Reasoning PYQs

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

Conclusions:

I. All Headset are Earphones → False (As only a few Earphones are Headset, some Headset are already Earphones but another portion of Headset is Earphones or not is possible but not definite, So it is false).

II. No Backpack is Headset → False (As some Headset are Backpack, some portion of Headset is already Backpack so No Backpack is Headset will be false).

Hence, Neither I nor II follows​.

Given Word: SANITY

We can use the positional values of letters/alphabets in the English alphabets series so as to map the letters in an easier manner.

According to the English alphabetical series, the positional values of letters are given as follows:

The letters are mapped as shown:

Hence, no such pairs of letters in the word "SANITY" which have as many letters between them as English alphabetical series (Both in the backward and forward direction)

From the given information,

S is married to A who is mother of G.

R is sister- in- law of A. K is grandson of P and only son of R. P has one granddaughter. 

As gender of P is not defined we cannot find its relation with anyone. 

Hence, P is grandfather of G is incorrect.

From the given information,

S is married to A who is mother of G.

R is sister- in- law of A. K is grandson of P and only son of R. P has one granddaughter. 

As gender of P cannot be define hence number of male member in the organisation cannot be determined.

From the given information,

S is married to A who is mother of G.

R is sister- in- law of A. K is grandson of P and only son of R. P has one granddaughter. 

S is son of P.

Boxes: A, B, C, D, E, F, and G.

1. There are  3 boxes between D and E.

2. Box E is kept above box D.

3. Box C is kept just above the box E.

 

4. There are 2 boxes between box G and box F.

 

5. Box A is kept in an odd number place.

6. Box B is kept one box above the box G.

Now case 2 did not follow the above-mentioned information, so terminated, as B can't be placed above G as no space left for B to be placed.

Box B will be place at 3^rd position as It is given Box B is placed one box above the box G, which means Box B will be placed immediately above box G, which is possible only when Box B is placed at 3^rd position.

Now only position is left that is 5^th position so Box A will be placed on 5^th position and also 5 is a odd number so this condition is also satisfied here.

 

Hence, box C is kept in the first position

Boxes: A, B, C, D, E, F, and G.

1. There are  3 boxes between D and E.

2. Box E is kept above box D.

3. Box C is kept just above the box E.

 

4. There are 2 boxes between box G and box F.

 

5. Box A is kept in an odd number place.

6. Box B is kept one box above the box G.

Now case 2 did not follow the above-mentioned information, so terminated, as B can't be placed above G as no space left for B to be placed.

Box B will be place at 3^rd position as It is given Box B will be placed immediately above box G, which is possible only when Box B is placed at 3^rd position.

Now only position is left that is 5^th position so Box A will be placed on 5^th position and also 5 is a odd number so this condition is also satisfied here.

 

Hence, 4 boxes kept above box A.

Boxes: A, B, C, D, E, F, and G.

1. There are  3 boxes between D and E.

2. Box E is kept above box D.

3. Box C is kept just above the box E.

 

4. There are 2 boxes between box G and box F.

 

 

5. Box A is kept in an odd number place.

6. Box B is kept one box above the box G.

Now case 2 did not follow the above-mentioned information, so terminated, as B can't be placed above G as no space left for B to be placed.

Box B will be place at 3^rd position as It is given Box B is placed one box above the box G, which means Box B will be placed immediately above box G, which is possible only when Box B is placed at 3^rd position.

Now only position is left that is 5^th position so Box A will be placed on 5^th position and also 5 is a odd number so this condition is also satisfied here.

Hence, 2 boxes are present between box C and G.

Boxes: A, B, C, D, E, F, and G.

1. There are  3 boxes between D and E.

2. Box E is kept above box D.

3. Box C is kept just above the box E.

 

4. There are 2 boxes between box G and box F.

 

 

5. Box A is kept in an odd number place.

6. Box B is kept one box above the box G.

Now case 2 did not follow the above-mentioned information, so terminated, as B can't be placed above G as no space left for B to be placed.

Box B will be place at 3^rd position as It is given Box B is placed one box above the box G, which means Box B will be placed immediately above box G, which is possible only when Box B is placed at 3^rd position.

Now only position is left that is 5^th position so Box A will be placed on 5^th position and also 5 is a odd number so this condition is also satisfied here.

 

Hence, the Zero box kept below box F.

Boxes: A, B, C, D, E, F, and G.

