Data Interpretation

69 questions

Question 1Question

A state tourism board recorded a total of 8,5008,500 visitors across three heritage sites—Site Alpha, Site Beta, and Site Gamma—during a cultural festival. The number of visitors to Site Alpha accounted for 40%40\% of the total visitors. Out of the visitors to Site Alpha, the ratio of domestic to international tourists was 3:23:2. Site Beta received 1,5001,500 fewer visitors than Site Alpha. At Site Beta, the number of international tourists was equal to the number of international tourists at Site Alpha. The remaining visitors went to Site Gamma, where 70%70\% were domestic tourists.

What is the difference between the total number of domestic tourists and the total number of international tourists across all three sites?

Show answer & explanation

Answer: 1,1401,140

Answer

The difference between the total number of domestic tourists and the total number of international tourists is 1,1401,140.
To find the final difference, the data must be extracted step-by-step for each site. Site Alpha has 3,4003,400 visitors (40%40\% of 8,5008,500), splitting into 2,0402,040 domestic and 1,3601,360 international based on the 3:23:2 ratio. Site Beta has 1,9001,900 visitors (3,4001,5003,400 - 1,500), with 1,3601,360 international tourists (matching Alpha) and the remaining 540540 being domestic. Site Gamma receives the leftover 3,2003,200 visitors, of which 70%70\% (2,2402,240) are domestic and 960960 are international. Summing the categories gives 4,8204,820 domestic and 3,6803,680 international tourists. The final difference is 4,8203,680=1,1404,820 - 3,680 = 1,140.

Step-by-Step Solution

1
Calculate the tourist breakdown for Site Alpha.
Total = 3,4003,400. Domestic = 2,0402,040. International = 1,3601,360.
Site Alpha has 40%40\% of 8,5008,500 total visitors, which is 3,4003,400. The ratio of domestic to international is 3:23:2. Domestic = 35×3,400=2,040\frac{3}{5} \times 3,400 = 2,040. International = 25×3,400=1,360\frac{2}{5} \times 3,400 = 1,360.
2
Calculate the tourist breakdown for Site Beta.
Total = 1,9001,900. International = 1,3601,360. Domestic = 540540.
Site Beta received 1,5001,500 fewer visitors than Site Alpha, so 3,4001,500=1,9003,400 - 1,500 = 1,900. The number of international tourists equals that of Site Alpha (1,3601,360). The remaining 1,9001,360=5401,900 - 1,360 = 540 are domestic tourists.
3
Calculate the tourist breakdown for Site Gamma.
Total = 3,2003,200. Domestic = 2,2402,240. International = 960960.
Site Gamma received the remaining visitors: 8,500(3,400+1,900)=3,2008,500 - (3,400 + 1,900) = 3,200. Since 70%70\% are domestic, Domestic = 0.70×3,200=2,2400.70 \times 3,200 = 2,240. International = 3,2002,240=9603,200 - 2,240 = 960.
4
Calculate overall totals and their difference.
Total Domestic = 4,8204,820. Total International = 3,6803,680. Difference = 1,1401,140.
Total Domestic = 2,040+540+2,240=4,8202,040 + 540 + 2,240 = 4,820. Total International = 1,360+1,360+960=3,6801,360 + 1,360 + 960 = 3,680. The difference is 4,8203,680=1,1404,820 - 3,680 = 1,140.

Key Concept

Data Extraction and Sequential Problem Solving in Caselets
Question 2Question

The table below presents the quarterly administrative software licensing costs (in ₹ Lakhs) for four governmental departments during the fiscal year 2025-26:

DepartmentQ1Q2Q3Q4
Dept A45506055
Dept B30354045
Dept C807590120
Dept D25303540

What is the absolute difference in Rupees between the average quarterly cost of Dept C and the combined cost of Dept B in the first half of the fiscal year (Q1 and Q2)?

Show answer & explanation

Answer: ₹ 26,25,000

Answer

₹ 26,25,000
The average quarterly cost for Dept C is calculated by adding its four quarters and dividing by 4: (80 + 75 + 90 + 120) / 4 = 365 / 4 = 91.25 Lakhs. The combined cost for Dept B in Q1 and Q2 is 30 + 35 = 65 Lakhs. The absolute difference is 91.25 - 65 = 26.25 Lakhs. Since the table scale is in Lakhs (where 1 Lakh = 1,00,000), the absolute difference in Rupees is 26.25 * 1,00,000 = ₹ 26,25,000.

