Staff Selection Commission Combined Higher Secondary Level Exam
Question : Directions: In the following question below are given some statements followed by some conclusions based on those statements. Taking the given statements to be true even if they seem to be at variance from commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follows the given statements. Statements: I. All T is P. II. All P is L. Conclusions: I. All T is L. II. All L is P. III. Some P is T.
Option 1: Both conclusions II and III follow
Option 2: Both conclusions I and II follow
Option 3: Both conclusions I and III follow
Option 4: All conclusions follow
Correct Answer: Both conclusions I and III follow
Solution : The possible Venn diagram according to the given statements is as follows –
Let's analyse the conclusions – Conclusion (I): All T is L – From the Venn diagram, it is evident that the circle representing T completely lies inside
Question : Directions: Select the correct alternative to indicate the arrangement of the following terms in a logical and meaningful order. 1. Post-Graduation 2. Graduation 3. Nursery 4. Matriculation 5. Higher Secondary
Option 1: 1, 2, 4, 3, 5
Option 2: 3, 1, 2, 4, 5
Option 3: 3, 4, 5, 2,1
Option 4: 1, 2, 3, 4, 5
Correct Answer: 3, 4, 5, 2,1
Solution : Given: 1. Post-Graduation 2. Graduation 3. Nursery 4. Matriculation 5. Higher Secondary
The stages of education involve – The first Initial stage of education is nursery, then comes matriculation which refers to secondary education or High school, which is followed by higher
Question : The highest waterfall in India is in the state of
Option 1: Andhra Pradesh
Option 2: Assam
Option 3: Maharashtra
Option 4: Karnataka
Correct Answer: Karnataka
Solution : The correct answer is Karnataka.
Kunchikal Falls, in the Shimoga district of Karnataka, is the highest waterfall in India. Formed by the Varahi River, it has a height of approx 455 m. The Varahi River is a tributary of the Souparnika River, one of the
Question : If $\sin (x - y) = \frac{1}2$ and $\cos (x + y) = \frac{1}2$, then what is the value of $\sin x \cos x + 2\sin^2x + cos^3x \sec x$?
Option 1: $2$
Option 2: $\sqrt{2}+1$
Option 3: $1$
Option 4: $\frac{3}{4}$
Correct Answer: $2$
Solution : $\sin (x - y) = \frac{1}2=\sin30^\circ$ $⇒(x-y)=30^\circ$---(1) $\cos (x + y) = \frac{1}2=\cos60^\circ$ $⇒(x+y)=60^\circ$---(2) Solving equation 1 and 2, we get, ⇒ $x=45^\circ$ Now, Putting the value of $x$, we get: $\sin x\cos x + 2\sin^2x + \cos^3 x \sec x$ $=\sin 45^\circ\cos45^\circ + 2\sin^2
Question : If $4(2x+3)>5-x$ and $5x-3(2x-7)>3x-1,$ then $x$ can take which of the following values?
Option 1: 6
Option 2: –1
Option 3: 5
Option 4: –6
Correct Answer: 5
Solution : The first inequality: $4(2x+3)>5-x$ $⇒8x+12>5-x$ $⇒9x>-7$ $⇒x>-\frac{7}{9}$ The second inequality: $5x-3(2x-7)>3x-1$ $⇒5x-6x+21>3x-1$ $⇒-4x> -22$ $⇒4x<22$ $⇒x<\frac{22}{4}$ $⇒x<5.5$ $\therefore$ The solution to the inequalities is $-\frac{7}{9} < x < 5.5$. Hence, the correct answer is 5.
