Staff Selection Commission Sub Inspector Exam
Question : The blood vessels that carry blood from the heart to the various parts of the body are called:
Option 1: Arteries
Option 2: Veins
Option 3: Septum
Option 4: Capillaries
Correct Answer: Arteries
Solution : The correct option is Arteries.
The blood vessels that carry blood from the heart to various parts of the body are called arteries. Arteries are responsible for transporting oxygenated blood (except for the pulmonary artery, which carries deoxygenated blood to the lungs for oxygenation) and
Question : PA and PB are two tangents from a point P outside the circle with centre O at the points A and B on it. If $\angle A P B=130^{\circ}$, then $\angle O A B$ is equal to:
Option 1: 45°
Option 2: 50°
Option 3: 35°
Option 4: 65°
Correct Answer: 65°
Solution : Given: PA and PB are tangents $\angle OAP = 90^\circ $ $\angle OBP = 90^\circ $ As, OAPB is a quadrilateral $\angle OAP + \angle APB + \angle PBO + \angle BOA = 360^\circ $ ⇒ $90^\circ + 130^\circ + 90^\circ + \angle BOA =
Question : If $\cot^2θ = 1 - e^2$, then the value of $\operatorname{cosec} θ + \cot^3θ \sec θ$ is:
Option 1: $\left(2-{e}^2\right)^ \frac{1}{2}$
Option 2: $\left(1-{e}^2\right)^ \frac{3}{2}$
Option 3: $\left(1-{e}^2\right)$
Option 4: $\left(2-{e}^2\right) ^\frac{3}{2}$
Correct Answer: $\left(2-{e}^2\right) ^\frac{3}{2}$
Solution : Given, $\cot^2θ = 1 - e^2$ Consider, $\operatorname{cosec} θ + \cot^3θ \sec θ$ $=\frac{1}{\sinθ} + \frac{\cos^3θ}{\sin^3θ}\frac1{\cosθ}$ $=\frac{\sin^2θ+\cos^2θ}{\sin^3θ}$ $=\frac{1}{\sin^3θ}$ [As $\sin^2θ+\cos^2θ=1$] $=\operatorname{cosec^3}θ$ Also, we know that, $\operatorname{cosec^2}θ=1+\cot^2θ$ ⇒ $\operatorname{cosec^2}θ=1+1-e^2$ ⇒ $\operatorname{cosec^2}θ=2-e^2$ ⇒ $\operatorname{cosec}θ=(2-e^2)^{\frac12}$ ⇒ $\operatorname{cosec^3}θ=(2-e^2)^{\frac32}$ Hence, the correct answer is $(2-e^2)^{\frac32}$.
Question : Directions: A scientist is related to the laboratory in the same way as a teacher is related to ______.
Option 1: School
Option 2: Research
Option 3: Job
Option 4: Students
Correct Answer: School
Solution : Given: A scientist is related to the laboratory.
The first term is a professional and the second term is the workplace – A scientist is a professional and a laboratory is a workplace or environment. Similarly, a teacher is a professional and a school is
Question : If 42 persons consume 144 kg of wheat in 15 days, how many days will 30 persons consume 48 kg of wheat?
Option 1: 8 days
Option 2: 7 days
Option 3: 12 days
Option 4: 6 days
Correct Answer: 7 days
Solution : Given: 42 persons can consume 144 kg of wheat in 15 days. Let in $x$ days 30 persons consume 48 kg of wheat. We know that $\frac{M1D1}{W1}=\frac{M2D2}{W2}$ According to the question, $\frac{42×15}{144}=\frac{30×x}{48}$ ⇒ $x=7$ Hence, the correct answer is 7 days.
Question : If $4\left(\operatorname{cosec}^2 57^{\circ}-\tan ^2 33^{\circ}\right)-\cos 90^{\circ}-y \tan ^2 66^{\circ} \tan ^2 24^{\circ}=\frac{y}{2}$, the value of $y$ is:
Option 1: $\frac{8}{3}$
Option 2: $\frac{3}{8}$
Option 3: $8$
Option 4: $\frac{1}{3}$
Correct Answer: $\frac{8}{3}$
Solution : $\operatorname{cosec}(90^{\circ} - \theta) = \sec \theta$ $\tan\theta = \frac{1}{\cot\theta}$ So, $4\left(\operatorname{cosec}^2 57^{\circ}-\tan ^2 33^{\circ}\right)-\cos 90^{\circ}-y \tan ^2 66^{\circ} \tan ^2 24^{\circ}=\frac{y}{2}$ ⇒ $(\operatorname{cosec}^{2}(90-33)^{\circ} - \tan^{2}33^{\circ}) - 0- y×\tan^{2}66^{\circ} × \tan^{2}(90-66)^{\circ}$ = $\frac{y}{2}$ ⇒ $4(\sec^{2}57^{\circ}- \tan^{2}33^{\circ}) - y × \tan^{2}66^{\circ} × \cot^{2}66^{\circ} = \frac{y}{2}$ ⇒ $4
Question : A person has to travel a distance of 30 km. He finds that he has covered $\frac{5}{6}$th of the distance in 3 hours and 20 minutes. what is his speed in km/hr?
Option 1: 6
Option 2: 5.4
Option 3: 6.3
Option 4: 7.5
Correct Answer: 7.5
Solution : Given: Total distance = 30 km 3 hr 20 min $= 3 + \frac{20}{60} = 3 + \frac{1}{3} = \frac{10}{3}$ $\frac{5}{6}$th of the distance $= 30 × \frac{5}{6}= 25$ km Speed of the person = $\frac{25}{\frac{10}{3}}= 7.5$ km/hr Hence, the correct answer is 7.5 km/hr.
Question : Which of the following painters of Jahangir's reign was conferred the title of Nadir-Ul-Asra?
Option 1: Bishan Das
Option 2: Mansur
Option 3: Manohar
Option 4: Daulat
Correct Answer: Mansur
Solution : The correct option is Mansur
Mansur was in charge of recording plants and animals during Jahangir's reign. He adopted the name Ustad Mansur and the emperor gave him the honorific title Nadir-ul- 'Asr (meaning "Miracle of the Age"). Because of his skills and expertise, he
Question : Ramesh borrowed INR 15,000 at a certain rate of simple interest from a bank. He repaid (the principal but not interest) INR 6,000 after 5 years, and after 5 more years he paid INR 18,600 and settled his account. Find the rate of simple interest per annum.
Option 1: 12%
Option 2: 10%
Option 3: 8%
Option 4: 6%
Correct Answer: 8%
Solution : Total amount = 6000 + 18600 = 24600 Principal sum, $P = 15000$ Simple interest = Total amount - principal sum = 24600 - 15000 = 9600 $n_1 = 5$ years $n_2 = 5$ years $P_2$ = 15000 - 6000 = 9000 Total simple interest
Question : Which among the following festivals is celebrated in Jharkhand during the spring when the saal trees wear new leaves?
Option 1: Karam
Option 2: Sarhul
Option 3: Jawa
Option 4: Sohrai
Correct Answer: Sarhul
Solution : The correct answer is Sarhul.
Sarhul is a spring festival celebrated in Jharkhand, India. During this season, the Saal trees blossom with new flowers, marking a significant natural event. The festival involves worshipping the village deity, considered the tribes' protector. People engage in lively
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