Step-by-Step Solution

1
Calculate the average quarterly cost for Dept C.
91.25 Lakhs
The average is the total of all four quarters divided by 4: (80 + 75 + 90 + 120) / 4 = 365 / 4 = 91.25.
2
Calculate the combined cost for Dept B in the first half of the year (Q1 and Q2).
65 Lakhs
The combined cost is the direct sum of Q1 and Q2 values: 30 + 35 = 65.
3
Find the absolute numeric difference between the two calculated values.
26.25
91.25 - 65 = 26.25.
4
Convert the numeric difference into actual Rupees using the scale multiplier.
₹ 26,25,000
The table header specifies the values are in '₹ Lakhs'. 1 Lakh is equal to 1,00,000. Multiplying 26.25 by 1,00,000 gives 26,25,000.

Key Concept

Extracting targeted data points from a grid, calculating averages and sums, and applying chart scale multipliers accurately.
Question 3Question

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 4Question

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 5Question

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 6Question

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 7Question

In a administrative evaluation conducted for 250250 government officers, data was collected regarding the completion of three specialized training modules: Cyber Security (CC), Data Analytics (DD), and Public Procurement (PP).

- 130130 officers completed Cyber Security
- 110110 officers completed Data Analytics
- 100100 officers completed Public Procurement
- 5555 officers completed both Cyber Security and Data Analytics
- 4040 officers completed both Data Analytics and Public Procurement
- 4545 officers completed both Cyber Security and Public Procurement
- 2020 officers completed all three modules

Based on the information provided, how many officers completed exactly two training modules?

Show answer & explanation

Answer: 8080

Answer

The number of officers who completed exactly two training modules is 8080.
To find the number of officers who completed exactly two modules, we isolate the three disjoint regions that represent taking two modules but not the third. Subtraction of the three-module intersection (2020) from each pair overlap yields: (5520)+(4020)+(4520)=35+20+25=80(55 - 20) + (40 - 20) + (45 - 20) = 35 + 20 + 25 = 80.

Step-by-Step Solution

1
Identify the given set values and intersections
N(C)=130N(C) = 130, N(D)=110N(D) = 110, N(P)=100N(P) = 100, N(CD)=55N(C \cap D) = 55, N(DP)=40N(D \cap P) = 40, N(CP)=45N(C \cap P) = 45, N(CDP)=20N(C \cap D \cap P) = 20
Extracting all cardinalities provided in the stem.
2
Calculate the number of officers in each exclusive two-set intersection region
Cyber & Data only =5520=35= 55 - 20 = 35; Data & Procurement only =4020=20= 40 - 20 = 20; Cyber & Procurement only =4520=25= 45 - 20 = 25
The intersection N(XY)N(X \cap Y) contains people taking all three modules as well. Subtracting N(CDP)N(C \cap D \cap P) isolates those taking ONLY two modules.
3
Sum the exclusive two-module regions
35+20+25=8035 + 20 + 25 = 80
Adding the three disjoint regions corresponding to exactly two training modules.

Key Concept

3-Set Venn Diagram Region Isolation and Inclusion-Exclusion
Estimated Time:1m 30s
Question 8Question

A survey was conducted among 400400 administrative officers in a state secretariat regarding their operational oversight of three major public welfare projects: Project Alpha, Project Beta, and Project Gamma. The survey collected the following data:

- 190190 officers oversee Project Alpha.
- 180180 officers oversee Project Beta.
- 170170 officers oversee Project Gamma.
- 8585 officers oversee both Project Alpha and Project Beta.
- 7575 officers oversee both Project Beta and Project Gamma.
- 7070 officers oversee both Project Alpha and Project Gamma.
- 4040 officers do not oversee any of these three projects.

Based on the given information, how many administrative officers oversee exactly two projects?

Show answer & explanation

Answer: 8080

Answer

The number of administrative officers who oversee exactly two projects is 8080.
The value 8080 is correct because subtracting those in no projects from the total of 400400 gives 360360 officers in at least one project. Using the inclusion-exclusion principle formula yields 5050 officers in all three projects. Subtracting 5050 from each pairwise intersection gives 3535, 2525, and 2020 for the three exclusive two-project regions, summing to 8080.

