It's a Question of your Career!

Correct Answer:

Solution : According to the given figure series –
1. The triangle at the sides of the rectangle increases by one in a clockwise direction as we move from one box to another.
2. Only two triangles are formed on each side of the rectangle.

So, following the

Correct Answer: $\frac{3}{80}$

Solution : Given numbers are $\frac{3}{4},\frac{15}{16}$ and $\frac{18}{5}$
So, HCF of $\frac{3}{4},\frac{15}{16},$ and $\frac{18}{5}$
= $\frac{\text{HCF of 3, 15, and 18}}{\text{LCM of 4, 16, and 5}}$
= $\frac{3}{80}$
Hence, the correct answer is $\frac{3}{80}$.

Correct Answer: To commit a course of action

Solution : The fourth option is correct.

The idiom to pull the trigger typically means to take action or make a decision that leads to a significant event or outcome. It refers to the moment of initiating or starting something that will

Correct Answer: (3)

Solution : The correct answer is the third option.

To should be used in the sentence to convey that the speaker took the shortest possible road through a park that was near a palace.

The preposition to is used to indicate the location of the park concerning

Correct Answer: Mohandas Karamchand Gandhi

Solution : The correct answer is Mohandas Karamchand Gandhi.

After being overseas for twenty years, Mohandas Karamchand Gandhi returned to India in January 1915. He had spent most of these years practising law in South Africa, where he eventually rose to prominence as the

Correct Answer: 17 – 18 years

Solution : The correct answer is 17 – 18 years.

Sub-junior boxers are those boys and girls who are 12 to 16 years old throughout the boxing season. A boxer's age is calculated from the year of birth.

Junior (youth) boxers are boys

Correct Answer: Bar

Solution : The correct option is Bar.

The metric unit of pressure that is equal to 0.986923 atmospheres (atm) is the bar. Therefore, 0.986923 atm is equivalent to 0.986923 bars. The conversion factor between atmospheres and bars is 1 atm = 1.01325 bar.

Pressure is a

Correct Answer: $\frac{127}{5}$

Solution : Given expression, $\frac14\text{ of }\frac45+\frac92\text{ of }\frac{17}{3}-\frac{3}{10}=p$
⇒ $p=\frac{1}{5}+\frac{51}{2}-\frac{3}{10}$
⇒ $p=\frac{2+255-3}{10}$
⇒ $p=\frac{254}{10}=\frac{127}{5}$
Hence, the correct answer is $\frac{127}{5}$.

Correct Answer: $x=y$

Solution : $\cos ^{2}\theta=\frac{(x+y)^{2}}{4xy}$
We know that $(x+y)\geq 2\sqrt{xy}$ (Since the arithmetic mean is always greater than or equal to the geometric mean)
$⇒(x+y)^2\geq 4xy$
$⇒\frac{(x+y)^2}{4xy}\geq 1$
Since $0\leq \cos ^{2}\theta\leq 1$, solution exists when $\frac{(x+y)^2}{4xy}=1$
$\therefore \frac{(x+y)^2}{4xy}=1$
$⇒4xy =x^2+y^2 + 2xy$
$⇒x^2+y^2 - 2xy=0$
$⇒(x-y)^2 =

Correct Answer: 2716

Solution : Given, $p=7+4\sqrt3$
⇒ $\frac1p=\frac{1}{7+4\sqrt3}$
⇒ $\frac1p=\frac{1}{7+4\sqrt3}\times\frac{7-4\sqrt3}{7-4\sqrt3}$
⇒ $\frac{1}{p}=\frac{7-4\sqrt3}{(7+4\sqrt3)(7-4\sqrt3)}$
We know $(a+b)(a-b)=a^2-b^2$
⇒ $\frac1p=\frac{7-4\sqrt3}{7^2-(4\sqrt3)^2}$
⇒ $\frac1p=\frac{7-4\sqrt3}{49-48}$
⇒ $\frac1p=7-4\sqrt3$
⇒ $p+\frac1p=7+4\sqrt3+7-4\sqrt3=14$
Now, consider $\frac{p^6+p^4+p^2+1}{p^3}$
$=\frac{p^6+1}{p^3}+\frac{p^4+p^2}{p^3}$
$=(p^3+\frac{1}{p^3})+(p+\frac{1}{p})$
We know, $(a+b)^3=a^3+b^3+3ab(a+b)$
So, $=(p^3+\frac{1}{p^3})+(p+\frac{1}{p})=(p+\frac1p)^3-3(p+\frac{1}{p})+(p+\frac1p)$
$=(p+\frac1p)^3-2(p+\frac{1}{p})$
$=14^3-2\times14$
$=2744-28$
$=2716$
Hence, the correct answer is 2716.