Staff Selection Commission Sub Inspector Exam
Question : In Pocket Veto, the President of India can keep a bill for how much duration?
Option 1: 1 month
Option 2: 6 months
Option 3: 12 months
Option 4: Indefinite period
Correct Answer: Indefinite period
Solution : The correct answer is indefinite period.
This technique for blocking bills has been referred to as a pocket veto. The veto power only allows the President to accept or reject an entire act that the Parliament has passed. It does not give the
Question : The ratio of the curved surface area of two cones is 1 : 4 and the ratio of slant height of the two cones is 2 : 1. What is the ratio of the radius of the two cones?
Option 1: 1 : 2
Option 2: 1 : 4
Option 3: 1 : 8
Option 4: 1 : 1
Correct Answer: 1 : 8
Solution : The curved surface area of a cone $ = \pi rl$ where \(r\) is the radius and \(l\) is the slant height of the cone. Given that the ratio of the curved surface areas of two cones = 1 : 4 $⇒\frac{\pi r_1
Question : If $\tanθ + \cotθ = 2$, $\theta$ is an acute angle, then find the value of $2 \tan^{25}θ + 3 \cot^{20}θ + 5 \tan^{30}θ \cot^{15}θ$.
Option 1: 8
Option 2: 6
Option 3: 10
Option 4: 12
Correct Answer: 10
Solution : Given: The value of the trigonometric equation is $\tanθ + \cotθ = 2$ where $θ$ is an acute angle. $\tanθ + \cotθ = 2$ ⇒ $\tan 45^{\circ} + \cot45^{\circ} = 2$ ⇒ $\theta = 45^{\circ}$ The value of the trigonometric expression $2 \tan^{25}45^{\circ} + 3
Question : Directions: Which of the following numbers will replace the question mark (?) and complete the given number series? 2, 12, 144, 2592, ?
Option 1: 62108
Option 2: 62208
Option 3: 62204
Option 4: 63408
Correct Answer: 62208
Solution : Given: 2, 12, 144, 2592, ?
The pattern is as follows –
So, 62208 is the missing number of the series. Hence, the second option is correct.
Question : Comprehension:
Read the given passage and answer the questions that follow.
On May 2, 2018, severe dust storms and thunderstorms hit parts of Rajasthan, Uttar Pradesh, and other adjoining regions. Explaining the reasons behind such a severe weather system, M. Mohapatra, a senior scientist with the India Meteorological Department (IMD), said that dust storms and thunderstorms are a result of nearly similar weather conditions, like intense heat. Areas that have moisture in the air experience thunderstorms, while those which don't have moisture experience dust storms. "All such conditions were being fulfilled on that day. The region had moist easterly winds coming from the Bay of Bengal, and there was a western disturbance system too. All this together triggered the events on May 2," he added.
In simple terms, a dust storm can be explained as a phenomenon when strong winds carry dust over an extensive area.
As far as the timing of their occurrence is concerned, Mohapatra stated that there is no deviation in the time of occurrence of a dust storm or thunderstorm activity, as they usually peak in the pre-monsoon period.
These events happen between March and May only. In the pre-monsoon period, temperatures are very high, around 44–45 degrees Celsius, which leads to such activities.
Even though dust storms and thunderstorms are a common feature every year, there has been no focused work on studying the trends related to them. "In the coming years, there could be more instances of intense thunderstorms and sand storms, but they have to be studied in detail, looking at the data from over 30-40 years to notice a trend. Work has been more or less confined to cyclones and monsoon systems, but not much on dust storms or thunderstorms. It needs study," admitted Mohapatra.
The extent of damage caused by thunderstorm activity in India can be gauged from the data of India's National Crime Records Bureau (NCRB). As per the NCRB data, compared to other natural disasters like cyclones, floods, or heat waves, lightning kills more people in India. For instance, at least 25 per cent of the 10,510 accidental deaths attributable to forces of nature in 2015 were due to lightning. The number of deaths due to lightning has constantly remained over 2,000 every year since 2005.