Step-by-Step Solution

1
Calculate the total number of officers supervising at least one project.
ABC=40040=360|A \cup B \cup C| = 400 - 40 = 360
Officers who oversee at least one project are found by subtracting those overseeing none from the total surveyed officers.
2
Apply the Principle of Inclusion-Exclusion for three sets to find the triple intersection (ABC|A \cap B \cap C|).
360=190+180+170(85+75+70)+ABC    360=540230+ABC    ABC=50360 = 190 + 180 + 170 - (85 + 75 + 70) + |A \cap B \cap C| \implies 360 = 540 - 230 + |A \cap B \cap C| \implies |A \cap B \cap C| = 50
The standard set formula accounts for single set totals, pairwise intersections, and the three-set intersection.
3
Determine the number of officers in each exclusive two-project intersection.
Only Alpha & Beta = 8550=3585 - 50 = 35; Only Beta & Gamma = 7550=2575 - 50 = 25; Only Alpha & Gamma = 7050=2070 - 50 = 20
Subtracting the triple intersection from each pairwise intersection yields the count of officers managing exactly two projects.
4
Sum the exclusive two-project regions.
35+25+20=8035 + 25 + 20 = 80
Adding these three disjoint set regions gives the total number of officers handling exactly two projects.

Key Concept

Three-Set Inclusion-Exclusion Principle and Set Region Decomposition
Estimated Time:2m 30s
Question 9Question

The following table shows the number of employees across three departments in a company over three years, with some missing data indicated by dashes:

Department202120222023Total
HR4050140
IT8090270
Finance6070180
Total170200220590

What is the number of employees in the HR department in the year 2022?

Show answer & explanation

Answer: 50

Answer

The number of employees in the HR department in 2022 is 50.
The row total for HR is 140. Subtracting the known figures for 2021 (40) and 2023 (50) from 140 yields 50 employees for 2022.

Step-by-Step Solution

1
Examine the row corresponding to the HR department.
Known HR employees in 2021 = 40, 2023 = 50, and Total HR employees = 140.
The row total equals the sum of values across all three years.
2
Calculate the missing 2022 HR value.
140 - (40 + 50) = 50 employees.
Subtracting the sum of 2021 and 2023 values from the total gives the missing value for 2022.

Key Concept

Deriving missing data elements using row sums
Estimated Time:45s
Question 10Question

During an urban mobility audit conducted among 400400 daily commuters in a metropolitan city, data was recorded regarding their regular use of three transit modes: Metro Rail (MM), Electric Bus (EE), and Shared Bicycle (BB). The survey revealed that 190190 commuters use Metro Rail, 160160 use Electric Bus, and 120120 use Shared Bicycle. Furthermore, 6060 commuters use both Metro Rail and Electric Bus, 4040 use both Electric Bus and Shared Bicycle, and 5050 use both Metro Rail and Shared Bicycle. If 2020 commuters utilize all three modes of transport, how many commuters do not use any of these three transit modes?

Show answer & explanation

Answer: 60

Answer

The number of commuters who do not use any of the three transit modes is 60.
Using the 3-set inclusion-exclusion principle, the total number of commuters using at least one mode is given by MEB=(190+160+120)(60+40+50)+20=340|M \cup E \cup B| = (190 + 160 + 120) - (60 + 40 + 50) + 20 = 340. Subtracting this from the total sample of 400 yields 400340=60400 - 340 = 60 commuters who use none of the three modes.

Step-by-Step Solution

1
Identify individual set cardinalities and intersections from the problem statement.
Total universe N=400N = 400, M=190|M| = 190, E=160|E| = 160, B=120|B| = 120, ME=60|M \cap E| = 60, EB=40|E \cap B| = 40, MB=50|M \cap B| = 50, and MEB=20|M \cap E \cap B| = 20.
Establishing accurate set values is required before applying set formulas.
2
Apply the Principle of Inclusion-Exclusion formula for three overlapping sets to find the union MEB|M \cup E \cup B|.
MEB=190+160+120604050+20=340|M \cup E \cup B| = 190 + 160 + 120 - 60 - 40 - 50 + 20 = 340.
Pairwise intersections are double-counted when summing individual sets and must be subtracted, while the triple intersection is subtracted thrice and must be added back.
3
Compute the complement of the union to determine commuters using none of the modes.
Neither mode =NMEB=400340=60= N - |M \cup E \cup B| = 400 - 340 = 60.
The complement set represents all members of the universe outside the three-set union.

