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Staff Selection Commission Sub Inspector Exam

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Question : If a + b + c = 6 and a2 + b2 + c2 = 38, then what is the value of a(b2 + c2) + b(c2 + a2) + c(a2 + b2) + 3abc?

Option 1: 3

Option 2: 6

Option 3: –6

Option 4: –3

Team Careers360 14th Jan, 2024

Correct Answer: –6


Solution :

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
⇒ 62 = 38 + 2ab + 2bc + 2ca
⇒ (ab + bc + ca) = $\frac{(36 – 38)}{2}$
⇒ (ab +

25 Views

Question : Directions: Four letter clusters have been given, out of which three are alike in some manner and one is different. Select the odd letter cluster.

Option 1: KNJZ

Option 2: FIEU

Option 3: MJLO

Option 4: SVRH

Team Careers360 15th Jan, 2024

Correct Answer: MJLO


Solution : Let's check each option –
First option: KNJZ; K + 3 = N; N – 4 = J; J – 10 = Z
Second option: FIEU; F + 3 = I; I – 4 = E; E – 10 = U
Third option: MJLO; M

28 Views

Question : Who was/were the founder(s) of the Theosophical Society?

Option 1: H. P. Blavatsky and Colonel Olcott

Option 2: Madam Bhikaji Cama

Option 3: Annie Besant

Option 4: Charles W. Leadbeater and Emily Lutyens

Team Careers360 20th Jan, 2024

Correct Answer: H. P. Blavatsky and Colonel Olcott


Solution : The correct answer is H. P. Blavatsky and Colonel Olcott.

The Theosophical Society, established in 1875 in New York City, is the organisational body of Theosophy, an esoteric new religious movement. Co-founded by Helena Blavatsky, a Russian mystic and

307 Views

Question : A person has three iron bars whose lengths are 20, 30 and 40 metres, respectively. He wants to cut pieces of the same length from each of the three bars. What is the least number of total pieces if he cuts without any wastage?

Option 1: 8

Option 2: 10

Option 3: 9

Option 4: 11

Team Careers360 16th Jan, 2024

Correct Answer: 9


Solution : HCF of 20, 30 and 40 = 10
Number of pieces from 20m long iron bar = $\frac{20}{10} = 2$
Number of pieces from 30m long iron bar = $\frac{30}{10} = 3$
Number of pieces from 40m long iron bar = $\frac{40}{10} = 4$
Total

21 Views

Question : Select the INCORRECTLY spelt word from the underlined words in the following sentence.
This was a blatent disregard to the advisory by the host against it and in violation of the norms of the meeting.

Option 1: advisory

Option 2: blatent

Option 3: disregard

Option 4: violation

Team Careers360 18th Jan, 2024

Correct Answer: blatent


Solution : The correct answer is the second option.

Explanation: The correct spelling is "blatant", not "blatent". The word "blatant" is an adjective used to describe something obvious, flagrant, or done without any attempt to conceal it. In the context of the sentence, it describes a disregard

28 Views

Question : The Rolling plan was implemented between which two five-year plans?

Option 1: Fourth and Fifth

Option 2: Sixth and Seventh

Option 3: Second and Third

Option 4: Fifth and Sixth

Team Careers360 24th Jan, 2024

Correct Answer: Fifth and Sixth


Solution : The correct option is the Fifth and Sixth

The term "rolling plan" typically refers to an economic planning approach where plans are continuously updated and modified rather than being fixed for a specific period, such as a five-year plan. The concept of the

119 Views

Question : Which of the following planets is not a Jovian planet?

Option 1: Saturn

Option 2: Uranus

Option 3: Mars

Option 4: Jupiter

Team Careers360 17th Jan, 2024

Correct Answer: Mars


Solution : The correct answer is Mars.

Mars is not a Jovian planet. Mars is a terrestrial planet. The Jovian planets are Jupiter, Saturn, Uranus and Neptune. They orbit far from the sun. These planets have no solid surfaces and are essentially large balls of gas

8 Views

Question : If $\frac{x}{(b–c)(b+c–2a)}=\frac{y}{(c–a)(c+a–2b)}=\frac{z}{(a–b)(a+b–2c)}$, then $(x+y+z)$ is:

Option 1: $a+b+c$

Option 2: $0$

Option 3: $a^2+b^2+c^2$

Option 4: $2$

Team Careers360 15th Jan, 2024

Correct Answer: $0$


Solution : Given: $\frac{x}{(b–c)(b+c–2a)}=\frac{y}{(c–a)(c+a–2b)}=\frac{z}{(a–b)(a+b–2c)}$
Let $\frac{x}{(b–c)(b+c–2a)}=\frac{y}{(c–a)(c+a–2b)}=\frac{z}{(a–b)(a+b–2c)}=k$.
${x}=k{(b–c)(b+c–2a)}$ (equation 1)
${y}=k{(c–a)(c+a–2b)}$ (equation 2)
${z}=k{(a–b)(a+b–2c)}$ (equation 3)
Adding equations 1, 2, and 3 respectively, we get,
$x+y+z=k{(b–c)(b+c–2a)}+k{(c–a)(c+a–2b)}+k{(a–b)(a+b–2c)}$
$x+y+z=k[(b–c)(b+c–2a)+(c–a)(c+a–2b)+(a–b)(a+b–2c)]$
$x+y+z= k[b^2+bc–2ab–bc–c^2+2ac+c^2+ac–2bc–ac–a^2+2ab+a^2+ab–2ac–ab–b^2+2bc)$
$x+y+z$ = 0
Hence, the correct answer is 0.

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