Given series:

1 @ € C D 6 & E 2 F G H 9 $ I @ € J K L 4 # 1 8 9 © H O V J 5 Y 6 ¥ Y K A %

As, Left - Left = Left

24^th from the Left - 7^th from the Left = 17^th from the Left

Clearly, 17^th from the Left is €$

Given series:

Left Side 1 @ € C D 6 & E 2 F G H 9 $ I @ € J K L 4 # 1 8 9 © H O V J 5 Y 6 ¥ Y K A % Right Side

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

Hence, 4 # 1 does not belong to the group$.

Given series:

Left Side 1 @ € C D 6 & E 2 F G H 9 $ I @ € J K L 4 # 1 8 9 © H O V J 5 Y 6 ¥ Y K A % Right Side

Vowels which are immediately followed by a consonant but not immediately preceded by a number:

required order is: Not number → Vowel → Consonant

1 @ € C D 6 & E 2 F G H 9 $ I @ € J K L 4 # 1 8 9 © H O V J 5 Y 6 ¥ Y K A %

Hence, there is only one vowel which is immediately followed by a consonant but not immediately preceded by a number: H O V

Hence, the correct answer is one.

Given series:

Left Side 1 @ € C D 6 & E 2 F G H 9 $ I @ € J K L 4 # 1 8 9 © H O V J 5 Y 6 ¥ Y K A % Right Side

1. If all the odd numbers are dropped:

@ € C D 6 & E 2 F G H $ I @ € J K L 4 # 8 © H O V J Y 6 ¥ Y K A %

2. 12^th element from the right end is ©.

Then, element that is twelfth from the right end is ‘©’.

Given series:

Left Side 1 @ € C D 6 & E 2 F G H 9 $ I @ € J K L 4 # 1 8 9 © H O V J 5 Y 6 ¥ Y K A % Right Side

Symbols which are immediately preceded by an odd number and followed by an alphabet:

Required order is: Odd number → Symbol → Alphabet

1 @ € C D 6 & E 2 F G H 9 $ I @ € J K L 4 # 1 8 9 © H O V J 5 Y 6 ¥ Y K A %

Hence, there are 2 symbols which is immediately preceded by an odd number and followed by an alphabet: 9 $ I and 9 © H$.

1. D sits second to the right of the person who faces the immediate neighbor of S. 2. D doesn't sit at the end.

Here, we have 3 Cases, i.e., Case 1, Case 2, and Case 3. 3. Only one person sits between S and the one who faces B. 4. There are two persons sitting between B and the one who faces P. 

Now, we can eliminate Case 3. 5. More than one person sits between R and Q, and Q does not sit at any end. 

Now, we can eliminate Case 2, but we have a subcase of Case 1, i.e., Case 1a. 6. T faces the person who sits third from C.

Now, we can eliminate Case 1. 7. A does not sit at any end.

This is our final arrangement,

Hence, the correct answer is S.

1. D sits second to the right of the person who faces the immediate neighbor of S. 2. D doesn't sit at the end.

Here, we have 3 Cases, i.e., Case 1, Case 2, and Case 3. 3. Only one person sits between S and the one who faces B. 4. There are two persons sitting between B and the one who faces P. 

Now, we can eliminate Case 3. 5. More than one person sits between R and Q, and Q does not sit at any end. 

Now, we can eliminate Case 2, but we have a subcase of Case 1, i.e., Case 1a. 6. T faces the person who sits third from C.

Now, we can eliminate Case 1. 7. A does not sit at any end.

This is our final arrangement,

Hence, the correct answer is The person who faces B, i.e., T.

1. D sits second to the right of the person who faces the immediate neighbor of S. 2. D doesn't sit at the end.

Here, we have 3 Cases, i.e., Case 1, Case 2, and Case 3. 3. Only one person sits between S and the one who faces B. 4. There are two persons sitting between B and the one who faces P. 

Now, we can eliminate Case 3. 5. More than one person sits between R and Q, and Q does not sit at any end. 

Now, we can eliminate Case 2, but we have a subcase of Case 1, i.e., Case 1a. 6. T faces the person who sits third from C.

Now, we can eliminate Case 1. 7. A does not sit at any end.

This is our final arrangement,

Except for S, all others sit at the ends.

Hence, the correct answer is S.

1. D sits second to the right of the person who faces the immediate neighbor of S. 2. D doesn't sit at the end.

Here, we have 3 Cases, i.e., Case 1, Case 2, and Case 3. 3. Only one person sits between S and the one who faces B. 4. There are two persons sitting between B and the one who faces P. 

Now, we can eliminate Case 3. 5. More than one person sits between R and Q, and Q does not sit at any end. 

Now, we can eliminate Case 2, but we have a subcase of Case 1, i.e., Case 1a. 6. T faces the person who sits third from C.

Now, we can eliminate Case 1. 7. A does not sit at any end.

This is our final arrangement,

A is related to R, and in the same way, C is related to Q, then D is related to P because a one-person gap is there.

Hence, the correct answer is P.