Key Concept

Principle of Inclusion-Exclusion for Three Sets
Question 11Question

The following table shows the number of students who appeared, passed, and failed in an annual examination across four different schools. Some data entries are missing and represented by a dash (--).

SchoolTotal StudentsPassed StudentsFailed Students
School A500350--
School B600--150
School C500400100
School D400280--

What is the total number of students who passed the examination across all four schools combined?

Show answer & explanation

Answer: 1480

Answer

1480
The total number of students who passed across all four schools is obtained by first calculating the missing entry for School B (600 Total150 Failed=450 Passed600 \text{ Total} - 150 \text{ Failed} = 450 \text{ Passed}) and then adding the passed student figures of all schools (350+450+400+280=1480350 + 450 + 400 + 280 = 1480).

Step-by-Step Solution

1
Find the missing number of passed students for School B
Passed students in School B = 600150=450600 - 150 = 450
Total students is the sum of passed and failed students.
2
Sum the passed students across all four schools
Total passed students = 350+450+400+280=1480350 + 450 + 400 + 280 = 1480
To find the combined total, add the passed student count of each individual school.

Key Concept

Missing Data Deduction and Summation
Estimated Time:45s
Question 12Question

The following table provides the annual expenditure (in ₹ Crores) of four administrative departments in a state over three fiscal years, with some values missing (indicated by dashes '—').

Department2021-222022-232023-24Total Expenditure
Agriculture4505201450
Education6807402000
Healthcare3504201250
Infrastructure8209701110
Total22002850

Based on the data provided, what is the difference (in ₹ Crores) between the average department expenditure in 2022-23 and the average department expenditure in 2021-22?

Show answer & explanation

Answer: 87.5

Answer

The difference between the average department expenditure in 2022-23 and 2021-22 is 87.5 ₹ Crores.
To find the difference between the average expenditures, we calculate the missing values first. For 2021-22, Education expenditure is 2000(680+740)=5802000 - (680 + 740) = 580 ₹ Crores. Total 2021-22 expenditure is 450+580+350+820=2200450 + 580 + 350 + 820 = 2200 ₹ Crores, making the average 2200/4=5502200 / 4 = 550 ₹ Crores. For 2022-23, Agriculture expenditure is 1450(450+520)=4801450 - (450 + 520) = 480 ₹ Crores. Total 2022-23 expenditure across all four departments is 480+680+420+970=2550480 + 680 + 420 + 970 = 2550 ₹ Crores, making the average 2550/4=637.52550 / 4 = 637.5 ₹ Crores. The difference between the averages is 637.5550=87.5637.5 - 550 = 87.5 ₹ Crores.

Step-by-Step Solution

1
Find missing 2021-22 Education expenditure
580 ₹ Crores
Subtracting known expenditures (680 and 740) from the Education row total (2000 gives 20001420=5802000 - 1420 = 580.
2
Compute 2021-22 average department expenditure
550 ₹ Crores
The sum for 2021-22 is 2200 ₹ Crores. Dividing by 4 departments gives 2200/4=5502200 / 4 = 550.
3
Find missing 2022-23 Agriculture expenditure
480 ₹ Crores
Subtracting known expenditures (450 and 520) from Agriculture row total (1450) gives 1450970=4801450 - 970 = 480.
4
Compute 2022-23 average department expenditure
637.5 ₹ Crores
Summing 2022-23 department values (480+680+420+970=2550480 + 680 + 420 + 970 = 2550) and dividing by 4 gives 2550/4=637.52550 / 4 = 637.5.
5
Subtract 2021-22 average from 2022-23 average
87.5 ₹ Crores
Calculating 637.5550=87.5637.5 - 550 = 87.5 ₹ Crores.

Key Concept

Deriving missing grid values using row totals and computing comparative averages.
Estimated Time:2m 30s
Question 13Question

Three numbers xx, yy, and zz are positive integers. What is the exact value of xx?

