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Question 241Question

Under the Registration of Births and Deaths (Amendment) Act, enacted to streamline digital public service delivery and create a centralized civil database, which document has been statutorily designated as the single conclusive proof of date and place of birth for school admissions, voter registration, government appointments, and issuance of passports and driving licences?

Show answer & explanation

Answer: Digital Birth Certificate issued through the Civil Registration System portal

Answer

Digital Birth Certificate issued through the Civil Registration System portal
The Registration of Births and Deaths (Amendment) Act modernized the Civil Registration System by creating a digital, interconnected national and state database. Under this legislation, the digital birth certificate is statutorily established as the single comprehensive document to prove the date and place of birth for multiple civic processes, including school admissions, passport issuance, driving licence applications, voter registration, marriage registrations, and recruitment to public posts.

Step-by-Step Solution

1
Examine the statutory objective of the Registration of Births and Deaths (Amendment) Act.
The Act amends the parent 1969 legislation to build a centralized national and state database of civil registrations under the Registrar General of India.
Understanding the legislative framework is necessary to identify which document was granted statutory primacy.
2
Analyze the specific provision concerning documentary proof for public governance services.
Section 3 of the amended Act mandates that the digital birth certificate alone shall serve as conclusive single proof of age and birth for educational admissions, voter lists, driving licences, passports, and public employment.
This establishes the legal standard that replaces multiple disparate age-verification documents.
3
Differentiate from alternative identity and educational credentials.
Documents like Aadhaar, EPIC, and matriculation certificates serve distinct identity or educational functions and do not possess statutory designation as the primary civil record of birth under this governance act.
Eliminating functional overlaps prevents conflation of identity proof with birth registration certification.

Key Concept

Statutory modernization of civil registration and the single-document digital governance framework under the Registration of Births and Deaths (Amendment) Act.
Estimated Time:1m 0s
Question 242Question

Regarding the eligibility criteria and benefit structure of the 'Yuva Nidhi' scheme introduced by the Government of Karnataka for educated unemployed youth, which of the following statements is accurate?

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Answer: It extends a monthly allowance of ₹3,000 to unemployed degree holders and ₹1,500 to diploma holders for a maximum period of two years.

Answer

The scheme extends a monthly allowance of ₹3,000 to unemployed degree holders and ₹1,500 to diploma holders for a maximum period of two years.
The 'Yuva Nidhi' scheme is a flagship social security initiative launched by the Government of Karnataka that provides direct financial assistance of ₹3,000 per month to eligible unemployed graduates and ₹1,500 per month to unemployed diploma holders for a maximum period of two years or until the beneficiary secures employment.

Step-by-Step Solution

1
Identify the target beneficiaries and core objective of Karnataka's Yuva Nidhi Scheme
The scheme assists graduates and diploma holders who have remained unemployed for at least six months following completion of their studies.
Establishing the policy focus confirms the baseline educational qualification requirements.
2
Evaluate the financial entitlement and temporal duration under the scheme
Unemployed degree holders receive ���3,000 per month, while diploma holders receive ₹1,500 per month for a maximum cap of two years (or until securing employment/higher education).
Direct benefit transfer parameters are explicitly defined in the state government operational guidelines.
3
Examine exclusion and disqualification criteria
Beneficiaries gainful in private/government sectors or those who enroll in higher education are rendered ineligible.
Ensures targeted allocation of welfare resources only to actively job-seeking youth.

Key Concept

Karnataka Yuva Nidhi Scheme Eligibility and Financial Assistance Provisions
Estimated Time:1m 0s
Question 243Question

A local health department issues the following public advisory: "Avoid consuming pre-cut fruits sold by roadside vendors to prevent the spread of stomach infections during the rainy season."

Below are two assumptions based on the advisory:
Assumption I: Pre-cut fruits sold by roadside vendors during the rainy season are likely to be contaminated and can cause stomach infections.
Assumption II: People do not consume pre-cut fruits during other seasons of the year.

Which of the given options correctly identifies the implicit assumption(s)?

Show answer & explanation

Answer: Only assumption I is implicit

Answer

The correct answer is that only assumption I is implicit.
Assumption I is implicit because the health advisory's warning is directly based on the premise that pre-cut fruits are a source of contamination and stomach infections during the rainy season. Without this premise, the advisory would be meaningless. Assumption II is not implicit because issuing a warning for a specific high-risk period does not imply that people do not consume the item at other times; it merely highlights when the risk is most significant.

Step-by-Step Solution

1
Analyze the core message of the given statement.
The statement is a health advisory warning people not to eat pre-cut fruits from roadside vendors during the rainy season to prevent stomach infections.
Understanding the core message is necessary to identify the unstated premises (assumptions) behind the warning.
2
Evaluate Assumption I.
Assumption I states that these pre-cut fruits are likely to be contaminated and cause infections. This is a direct underlying premise; if this were not assumed to be true, the health department would have no reason to issue the specific warning.
An assumption is implicit if the statement cannot logically stand without it.
3
Evaluate Assumption II.
Assumption II claims that people do not consume pre-cut fruits during other seasons. The statement only addresses the elevated risk during the rainy season and makes no claims or assumptions about consumption habits during the rest of the year.
Assumptions must be strictly derived from the context of the statement without adding external generalizations.

