Staff Selection Commission Sub Inspector Exam
Question : In a triangle ABC, AD, BE, and CF are three medians. The perimeter of ABC is:
Option 1: equal to $(\overline {AD}+\overline{BE}+\overline{CF})$
Option 2: greater than $(\overline {AD}+\overline{BE}+\overline{CF})$
Option 3: less than $(\overline {AD}+\overline{BE}+\overline{CF})$
Option 4: none of these
Correct Answer: greater than $(\overline {AD}+\overline{BE}+\overline{CF})$
Solution : We know that the sum of any two sides of a triangle is greater than twice the median bisecting the third side. Here, D, E, and F are the midpoints of the sides BC, AC, and AB respectively. ∴ AB + AC
Question : If $a+\frac{1}{a}=P^2$, then find $a^2+\frac{1}{a^2}$.
Option 1: $p^2 - 2$
Option 2: $p^2 + 2$
Option 3: $p^4 - 2$
Option 4: $p^4 + 2$
Correct Answer: $p^4 - 2$
Solution : Given: $a+\frac{1}{a}=P^2$ Squaring both sides, we get: ⇒ $(a+\frac{1}{a})^2=(P^2)^2$ ⇒ $a^2+\frac{1}{a^2}+2\times a\times\frac{1}{a}=P^4$ $\therefore a^2+\frac{1}{a^2}=P^4-2$ Hence, the correct answer is $P^4-2$.
Question : With which sport do you associate the name of Koneru Humpy?
Option 1: Basketball
Option 2: Chess
Option 3: Volleyball
Option 4: Table Tennis
Correct Answer: Chess
Solution : The correct option is Chess
Koneru Humpy is connected to the chess game. She is a grandmaster of Indian chess and is regarded as one of the best female players in the world. Koneru Humpy is a notable chess player who has represented India in
Question : Simplify the given expression. $\frac{a^2-b^2-2 b c-c^2}{a^2+b^2+2 a b-c^2}$
Option 1: $\frac{a+b+c}{a-b-c}$
Option 2: $\frac{a-b-c}{a+b-c}$
Option 3: $\frac{a+b-c}{a-b-c}$
Option 4: $\frac{a-b+c}{a+b-c}$
Correct Answer: $\frac{a-b-c}{a+b-c}$
Solution : Given: $\frac{a^2-b^2-2 b c-c^2}{a^2+b^2+2 a b-c^2}$ = $\frac{a^2-(b^2+2 b c+c^2)}{(a^2+b^2+2 a b)-c^2}$ = $\frac{a^2-(b+c)^2}{(a+b)^2-c^2}$ = $\frac{(a-b-c)(a+b+c)}{(a+b-c)(a+b+c)}$ = $\frac{a-b-c}{a+b-c}$ Hence, the correct answer is $\frac{a-b-c}{a+b-c}$.
Question : Solve the following equation. 123 × (162 − 142 − 40) ÷ 2 − 94 = ________.
Option 1: 17280
Option 2: 6561
Option 3: 10719
Option 4: 9865
Correct Answer: 10719
Solution : 123 × (162 − 142 − 40) ÷ 2 − 94 = 1728 × (256 − 196 − 40) ÷ 2 − 6561 = 1728 × 20 ÷ 2 − 6561 = 1728 × 10 − 6561 = 17280 − 6561
Question : A tradesman marks his goods at 26% above the cost price. He allows his customers a discount of 12% on the marked price. His profit percentage is:
Option 1: 10.88%
Option 2: 11.08%
Option 3: 10.50%
Option 4: 10%
Correct Answer: 10.88%
Solution : Let the cost price be 100. So, marked price = 100 + 26 = 126 He allows his customers a discount of 12% on the marked price. $\therefore$ Selling price = $\frac{100-12}{100}×126=110.88$ $\therefore$ Profit percentage $=\frac{110.88-100}{100}×100=10.88\%$ Hence, the correct answer is 10.88%.
Question : If $(\cos \theta+\sin \theta):(\cos \theta-\sin \theta)=(\sqrt{3}+1):(\sqrt{3}-1), 0^{\circ}<\theta<90^{\circ}$, then what is the value of $\sec \theta$?
Option 1: 2
Option 2: $\sqrt{2}$
Option 3: 1
Option 4: $\frac{2 \sqrt{3}}{3}$
Correct Answer: $\frac{2 \sqrt{3}}{3}$
Solution : $(\cos \theta+\sin \theta):(\cos \theta-\sin \theta)=(\sqrt{3}+1):(\sqrt{3}-1)$ $\Rightarrow \frac{\cos\ \theta\ +\ \sin\ \theta}{\cos\ \theta\ -\ \sin\ \theta}\ =\ \frac{\sqrt{3}+\ 1}{\sqrt{3}-\ 1}$ By componendo and dividendo $\Rightarrow \frac{\cos\ \theta}{\sin\ \theta}\ =\ \frac{\sqrt{3}}{1}$ $\Rightarrow \cot\ \theta = \sqrt{3}=\cot30^\circ$ $\Rightarrow \theta = 30^\circ$ $\therefore \sec\ 30^\circ=\frac{2}{\sqrt{3}}=\frac{2 \sqrt{3}}{3}$ Hence, the
Question : Find out the wrong number in the sequence 169, 144, 121, 100, 82, 64, 49.
Option 1: 144
Option 2: 49
Option 3: 64
Option 4: 82
Correct Answer: 82
Solution : Given: The sequence 169, 144, 121, 100, 82, 64, 49 can be written as 132, 122, 112, 102, 82, 82, 72 in the same order. Observe that all the numbers are perfect squares except
Question : Select the most appropriate phrasal verb to fill in the blank. Arunima _______________ a lot during her vacation time.
Option 1: gets on
Option 2: gets after
Option 3: gets about
Option 4: gets along
Correct Answer: gets about
Solution : The correct answer is the third option.
Explanation: The phrasal verb "gets about" is commonly used to describe someone moving around or travelling to various places. It suggests a level of activity and mobility. In the context of the sentence, using "gets about" implies
Question : If $\cos A = \sin ^2A$, and $a \sin ^{12}A+b \sin ^{10}A+c \sin ^{8}A+ \sin ^{6}A=1$, then $a + b + c$?
Option 1: 7
Option 2: 8
Option 3: 9
Option 4: 6
Correct Answer: 7
Solution : $\cos A = \sin^2A$ Square both sides, $⇒\cos^2A = \sin^4A$ $⇒1 - \sin^2A = \sin^4A$ $⇒1 = \sin^4A + \sin^2A$ Cubing both sides, $⇒1 =(\sin^4A + \sin^2A)^3$ $⇒1 = \sin^{12}A + \sin^6A + 3\sin^8A + 3\sin^{10}A$ Comparing this with the given expression, $a \sin^{12}A +
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