Statement I: The arithmetic mean of xx, yy, and zz is equal to x+2x + 2, where yzy \le z.
Statement II: The least common multiple (LCM) of yy and zz is 1212, and their greatest common divisor (GCD) is 22.

Which of the following options correctly describes the sufficiency of the statements?

Show answer & explanation

Answer: Neither Statement I nor Statement II is sufficient to answer the question, even when taken together.

Answer

Neither Statement I nor Statement II is sufficient to answer the question, even when taken together.
The correct answer states that neither statement is sufficient, even when combined. Statement I reduces to y+z=2x+6y + z = 2x + 6, which leaves xx dependent on the sum y+zy + z. Statement II yields two possible pairs for (y,z)(y, z) with GCD=2\text{GCD}=2 and LCM=12\text{LCM}=12, namely (2,12)(2, 12) with sum 1414, and (4,6)(4, 6) with sum 1010. Substituting y+z=14y + z = 14 gives x=4x = 4, while substituting y+z=10y + z = 10 gives x=2x = 2. Because both x=4x = 4 and x=2x = 2 are valid positive integer solutions, a single unique value for xx cannot be determined.

Step-by-Step Solution

1
Analyze Statement I individually.
The arithmetic mean formula gives x+y+z3=x+2    x+y+z=3x+6    y+z=2x+6\frac{x + y + z}{3} = x + 2 \implies x + y + z = 3x + 6 \implies y + z = 2x + 6.
Since yy and zz are unknown positive integers, xx can take multiple values (e.g., if y+z=8y+z = 8, x=1x=1; if y+z=10y+z = 10, x=2x=2). Thus, Statement I alone is insufficient.
2
Analyze Statement II individually.
Find pairs (y,z)(y, z) such that yzy \le z, GCD(y,z)=2\text{GCD}(y, z) = 2, and LCM(y,z)=12\text{LCM}(y, z) = 12.
Since y×z=GCD×LCM=24y \times z = \text{GCD} \times \text{LCM} = 24, the valid pairs of positive integers with GCD=2\text{GCD}=2 and LCM=12\text{LCM}=12 are (2,12)(2, 12) and (4,6)(4, 6). Statement II makes no reference to xx, so it is insufficient.
3
Evaluate Statements I and II together.
Test both valid pairs from Statement II in the relation 2x=y+z62x = y + z - 6 from Statement I.
Case 1: For (y,z)=(2,12)(y, z) = (2, 12), y+z=14    2x=146=8    x=4y + z = 14 \implies 2x = 14 - 6 = 8 \implies x = 4.
Case 2: For (y,z)=(4,6)(y, z) = (4, 6), y+z=10    2x=106=4    x=2y + z = 10 \implies 2x = 10 - 6 = 4 \implies x = 2.
Since two distinct values of xx (x=4x = 4 and x=2x = 2) satisfy all conditions, both statements together are NOT sufficient to determine a unique value for xx.

Key Concept

Data Sufficiency with Multiple Discrete Solutions in Number Theory
Question 14Question

A survey was conducted among 500500 rural households in a district regarding their adoption of three e-governance services: Digital Payments (PP), E-Health Cards (HH), and Online Land Records (LL). The survey revealed that 210210 households use Digital Payments, 190190 use E-Health Cards, and 220220 use Online Land Records. Additionally, 8080 households use both Digital Payments and E-Health Cards, 7070 use both E-Health Cards and Online Land Records, and 9090 use both Digital Payments and Online Land Records. If 4040 households use all three e-governance services, how many households use at least two of these services?

Show answer & explanation

Answer: 160160

Answer

The number of households using at least two e-governance services is 160160.
To find the number of households using at least two services, we sum the households using exactly two services and those using all three services. The households using only Digital Payments and E-Health Cards is 8040=4080 - 40 = 40. The households using only E-Health Cards and Online Land Records is 7040=3070 - 40 = 30. The households using only Digital Payments and Online Land Records is 9040=5090 - 40 = 50. Adding these to the 4040 households using all three services yields 40+30+50+40=16040 + 30 + 50 + 40 = 160.