Key Concept

An assumption is an unstated premise that the author of a statement takes for granted. A valid assumption must be essential for the statement to hold true.
Question 244Question

During gametogenesis in a diploid organism, non-disjunction of a homologous chromosome pair occurs during Anaphase I of meiosis. Assuming normal segregation in Meiosis II, which of the following chromosome distributions will be observed in the four resulting gametes?

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Answer: Two gametes with n+1n + 1 chromosomes and two gametes with n1n - 1 chromosomes

Answer

Two gametes with n+1n + 1 chromosomes and two gametes with n1n - 1 chromosomes
When non-disjunction takes place in Anaphase I of meiosis, the failure of homologous chromosomes to segregate properly yields one daughter cell with an extra chromosome and another daughter cell missing that chromosome. When these cells undergo normal Meiosis II, the resulting four gametes consist of two (n+1)(n + 1) gametes and two (n1)(n - 1) gametes, leading to complete aneuploidy among all four products.

Step-by-Step Solution

1
Analyze the meiotic stage and event specified in the scenario
Non-disjunction occurs at Anaphase I, meaning a homologous chromosome pair fails to separate to opposite poles.
Meiosis I is the reductional division where homologous pairs segregate.
2
Determine the chromosomal content of the two intermediate secondary gametocytes after Meiosis I
One secondary cell receives both homologous chromosomes (an extra chromosome, n+1n + 1), while the other cell receives neither (lacking a chromosome, n1n - 1).
The entire pair moves to one daughter cell instead of dividing equally.
3
Trace the chromosome distribution through normal Meiosis II equational division
The (n+1)(n + 1) cell divides sister chromatids into two (n+1)(n + 1) gametes, and the (n1)(n - 1) cell divides into two (n1)(n - 1) gametes.
Meiosis II separates sister chromatids normally, generating two (n+1)(n + 1) and two (n1)(n - 1) gametes from the four total products.

Key Concept

Meiotic Non-Disjunction and Aneuploidy Formation
Estimated Time:1m 0s
Question 245Question

What is the unit digit of the expression 743+9277^{43} + 9^{27}?

Show answer & explanation

Answer: 2

Answer

The unit digit of the given expression 743+9277^{43} + 9^{27} is 2.
The unit digit of 7437^{43} is determined by dividing 43 by 4, giving a remainder of 3, so 737^3 ends in 3. The unit digit of 9279^{27} is determined by taking 9 to an odd power, which ends in 9. Adding 3+9=123 + 9 = 12 gives a final unit digit of 2.

Step-by-Step Solution

1
Determine the unit digit of 7437^{43} using cyclicity rules.
The unit digit of 7437^{43} is 3.
The base 7 follows a 4-step cyclicity pattern (7, 9, 3, 1). Dividing exponent 43 by 4 leaves remainder 3, giving 73(mod10)=37^3 \pmod{10} = 3.
2
Determine the unit digit of 9279^{27} using cyclicity rules.
The unit digit of 9279^{27} is 9.
The base 9 follows a 2-step cyclicity pattern (9 for odd powers, 1 for even powers). Since 27 is odd, the unit digit is 9.
3
Combine the resulting unit digits.
The unit digit of the sum is 2.
Summing the individual unit digits gives 3+9=123 + 9 = 12, whose unit digit is 2.

Key Concept

Unit Digit and Cyclicity
Question 246Question

In a civil services training academy of 120120 officers, 7070 officers speak English, 6060 officers speak Hindi, and 3030 officers speak both English and Hindi. Based on this set data, which of the following statements are correct?

Select all that apply

Show answer & explanation

Answer: The number of officers who speak at least one of the two languages is 100100.; The number of officers who speak only English is 4040.; The number of officers who speak exactly one language is 7070.

Answer

The correct statements are: the number of officers who speak at least one of the two languages is 100100; the number of officers who speak only English is 4040; and the number of officers who speak exactly one language is 7070.
Statements asserting that 100100 officers speak at least one language, 4040 speak only English, and 7070 speak exactly one language are mathematically correct based on standard set operations: EH=70+6030=100|E \cup H| = 70 + 60 - 30 = 100, Only E=7030=40E = 70 - 30 = 40, and Exactly One =40+30=70= 40 + 30 = 70.

Step-by-Step Solution

1
Calculate the number of officers speaking at least one language using Inclusion-Exclusion Principle.
EH=E+HEH=70+6030=100|E \cup H| = |E| + |H| - |E \cap H| = 70 + 60 - 30 = 100.
The intersection must be subtracted once to avoid double counting.
2
Find the number of officers speaking only English and only Hindi.
Only English = 7030=4070 - 30 = 40; Only Hindi = 6030=3060 - 30 = 30.
Subtracting the overlap from each set leaves elements exclusive to that set.
3
Calculate officers speaking neither language and officers speaking exactly one language.
Neither = 120100=20120 - 100 = 20; Exactly one = 40+30=7040 + 30 = 70.
Neither language is the complement of the union. Exactly one language is the sum of disjoint set regions.

