Correct Answer: 3

Solution : Given: $x+y+z=0$
Cubing both sides, we get,
$⇒(x+y+z)^3=0$
$⇒x^3 + y^3 + z^3 + 3(x+y)(y+z)(z+x)=0$
$⇒x^3+y^3+z^3 = -3(x+y)(y+z)(z+x)$ ......(1)
From the given equation ($x+y+z=0$), we can get,
$x+y=-z,$ $y+z=-x$ and $z+x=-y$
Putting in (1), we get,
$x^3+y^3+z^3 = -3(-z)(-x)(-y)=3xyz$
Consider, $\frac{x^2}{(y z)}+\frac{y^2}{(x z)}+\frac{z^2}{(x y)}$
$=\frac{x^2(xz)(xy)+y^2(yz)(xy)+z^2(yz)(xz)}{(yz)(xz)(xy)}$

Correct Answer: MOQX

Solution : Let's check the options –
First option: EILN; E + 4 = I; I + 3 = L; L + 2 = N
Second option: AEHJ; A + 4 = E; E + 3 = H; H + 2 = J
Third option: MOQX; M

Correct Answer: $1-\sin \theta$

Solution : $\frac{(1+\sec \theta \operatorname{cosec} \theta)^2(\sec \theta-\tan \theta)^2(1+\sin \theta)}{(\sin \theta+\sec \theta)^2+(\cos \theta+\operatorname{cosec} \theta)^2}$
Since $ 0^{\circ}<\theta<90^{\circ}$.
$=\frac{(1+\frac{1}{\cos\theta.\sin\theta})^2(\frac{1}{cos\theta}-\frac{\sin\theta}{\cos\theta})^2 (1+\sin\theta)}{(\sin\theta+\frac{1}{\cos\theta})^2+(\cos\theta+\frac{1}{\sin\theta})^2}$
$=\frac{(\frac{\cos\theta.\sin\theta+1}{\cos\theta.\sin\theta})^2(\frac{1-\sin\theta}{\cos\theta})^2 (1+\sin\theta)}{(\frac{\sin\theta.\cos\theta+1}{\cos\theta})^2+(\frac{\cos\theta\sin\theta+1}{\sin\theta})^2}$
$=\frac{(\frac{1}{\cos\theta.\sin\theta})^2(\frac{1-\sin\theta}{\cos\theta})^2 (1+\sin\theta)}{\frac{\sin^2\theta+\cos^2\theta}{\sin^2\theta.\cos^2\theta}}$
$=\frac{(1-\sin\theta)(1-\sin\theta)(1+\sin\theta)}{\cos^2\theta}$
$=\frac{(1-\sin\theta)(1-\sin^2\theta)}{\cos^2\theta}$
$=\frac{(1-\sin\theta)(\cos^2\theta)}{\cos^2\theta}$
$=1-\sin\theta$
Hence, the correct answer is $1-\sin \theta$.

Correct Answer: 4, 3, 2, 1

Solution : Given:
1. Nest 2. Neck 3. Neat 4. Near

Step 1: Compare the first and second letters of each word. Since all the words have the same letter N and e, so move on to the next letter.
Step 2: The

Correct Answer: 366 m

Solution :
Given:
AB = Height of mountain = 500 m
$\angle$ACB = 30°; $\angle$ADB = 45°
C and D ⇒ Positions of boats
Let CD = $x$ m
Solution:
From $\triangle$ABD,
$\tan 45° = \frac{AB}{BD}$ 
⇒ AB = BD
⇒ AB = BD = 500

Correct Answer: P ÷ Q + R

Solution : Given:
A ÷ B means that A is the mother of B, A – B means that A is the father of B, and A + B means that A is the sister of B.

Let's check the options –
First

Correct Answer: MXO

Solution : Given:
CNY : GRC :: FRL : JVP :: ITK : ?

Add 4 to the place value of each letter of CNY to get the required code –
C + 4 = G; N + 4 = R; Y + 4 = C
Thus,

Correct Answer: (1)

Solution : The error lies in the first part of the sentence.

"On last Sunday" should be replaced with ''Last Sunday'', as it is more widely used to refer to the specific day and the use of the preposition "on" is not required. "Last Sunday" is a

Correct Answer: –1 or 2

Solution : Given: $2\left [x^{2} +\frac{1}{x^{2}} \right]-2\left [ x-\frac{1}{x} \right]-8=0$
$⇒2\left [(x -\frac{1}{x})^{2}+2 \right]-2\left [ x-\frac{1}{x} \right ]-8=0$
Let $(x -\frac{1}{x})=t$
The equation becomes $2(t^{2}+2)-2t-8=0$
$⇒2t^{2}-2t-4=0$
$⇒t^{2}-2t+t-1=0$
$⇒(t-2)(t+1)=0$
$⇒t=-1,2$
So, $(x -\frac{1}{x})=-1,2$
Hence, the correct answer is –1 or 2.

Correct Answer: Abhinav Bindra

Solution : The correct option is - Abhinav Bindra.

In his final Olympic appearance, Abhinav Bindra, the nation's first individual gold medalist, carried the Indian flag during the 2016 Summer Olympics in Rio. He has received the Padma Bhushan award from the Indian government.