Step-by-Step Solution

1
Identify given set parameters and triple intersection.
Total N=500N = 500, PH=80|P \cap H| = 80, HL=70|H \cap L| = 70, PL=90|P \cap L| = 90, and PHL=40|P \cap H \cap L| = 40.
Break down pairwise intersections into mutually exclusive regions.
2
Calculate households using exactly two services for each pair.
Only PP and H=8040=40H = 80 - 40 = 40; Only HH and L=7040=30L = 70 - 40 = 30; Only PP and L=9040=50L = 90 - 40 = 50.
Subtract the triple intersection count (4040) from each pairwise intersection to isolate households using exactly two services.
3
Sum households using exactly two services and households using all three services.
At least two services =40+30+50+40=160= 40 + 30 + 50 + 40 = 160.
'At least two' encompasses both 'exactly two' and 'all three' regions of the Venn diagram.

Key Concept

Principle of Inclusion-Exclusion for Three Sets and Mutually Exclusive Region Calculation
Estimated Time:1m 30s
Question 15Question

What is the value of the positive integer nn?

Statement I: n+8=15n + 8 = 15
Statement II: 3n=213n = 21

Which of the following options correctly describes the sufficiency of the statements to answer the question?

Show answer & explanation

Answer: Either Statement I alone or Statement II alone is sufficient to answer the question.

Answer

Either Statement I alone or Statement II alone is sufficient to answer the question.
Evaluating Statement I alone gives n=158=7n = 15 - 8 = 7, which uniquely determines nn. Evaluating Statement II alone gives n=213=7n = \frac{21}{3} = 7, which also uniquely determines nn. Since each statement independently provides a unique answer, either statement alone is sufficient.

Step-by-Step Solution

1
Evaluate Statement I independently
From n+8=15n + 8 = 15, subtracting 88 from both sides gives n=7n = 7.
This yields a unique value for nn, so Statement I alone is sufficient.
2
Evaluate Statement II independently
From 3n=213n = 21, dividing both sides by 33 gives n=7n = 7.
This also yields a unique value for nn, so Statement II alone is sufficient.
3
Determine overall sufficiency choice
Since both Statement I alone and Statement II alone yield the exact unique answer, either statement alone is sufficient.
This matches the standard option where either statement independently suffices.

Key Concept

Single-Variable Linear Data Sufficiency
Estimated Time:45s
Question 16Question

What is the average age of a class of 20 students?

Statement I: The sum of the ages of all 20 students in the class is 300 years.
Statement II: The age of the youngest student in the class is 12 years.

Show answer & explanation

Answer: Statement I alone is sufficient to answer the question, while Statement II alone is not sufficient.

Answer

Statement I alone is sufficient to answer the question, while Statement II alone is not sufficient.
The correct choice highlights that Statement I alone provides the total sum of ages for the 20 students. Since average equals total sum divided by the number of items (Average=30020=15\text{Average} = \frac{300}{20} = 15), Statement I alone is sufficient. Statement II only tells us the age of one specific student, which does not allow us to compute the average for all 20 students.

Step-by-Step Solution

1
Analyze the core formula needed to solve the question
The average age of a group is calculated using the formula: Average Age=Sum of ages of all studentsTotal number of students\text{Average Age} = \frac{\text{Sum of ages of all students}}{\text{Total number of students}}. The total number of students is given as 2020.
Understanding the required parameters helps evaluate statement sufficiency.
2
Evaluate Statement I individually
Statement I gives the sum of ages =300= 300 years. Average Age=30020=15\text{Average Age} = \frac{300}{20} = 15 years. Thus, Statement I alone yields a unique answer and is sufficient.
Determines if Statement I provides enough data on its own.
3
Evaluate Statement II individually
Statement II gives the age of the youngest student =12= 12 years. This tells us nothing about the remaining 1919 students, so we cannot determine the sum or the average age. Thus, Statement II alone is not sufficient.
Determines if Statement II provides enough data on its own.

Key Concept

Data Sufficiency - Evaluation of Individual Statements in Average Calculations
Question 17Question

A rectangular garden is surrounded on all four sides by a concrete walking path of uniform width ww meters. What is the area of the concrete walking path in square meters?

Statement I: The perimeter of the outer boundary of the walking path is 80 meters80\text{ meters}.
Statement II: The area of the rectangular garden is 300 square meters300\text{ square meters}, and its length is 5 meters5\text{ meters} greater than its width.

Which of the following options correctly describes the sufficiency of the statements to answer the question?