Key Concept

Two-set inclusion-exclusion principle and disjoint region classification in set theory
Estimated Time:50s
Question 247Question

As the Commandant of the State Disaster Response Force (SDRF) responding to a major freight train derailment involving breached hazardous chemical tankers near a populated area, you must execute standard incident command operational procedures. Arrange the following crisis management measures in their correct operational sequence from first to last:

Drag items to arrange them in the correct order

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Answer

The correct operational sequence begins with establishing perimeter control, followed by hazard assessment/reconnaissance, targeted downwind evacuation, technical leak containment, and finally decontamination with handover to environmental restoration authorities.
In standard crisis management protocols, response follows a strict priority logic: protecting life via scene isolation, evaluating the specific hazard (reconnaissance), evacuating vulnerable populations based on empirical dispersion data, mitigating the chemical source (containment), and finally conducting decontamination prior to site transition.

Step-by-Step Solution

1
Isolate the incident area and secure perimeters.
Unauthorised public entry is prevented and immediate civilian exposure risks are contained.
Safety protocols dictate that securing the scene perimeter is the indispensable first step in crisis response.
2
Perform hazmat hazard identification and dispersion analysis.
The exact chemical agent, toxicity thresholds, and wind dispersion directions are determined.
Operational decisions regarding personnel PPE and evacuation parameters depend strictly on empirical hazard data.
3
Evacuate at-risk populations in the downwind sector.
Civilians directly in the plume path are systematically moved to safe assembly zones.
Evacuation must be tailored to actual wind vectors and toxicity profiles to prevent panic and mass exposure.
4
Execute source containment and valve capping.
The chemical leak is plugged and active release of hazardous material is terminated.
Source containment stops ongoing environmental contamination once responder safety and public evacuation are established.
5
Decontaminate teams and hand over the site.
Emergency responders are safely decontaminated, and long-term site restoration is assigned to environmental agencies.
Decontamination eliminates secondary transport of hazardous residues outside the hot zone.

Key Concept

Standard Incident Command System (ICS) operational sequence in hazardous material emergency management
Question 248Question

Four committee members—Aman, Bina, Charan, and Divya—are to be assigned to two discussion groups, Group X and Group Y, such that each group contains exactly two members.

The group assignments must follow these conditions:
1. Aman and Bina cannot be placed in the same group.
2. Charan is assigned to Group X.
3. Divya is assigned to the same group as Bina.

Which member is assigned to Group X along with Charan?

Show answer & explanation

Answer: Aman

Answer

Aman is assigned to Group X along with Charan.
Charan is already assigned to Group X. Since Bina and Divya must be in the same group, placing them in Group X would result in three members in Group X, violating the condition that each group has exactly two members. Therefore, Bina and Divya must be in Group Y. Since Aman cannot be in the same group as Bina, Aman must be assigned to Group X alongside Charan.

Step-by-Step Solution

1
Determine the capacity limit per group
Group X must have 2 members, and Group Y must have 2 members.
There are 4 members total divided equally between 2 groups.
2
Analyze the placement of Bina and Divya based on Group X capacity
Bina and Divya must both be in Group Y.
Charan is already in Group X. If Bina were in Group X, Divya would also join Group X, making 3 members (Charan, Bina, Divya), which violates the capacity constraint.
3
Determine Aman's placement
Aman is placed in Group X with Charan.
Aman and Bina cannot be in the same group. Since Bina is in Group Y, Aman must be in Group X.

Key Concept

Group Allocation and Capacity Constraints
Estimated Time:45s
Question 249Question

A land surveyor begins at a base camp OO. She walks 12 m12\text{ m} due East to reach point AA. From point AA, she turns 9090^\circ clockwise and walks 16 m16\text{ m} to point BB. Facing South at point BB, she turns 135135^\circ anti-clockwise and walks 102 m10\sqrt{2}\text{ m} to point CC. Finally, she turns due West and walks 14 m14\text{ m} to stop at point DD. What is the shortest distance (in meters) between the base camp OO and point DD?

Show answer & explanation

Answer: 10

Answer

The shortest distance between the base camp OO and point DD is 10 meters10\text{ meters}.
Tracking relative coordinates gives point A(12,0)A(12,0), point B(12,16)B(12,-16), point C(22,6)C(22,-6), and final point D(8,6)D(8,-6). The straight-line distance from the origin O(0,0)O(0,0) to D(8,6)D(8,-6) is 82+(6)2=10 meters\sqrt{8^2 + (-6)^2} = 10\text{ meters}.

