UPSC CISF-AC-EXE Archives

42. Which one of the following statements about the Directive Principles o

Which one of the following statements about the Directive Principles of State Policy is not correct ?

The Directive Principles constitute certain ideals for social justice.

The Directive Principles are not confined to Part IV of the Constitution of India.

The sanction behind the Directive Principles is legal.

It aims to fulfil the objective of socio-economic justice.

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UPSC CISF-AC-EXE – 2018

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The correct answer is C) The sanction behind the Directive Principles is legal.

Directive Principles of State Policy (DPSPs) are contained primarily in Part IV of the Constitution of India (Articles 36-51). They are fundamental in the governance of the country and it shall be the duty of the State to apply these principles in making laws.
Statement A is correct as DPSPs aim to promote social and economic justice and welfare.
Statement B is correct as principles similar to DPSPs are found outside Part IV (e.g., promotion of Hindi language, claims of SC/STs to services).
Statement D is correct as DPSPs are intended to achieve the objective of socio-economic justice, contributing to the establishment of a welfare state.
Statement C is incorrect because DPSPs are non-justiciable, meaning they cannot be enforced by courts. Their sanction is primarily moral and political, dependent on public opinion and the ruling government’s commitment.

The justiciability of Fundamental Rights (Part III) and the non-justiciability of Directive Principles (Part IV) is a key distinction in the Indian Constitution. While not legally enforceable, DPSPs serve as guidelines for the government in formulating policies and laws.

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43. Which one among the following is responsible for the command, control

Which one among the following is responsible for the command, control and operational decisions of nuclear weapons in India ?

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Nuclear Commission of India

Nuclear Command Authority

The Ministry of Defence

The Cabinet Committee on Security

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UPSC CISF-AC-EXE – 2018

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The correct answer is B) Nuclear Command Authority.

In India, the Nuclear Command Authority (NCA) is the apex body responsible for the command, control, and operational decisions regarding nuclear weapons. It comprises a Political Council, chaired by the Prime Minister, which is the sole body authorized to order the use of nuclear weapons, and an Executive Council, chaired by the National Security Advisor, which provides inputs and carries out the directives of the Political Council.

India adopted its nuclear doctrine in 2003, which includes the principles of ‘No First Use’ and ‘Credible Minimum Deterrence’. The establishment of the NCA is a key component of this doctrine, ensuring civilian political leadership maintains control over nuclear weapons decisions.

44. Which among the following countries in not a founder member of Shangha

Which among the following countries in not a founder member of Shanghai Cooperation Organization ?

Kazakhstan

Kyrgyzstan

Tajikistan

Afghanistan

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UPSC CISF-AC-EXE – 2018

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The correct answer is D) Afghanistan.

The Shanghai Cooperation Organization (SCO) was founded in 2001. The original six founder members are China, Kazakhstan, Kyrgyzstan, Russia, Tajikistan, and Uzbekistan. Afghanistan is not one of the founder members.

Afghanistan holds Observer Status in the SCO. India and Pakistan became full members of the SCO in 2017. Other Observer States include Belarus, Iran (full member from 2023), and Mongolia.

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45. 25 writers, 20 doctors, 18 dentists and 12 bank employees spent altoge

25 writers, 20 doctors, 18 dentists and 12 bank employees spent altogether ₹ 1,330 in a hotel. However, it was found that 5 writers spent as much as 4 doctors, that 12 doctors spent as much as 9 dentists, and that 6 dentists spent as much as 8 bank employees. Out of the four professional groups, which group spent the maximum amount ?

Bank employees

Dentists

Doctors

Writers

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UPSC CISF-AC-EXE – 2018

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The correct answer is B) Dentists.

