Given:

A team of five persons is to be selected from amongst:

Five girls: A, B, C, D, and E.

Four boys: P, Q, R, and S.

Q is already a selected member.

1. S and Q cannot be together

⇒ As Q is already selected, S cannot be there in the team.

Therefore, 

Option 1: P, S, C, A → Cannot form a team.

Option 2: R, S, C, A → Cannot form a team.

Option 3: S, R, D, C → Cannot form a team.

2. E and P have to be together.

3. B and S cannot be together.

⇒ As E and P have to be together and B and S cannot be together.

Therefore,

Option 4: E, P, B, R → Can form a possible team.

Therefore, after Q, the other 4 selected team members are E, P, B, R

Hence, the correct answer is "E, P, B, R".

Here, we can see Sumita is common in  brilliant, obedient, regular and studious 

Hence, Sumita is correct answer.

Logic: We have to replace the symbols as per the given operators then apply the "BODMAS" rule to solve the equation.

Now, the equation will becomes:

⇒ 36 - 12 × 2 + 18 ÷ 9.

⇒ 36 - 12 × 2 + 2.

⇒ 36 - 24 + 2.

⇒ 38 - 24.

⇒ 14.

Hence, the correct answer is "14".

People: A, B, C, D, E and F

Married Couples = 2

Professions: Not working (2), Lecturer (1), Architect (1), Accountant (1), Lawyer (1) No lady in the group is either an architect or an accountant. ⇒ The women in the group are either not working or Lecturer or Lawyer.

1. The lawyer is married to D, D is a housewife ⇒ The lawyer is a male and D is not working.

C, the accountant, is married to F who is a lecturer. Therefore, C is a male as he is an Accountant and F is a female

2. A is married to D.

B is not working and either E or B is a female ⇒ E is an architect, hence male. B is female.

People: A, B, C, D, E and F

Married Couples = 2

Professions: Not working (2), Lecturer (1), Architect (1), Accountant (1), Lawyer (1) No lady in the group is either an architect or an accountant. ⇒ The women in the group are either not working or Lecturer or Lawyer.

1. The lawyer is married to D, D is a housewife ⇒ the lawyer is a male and D is not working.

C, the accountant, is married to F who is a lecturer. Therefore, C is a male as he is an Accountant and F is a female.

2. A is married to D.

B is not working and either E or B is a female ⇒ E is an architect, hence male. B is female.

Clearly, E is the architect.

(1) 2 Wolves in front of a Wolf.

(2) 2 behind a Wolf.

(3) 1 Wolf in between 2 Wolves.

So, there can be minimum 3 Wolves.

Hence, 3 is the correct answer.

1) D and G have to be together.

2) E and D have to be together.

Implies, E, D, and G are in the same team.

3) C cannot be with E.

Implies, C is not in the same team as E, D, and G.

4) F and D cannot be together.

Implies, F is not in the same team as E, D, and G

Team 1 - E, D, and G

Team 2 - C and F

5) A and B cannot be in the same team.

Implies, either A or B is in team 1.

Possibilities for team 1: AEDG or BEDG.

6) A has to be with F.

Therefore, AEDG can be eliminated.

Hence, EDGB can be formed using the given data.

1) 18 + 843 + 288 + 980 + 82 = 1818

LHS = 18 + 843 + 288 + 980 + 82 = 2211 ≠ RHS

2) 14 + 443 + 244 + 940 + 42 = 1814

LHS =  14 + 443 + 244 + 940 + 42 = 1683 ≠ RHS 

3) 15 + 543 + 255 + 950 + 52 = 1815

LHS = 15 + 543 + 255 + 950 + 52 = 1815 = RHS

4) 16 + 643 + 266 + 960 + 62 = 1816

LHS = 16 + 643 + 266 + 960 + 62 = 1947 ≠ RHS 

Hence, option 3 is the correct answer 

Let the number of boys be ‘x’ and number of girls be ‘y’.

So, total number of students in the class = x + y

If each boy gets 5 apples, then each girl will get 7 apples.

According to question:

5x + 7y = 85 ______ (i)

And, if each boy gets 4 apples, then each girl will get 9 apples.

4x + 9y = 85 ______ (ii)

Equating equation (i) and (ii), we get:

5x + 7y = 4x + 9y

⇒ 5x – 4x = 9y – 7y

⇒ x = 2y

Putting the value of ‘x’ in equation (i), we get:

5(2y) + 7y = 85

⇒ 10y + 7y = 85

⇒ 17y = 85

⇒ y = 85 ÷ 17

⇒ y = 5

As x = 2y and y = 5

The value of x = 2 × 5 = 10

Therefore, total number of students in the class = x + y = 10 + 5 = 15

Hence, ‘15’ is the correct answer.

Six persons - Seema, Vaibhav, Ajay, Manisha, Tulika and Ananya.

States - Assam, Gujarat, Madhya Pradesh, Punjab, Bihar and Rajasthan.

Games - Chess, Football, Hockey, Ludo, Badminton and Cricket.

(1) Ananya was born in Gujarat and she plays Cricket.

(2) The person who was born in Bihar plays Football.

(3) Seema plays Hockey and she was not born in Assam or Madhya Pradesh. Therefore she was born in Punjab, 

(4) Manisha was born in Rajasthan and she plays badminton.