Step-by-Step Solution

1
Establish coordinate system
Base camp O=(0,0)O = (0, 0)
Assigning the origin to the start point allows vector tracking.
2
Calculate displacement to point A and B
Point A=(12,0)A = (12, 0), Point B=(12,16)B = (12, -16)
East increases x-coordinate by 12; South decreases y-coordinate by 16.
3
Decompose angular movement from B to C
Point C=(22,6)C = (22, -6)
Turning 135135^\circ anti-clockwise from South points to North-East. Δx=102×12=10 m\Delta x = 10\sqrt{2} \times \frac{1}{\sqrt{2}} = 10\text{ m} East, Δy=102×12=10 m\Delta y = 10\sqrt{2} \times \frac{1}{\sqrt{2}} = 10\text{ m} North.
4
Determine final coordinates at point D
Point D=(8,6)D = (8, -6)
Walking 14 m14\text{ m} West reduces the x-coordinate from 22 to 8.
5
Apply distance formula for straight-line displacement
Distance OD=10 mOD = 10\text{ m}
OD=(80)2+(60)2=64+36=10 mOD = \sqrt{(8 - 0)^2 + (-6 - 0)^2} = \sqrt{64 + 36} = 10\text{ m}.

Key Concept

Vector displacement decomposition and 2D Cartesian plane distance calculation
Question 250Question
What is the numerical value of the following mathematical expression evaluated using the BODMAS rule?
12.5% of 480[334÷{1.25+12 of (3.61.80.7)}]×81512.5\% \text{ of } 480 - \left[ 3 \frac{3}{4} \div \left\{ 1.25 + \frac{1}{2} \text{ of } \left( 3.6 - \overline{1.8 - 0.7} \right) \right\} \right] \times \frac{8}{15}
Show answer & explanation

Answer: 59.2

Answer

The evaluated numerical value of the expression is 59.2.
Following strict BODMAS hierarchy—evaluating the vinculum (1.1), parentheses (2.5), 'of' operation (1.25), curly brackets (2.5), square bracket division (1.5), multiplication (0.8), percentage term (60), and final subtraction—gives 60 - 0.8 = 59.2.

Step-by-Step Solution

1
Evaluate the expression under the vinculum bar.
1.8 - 0.7 = 1.1
The vinculum has the highest priority and acts as an implicit innermost bracket.
2
Evaluate the terms inside the round brackets (parentheses).
3.6 - 1.1 = 2.5
Round brackets must be resolved after the vinculum.
3
Evaluate the 'of' operation inside the curly brackets.
(1/2) * 2.5 = 1.25
'Of' precedes addition and multiplication in the standard BODMAS order.
4
Complete the addition within the curly brackets.
1.25 + 1.25 = 2.5 = 5/2
Brackets must be fully simplified from inside out.
5
Perform division inside the square brackets.
(15/4) / (5/2) = (15/4) * (2/5) = 3/2 = 1.5
Square brackets are evaluated after inner curly brackets.
6
Multiply the square bracket result by 8/15.
1.5 * (8/15) = (3/2) * (8/15) = 4/5 = 0.8
Multiplication takes precedence over final subtraction.
7
Calculate the initial percentage 'of' term.
12.5% of 480 = 0.125 * 480 = 60
'Of' operation on percentage takes precedence before subtraction.
8
Subtract the bracketed product from the initial term.
60 - 0.8 = 59.2
Subtraction is performed last according to BODMAS.

Key Concept

BODMAS Order of Operations with Vinculum and Nested Brackets
Estimated Time:2m 0s
Question 251Question

What is the unit digit of the expression S=(1!+2!+3!++99!)+399S = (1! + 2! + 3! + \dots + 99!) + 3^{99}?

Show answer & explanation

Answer: 0

Answer

The unit digit of the given expression is 0.
The unit digit of a sum is determined by the sum of the unit digits of its individual terms. For the factorial sum (1!+2!+3!++99!)(1! + 2! + 3! + \dots + 99!), terms from 5!5! onward end in 0 because 5!=1205! = 120. Summing the first four terms yields 1+2+6+24=331 + 2 + 6 + 24 = 33, contributing a unit digit of 3. For 3993^{99}, the cyclicity sequence of base 3 is 3, 9, 7, 1 (length 4). Dividing 99 by 4 leaves a remainder of 3, corresponding to 33=273^3 = 27, which contributes a unit digit of 7. Summing the two unit digits yields 3+7=103 + 7 = 10, giving a final unit digit of 0.

Step-by-Step Solution

1
Evaluate the unit digit contribution of the factorial terms
Unit digit of (1!+2!+3!++99!)(1! + 2! + 3! + \dots + 99!) is 3
Since n!n! ends in 0 for every n5n \ge 5, only 1!+2!+3!+4!=331! + 2! + 3! + 4! = 33 affects the unit digit.
2
Find the unit digit of 3993^{99} using pattern cyclicity
Unit digit of 3993^{99} is 7
The cyclicity of base 3 is 4. Since 99(mod4)=399 \pmod 4 = 3, the unit digit corresponds to 33=273^3 = 27.
3
Combine the unit digits of both parts
Unit digit of SS is 0
Adding the unit digits gives 3+7=103 + 7 = 10, making the final unit digit 0.

Key Concept

Unit digit evaluation using factorial terminal zeros and exponential cyclicity
Question 252Question

A district administration conducted a specialized training audit for a cadre of 300300 officers across three administrative domains: E-Governance (EE), Disaster Management (DD), and Financial Administration (FF). The data gathered from the audit is as follows:

- 140140 officers are trained in E-Governance.
- 130130 officers are trained in Disaster Management.
- 120120 officers are trained in Financial Administration.
- 5050 officers are trained in both E-Governance and Disaster Management.
- 4040 officers are trained in both Disaster Management and Financial Administration.
- 4545 officers are trained in both E-Governance and Financial Administration.
- 2020 officers are trained in all three domains.