Question : Directions: After interchanging the given two numbers (not digits), what will be the value of the given equation? 13 and 11 48 ÷ 6 + 9 – 11 × 11 + 13
Option 1: –121
Option 2: –113
Option 3: –141
Option 4: –137
Correct Answer: –141
Solution : Given: 48 ÷ 6 + 9 – 11 × 11 + 13
On interchanging the numbers, we get – = 48 ÷ 6 + 9 – 13 × 13 + 11 = 8 + 9 – 13 × 13 + 11 = 8 + 9
Question : If $x=\frac{4\sqrt{15}}{\sqrt{5}+\sqrt{3}}$, the value of $\frac{x+\sqrt{20}}{x–\sqrt{20}}+\frac{x+\sqrt{12}}{x–\sqrt{12}}$ is:
Option 1: $1$
Option 2: $2$
Option 3: $\sqrt{3}$
Option 4: $\sqrt{5}$
Solution : Given: $x=\frac{4\sqrt{15}}{\sqrt{5}+\sqrt{3}}=\frac{4\sqrt{5}\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ Now, $\frac{x}{\sqrt{20}}=\frac{4\sqrt{5}\sqrt{3}}{\sqrt{20}(\sqrt{5}+\sqrt{3})} = \frac{2\sqrt{3}}{(\sqrt{5}+\sqrt{3})}$ Using componendo and dividendo $\frac{x+\sqrt{20}}{x-\sqrt{20}}=\frac{\sqrt{5}+3\sqrt{3}}{(\sqrt{3}-\sqrt{5})}$ Similarly, we can find $\frac{x+\sqrt{12}}{x-\sqrt{12}}=\frac{\sqrt{3}+3\sqrt{5}}{(\sqrt{5}-\sqrt{3})}$ Now, $\frac{x+\sqrt{20}}{x–\sqrt{20}}+\frac{x+\sqrt{12}}{x–\sqrt{12}}=\frac{\sqrt{5}+3\sqrt{3}}{(\sqrt{3}-\sqrt{5})}+\frac{\sqrt{3}+3\sqrt{5}}{(\sqrt{5}-\sqrt{3})} =\frac{2\sqrt{3}-2\sqrt{5}}{(\sqrt{3}-\sqrt{5})}$ =2 Hence, the correct answer is $2$.
Question : If a : b : c = 2 : 3 : 5 and 5b – a + 2c = 115, then what is the value of b?
Option 1: 15
Option 2: 12
Option 3: 6
Option 4: 24
Correct Answer: 15
Solution : a ∶ b ∶ c = 2 ∶ 3 ∶ 5 Let the value of a, b, and c be 2x, 3x, and 5x, respectively. So, ⇒ 5b - a + 2c = 115 ⇒ 5 × (3x) – (2x) + 2 × (5x) =
Question : A computer is available for INR 39,000 on cash payment or INR 19,000 as cash payment followed by five monthly instalments of INR 4,200 each. What is the rate of interest per annum under the instalment plan?
Option 1: $20 \frac{19}{29} \%$
Option 2: $20 \frac{17}{29} \%$
Option 3: $20 \frac{20}{29} \%$
Option 4: $20 \frac{18}{29} \%$
Correct Answer: $20 \frac{20}{29} \%$
Solution : Total cost of the computer = Rs. 39000 Down payment = Rs. 19000 Balance = Rs. 20000 Let the rate of interest be $r$ Here simple interest is denoted by S.I. Hence, the amount of Rs. 20000 for 5 months = $20000+ \frac{20000\times
Question : If $x+\frac{1}{x}=0$, then the value of $x^{5}+\frac{1}{x^{5}}$ is:
Option 1: 2
Option 3: 1
Option 4: 0
Correct Answer: 0
Solution : Given: $x+\frac{1}{x}=0$............................................ $(i)$ Now, $(x^2+\frac{1}{x^2})(x+\frac{1}{x})$ = 0 ⇒ $x^3+\frac{1}{x^3} + x+\frac{1}{x}$= 0 ⇒ $x^3+\frac{1}{x^3}=0$ Also, $(x^4+\frac{1}{x^4})(x+\frac{1}{x})$ = 0 ⇒ $x^5+\frac{1}{x^5} + x^3+\frac{1}{x^3}$= 0 ⇒ $ x^5+\frac{1}{x^5} + 0 = 0$ ⇒ $x^5+\frac{1}{x^5} = 0$ Hence, the correct answer is 0.
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