Staff Selection Commission Combined Graduate Level Exam
Question : If $x=\frac{4\sqrt{ab}}{\sqrt a+ \sqrt b}$, then what is the value of $\frac{x+2\sqrt{a}}{x-2\sqrt a}+\frac{x+2\sqrt{b}}{x-2\sqrt b}$(when $a\neq b$)?
Option 1: 0
Option 2: 2
Option 3: 4
Option 4: $\frac{(\sqrt a+\sqrt b)}{(\sqrt a - \sqrt b)}$
Correct Answer: 2
Solution : Given: $x=\frac{4\sqrt ab}{\sqrt a+\sqrt b}$ Equation $=\frac{x+2\sqrt a}{x-2\sqrt a}+\frac{x+2\sqrt b}{x-2\sqrt b}$ Put the value of $x$ in equation: $=\frac{\frac{4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}+2\sqrt{a}}{\frac{4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-2\sqrt a}+\frac{\frac{4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}+2\sqrt{b}}{\frac{4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-2\sqrt b}$ $=\frac{\frac{4\sqrt{ab}+2a+2\sqrt {ab}}{\sqrt{a}+\sqrt{b}}}{\frac{4\sqrt{ab}-2a-2\sqrt {ab}}{\sqrt{a}+\sqrt{b}}}+\frac{\frac{4\sqrt{ab}+2\sqrt {ab}+2b}{\sqrt{a}+\sqrt{b}}}{\frac{4\sqrt{ab}-2\sqrt {ab}-2b}{\sqrt{a}+\sqrt{b}}}$ $=\frac{\frac{4\sqrt{ab}+2a+2\sqrt {a}b}{\sqrt{a}+\sqrt{b}}}{\frac{4\sqrt{ab}-2a-2\sqrt {a}b}{\sqrt{a}+\sqrt{b}}}+\frac{\frac{4\sqrt{ab}+2\sqrt {ab}+2b}{\sqrt{a}+\sqrt{b}}}{\frac{4\sqrt{ab}-2\sqrt {a}b-2b}{\sqrt{a}+\sqrt{b}}}$ $=\frac{4\sqrt{ab}+2a+2\sqrt {a}b}{4\sqrt{ab}-2a-2\sqrt {ab}}+\frac{4\sqrt{ab}+2\sqrt {ab}+2b}{4\sqrt{ab}-2\sqrt {ab}-2b}$ $=\frac{2}{2}\left [\frac{2\sqrt{ab}+a+\sqrt {a}b}{2\sqrt{ab}-a-\sqrt {a}b} \right ]+\frac{2}{2}\left[\frac{2\sqrt{ab}+\sqrt {ab}+b}{2\sqrt{ab}-\sqrt {a}b-b}\right]$ $=\frac{3\sqrt{ab}+a}{\sqrt{ab}-a}+\frac{3\sqrt{ab}+b}{\sqrt{ab}-b}$
Question : If $x+\frac{1}{x}=\sqrt{3}$, then the value of $x^{3}+\frac{1}{x^{3}}$ is equal to:
Option 1: $1$
Option 2: $3\sqrt{3}$
Option 3: $0$
Option 4: $3$
Correct Answer: $0$
Solution : Given: $x+\frac{1}{x}=\sqrt{3}$ Cubing both sides we get $(x+\frac{1}{x})^3=(\sqrt{3})^3$ ⇒ $x^3+\frac{1}{x^3}+3×x×\frac{1}{x}(x+\frac{1}{x})=(3\sqrt{3})$ ⇒ $x^3+\frac{1}{x^3}=3\sqrt{3}–3(x+\frac{1}{x})$ $\because x+\frac{1}{x}=\sqrt{3}$ Thus, $x^3+\frac{1}{x^3}=3\sqrt{3}–3\sqrt{3} = 0$ Hence, the correct answer is $0$.
Question : Neeraj Chopra is associated with which sports ?