Based on the data provided, which of the following statements are correct?

Select all that apply

Show answer & explanation

Answer: The number of officers trained in exactly two subjects is 7575.; The number of officers trained in at most one subject is 205205.

Answer

The correct statements are that the number of officers trained in exactly two subjects is 75, and the number of officers trained in at most one subject is 205.
The statement declaring that 75 officers are trained in exactly two subjects is accurate because summing the disjoint exclusive intersections (5020)+(4020)+(4520)(50-20) + (40-20) + (45-20) yields 30+20+25=7530 + 20 + 25 = 75. Furthermore, the statement claiming that 205 officers are trained in at most one subject is correct because adding the single-domain officers (65+60+55=18065 + 60 + 55 = 180) to those trained in none of the three domains (300275=25300 - 275 = 25) gives exactly 205205.

Step-by-Step Solution

1
Calculate the region cardinalities for exactly two domains.
E-Governance and Disaster Management only = 5020=3050 - 20 = 30; Disaster Management and Financial Administration only = 4020=2040 - 20 = 20; E-Governance and Financial Administration only = 4520=2545 - 20 = 25. Total in exactly two domains = 30+20+25=7530 + 20 + 25 = 75.
The intersection of any two sets includes elements present in all three sets; subtracting the three-set intersection yields the count for 'only two'.
2
Calculate single-domain region cardinalities.
Only E-Governance = 140(30+25+20)=65140 - (30 + 25 + 20) = 65; Only Disaster Management = 130(30+20+20)=60130 - (30 + 20 + 20) = 60; Only Financial Administration = 120(25+20+20)=55120 - (25 + 20 + 20) = 55.
Each total set count contains its single-domain region plus three overlapping sub-regions.
3
Calculate total officers trained in at least one domain and those trained in none.
Total trained in at least one domain = 65+60+55+30+20+25+20=27565 + 60 + 55 + 30 + 20 + 25 + 20 = 275. Officers trained in none = 300275=25300 - 275 = 25.
Summing all 7 disjoint set regions gives the union cardinality; subtracting from total population yields the complement.
4
Evaluate the statement regarding 'at most one subject'.
Officers in at most one domain = (Only E + Only D + Only F) + None = 65+60+55+25=20565 + 60 + 55 + 25 = 205.
'At most one' includes both the zero-domain category and single-domain categories.

Key Concept

Three-Set Principle of Inclusion-Exclusion and Region Partitioning
Question 253Question
Evaluate the following numerical expression using the correct order of operations (BODMAS):
404×(62)÷2+540 - 4 \times (6 - 2) \div 2 + 5
What is the final result?
Show answer & explanation

Answer: 37

Answer

The final result of the expression is 37.
Following the BODMAS rule: first simplify the bracket (62)=4(6 - 2) = 4. Next, execute multiplication and division from left to right: 4×4=164 \times 4 = 16 and 16÷2=816 \div 2 = 8. The expression reduces to 408+540 - 8 + 5. Finally, carry out subtraction and addition from left to right: 408=3240 - 8 = 32, and 32+5=3732 + 5 = 37.

Step-by-Step Solution

1
Evaluate the expression inside the brackets
62=46 - 2 = 4
According to BODMAS, operations inside brackets (B) take highest priority.
2
Perform multiplication and division from left to right
4×4=164 \times 4 = 16, then 16÷2=816 \div 2 = 8
Multiplication and Division (DM) carry equal priority and are evaluated in order from left to right.
3
Perform addition and subtraction from left to right
408+5=32+5=3740 - 8 + 5 = 32 + 5 = 37
Addition and Subtraction (AS) carry equal priority and are evaluated sequentially from left to right.

Key Concept

BODMAS Order of Operations
Question 254Question

In a rural development assessment of 200200 villages in a district, data regarding access to three basic infrastructure facilities—Piped Water, Solar Power, and Broadband Internet—was recorded as follows:
- 110110 villages have access to Piped Water.
- 100100 villages have access to Solar Power.
- 9090 villages have access to Broadband Internet.
- 5050 villages have access to both Piped Water and Solar Power.
- 4040 villages have access to both Solar Power and Broadband Internet.
- 4545 villages have access to both Piped Water and Broadband Internet.
- 2020 villages have access to all three facilities.

Which of the following statements are correct?

Select all that apply

Show answer & explanation

Answer: Exactly 7575 villages have access to exactly two infrastructure facilities.; The number of villages having access to at least two infrastructure facilities is 9595.; Exactly 3535 villages have access to Piped Water only.

Answer

The correct statements are those indicating that 7575 villages have access to exactly two facilities, 9595 villages have access to at least two facilities, and 3535 villages have access to Piped Water only.
Evaluating each region using set theory shows: exactly two facilities = 30+20+25=7530 + 20 + 25 = 75; at least two facilities = 75+20=9575 + 20 = 95; Piped Water only = 110(30+25+20)=35110 - (30 + 25 + 20) = 35. All three statements accurately reflect the set cardinality.