Let $w_s, d_s, t_s, b_s$ be the average amount spent by one writer, doctor, dentist, and bank employee, respectively. The number of people in each group is 25 writers, 20 doctors, 18 dentists, and 12 bank employees.
We are given the following relationships:
1. 5 writers spend as much as 4 doctors: $5w_s = 4d_s \implies w_s = \frac{4}{5}d_s$
2. 12 doctors spend as much as 9 dentists: $12d_s = 9t_s \implies 4d_s = 3t_s \implies t_s = \frac{4}{3}d_s$
3. 6 dentists spend as much as 8 bank employees: $6t_s = 8b_s \implies 3t_s = 4b_s$
Substitute $t_s$ from (2) into (3): $3(\frac{4}{3}d_s) = 4b_s \implies 4d_s = 4b_s \implies d_s = b_s$
So, we have $w_s = \frac{4}{5}d_s$, $t_s = \frac{4}{3}d_s$, and $b_s = d_s$.
The total amount spent is ₹ 1330: $25w_s + 20d_s + 18t_s + 12b_s = 1330$
Substitute the values in terms of $d_s$: $25(\frac{4}{5}d_s) + 20d_s + 18(\frac{4}{3}d_s) + 12d_s = 1330$
$20d_s + 20d_s + 24d_s + 12d_s = 1330$
$76d_s = 1330 \implies d_s = \frac{1330}{76} = 17.5$
Now calculate the amount spent per person for each group:
$d_s = 17.5$
$w_s = \frac{4}{5} \times 17.5 = 14$
$t_s = \frac{4}{3} \times 17.5 = \frac{70}{3} \approx 23.33$
$b_s = 17.5$
Total amount spent by each group:
Writers: $25 \times 14 = 350$
Doctors: $20 \times 17.5 = 350$
Dentists: $18 \times \frac{70}{3} = 6 \times 70 = 420$
Bank Employees: $12 \times 17.5 = 210$
Comparing the total amounts, the Dentists group spent the maximum amount (₹ 420).

This problem requires setting up equations based on the given relationships and solving for the per-person cost of one group, then using that to find the costs for others and finally the total expenditure per group.

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46. Find the missing letter. – The question contains a table which is not

Find the missing letter.
– The question contains a table which is not rendered here.

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This question was previously asked in

UPSC CISF-AC-EXE – 2018

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The question refers to a missing table which is essential to identify the pattern and find the missing letter. Without the table or the context of the pattern, it is impossible to determine the correct answer from the given options. Therefore, a definitive correct option cannot be provided based solely on the text of the question.

Questions involving missing letters or numbers typically require identifying a sequence, pattern, or relationship within a given set of data, often presented in a table, matrix, series, or diagram. The pattern could be based on alphabetical position, arithmetic operations, visual arrangement, or a combination thereof.

To solve such problems, one would need to carefully observe the provided data (the table, in this case) and test potential patterns (e.g., row-wise, column-wise, diagonal, sum, difference, product, letter values, etc.) to find the rule governing the arrangement. Once the rule is found, it is applied to determine the missing element. As the table is unavailable, this step cannot be performed.

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47. Find the odd one out of the following items : EF22, JK42, GH24, VW90,

Find the odd one out of the following items :
EF22, JK42, GH24, VW90, IJ38

EF22

GH24

IJ38

VW90

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UPSC CISF-AC-EXE – 2018

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The correct answer is B) GH24.

The pattern for most items is that the sum of the alphabetical positions of the two letters, when multiplied by 2, equals the number.
– EF: E is the 5th letter, F is the 6th. 5 + 6 = 11. 11 * 2 = 22. (Matches EF22)
– JK: J is the 10th letter, K is the 11th. 10 + 11 = 21. 21 * 2 = 42. (Matches JK42)
– GH: G is the 7th letter, H is the 8th. 7 + 8 = 15. 15 * 2 = 30. The number given is 24. (Does not match GH24)
– VW: V is the 22nd letter, W is the 23rd. 22 + 23 = 45. 45 * 2 = 90. (Matches VW90)
– IJ: I is the 9th letter, J is the 10th. 9 + 10 = 19. 19 * 2 = 38. (Matches IJ38)
All items except GH24 follow this pattern.

This is a common type of reasoning question that tests pattern recognition based on alphabetical order and arithmetic operations. Always check for simple relationships between letter positions and the given numbers, such as sums, differences, products, or sequences.

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48. Delhi is bigger than Pune, Kanpur is bigger than Raipur. Jhansi is not

Delhi is bigger than Pune, Kanpur is bigger than Raipur. Jhansi is not as big as Pune, but is bigger than Kanpur. Which one of the following is the smallest city ?

Delhi

Pune

Kanpur

Raipur

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UPSC CISF-AC-EXE – 2018

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The smallest city is Raipur.

We are given the following comparisons regarding size:
1. Delhi is bigger than Pune: Delhi > Pune
2. Kanpur is bigger than Raipur: Kanpur > Raipur
3. Jhansi is not as big as Pune (i.e., smaller than Pune): Jhansi < Pune 4. Jhansi is bigger than Kanpur: Jhansi > Kanpur

Let’s combine these inequalities:
From (3) and (4): Pune > Jhansi and Jhansi > Kanpur. Combining these gives Pune > Jhansi > Kanpur.
From (2) and the result above (Jhansi > Kanpur): Jhansi > Kanpur > Raipur.
From (1) and the result Pune > Jhansi: Delhi > Pune > Jhansi.

