Correct Answer: 5364

Solution : Given: $a=299, b=298, c=297$
We know that the algebraic identity is $(a^3+b^3+c^3-3abc)=\frac{1}{2} \times(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$
$2a^3+2b^3+2c^3-6abc$
$=2(a^3+b^3+c^3-3abc)$
$=2\times\frac{1}{2} \times(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$
$=2\times\frac{1}{2} \times(299+298+297)[(299-298)^2+(298-297)^2+(297-299)^2]$
$=894[1^2+1^2+2^2]=894\times6$
$= 5364$
Hence, the correct answer is 5364.

Correct Answer: Mohiniyattam

Solution : The correct answer is Mohiniyattam.

Sunanda Nair is a renowned Indian classical dancer known for performing the dance form Mohiniyattam. It is a traditional dance style from the southern Indian state of Kerala. It is characterised by its graceful and swift movements, expressive storytelling,

Correct Answer: DACB

Solution : The correct choice is the fourth option: DACB.

The logical flow of sentences is as follows:

• Sentence D (Henry D. Penrose was a dog with a pedigree): This is the introductory sentence introducing the main character, Henry D. Penrose.
• Sentence A (Each morning he was

Correct Answer: 11

Solution : Let the original fraction be $\frac{x}{y}$ 
According to the question,
$y = 2x + 4$ .............(1)
When 3 is added to the numerator and 3 is subtracted from the denominator the fraction becomes $\frac{2}{3}$
⇒ new fraction = $\frac{x + 3}{y – 3} = \frac{2}{3}$

Correct Answer: A. G. Tansley

Solution : The correct option is A. G. Tansley.

The term ecosystem was proposed by Sir Arthur Tansley, a British botanist and ecologist, in 1935. In addition to coining the term, Tansley made several significant contributions to ecology, including developing the concept of the

Correct Answer: 1

Solution : Given: $\sin A+\sin^{2} A=1$ ----(1)
⇒ $\sin A=1–\sin^{2} A$
⇒ $\sin A=\cos^{2} A$ -----(2)
⇒ $\sin^{2} A=\cos^{4} A$ -----(3)
From equation (2) and (3), we get:
So, $\cos^{2} A+\cos^{4} A=\sin A+\sin^{2} A$
⇒ $\cos^{2} A+\cos^{4} A=1$
Hence, the correct answer is 1.

Correct Answer: 970

Solution : Given: $x^{2}+\frac{1}{x^{2}} = 98$
We know, $(x+y)^{2}= x^{2}+y^{2}+2xy$
⇒ $(x+\frac{1}{x})^{2}= x^{2}+\frac{1}{x^{2}}+2(x)(\frac{1}{x})$
⇒ $(x+\frac{1}{x})^{2}= 98+2$
⇒ $(x+\frac{1}{x})^{2}= 100$
⇒ $ (x+\frac{1}{x})= 10$
Now,
$x^{3}+\frac{1}{x^{3}} = (x+\frac{1}{x})^{3}-3(x)(\frac{1}{x})(x+\frac{1}{x})$
= $10^{3}-3(10)$
= $1000-30$
= $970$
Hence, the correct answer is 970.

Correct Answer: 198

Solution : Given:
$\mathrm{p}=\frac{\sqrt{2}+1}{\sqrt{2}-1}$, $\mathrm{q}=\frac{\sqrt{2}-1}{\sqrt{2}+1}$
Rationalising $p$, we get:
$p= 3+2\sqrt{2}$
⇒ $p^2=17+12\sqrt{2}$
$\mathrm{q}=\frac{\sqrt{2}-1}{\sqrt{2}+1}$
Rationalising $q$, we get:
$q= 3-2\sqrt{2}$
⇒ $q^2=17-12\sqrt{2}$
$\therefore$ $\frac{\mathrm{p}^2}{\mathrm{q}}+\frac{\mathrm{q}^2}{\mathrm{p}}=(\frac{17+12\sqrt{2}}{3-2\sqrt{2}})+(\frac{17-12\sqrt{2}}{3+2\sqrt{2}})$
⇒ $\frac{\mathrm{p}^2}{\mathrm{q}}+\frac{\mathrm{q}^2}{\mathrm{p}}=\frac{(17+12\sqrt{2})(3+2\sqrt{2})+(17-12\sqrt{2})(3-2\sqrt{2})}{(3+2\sqrt{2})(3-2\sqrt{2})}$
⇒ $\frac{\mathrm{p}^2}{\mathrm{q}}+\frac{\mathrm{q}^2}{\mathrm{p}}=\frac{2\times 17\times 3+2\times 12\times2\times 2}{(3+2\sqrt{2})(3-2\sqrt{2})}$
⇒ $\frac{\mathrm{p}^2}{\mathrm{q}}+\frac{\mathrm{q}^2}{\mathrm{p}}=198$
Hence, the correct answer is 198.

Correct Answer: go bankrupt

Solution : The correct option is the second option.

Go bankrupt is a commonly used collocation in English that describes the financial state of a company or organisation when it is unable to pay its debts and is declared legally insolvent.
In the sentence context, the

Correct Answer: Criterion

Solution : The fourth option is the correct answer.

In the context of the passage, the sentence discusses how remote working has become a significant factor in the hiring process. The term that best fits here is criterion, as it refers to a principle or standard by