Step-by-Step Solution

1
Determine the exclusive two-set overlap regions.
Piped Water & Solar Power only = 5020=3050 - 20 = 30; Solar Power & Broadband Internet only = 4020=2040 - 20 = 20; Piped Water & Broadband Internet only = 4520=2545 - 20 = 25.
The given pairwise intersections include the 2020 villages that have access to all three facilities.
2
Determine the single-set only regions.
Piped Water only = 110(30+25+20)=35110 - (30 + 25 + 20) = 35; Solar Power only = 100(30+20+20)=30100 - (30 + 20 + 20) = 30; Broadband Internet only = 90(25+20+20)=2590 - (25 + 20 + 20) = 25.
Subtract all overlapping regions containing that facility from its total.
3
Calculate total villages with at least one facility using the Principle of Inclusion-Exclusion.
Total with at least one facility = 35+30+25+30+20+25+20=18535 + 30 + 25 + 30 + 20 + 25 + 20 = 185.
Summing all 7 disjoint regions yields the union of the three sets.
4
Find villages with none of the facilities.
Villages with none = 200185=15200 - 185 = 15.
Subtract total union from the universe of 200200 villages.

Key Concept

Three-Set Principle of Inclusion-Exclusion and Venn Diagram Region Cardinality
Question 255Question

Let NN be the smallest positive integer that leaves a remainder of 33 when divided by 1111, a remainder of 55 when divided by 1313, and is exactly divisible by 77. What is the remainder when N4+N2+1N^4 + N^2 + 1 is divided by 1717?

Show answer & explanation

Answer: 3

Answer

The remainder when N4+N2+1N^4 + N^2 + 1 is divided by 1717 is 3.
Using the Chinese Remainder Theorem, the smallest integer meeting all divisibility constraints is determined to be N=707N = 707. Reducing 707707 modulo 1717 gives 1010. Substituting 1010 into the expression N4+N2+1N^4 + N^2 + 1 modulo 1717 yields (104+102+1)(mod17)=(4+15+1)(mod17)=20(mod17)=3(10^4 + 10^2 + 1) \pmod{17} = (4 + 15 + 1) \pmod{17} = 20 \pmod{17} = 3.

Step-by-Step Solution

1
Express the given conditions as a system of linear congruences
N3(mod11)N \equiv 3 \pmod{11}, N5(mod13)N \equiv 5 \pmod{13}, and N0(mod7)N \equiv 0 \pmod{7}.
Translates the problem statement into standard modular arithmetic notation.
2
Solve the simultaneous congruences for modulo 11 and modulo 13
Let N=11k+3N = 11k + 3. Substituting into 11k+35(mod13)11k + 3 \equiv 5 \pmod{13} gives 11k2(mod13)    2k2(mod13)    k12(mod13)11k \equiv 2 \pmod{13} \implies -2k \equiv 2 \pmod{13} \implies k \equiv 12 \pmod{13}. Thus, N=11(13m+12)+3=143m+135N = 11(13m + 12) + 3 = 143m + 135.
Combines two modular constraints into a single general expression using the Chinese Remainder Theorem principle.
3
Apply the divisibility condition for 7 to determine the smallest value of NN
143m+1350(mod7)    3m+20(mod7)    3m5(mod7)    m4(mod7)143m + 135 \equiv 0 \pmod{7} \implies 3m + 2 \equiv 0 \pmod{7} \implies 3m \equiv 5 \pmod{7} \implies m \equiv 4 \pmod{7}. For the smallest positive integer, set m=4m = 4, so N=143(4)+135=707N = 143(4) + 135 = 707.
Finds the specific smallest integer NN that satisfies all three divisor requirements.
4
Reduce NN modulo 17
707=17×41+10    N10(mod17)707 = 17 \times 41 + 10 \implies N \equiv 10 \pmod{17}.
Simplifies the base before evaluating higher powers.
5
Calculate the polynomial expression (N4+N2+1)(mod17)(N^4 + N^2 + 1) \pmod{17}
N2102=100152(mod17)N^2 \equiv 10^2 = 100 \equiv 15 \equiv -2 \pmod{17}, and N4(2)2=4(mod17)N^4 \equiv (-2)^2 = 4 \pmod{17}. Therefore, N4+N2+14+15+1=203(mod17)N^4 + N^2 + 1 \equiv 4 + 15 + 1 = 20 \equiv 3 \pmod{17}.
Uses modular exponentiation properties to efficiently evaluate the final remainder.

Key Concept

Chinese Remainder Theorem and Modular Exponentiation
Question 256Question

In a state administrative academy, a batch of 300300 probationary officers were surveyed regarding their enrolment in three specialized training modules: Cyber Security (CC), Public Policy (PP), and Financial Management (FF). The survey revealed the following data:
- Total officers enrolled in Cyber Security: 160160
- Total officers enrolled in Public Policy: 140140
- Total officers enrolled in Financial Management: 130130
- Officers enrolled in both Cyber Security and Public Policy: 6565
- Officers enrolled in both Public Policy and Financial Management: 5555
- Officers enrolled in both Cyber Security and Financial Management: 5050
- Officers enrolled in all three modules: 2020

Based on the data provided above, which of the following statements are correct?

