusions:
I. All pets are bad is a possibility.
II. Some tricks are good.
A) only I follows
B) only II follows
C) either I or II follows
D) neither I nor II follow
E) both I and II follow

Option D

Directions (1-3): In each of the questions below, statements are given followed by two conclusions numbered I and II. Read all the conclusions and then decide which of the given conclusions DEFINITELY DO NOT follow from the given statements.

1. Statements: All chairs are tables. Some tables are pens. No table is pencil. Some pencils are erasers.
Conclusions:
I. Some chairs are pencils
II. All pens are pencils
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option E
2. Statements: All pigs are cats. Some pigs are dogs. No dog is tiger. All lions are tigers
Conclusions:
I. No cat is lion
II. All lions are pigs
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option D
3. Statements: Some chairs are tables. All tables are pens. Some pens are pencils. No pen is eraser.
Conclusions:
I. Some tables are erasers
II. All pencil are eraser
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option E

Directions (4-7): In each of the questions below, statements are given followed by two conclusions numbered I and II. Read all the conclusions and then decide which of the given conclusions follow from the given statements.

1. Statements: All green are orange. Some orange are pink. All pink are white. Some red are pink.
Conclusions:
I. Some orange are white.
II. No green is red.
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option A
2. Statements: Some copy are pen. All pen are erasers. No book is eraser. Some pencils are books.
Conclusions:
I. Some pens are books
II. All pencils are pens is a possibility.
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option D
3. Statements: All chairs are tables. Some pens are chairs. No pencil is pen. All pencils are erasers.
Conclusions:
I. All chairs are erasers is a possibility.
II. Some pencils are chairs.
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option A
4. Statements: All green are white. Some green are black. No blue is green. All red are blue.
Conclusions:
I. No white is red.
II. All black are red is a possibility.
A) only I
B) only II
C) either I or II
D) neither I nor II
E) both I and II

Option D

Direction (8-10): Each option contains some statements. Determine from which statements the given two conclusions follow.

1. Conclusions:
I. Some pens are pencils is a possibility
II. All copy are pencils is a possibility
A) Some copy are pen. All pen are eraser. No pencil is eraser. Some pencils are books.
B) All erasers are copy. Some eraser are pen. No pencil is eraser. All books are pencils.
C) All copy are pen. Some eraser are pen. No book is eraser. All pencils are books.
D) Some copy are pen. Some eraser are pen. No book is pen. All pencils are books.
E) both I and II

Option C
2. Conclusions:
I. All white are blue is a possibility.
II. Some black are not red.
A) All white are black.Some green are black. Some red are green. No black is blue.
B) Some white are black. All black are green. No red is green. Some red are blue
C) Some white are black. All black are green. Some red are green. Some red are blue
D) All white are black. Some white are green. No blue is green. Some red are blue.
E) both I and II

Option B
3. Conclusions:
I. No cat is pig.
II. All pig are rabbit is a possibility.
A) All monkey are cats. Some cats are lions. Some lions are rabbits. No pig is lion.
B) All monkey are cats. No monkey is lion. All lions are rabbits. Some pigs are lion.
C) Some monkey are cats. Some lions are cats. All rabbits are lions. No pig is lion.
D) Some monkey are cats. All cats are lions. Some lions are rabbits. No pig is lion.
E) both I and II

Option D

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