Correct Answer: Q

Solution : Given:

(i) T is sitting exactly in the middle of the row.

(ii) Q is sitting to the immediate right and immediate left of P and T, respectively. S is not sitting at the extreme end.

From the final seating arrangement, Q is sitting third

Correct Answer: 1

Solution : The correct answer is the first option.

"Last night" is an appropriate phrase to use in this context to specify the time when the action took place i.e., yesternight.

So, the correct sentence should be: I walked back home after the movie last night.

Correct Answer: 3080

Solution : Given: The series is $1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots \ldots$.
The sum of the first $n$ consecutive number = $\frac{n(n+1)}{2}$
The sum of the square of the first $n$ consecutive number = $\frac{n(n + 1) (2n + 1)}{6}$
The sum of

Correct Answer: much happier

Solution : The correct choice is the second option.

The adjective happier is already a comparative form, so adding more before it is redundant and grammatically incorrect. 

Therefore, the correct sentence is: I think she would be much happier in her hometown.

Correct Answer: Kathak

Solution : The correct answer is Kathak.

Rani Karnaa Nayak performed Kathak. In 1927, she was born in Hyderabad, Sindh. Pandit Sohanlal Nayak, her father, began her Kathak instruction when she was five years old. She then studied under other great Kathak dancers, including Shambhu Maharaj

Correct Answer: 500%

Solution : 20% of (A + B) = 30% of (A − B)
⇒ 20% of A + 20% of B = 30% of A – 30% of B
⇒ 10% of A = 50% of B
$\therefore$ A = 500% of B
Hence, the correct answer

Correct Answer: 2

Solution : Given:
$x=\frac{4\sqrt ab}{\sqrt a+\sqrt b}$
Equation $=\frac{x+2\sqrt a}{x-2\sqrt a}+\frac{x+2\sqrt b}{x-2\sqrt b}$
Put the value of $x$ in equation:
$=\frac{\frac{4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}+2\sqrt{a}}{\frac{4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-2\sqrt a}+\frac{\frac{4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}+2\sqrt{b}}{\frac{4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-2\sqrt b}$
$=\frac{\frac{4\sqrt{ab}+2a+2\sqrt {ab}}{\sqrt{a}+\sqrt{b}}}{\frac{4\sqrt{ab}-2a-2\sqrt {ab}}{\sqrt{a}+\sqrt{b}}}+\frac{\frac{4\sqrt{ab}+2\sqrt {ab}+2b}{\sqrt{a}+\sqrt{b}}}{\frac{4\sqrt{ab}-2\sqrt {ab}-2b}{\sqrt{a}+\sqrt{b}}}$
$=\frac{\frac{4\sqrt{ab}+2a+2\sqrt {a}b}{\sqrt{a}+\sqrt{b}}}{\frac{4\sqrt{ab}-2a-2\sqrt {a}b}{\sqrt{a}+\sqrt{b}}}+\frac{\frac{4\sqrt{ab}+2\sqrt {ab}+2b}{\sqrt{a}+\sqrt{b}}}{\frac{4\sqrt{ab}-2\sqrt {a}b-2b}{\sqrt{a}+\sqrt{b}}}$
$=\frac{4\sqrt{ab}+2a+2\sqrt {a}b}{4\sqrt{ab}-2a-2\sqrt {ab}}+\frac{4\sqrt{ab}+2\sqrt {ab}+2b}{4\sqrt{ab}-2\sqrt {ab}-2b}$
$=\frac{2}{2}\left [\frac{2\sqrt{ab}+a+\sqrt {a}b}{2\sqrt{ab}-a-\sqrt {a}b} \right ]+\frac{2}{2}\left[\frac{2\sqrt{ab}+\sqrt {ab}+b}{2\sqrt{ab}-\sqrt {a}b-b}\right]$
$=\frac{3\sqrt{ab}+a}{\sqrt{ab}-a}+\frac{3\sqrt{ab}+b}{\sqrt{ab}-b}$

Correct Answer: $0$

Solution : Given: $x+\frac{1}{x}=\sqrt{3}$
Cubing both sides we get
$(x+\frac{1}{x})^3=(\sqrt{3})^3$
⇒ $x^3+\frac{1}{x^3}+3×x×\frac{1}{x}(x+\frac{1}{x})=(3\sqrt{3})$
⇒ $x^3+\frac{1}{x^3}=3\sqrt{3}–3(x+\frac{1}{x})$
$\because x+\frac{1}{x}=\sqrt{3}$
Thus, $x^3+\frac{1}{x^3}=3\sqrt{3}–3\sqrt{3} = 0$
Hence, the correct answer is $0$.

Correct Answer: Javelin Throw 

Solution : The correct option is - Javelin Throw .

Neeraj Chopra is an Indian javelin thrower, born on 24 December 1997. He now holds the title of Olympic champion in the javelin throw and has won silver at the World Championships and the Diamond League.

Correct Answer: are visiting

Solution : The most appropriate choice is the fourth option.

The original sentence is incorrect as per the subject-verb agreement. Since "My friend Meera and her mother" is a compound subject involving more than one person, the verb should also be in the plural form to