Question:
When do dust storms and thunderstorms generally occur?
Option 1: In the pre-monsoon period
Option 2: After the rains
Option 3: During monsoon
Option 4: During winters
Correct Answer: In the pre-monsoon period
Solution : The first option is correct.
As stated by M. Mohapatra in the passage, the timing of the occurrence of dust storms and thunderstorm activity usually peaks in the pre-monsoon period. This period falls between March and May and is characterised by very
Question : Directions: Which of the option figures is the exact mirror image of the given problem figure when the mirror is held to the right side of the problem figure?
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution : As per the mirror image properties, closer things appear close to the mirror in the reflection. Here, according to the information provided, the mirror is placed on the right side of the figure. So, the left side of the reflected image will appear as the right
Question : Directions: A person walks towards his house at 8:00 a.m. and observes his shadow to his right. In which direction he is walking?
Option 1: North
Option 2: South
Option 3: East
Option 4: West
Correct Answer: South
Solution : As the sun rises in the East in the morning i.e. 8:00 a.m., the shadow will be formed in the opposite direction of the sun.
Here, the shadow is formed to his right, so the left side of the person is facing the sun.
So,
Question : What is the HCF of $\frac{4}{5}, \frac{6}{8}, \frac{8}{25}?$
Option 1: $\frac{1}{100}$
Option 2: $\frac{1}{5}$
Option 3: $\frac{1}{50}$
Option 4: $\frac{1}{200}$
Correct Answer: $\frac{1}{100}$
Solution : HCF of $\frac{4}{5}, \frac{6}{8}, \frac{8}{25} =\frac{\text{HCF of numerator}}{\text{LCM of denominator}}$ HCF of numerator (i.e. 4, 6 and 8) = 2 LCM of denominator (i.e. 5, 8 and 25) = 200 So, the HCF of $\frac{4}{5}, \frac{6}{8}, \frac{8}{25} =\frac{2}{200} =\frac{1}{100}$ Hence, the correct answer is $\frac{1}{100}$.
Question : When a host is exposed to antigens, which may be in the form of living or dead microbes or other proteins, antibodies are produced in the host body. This type of immunity is called ______ immunity.
Option 1: acquired
Option 2: active
Option 3: passive
Option 4: strong
Correct Answer: active
Solution : The correct answer is active.
When a host encounters antigens, which could be living or dead microbes or other proteins, the host's body produces antibodies in response. This form of immunity is termed "active immunity." Active immunity is a gradual process and can be either
Question : If $\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}=4,0^{\circ}<\theta<90^{\circ}$, then what is the value of $(\sec \theta+\operatorname{cosec} \theta+\cot \theta) ?$
Option 1: $1+2 \sqrt{3}$
Option 2: $\frac{1+2 \sqrt{3}}{3}$
Option 3: $\frac{2+\sqrt{3}}{3}$
Option 4: $2+\sqrt{3}$
Correct Answer: $2+\sqrt{3}$
Solution : Given, $\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}=4$ ⇒ $\frac{\cos\theta(1+\sin\theta)+\cos\theta(1-\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}=4$ ⇒ $\frac{\cos\theta+\sin\theta\cos\theta+\cos\theta-\sin\theta\cos\theta}{(1+\sin\theta)(1-\sin\theta)}=4$ ⇒ $\frac{2\cos\theta}{1-\sin^2\theta}=4$ [As $\sin^2\theta+\cos^2\theta=1$] ⇒ $\frac{\cos\theta}{\cos^2\theta}=2$ ⇒ $\cos\theta=\frac{1}{2}$ ⇒ $\theta=60°$ Now, $\sec \theta+\operatorname{cosec} \theta+\cot \theta=\sec60°+\operatorname{cosec}60°+\cot60°$ $=2+\frac{2}{\sqrt3}+\frac1{\sqrt3}$ $=\frac{2\sqrt3+2+1}{\sqrt3}$ $=\frac{2\sqrt3+3}{\sqrt3}$ $=2+\sqrt3$ Hence, the correct answer is $2+\sqrt3$.
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