Option 1: Kabaddi
Option 2: Cricket
Option 3: Javelin Throw
Option 4: Wrestling
Correct Answer: Javelin Throw
Solution : The correct option is - Javelin Throw .
Neeraj Chopra is an Indian javelin thrower, born on 24 December 1997. He now holds the title of Olympic champion in the javelin throw and has won silver at the World Championships and the Diamond League.
Question : Select the most appropriate option that can substitute the underlined segment in the given sentence.
My friend Meera and her mother is visiting me this weekend.
Option 1: have visiting
Option 2: am visiting
Option 3: was visiting
Option 4: are visiting
Correct Answer: are visiting
Solution : The most appropriate choice is the fourth option.
The original sentence is incorrect as per the subject-verb agreement. Since "My friend Meera and her mother" is a compound subject involving more than one person, the verb should also be in the plural form to
Question : Directions: A – B means A is the mother of B; A * B means A is the sister of B; A % B means A is the husband of B; A & B means A is the son of B. If Z * A & B – C * D % E, then how is Z related to E?
Option 1: Sister
Option 2: Wife
Option 3: Husband’s sister
Option 4: Mother-in-law
Correct Answer: Husband’s sister
Solution : Given: A – B ⇒ A is the mother of B A * B ⇒ A is the sister of B A % B ⇒ A is the husband of B A & B ⇒ A is the son of B
As per the
Question : In the figure, AB = AD = 7 cm, and AC = AE, and BC = 11 cm, then find the length of ED.
Option 1: 12
Option 2: 10
Option 3: 11
Option 4: 2
Correct Answer: 11
Solution : Given, AB = AD = 7 cm, and AC = AE, and BC = 11 cm $\angle BAC = \angle DAE$ (vertically opposite angles) In the given figure, $\frac{AB}{BC}=\frac{AD}{ED}$ So, $\triangle ABC \cong \triangle ADE$ [by Side-Angle-Side criteria] ⇒ $ED = BC = 11$ cm
Question : Directions: Select the option that is related to the fifth number in the same way as the second number is related to the first number and the fourth number is related to the third number. 7 : 13 :: 16 : 31 :: 46 : ?
Option 1: 81
Option 2: 83
Option 3: 91
Option 4: 97
Correct Answer: 91
Solution : Given: 7 : 13 :: 16 : 31 :: 46 : ?
Like, 7 : 13→7 × 2 = 14; 14 – 1 = 13 16 : 31→16 × 2 = 32; 32 – 1 = 31 Similarly, for 46 : ?→(46 × 2) =
Question : Directions: In the following question, find the odd number pair from the given alternatives.
Option 1: 73 – 61
Option 2: 57 – 69
Option 3: 47 – 59
Option 4: 42 – 29
Correct Answer: 42 – 29
Solution : Let's check the options – First option: 73 – 61→73 – 61 = 12 Second option: 57 – 69→69 – 57 = 12 Third option: 47 – 59→59 – 47 = 12 Fourth option: 42 – 29
Question : What is the possible value of (a + b + c) – 3, if a2 + b2 + c2 = 9 and ab + bc + ca = 8?
Option 1: 5
Option 2: 3
Option 3: 9
Solution : Given: $a^2 + b^2 + c^2 = 9$ and $ab + bc + ca = 8$ We know, $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ ⇒ $(a+b+c)^2=9+2\times 8$ ⇒ $(a+b+c)^2=9+16$ ⇒ $a+b+c=\sqrt{25}=5$ $\therefore$ $(a+b+c)-3 = 5-3 = 2$ Hence, the correct answer is 2.
Question : Who was the first to use the term 'State'?
Option 1: Hobbes
Option 2: Plato
Option 3: Aristotle
Option 4: Machiavelli
Correct Answer: Machiavelli
Solution : The correct option is Machiavelli.
Niccol Machiavelli (1469-1527), an Italian philosopher, is credited with coining the term "state" in its contemporary connotation. He describes the state in his book The Prince as "a political unit that is sovereign and has a monopoly on the legitimate
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