Select all that apply

Show answer & explanation

Answer: The total number of officers enrolled in exactly one training module is 150150.; The ratio of officers enrolled in Cyber Security only to those enrolled in Financial Management only is 13:913 : 9.

Answer

The correct statements are that the total number of officers enrolled in exactly one training module is 150150, and the ratio of officers enrolled in Cyber Security only to those enrolled in Financial Management only is 13:913 : 9.
The statements confirming that 150150 officers are enrolled in exactly one module and that the ratio of Cyber Security only to Financial Management only is 13:913 : 9 are both mathematically true based on region decomposition.

Step-by-Step Solution

1
Identify the 3-set intersection region
The number of officers in all three modules (CPFC \cap P \cap F) is given as 2020.
The central intersection is the foundation for calculating all non-overlapping regions in a 3-set Venn diagram.
2
Calculate the regions corresponding to exactly two modules
CP only=6520=45C \cap P \text{ only} = 65 - 20 = 45; PF only=5520=35P \cap F \text{ only} = 55 - 20 = 35; CF only=5020=30C \cap F \text{ only} = 50 - 20 = 30. Total in exactly two modules = 45+35+30=11045 + 35 + 30 = 110.
Subtracting the 3-set intersection from each 2-set intersection isolates the elements belonging exclusively to two sets.
3
Calculate the regions corresponding to exactly one module
C only=160(45+30+20)=65C \text{ only} = 160 - (45 + 30 + 20) = 65; P only=140(45+35+20)=40P \text{ only} = 140 - (45 + 35 + 20) = 40; F only=130(30+35+20)=45F \text{ only} = 130 - (30 + 35 + 20) = 45. Total in exactly one module = 65+40+45=15065 + 40 + 45 = 150.
Subtracting all double-counted and triple-counted intersections from total set counts yields single-set cardinalities.
4
Evaluate the complement (neither set) and statement conditions
Total in at least one module = 150+110+20=280150 + 110 + 20 = 280. Neither = 300280=20300 - 280 = 20. At least two modules = 110+20=130110 + 20 = 130. Ratio C only:F only=65:45=13:9C \text{ only} : F \text{ only} = 65 : 45 = 13 : 9.
Verifying each statement against calculated set region cardinalities confirms which statements are true.

Key Concept

Principle of Inclusion-Exclusion for Three Sets
Question 257Question

Determine the unit digit of the composite exponential expression E=171819+181920192021E = 17^{18^{19}} + 18^{19^{20}} - 19^{20^{21}}.

Show answer & explanation

Answer: 8

Answer

8
The unit digit of 17181917^{18^{19}} is 1 because 181918^{19} is divisible by 4, which corresponds to the 4th power 741(mod10)7^4 \equiv 1 \pmod{10}. The unit digit of 18192018^{19^{20}} is 8 because 19201(mod4)19^{20} \equiv 1 \pmod 4, giving 81=88^1 = 8. The unit digit of 19202119^{20^{21}} is 1 because 202120^{21} is an even exponent, giving 9even1(mod10)9^{\text{even}} \equiv 1 \pmod{10}. Combining these gives 1+81=81 + 8 - 1 = 8.

Step-by-Step Solution

1
Find the unit digit of 17181917^{18^{19}}
The unit digit is 1
Base 7 has cyclicity 4. Exponent 1819=2190(mod4)18^{19} = 2^{19} \equiv 0 \pmod 4. When the remainder modulo 4 is 0, we take the 4th power in the cycle: 741(mod10)7^4 \equiv 1 \pmod{10}.
2
Find the unit digit of 18192018^{19^{20}}
The unit digit is 8
Base 8 has cyclicity 4. Exponent 1920(1)201(mod4)19^{20} \equiv (-1)^{20} \equiv 1 \pmod 4. The 1st power in the cycle gives 81=88^1 = 8.
3
Find the unit digit of 19202119^{20^{21}}
The unit digit is 1
Base 9 has cyclicity 2. The exponent 202120^{21} is even, so 9even1(mod10)9^{\text{even}} \equiv 1 \pmod{10}.
4
Evaluate the combined expression
The final unit digit is 8
Combine the unit digits of the individual terms: 1+81=81 + 8 - 1 = 8.

Key Concept

Cyclicity of numbers and modular arithmetic for tower exponents
Question 258Question

Four senior administrative officers—Ananya, Bikram, Chitra, and Dev—are to be assigned to head four different departments (Agriculture, Education, Transport, and Urban Development) across four city postings (Ranchi, Raipur, Shimla, and Dehradun), based on the following analytical conditions:

1. Ananya is not assigned to Ranchi or Dehradun, and does not head Agriculture or Transport.
2. The officer heading Urban Development is posted in Raipur.
3. Bikram heads the Education department, but is posted neither in Dehradun nor in Ranchi.
4. Chitra is posted in Ranchi, but does not head the Transport department.
5. Dev is not posted in Raipur.