Combining all the relationships, we get a clear order of size:
Delhi > Pune > Jhansi > Kanpur > Raipur.

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The city at the smaller end of this chain is Raipur.

The comparisons allow us to establish a strict linear ordering of the four cities by size: Delhi is the largest, followed by Pune, then Jhansi, then Kanpur, and finally Raipur is the smallest.
Delhi > Pune
Pune > Jhansi (from Jhansi < Pune) Jhansi > Kanpur
Kanpur > Raipur
Combining these gives Delhi > Pune > Jhansi > Kanpur > Raipur.

49. How many times in a day are the hour hand and the minute hand of a wal

How many times in a day are the hour hand and the minute hand of a wall clock straight (i.e., the angle between them is 180°)?

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UPSC CISF-AC-EXE – 2018

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The hour hand and the minute hand of a wall clock are straight (180° apart) 22 times in a day (24 hours).

In a 12-hour period, the minute hand completes 12 revolutions while the hour hand completes 1 revolution. The minute hand gains 11 full revolutions over the hour hand. During this process, the hands coincide (0° apart) 11 times and are opposite (180° apart) 11 times.
The hands are 180° apart once in every hour interval, except for the interval between 6 o’clock and 7 o’clock, where they are 180° apart exactly at 6 o’clock. The 6 o’clock position is counted in both the 5-6 interval and the 6-7 interval if we consider specific time points, but considering distinct occurrences in a 12-hour cycle, it happens 11 times.
For example, between 12 pm and 12 am, the hands are opposite at approximately 12:33, 1:38, 2:44, 3:49, 4:55, 6:00, 7:05, 8:11, 9:16, 10:22, 11:27. (These are approximate times, the exact times are fractions). This is 11 distinct times in a 12-hour period.
A full day is 24 hours, which consists of two 12-hour periods.
Therefore, in 24 hours, the hands will be straight (180° apart) 11 times + 11 times = 22 times.

The general formula for the time t (in minutes past H o’clock) when the hands are at an angle $\theta$ is $t = \frac{2}{11} (30H \pm \theta)$.
For the hands to be straight, $\theta = 180^\circ$.
$t = \frac{2}{11} (30H \pm 180)$.
Let’s check for H from 1 to 12.
For H=6, $t = \frac{2}{11} (180 \pm 180)$. $t = \frac{2}{11} (360) \approx 65.45$ min (past 6) or $t = \frac{2}{11} (0) = 0$ min (past 6). This confirms 6:00 is one time.
For H=5, $t = \frac{2}{11} (150 \pm 180)$. $t = \frac{2}{11} (330) = 60$ min (past 5, which is 6:00) or $t = \frac{2}{11} (-30)$ (not valid in this hour).
The hands are exactly opposite at 6:00. This instance is the boundary point that is counted only once in a 12-hour period when counting the intervals between hours.
The number of times the hands are 180° apart is 11 in 12 hours, and 22 in 24 hours.

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50. In a bag, there are notes of ₹ 10, ₹ 20 and ₹ 50 in the ratio of 1 : 2

In a bag, there are notes of ₹ 10, ₹ 20 and ₹ 50 in the ratio of 1 : 2 : 3. If the total money is ₹ 1,000, how many notes of ₹ 10 are there ?

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This question was previously asked in

UPSC CISF-AC-EXE – 2018

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There are 5 notes of ₹10 in the bag.

The ratio of the number of notes of ₹10, ₹20, and ₹50 is 1 : 2 : 3.
Let the number of ₹10 notes be x.
Then, the number of ₹20 notes is 2x.
And the number of ₹50 notes is 3x.

The total value of the money in the bag is the sum of the values of each type of note:
Value from ₹10 notes = x * ₹10 = 10x
Value from ₹20 notes = 2x * ₹20 = 40x
Value from ₹50 notes = 3x * ₹50 = 150x

Total money = 10x + 40x + 150x = 200x.
We are given that the total money is ₹1,000.
So, 200x = 1000.
Solving for x:
x = 1000 / 200
x = 5.

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The number of ₹10 notes is x, which is 5.

Number of ₹10 notes = 5 (Value = 5 * ₹10 = ₹50)
Number of ₹20 notes = 2 * 5 = 10 (Value = 10 * ₹20 = ₹200)
Number of ₹50 notes = 3 * 5 = 15 (Value = 15 * ₹50 = ₹750)
Total value = ₹50 + ₹200 + ₹750 = ₹1000. This confirms the calculation.