Reasoning PYQs

Given

Statements:

U > V ≥ D; K ≥ T ≤ D

On combining: U > V ≥ D ≥ T ≤ K

Conclusions:

I. T ≤ V → True (as V ≥ D ≥ T → V ≥ T)

II. K > U → False (as U > V ≥ D ≥ T ≤ K thus clear relation between K and U cannot be determined)

Hence, only I is True.

**Given

Statements:

** X < Q ≤ N = L; H > F > X

On combining: H > F > X < Q ≤ N = L

Conclusions:

I. H > N → False (as H > F > X < Q ≤ N there is opposite sign between H and N. So, It is false)

II. F ≤ L → False (as F > X < Q ≤ N = L there is opposite sign between F and L. So, It is false)

Hence, Neither I nor II is True.

Given Statements: A > B ≥ Q; G < C > D = Q

On combining: A > B ≥ Q = D < C > G

Conclusions:

I. A > C → False (as A > D < C → thus clear relation between A and C cannot be determined).

II. A > D → True ( as A > Q = D)

Therefore, only conclusion II is true

**Given

Statements:

** A < B ≥ C > D ; C < F ; E ≤ D

**On combining:**A < B ≥ C > D ≥ E ; C < F

Conclusions:

I. A < F → False ( As A < B ≥ C < F  →  thus clear relation between A and F cannot be determined)

II. E < F  →True (As F >  C > D  ≥  E  → E < F)

Therfore only conclusion II is True.

Given series is:

Left Side N # S 3 @ E B 7 $ 9 I £ O R ! 6 M > 3 % D 5 X & Z 8 < F ? 4 G A Right Side

Arrangement required: Symbol → Vowel → Consonant

Left Side N # S 3 @ E B 7 $ 9 I £ O R ! 6 M > 3 % D 5 X & Z 8 < F ? 4 G A Right Side

Clearly there are two such pairs they are**: @ E B** and £ O R

Given series is:

Left Side N # S 3 @ E B 7 $ 9 I £ O R ! 6 M > 3 % D 5 X & Z 8 < F ? 4 G A Right Side

Arrangement required: Symbol → Consonant → Symbol

Left Side N # S 3 @ E B 7 $ 9 I £ O R ! 6 M > 3 % D 5 X & Z 8 < F ? 4 G A Right Side

Clearly there is one such pair i.e., < F ?

Given series is:

Left Side N # S 3 @ E B 7 $ 9 I £ O R ! 6 M > 3 % D 5 X & Z 8 < F ? 4 G A Right Side

Arrangement required: Consonant → Number → Vowel

Clearly none of the number follows the given condition$.

Given series:

Left Side N # S 3 @ E B 7 $ 9 I £ O R ! 6 M > 3 % D 5 X & Z 8 < F ? 4 G A Right Side

If all the symbols are dropped: N S 3 E B 7 9 I O R 6 M 3 D 5 X Z 8 F 4 G A.

Hence, 7^th letter from the left end is 9.$

Given series:

Left Side N # S 3 @ E B 7 $ 9 I £ O R ! 6 M > 3 % D 5 X & Z 8 < F ? 4 G A Right Side

Here the group is formed in which second element is second next to the third element, and the third element is to the just next of the first element.

Hence, 9 I O does not belong to the group$.

Hence, three pairs are there in the word ‘ENLIGHTEN’, each of which has as many letters between them in the word (both forward and backward direction) as they have between them in the English Alphabet.

According to the question,

1. Only one person sits between A4 and A1. 

2. B3 sits immediate left of A4.

3. A2 sits third to the right of B3.

4. B4 is not a neighbor of A1 and A2.

5. A3 is not immediate neighbor of A2.

6. B1 is not a neighbor of B4.

Hence, B2 sits third to the right of A4.

According to the question,

1. Only one person sits between A4 and A1. 

2. B3 sits immediate left of A4.

3. A2 sits third to the right of B3.

4. B4 is not a neighbor of A1 and A2.

5. A3 is not immediate neighbor of A2.

6. B1 is not a neighbor of B4.

Here A3 sits second to the right of B2.

Hence, “None of the above” is the correct answer.

According to the question,

1. Only one person sits between A4 and A1. 

2. B3 sits immediate left of A4.

3. A2 sits third to the right of B3.

4. B4 is not a neighbor of A1 and A2.

5. A3 is not immediate neighbor of A2.

6. B1 is not a neighbor of B4.

Hence, “B2 and A3” are the neighbors of B4.

According to the question,

1. Only one person sits between A4 and A1. 

2. B3 sits immediate left of A4.

3. A2 sits third to the right of B3.

4. B4 is not a neighbor of A1 and A2.

5. A3 is not immediate neighbor of A2.

6. B1 is not a neighbor of B4.

Hence, A3 sits immediate right of B4.

According to the question,

1. Only one person sits between A4 and A1. 

2. B3 sits immediate left of A4.

3. A2 sits third to the right of B3.

4. B4 is not a neighbor of A1 and A2.

5. A3 is not immediate neighbor of A2.

6. B1 is not a neighbor of B4.

Hence, A4 sits immediate left with respect to B1.

The least possible diagram for the given statements is as follows

Conclusions:

I. Some winter is nights → False (It is possible but not definite)

II. No winter is night → False (It is possible but not definite)

Conclusion I and II is complementary pair.

Hence, Either conclusion I or conclusion II follows.

The least possible Venn diagram is shown below

Conclusions:

I. Some Purse are Bag → False (It is possible but not definite. As no direct relation between Bag and Purse is given).

So it is considered as Cannot be determined.

II. No Bag is Purse → False (It is possible but not definite. As no direct relation between Bag and Purse is given).

Part of Bag can go in Purse.

So, it is considered as Cannot be determined.

Thus conclusion I and II form a complementary pair.

Hence, Either conclusion I or II follow.

Additional Information:

Conditions for the Either Or:

1. Both the conclusion must be Cannot be determined.
2. Both the objects in both the conclusion must be same.
3. If one conclusion is positive, then other should be negative.

Complementary pairs:

1. All - Some not (Positions of the object should be same in the conclusion)
2. Some - No

The least possible Venn diagram for the given statements is as follows-

Conclusions:

I. Some Daisy are not Camelia → False (It is possible but not definite)

II. All Pansy being Poppy is a possibility → False (As ‘Only a few Pansy are Poppy’, some parts of Pansy will never be Poppy.)

Additional Information:

‘Only a few Pansy are Poppy’ → here the first element is Pansy and the second element is Poppy. So, the statement means ‘some Pansy are Poppy and some parts of Pansy will never be Poppy but all Poppy can be Pansy.

So, only a few = some + some not.

Hence, neither conclusion I nor II follow.

The least possible Venn diagram for the given statements is as follows

Conclusions:

I. Some Dogs are Zebra ⇒ False ( As it is possible but not definite, as no direct relation between the two is given).

II. All Zebra can be dog ⇒ True ( As there is no relation between Zebra and Dog, so the possibility is true).

Hence, Only conclusion II follows.

Given words are: BTS   WHN   CKU   POQ   RTN   ZGN

1. Replacing second letter of each word by immediate succeeding letter as per alphabetical series: BUS   WIN   CLU   PPQ   RUN   ZHN

2. Clearly, three English meaningful words BUS, WIN, RUN are formed.

Therefore, three English meaningful words are formed.