Match each officer on the left with their correct department and city posting combination on the right.

Click a left item, then click its matching right item

Items

Ananya
Bikram
Chitra
Dev

Matches

Show answer & explanation

Answer

Ananya matches with Urban Development (Raipur); Bikram matches with Education (Shimla); Chitra matches with Agriculture (Ranchi); Dev matches with Transport (Dehradun).
Through step-by-step logical elimination of city-department constraints: Chitra is posted in Ranchi. Bikram heads Education and cannot be in Ranchi, Dehradun, or Raipur (reserved for Urban Development), placing Bikram in Shimla. Ananya is not in Ranchi or Dehradun, placing her in Raipur with Urban Development. Dev receives the remaining city (Dehradun) and department (Transport), while Chitra takes Agriculture.

Step-by-Step Solution

1
Determine fixed city allocations for officers where direct or tight constraints exist.
Chitra is posted in Ranchi (Condition 4). Bikram cannot be in Ranchi or Dehradun (Condition 3). Ananya cannot be in Ranchi or Dehradun (Condition 1). Therefore, the only available city options for Bikram and Ananya are Raipur and Shimla.
Eliminating impossible city assignments narrows down the domain for Bikram and Ananya.
2
Analyze department-city linkages to deduce Bikram and Ananya's exact locations and departments.
Condition 2 states the Urban Development head is in Raipur. Condition 3 states Bikram heads Education. Hence, Bikram cannot be in Raipur. This forces Bikram to be posted in Shimla (Education department). Consequently, Ananya must be posted in Raipur, heading Urban Development.
A person cannot head Education while occupying a city reserved exclusively for the head of Urban Development.
3
Determine the city assignment for Dev.
The cities assigned so far are Ranchi (Chitra), Shimla (Bikram), and Raipur (Ananya). The only remaining city is Dehradun, which must be assigned to Dev.
Four distinct officers must map to four distinct cities.
4
Assign remaining departments (Agriculture and Transport) to Chitra and Dev.
Condition 4 specifies Chitra does not head Transport. Thus, Chitra must head Agriculture in Ranchi. This leaves Transport for Dev in Dehradun.
Process of elimination across remaining unassigned attributes.

Key Concept

Matrix Elimination and Analytical Grouping
Question 259Question

What is the next letter in the given alphabetical series: Z,X,V,T,R,Z, X, V, T, R, \dots?

Show answer & explanation

Answer: PP

Answer

The correct next letter in the sequence is PP.
Each term in the series moves backward by 2 steps in the standard English alphabet: Z(26)X(24)V(22)T(20)R(18)Z (26) \rightarrow X (24) \rightarrow V (22) \rightarrow T (20) \rightarrow R (18). Continuing this constant decrease of 2 leads to the 16th letter, which is PP.

Step-by-Step Solution

1
Determine the alphabetical positions of each given letter in the series
Z=26Z = 26, X=24X = 24, V=22V = 22, T=20T = 20, R=18R = 18
Converting letters to numerical positions helps reveal the mathematical pattern.
2
Identify the common difference between consecutive terms
2426=224 - 26 = -2, 2224=222 - 24 = -2, 2022=220 - 22 = -2, 1820=218 - 20 = -2
Each letter is obtained by moving 2 positions backward in the alphabet.
3
Apply the pattern to find the next position and letter
182=1618 - 2 = 16, which corresponds to the letter PP
Subtracting 2 positions from the 18th letter (RR) gives the 16th letter (PP).

Key Concept

Reverse Alphabetical Positional Shifts
Question 260Question

Following seasonal monsoon rains, a District Medical Officer notes a sudden spike in gastrointestinal infections in a specific residential sector, despite the central water treatment plant operating at full chlorination standards. A quick technical audit reveals that informal household connections have caused drinking water pipes to run directly through open stormwater drains, leading to localized cross-contamination during flooding. Which of the following is the most proportional and effective administrative intervention to address the root cause of this health crisis?

Show answer & explanation

Answer: Immediately isolate and repair the compromised water pipelines to segregate them from stormwater drains, while supplying temporary clean drinking water via municipal tankers.

Answer

Isolate and repair the compromised water pipelines to segregate them from stormwater drains, while supplying temporary clean drinking water via municipal tankers.
The correct response directly addresses the root cause of the contamination—the physical intersection of drinking water pipes and drainage lines—while maintaining public welfare through temporary water delivery. It is a targeted, proportional, and technically sound administrative measure.

Step-by-Step Solution

1
Identify the primary qualitative cause of the problem.
The contamination is caused by physical cross-contamination between drinking water pipes and open stormwater drains in the sector, not by failure of central treatment.
Effective problem solving requires addressing the structural cause rather than secondary symptoms.
2
Evaluate proposed administrative interventions for proportionality and efficacy.
Repairing and re-routing the pipes addresses the physical vulnerability, while temporary tanker distribution ensures continuity of essential services.
Public administration decisions must balance root-cause rectification with immediate civic duty of care.

Key Concept

Qualitative Cause-and-Effect Problem Solving in Public Administration
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