Correct Answer: 40

Solution : Given:
5 – 9 × 8 ÷ 12 + 4 = ?

After interchanging + and ÷, and × and –, the equation will be as follows –
= 5 × 9 – 8 + 12 ÷ 4
= 5 × 9 – 8 +

Correct Answer: $\frac{k^{32}-\frac{1}{k^{32}}}{k+\frac{1}{k}}$

Solution : Consider, $\left(k-\frac{1}{k}\right)\left(k^2+\frac{1}{k^2}\right)\left(k^4+\frac{1}{k^4}\right)\left(k^8+\frac{1}{k^8}\right)\left(k^{16}+\frac{1}{k^{16}}\right)$
Multiplying and dividing by $(k+\frac{1}{k})$
⇒ $\left(k-\frac{1}{k}\right)\left(k^2+\frac{1}{k^2}\right)\left(k^4+\frac{1}{k^4}\right)\left(k^8+\frac{1}{k^8}\right)\left(k^{16}+\frac{1}{k^{16}}\right)=\frac{(k-\frac{1}{k})(k+\frac{1}{k})(k^2+\frac{1}{k^2})(k^4+\frac{1}{k^4})(k^8+\frac{1}{k^8})(k^{16}+\frac{1}{k^{16}})}{k+\frac{1}{k}}$
Now using, $(a+b)(a-b)=a^2 - b^2$
⇒ $\left(k-\frac{1}{k}\right)\left(k^2+\frac{1}{k^2}\right)\left(k^4+\frac{1}{k^4}\right)\left(k^8+\frac{1}{k^8}\right)\left(k^{16}+\frac{1}{k^{16}}\right)=\frac{(k^2-\frac{1}{k^2})(k^2+\frac{1}{k^2})(k^4+\frac{1}{k^4})(k^8+\frac{1}{k^8})(k^{16}+\frac{1}{k^{16}})}{k+\frac{1}{k}}$
$=\frac{(k^4-\frac{1}{k^4})(k^4+\frac{1}{k^4})(k^8+\frac{1}{k^8})(k^{16}+\frac{1}{k^{16}})}{k+\frac{1}{k}}$
$=\frac{(k^8-\frac{1}{k^8})(k^8+\frac{1}{k^8})(k^{16} + \frac{1}{k^{16}})}{k+\frac{1}{k}}$
$=\frac{(k^{16} - \frac{1}{k^{16}})(k^{16} + \frac{1}{k^{16}})}{k+\frac{1}{k}}$
$=\frac{k^{32} - \frac{1}{k^{32}}}{k+\frac{1}{k}}$
Hence, the correct answer is $\frac{k^{32} - \frac{1}{k^{32}}}{k+\frac{1}{k}}$.

Correct Answer: had worked hard

Solution : The correct choice is the fourth option. 

"Had worked hard" should be used to fill in the blank since it uses the past perfect tense, which is appropriate for describing an action that was completed before another past action. In this sentence, the

Correct Answer: P

Solution : Subtract 3 from the place value of the letters of each row to obtain the next letter.
Row I: K – 3 = H; H – 3 = E
Row II: B – 3 = Y; Y – 3 = V
Row III: S –

Correct Answer: Ghiyas-ud-din Tughlaq

Solution : The correct option is Ghiyas-ud-din Tughlaq.

Ghiyas-ud-din Tughlaq raised the banner of revolt against Nasiruddin Khusrau. Nasiruddin Khusrau was the last ruler of the Khalji dynasty. He was a weak and incompetent ruler, and his reign was marked by corruption and mismanagement. This

Correct Answer: Mohiniyattam dance

Solution : The correct answer is the Mohiniyattam dance.

The dance forms Tevitichiyattam, Nangai Natakam, and Dasiyattam are all associated with Mohiniyattam, which is a traditional Indian classical dance form that originated in the state of Kerala.

Mohiniyattam is known for its graceful and expressive

Correct Answer: 5

Solution : Marked Price (MP) = Rs. 1600
First discount = 10%
Final selling price (SP) = Rs. 1368
Selling price after discount = Marked Price × (1 – Discount %) 
⇒ Selling Price after first discount = Rs. 1600 × (1 – $\frac{10}{100}$) = Rs. 1440

Correct Answer: $(200+\frac{200}{\sqrt{3}})$ metres

Solution :
Let BD be $x$ and CD be $y$.
In right $\triangle$ABD,
$\tan 45^{\circ}$ = $\frac{AD}{BD}$
$⇒1 = \frac{200}{x}$
$⇒x= 200$
Now, in right $\triangle$ACD,
$\tan 60^{\circ}= \frac{AD}{CD}$
$⇒\sqrt3= \frac{200}{y}$
$⇒y= \frac{200}{\sqrt3}$
$\therefore$ The width of the river $=x+y = 200+\frac{200}{\sqrt3}$ metres
Hence, the correct

Correct Answer: Decomposer - Bacteria

Solution : The correct answer is Decomposer - Bacteria.

• Producers are the autotrophs, which are the sole energy producers in an ecosystem. They are green plants, blue-green algae and phytoplankton, which can generate food with the help of photosynthesis.
• Consumers depend on producers for

Correct Answer: $a<b<c$

Solution : Given: $a=\sqrt{6}–\sqrt{5}$, $b=\sqrt{5}–2$, $c=2–\sqrt{3}$
Rationalising the given values,
$a=\sqrt{6}–\sqrt{5}×\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}$
⇒ $a=\frac{1}{\sqrt{6}+\sqrt{5}}$ -----(1)
$b=\sqrt{5}–2×\frac{\sqrt{5}+2}{\sqrt{5}+2}$
⇒ $b=\frac{1}{\sqrt{5}+2}$
⇒ $b=\frac{1}{\sqrt{5}+\sqrt{4}}$------(2)
$c=2–\sqrt{3}×\frac{2+\sqrt{3}}{2+\sqrt{3}}$
⇒ $c=\frac{1}{2+\sqrt{3}}$
⇒ $c=\frac{1}{\sqrt{4}+\sqrt{3}}$------(3)
By comparing $(a,b,c)$ we can find $a<b<c$.
Hence, the correct answer is $a<b<c$.