1. There are  3 boxes between D and E.

2. Box E is kept above box D.

3. Box C is kept just above the box E.

 

4. There are 2 boxes between box G and box F.

 

5. Box A is kept in an odd number place.

6. Box B is kept one box above the box G.

Now case 2 did not follow the above-mentioned information, so terminated, as B can't be placed above G as no space left for B to be placed.

Box B will be place at 3^rd position as It is given Box B is placed one box above the box G, which means Box B will be placed immediately above box G, which is possible only when Box B is placed at 3^rd position.

Now only one position is left that is 5^th position so Box A will be placed on 5^th position and also 5 is a odd number so this condition is also satisfied here.

 

Four of the following are placed in odd places so they form a group but box G is kept in an even place.

Hence, box G is an odd one.

Given Series

Q O & P 6 D 2 M E 9 R ! @ N # 5 1 U 8 A * V 7 T I % $ Y W 3

As per the given statement, the vowels and numbers are dropped then the series will be, 

Q & P D M R ! @ N # * V T % $ Y W

Therefore 13^th element from the right end is, 

Q & P D M R ! @ N # * V T % $ Y W

Hence, M is the correct answer.$

Given series

Q O & P 6 D 2 M E 9 R ! @ N # 5 1 U 8 A * V 7 T I % $ Y W 3

The number of symbols is between the fifth element from the right end and the sixth element from the left end is

Q O & P 6 D 2 M E 9 R ! @ N # 5 1 U 8 A * V 7 T I % $ Y W 3

The symbols are !, @,  #, and *

Hence, Four is the correct answer.

Given Series

Q O & P 6 D 2 M E 9 R ! @ N # 5 1 U 8 A * V 7 T I % $ Y W 3

 The series: 

Q&P, DME, R@N ___

Q O & P 6 D 2 M E 9 R ! @ N # 5 1 U 8 A * V 7 T I % $ Y W 3

In the same way,

5U8 is the next pair.

Given Series

Q O & P 6 D 2 M E 9 R ! @ N # 5 1 U 8 A * V 7 T I % $ Y W 3

As per the given information, numbers which are immediately followed by consonants is

Q O & P 6 D 2 M E 9 R ! @ N # 5 1 U 8 A * V 7 T I % $ Y W 3

6 + 2 + 9 + 7 = 24

Therefore, 24 is the sum of the numbers which are immediately followed by consonants.

Given Series

Q O & P 6 D 2 M E 9 R ! @ N # 5 1 U 8 A * V 7 T I % $ Y W 3

As per the given statement, 

the numbers are immediately preceded by an alphabet and followed by a number is none. 

Hence, none is the correct answer$.

All given options, except 'RSTY', have the logic of +1, +2, +2.

In RSTY logic is +1, +1, +5.

Hence, RSTY is the odd one out.

**Given

Statement:

** F ≥ T ≥ H = G ≥ W = Z ≥ L

Conclusions: 

I. L > H → False (As H = G ≥ W = Z ≥ L → H ≥ L. So, it is false).

II. F ≥ L → True (As F ≥ T ≥ H = G ≥ W = Z ≥ L → F ≥ L. So, it is true).

Hence, Only II follow.

**Given

Statements:

** O = P ≥ N > M; P ≥ X > A = Q

On combining: M < N ≤ P = O ≥ X > A = Q

Conclusions: 

I. O = Q → False (As O ≥ X > A = Q → O > Q. So, it is false).

II. P > A → True (As P = O ≥ X > A → P > A. So, it is true).

Hence, Only II follow.

**Given

Statement:

** E ≥ W = A < B ≤ S ≤ T < X

Conclusions: 

I. X > W → True (As W = A < B ≤ S ≤ T < X → X > W. So, it is true).

II. B > X → False (As B ≤ S ≤ T < X → B < X. So, it is false).

Hence, Only I follow.

1. A is sitting second from one of the ends.

(Here, we will have two cases, Case 1: when A is second from the left end and Case 2: when A is second from the right end).

2. There are two persons sitting between A and B.

3. Four persons are sitting between B and D.

4. D is sitting just left of C.

5. F sits three places away to the right of B.

(Here, we can eliminate case 2 as A is already three places away to the right of B).

6. E sits at the extreme right end of the row.

7. G sits to the immediate right of B.

8. Two persons are sitting between C and E.

Hence, the final arrangement will be as follows:

Hence, 14 persons are sitting in the row.