1. D sits second to the right of the person who faces the immediate neighbor of S. 2. D doesn't sit at the end.

Here, we have 3 Cases, i.e., Case 1, Case 2, and Case 3. 3. Only one person sits between S and the one who faces B. 4. There are two persons sitting between B and the one who faces P. 

Now, we can eliminate Case 3. 5. More than one person sits between R and Q, and Q does not sit at any end. 

Now, we can eliminate Case 2, but we have a subcase of Case 1, i.e., Case 1a. 6. T faces the person who sits third from C.

Now, we can eliminate Case 1. 7. A does not sit at any end.

This is our final arrangement,

Hence, the correct answer is Two.

**Given

Statements:

** P > A ≥ R; A = C ≤ H < U = T > E

On combining: P > A = C ≥ R; R ≤ A = C ≤ H < U = T > E

Conclusions:

I. R = H → False (as R ≤ A = C ≤ H; R ≤ H)

II. R < H → False (as R ≤ A = C ≤ H; R ≤ H)

Note:

conclusion I and II form a complimentary pair.

Hence, Either conclusion I or II is true.

Given

Statements:

O > U ≥ T ≤ G; R ≥ O > W

On combining: R ≥ O > U ≥ T ≤ G; R ≥ O > W

Conclusions:

I. R > G → False (as R ≥ O > U ≥ T ≤ G; as clear relation between R and G cannot be established)

II. W > U → False (as W < O > U; as clear relation between W and U cannot be established)

Hence, none of the conclusion is true.

**Given

Statements:

** P < E = N ≤ T; N ≤ S < U = L < A

On combining: P < E = N ≤ T; P < E = N ≤ S < U = L < A

Conclusions:

I. P < U → true (as P < E = N < U → P < U)

II. E ≤ S → true (as E = N ≤ S → E ≤ S)

Hence, Both the conclusions I and II are true.

The logic followed here is:

Option 1) QSU

Option 2) UWY

Option 3) CEG

Option 4) HJI

Here, 'HJI' is odd among the given options.

Hence, the correct answer is "Option 4".

Here, persons are N, O, P, Q, R, S and T.

1. Three persons went temple between T and N.

2. T went temple two days before O.

3. N went temple neither Friday nor Saturday.

 

4. As many persons went temple before P as many after Q.

5. Q went temple after R.

6. S went temple before R but not after P. Therefore, case 1 is eliminated.

Hence, the final arrangement is as follows

 

Since, R went temple on Thursday. Therefore, statement A is correct.

O went temple just after R. Therefore, statement B is correct.

As many persons went temple before R as many after him. Therefore, statement C is correct.

Hence, the incorrect statement is None of these.

Here, persons are N, O, P, Q, R, S and T.

1. Three persons went temple between T and N.

2. T went temple two days before O.

3. N went temple neither Friday nor Saturday.

 

4. As many persons went temple before P as many after Q.

5. Q went temple after R.

6. S went temple before R but not after P. Therefore, case 1 is eliminated. Hence, the final arrangement is as follows

 

Since, P went temple on Tuesday not on Wednesday. Therefore, statement first is incorrect.

Two persons went temple after O. Therefore, statement second is incorrect.

R went temple on Thursday not on Tuesday. Therefore, statement third is incorrect.

As many persons went temple before T as many after O not R. Therefore, statement fourth is incorrect.

Hence, the correct statement is None of these.

Here, persons are N, O, P, Q, R, S and T.

1. Three persons went temple between T and N.

2. T went temple two days before O.

3. N went temple neither Friday nor Saturday.

 

4. As many persons went temple before P as many after Q.

5. Q went temple after R.

6. S went temple before R but not after P. Therefore, case 1 is eliminated. Hence, the final arrangement is as follows

 

Hence, two persons went temple between P and O.

Here, persons are N, O, P, Q, R, S and T.

1. Three persons went temple between T and N.

2. T went temple two days before O.

3. N went temple neither Friday nor Saturday.

 

4. As many persons went temple before P as many after Q.

5. Q went temple after R.

6. S went temple before R but not after P. Therefore, case 1 is eliminated. Hence, the final arrangement is as follows

 

Hence, O went temple on Friday.

Here, persons are N, O, P, Q, R, S and T.

1. Three persons went temple between T and N.

2. T went temple two days before O.

3. N went temple neither Friday nor Saturday.

 

4. As many persons went temple before P as many after Q.

5. Q went temple after R.

6. S went temple before R but not after P. Therefore, case 1 is eliminated. Hence, the final arrangement is as follows

 

Since, S is the first person who went temple so no one went temple before him and N is the last person so no one went after him.

Hence, the correct answer is N.

1. D has fewer items than A and C, but more items than E and F.

A/C > C/A > D > E/F > F/E

2. Exactly two persons have more items than B.

_ > _ > B > _ > _ > _

3. Exactly one person has fewer items than E.

A/C > C/A > B > D > E > F

Either A or C has the highest number of items.

Hence, the correct answer is Can't be determined.