Correct Answer: Forty-second Amendment Act, 1976

Solution : The Correct option is the Forty-second Amendment Act of 1976.

The Forty-second Amendment Act of 1976 amended the Indian Constitution to include the Fundamental Duties. This modification, which became effective on November 3, 1976, included a new Part IVA that contains

Correct Answer: Netaji Subhas Chandra Bose

Solution : The correct answer is Netaji Subhash Chandra Bose.

The Samagra Shiksha Scheme is an integrated scheme for school education, and it ensures quality education for all by focusing on teachers and technology. Also, residential facilities are provided under the scheme for

Correct Answer: 1 km/h

Solution : Suppose he moves 4 km downstream in $x$ hours.
⇒ Speed downstream =$\frac{4}{x}$ km/hr
And, speed upstream =$\frac{3}{x}$ km/hr
According to the question,
$\frac{48}{\frac{4}{x}}+\frac{48}{\frac{3}{x}}=14$
⇒ $\frac{48x}{4}+\frac{48x}{3}=14$
⇒ $\frac{144x+192x}{12}=14$
⇒ $336x=168$
⇒ $x=\frac{168}{336}=\frac{1}{2}$
⇒ Speed downstream = 8 km/hr
And, speed upstream = 6 km/hr.

Correct Answer: 0

Solution : Given:
$x= {\frac{1}{\sqrt{2}+1}}$
Rationalising the denominator, we get,
⇒ $x ={\frac{{\sqrt{2}-1}}{(\sqrt{2}+1)×{(\sqrt{2}-1)}}}=\frac{\sqrt{2}-1}{2-1} = {\sqrt{2}-1}$
Putting this value in the expression $(x^{2}+2x-1)$, we get,
$=[(\sqrt{2}-1)^{2}+2(\sqrt{2}-1)-1]$
$=2-2\sqrt{2}+1+2\sqrt{2}-2-1 $
$= 0$
Hence, the correct answer is 0.

Correct Answer: ÷, +, ÷, +, =

Solution : Given:
12 * 4 * 15 * 3 * 16 * 24

Replace * with the mathematical signs and solve the equations one by one using BODMAS.
Let's check the given options –
First option: +, –, ÷, ×, =
⇒

Correct Answer: $\sqrt{2}+1$

Solution : Given: The expression is $\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}–\sqrt{5}} \div \frac{\sqrt{14}+\sqrt{10}}{\sqrt{14}–\sqrt{10}}+\frac{\sqrt{10}}{\sqrt{5}}$.
Take $\sqrt2$ common in the term $\frac{\sqrt{14}+\sqrt{10}}{\sqrt{14}–\sqrt{10}}$, we get,
$\frac{\sqrt{14}+\sqrt{10}}{\sqrt{14}–\sqrt{10}}=\frac{\sqrt2}{\sqrt2}\times\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}–\sqrt{5}}=\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}–\sqrt{5}}$
The expression can be written as $\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}–\sqrt{5}} \div\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}–\sqrt{5}} +\frac{\sqrt{10}}{\sqrt{5}}$
= $\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}–\sqrt{5}} \times\frac{\sqrt{7}–\sqrt{5}}{\sqrt{7}+\sqrt{5}} +\frac{\sqrt{10}}{\sqrt{5}}$
= $1+\sqrt2$
= ${\sqrt{2}}+1$
Hence, the correct answer is ${\sqrt{2}}+1$.

Correct Answer: Eratosthenese

Solution : The correct answer is Erastothenese.

Erastothenese was a Greek astronomer, mathematician and philosopher. He was the first to used the word geography which meant Earth writing in Greek. He is also known as father of geography. He invented the system of latitudes and longitudes and

Correct Answer: 182

Solution : Given:
382, 322, 272, 232, 202, ?

Subtract the multiples of 10 (in decreasing order), starting from 60 from the previous number to obtain the following number of the series.
382 – 60 = 322; 322 – 50 = 272; 272 – 40 = 232;

Correct Answer: Chandragupta II

Solution : The correct answer is Chandragupta II.

Chandragupta II, who was famously called Chandragupta Vikramaditya, served as a Gupta emperor. His royal assembly boasted the esteemed presence of both Aryabhata and Kalidasa. Among the nine pearls of Chandragupta's court was the celebrated Sanskrit writer Kalidasa.

Correct Answer: Dr. Kanak Rele

Solution : The correct answer is Dr. Kanak Rele.

Kanak Rele was a prominent Indian dancer, choreographer, and academic renowned for her expertise in Mohiniyattam. She gained recognition not only for her contributions as a performer but also for her role as an advisor on