Show answer & explanation

Answer: Both Statement I and Statement II together are sufficient, but neither statement alone is sufficient.

Answer

Both Statement I and Statement II together are sufficient, but neither statement alone is sufficient.
The correct answer states that both statements together are sufficient, but neither alone is sufficient. Statement I leaves the path width and garden dimensions undetermined, while Statement II gives the garden dimensions without the path width. Combining them allows the path width and consequently the exact path area to be calculated uniquely.

Step-by-Step Solution

1
Analyze Statement I independently
Let the length and width of the garden be LL and WW meters, and the uniform path width be ww meters. The outer dimensions are (L+2w)(L + 2w) and (W+2w)(W + 2w). The outer perimeter is 2[(L+2w)+(W+2w)]=2(L+W+4w)=802[(L + 2w) + (W + 2w)] = 2(L + W + 4w) = 80, which simplifies to L+W+4w=40L + W + 4w = 40. The area of the path is (L+2w)(W+2w)LW=2w(L+W+2w)(L + 2w)(W + 2w) - LW = 2w(L + W + 2w). Since L+WL + W and ww can take multiple values satisfying L+W+4w=40L + W + 4w = 40, the area of the path cannot be uniquely determined. Hence, Statement I alone is NOT sufficient.
Evaluating whether Statement I alone fixes a single numerical value for the path area.
2
Analyze Statement II independently
Statement II states that LW=300L \cdot W = 300 and L=W+5L = W + 5. Substituting gives (W+5)W=300(W + 5)W = 300, so W2+5W300=0W^2 + 5W - 300 = 0. Factoring gives (W+20)(W15)=0(W + 20)(W - 15) = 0, yielding W=15 mW = 15\text{ m} and L=20 mL = 20\text{ m}. However, no information is given about the path width ww. Thus, the path area 2w(35+2w)2w(35 + 2w) cannot be computed without ww. Hence, Statement II alone is NOT sufficient.
Evaluating whether Statement II alone fixes a single numerical value for the path area.
3
Analyze Statements I and II combined
From Statement II, L=20 mL = 20\text{ m} and W=15 mW = 15\text{ m}, so L+W=35 mL + W = 35\text{ m}. Substitute this into the relation from Statement I: 35+4w=404w=5w=1.25 meters35 + 4w = 40 \Rightarrow 4w = 5 \Rightarrow w = 1.25\text{ meters}. The area of the path is 2(1.25)(35+2.5)=2.5×37.5=93.75 square meters2(1.25)(35 + 2.5) = 2.5 \times 37.5 = 93.75\text{ square meters}, which is a unique value. Therefore, both statements together are sufficient.
Determining if combining both statements resolves all unknown variables uniquely.

Key Concept

Evaluating individual statement sufficiency for multi-variable geometric systems prior to combination.
Question 18Question

In a performance audit of 200200 public health centers across a administrative division, adoption of three digital portals was evaluated: Tele-consultation (TT), Electronic Health Records (EE), and Medicine Inventory Management (MM). The audit revealed the following findings:

- 105105 centers use Tele-consultation (TT)
- 9090 centers use Electronic Health Records (EE)
- 8585 centers use Medicine Inventory Management (MM)
- 4040 centers use both TT and EE
- 3535 centers use both EE and MM
- 3030 centers use both TT and MM
- 1515 centers use all three platforms

How many public health centers utilize exactly two of the three digital health platforms?

Show answer & explanation

Answer: 60

Answer

60 public health centers utilize exactly two of the three digital health platforms.
The number of centers using 'exactly two' platforms consists of three distinct regions: (Centers in TT and EE only) + (Centers in EE and MM only) + (Centers in TT and MM only). Since each given pairwise intersection includes the 15 centers that use all three portals, we subtract 15 from each pairwise value: (4015)+(3515)+(3015)=25+20+15=60(40 - 15) + (35 - 15) + (30 - 15) = 25 + 20 + 15 = 60.

