It's a Question of your Career!

Correct Answer: XUJBG

Solution : Given:
GOPAL is coded as MIVUR –

Add 6 and subtract 6 alternatively from the place value of each letter of the word GOPAL to obtain the required code –
G + 6 = M; O – 6 = I; P + 6 = V;

Correct Answer: Aquatic

Solution : Given:
1. Aquarium 2. Aquatic 3. Aquarial 4. Aquatint 5. Aquavit

Step 1: The first four letters of each word are the same – a, q, u, a; so move on to the next letter.
Step 2: The fifth letter of

Correct Answer: TVWY

Solution : Given:

KMNP : ACDF :: PRSU : ?

Here, the logic is that the second alphabet of each letter adds 2 letters ahead of the first letter, the third letter adds 1 ahead of the second letter and the fourth letter adds 2 ahead of

Correct Answer: Perishable

Solution : The correct option is Perishable.

Money does not have the quality of perishability. The widespread consensus is that money is durable, meaning it doesn't rot, corrode, or expire and doesn't lose value over time. Certain products or commodities that have a short shelf life and

Correct Answer:

Sect contemporary to the Buddha   

Solution : Correct Answer is Sect contemporary to the Buddha  

The Ajivikas is a prehistoric Indian ascetic religion that believed in freedom, fatalism, and karma. This sect was particularly influential during the reign of the Mauryan king Bindusara. King Bindusara and his wife

Correct Answer: 8 days

Solution : Total work = LCM (11, 20 and 55) = 220 units
Efficiency of A = $\frac{220}{11}$ = 20 units/day
Efficiency of B = $\frac{220}{20}$ = 11 units/day
Efficiency of C = $\frac{220}{55}$ = 4 units/day
Combined work of A and B = 20 +

Correct Answer: $\pm \frac{1}{2}(\sqrt{3}+1)$

Solution : Given: $\frac{2+\sqrt{3}}{2}$
$= \frac{1}{4}({4+2\sqrt{3}})$
$= \frac{1}{4}[1^2+(\sqrt{3})^2+2×1×\sqrt{3}]$
$= \frac{1}{4}(1+\sqrt{3})^2$
So, the square root of $\frac{2+\sqrt{3}}{2}$
$= \sqrt{\frac{2+\sqrt{3}}{2}}$
$= \sqrt{\frac{1}{4}(1+\sqrt{3})^2}$
$= \pm \frac{1}{2}(1+\sqrt{3})$
Hence, the correct answer is $\pm \frac{1}{2}(\sqrt{3}+1)$ .

Correct Answer: $8(\sqrt{3}-2)$

Solution : Given, $\sin \theta+\cos \theta=\frac{\sqrt{3}-1}{2 \sqrt{2}}$
Squaring both sides, we get,
⇒ $(\sin \theta+\cos \theta)^2=(\frac{\sqrt{3}-1}{2 \sqrt{2}})^2$
⇒ $\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=\frac{3+1-2\sqrt3}{8}$
⇒ $1+2\sin\theta\cos\theta=\frac{4-2\sqrt{3}}{8}$
⇒ $2\sin\theta\cos\theta=\frac{2-\sqrt3}{4}-1$
⇒ $2\sin\theta\cos\theta=\frac{2-\sqrt3-4}{4}$
⇒ $\sin\theta\cos\theta=\frac{-2-\sqrt3}{8}$
Now consider, $\tan \theta+\cot \theta$
$=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}$
$=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}$
$=\frac{1}{\sin\theta\cos\theta}$
$=\frac{1}{\frac{-2-\sqrt3}{8}}$
$=\frac{-8}{2+\sqrt3}$
Rationalizing it, we get,
$=\frac{-8}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}$
$=\frac{-8(2-\sqrt3)}{2^2-(\sqrt3)^2}$
$=\frac{8(\sqrt{3}-2)}{4-3}=8(\sqrt3-2)$
Hence, the correct

Correct Answer: WVHRTMVI

Solution : Given:
CURTAIN : XFIGZRM :: DESIGNER : ?

Opposite letter pairs of CURTAIN –

So, CURTAIN is related to XFIGZRM.

Similarly, follow the same pattern for DESIGNER –

Correct Answer: $2N^2-1$

Solution : Mean of $x$ and $\frac{1}{x}$ is $N$.
So, $\frac{x+\frac{1}{x}}{2}$ = $N$
Squaring both sides,
⇒ $\frac{(x+\frac{1}{x})^2}{2^2}$ = $N^2$
⇒ $\frac{x^2+\frac{1}{x^2}+2×x×\frac{1}{x}}{4}$ = $N^2$
⇒ $x^2+\frac{1}{x^2}$ = $4N^2-2$
So, the mean of $x^2$ and $\frac{1}{x^2}$ = $\frac{x^2+\frac{1}{x^2}}{2}$ = $2N^2-1$
Hence, the correct answer is $2N^2-1$.