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Staff Selection Commission Combined Graduate Level Exam

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Question : A man rows to a place 48 km distance and comes back in 14 hours. He finds that he can row 4 km with the stream at the same time as 3 km against the stream. The speed of the stream is:

Option 1: 1.5 km/h

Option 2: 3.5 km/h

Option 3: 1.8 km/h

Option 4: 1 km/h

Team Careers360 12th Jan, 2024

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.

19 Views

Question : If $x$ = ${\frac{1}{\sqrt{2}+1}}$, then the value of $(x^{2}+2x–1)$ is:

Option 1: $\sqrt[2]{2}$

Option 2: 4

Option 3: 0

Option 4: 2

Team Careers360 20th Jan, 2024

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.

10 Views

Question : Directions: Select the correct combination of mathematical signs to sequentially replace the * signs and balance the given equation.
12 * 4 * 15 * 3 * 16 * 24

Option 1: +, –, ÷, ×, =

Option 2: –, +, ÷, ×, =

Option 3: ÷, –, +, ×, =

Option 4: ÷, +, ÷, +, =

Team Careers360 12th Jan, 2024

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: +, –, ÷, ×, =

14 Views

Question : What is the value of $\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}–\sqrt{5}} \div \frac{\sqrt{14}+\sqrt{10}}{\sqrt{14}–\sqrt{10}}+\frac{\sqrt{10}}{\sqrt{5}}$?

Option 1: $\sqrt{2}+2$

Option 2: $2 \sqrt{2}+2$

Option 3: $\sqrt{2}+1$

Option 4: $2 \sqrt{2}+1$

Team Careers360 24th Jan, 2024

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 ${\sqrt{2}}+1$.

17 Views

Question : Who coined the word 'Geography'?

Option 1: Ptolemy

Option 2: Eratosthenese

Option 3: Hacatus

Option 4: Herodatus

Team Careers360 22nd Jan, 2024

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

120 Views

Question : Directions: Which of the following numbers will replace the question mark (?) in the given series?
382, 322, 272, 232, 202, ?

Option 1: 168

Option 2: 150

Option 3: 182

Option 4: 132

Team Careers360 18th Jan, 2024

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;

11 Views

Question : Aryabhata and Kalidasa were in the court of which Gupta Emperor?

Option 1: Kumaragupta I

Option 2: Chandragupta II

Option 3: Samudragupta

Option 4: Skandagupta

Team Careers360 14th Jan, 2024

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.

67 Views

Question : Who among the following is known as a pioneering dance educationist and a prominent Mohiniyattam exponent?

Option 1: Madhavi Mudgal

Option 2: Shagun Butani

Option 3: Mohanrao Kallianpurkar

Option 4: Dr. Kanak Rele

Team Careers360 18th Jan, 2024

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

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