Step-by-Step Solution

1
Identify given set cardinalities and intersections
Total universe N=200N = 200; n(TE)=40n(T \cap E) = 40, n(EM)=35n(E \cap M) = 35, n(TM)=30n(T \cap M) = 30, and n(TEM)=15n(T \cap E \cap M) = 15.
The given pairwise intersections include centers that use all three platforms.
2
Calculate centers using strictly two platforms (excluding the triple overlap)
Centers using ONLY TT and E=4015=25E = 40 - 15 = 25.
Centers using ONLY EE and M=3515=20M = 35 - 15 = 20.
Centers using ONLY TT and M=3015=15M = 30 - 15 = 15.
To find 'exactly two', centers using all three must be removed from each pairwise intersection.
3
Sum the exclusive two-set regions
Total = 25+20+15=6025 + 20 + 15 = 60.
Adding these mutually exclusive regions yields the exact total of centers using exactly two portals.

Key Concept

3-Set Venn Diagram Region Isolation (Inclusion-Exclusion Principle)
Estimated Time:1m 30s
Question 19Question

The table below presents the quarterly renewable energy generation (in Megawatts, MW) of four geographical zones of a state in 2025:

ZoneQ1 (MW)Q2 (MW)Q3 (MW)Q4 (MW)
North Zone120150180150
South Zone200220250210
East Zone90110130110
West Zone160180200160

Based on the table, what is the average quarterly renewable energy generation (in MW) for the South Zone in 2025?

Show answer & explanation

Answer: 220

Answer

The average quarterly renewable energy generation for the South Zone in 2025 is 220 MW220\text{ MW}.
The correct average is obtained by summing the four quarterly generation figures of the South Zone (200+220+250+210=880 MW200 + 220 + 250 + 210 = 880\text{ MW}) and dividing by 44, giving 220 MW220\text{ MW}.

Step-by-Step Solution

1
Locate the row for the South Zone and extract data for all four quarters.
Q1 = 200 MW200\text{ MW}, Q2 = 220 MW220\text{ MW}, Q3 = 250 MW250\text{ MW}, Q4 = 210 MW210\text{ MW}.
Data needs to be aggregated across the full year.
2
Calculate total annual generation for the South Zone.
Total = 200+220+250+210=880 MW200 + 220 + 250 + 210 = 880\text{ MW}.
Summing values is required before finding the average.
3
Divide total generation by the number of quarters.
Average = 8804=220 MW\frac{880}{4} = 220\text{ MW}.
An average across four periods requires dividing the total sum by 4.

Key Concept

Average calculation from a tabular dataset
Question 20Question

In an agricultural extension survey conducted among 250250 farmers in an administrative block, awareness of three welfare schemes was evaluated: Crop Insurance (CC), Soil Health Card (SS), and Kisan Credit Card (KK). The survey revealed that 120120 farmers are aware of CC, 110110 are aware of SS, and 130130 are aware of KK. Additionally, 4545 farmers are aware of both CC and SS, 5050 are aware of both SS and KK, and 4040 are aware of both CC and KK. If 1515 farmers are aware of all three schemes, how many farmers are aware of exactly two of these schemes?

Show answer & explanation

Answer: 90

Answer

The total number of farmers aware of exactly two schemes is 9090.
To find the number of farmers aware of exactly two schemes, we isolate the three two-set intersection regions that exclude the three-set intersection. Subtracting the 1515 farmers aware of all three schemes from each pairwise intersection yields 3030 (for CC and SS only), 3535 (for SS and KK only), and 2525 (for CC and KK only). Summing these mutually exclusive regions gives 30+35+25=9030 + 35 + 25 = 90.

Step-by-Step Solution

1
Calculate the count of farmers aware of only Crop Insurance and Soil Health Card
4515=3045 - 15 = 30
The pairwise intersection includes farmers aware of all three schemes, so subtracting the triple intersection isolates those aware of only these two schemes.
2
Calculate the count of farmers aware of only Soil Health Card and Kisan Credit Card
5015=3550 - 15 = 35
Subtract the triple intersection count from the pairwise intersection count of SS and KK.
3
Calculate the count of farmers aware of only Crop Insurance and Kisan Credit Card
4015=2540 - 15 = 25
Subtract the triple intersection count from the pairwise intersection count of CC and KK.
4
Sum the three region counts for exactly two schemes
30+35+25=9030 + 35 + 25 = 90
The regions representing 'only C and S', 'only S and K', and 'only C and K' are mutually exclusive.

Key Concept

3-Set Venn Diagram Region Isolation
Page 1 / 4Next