Venn Diagrams and Set-Based Data

16 soru

Soru 1Soru

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?

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Cevabı ve açıklamayı göster

Cevap: 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.

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Two-set inclusion-exclusion principle and disjoint region classification in set theory
Tahmini Süre:50s
Soru 2Soru

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?

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Cevabı ve açıklamayı göster

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

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Three-Set Principle of Inclusion-Exclusion and Region Partitioning
Soru 3Soru

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?

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Cevabı ve açıklamayı göster

Cevap: 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.

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Three-Set Principle of Inclusion-Exclusion and Venn Diagram Region Cardinality
Soru 4Soru

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?

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Cevabı ve açıklamayı göster

Cevap: 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.

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Principle of Inclusion-Exclusion for Three Sets
Soru 5Soru

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?

Cevabı ve açıklamayı göster

Cevap: 8080

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

3-Set Venn Diagram Region Isolation and Inclusion-Exclusion
Tahmini Süre:1m 30s
Soru 6Soru

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?

Cevabı ve açıklamayı göster

Cevap: 8080

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Three-Set Inclusion-Exclusion Principle and Set Region Decomposition
Tahmini Süre:2m 30s
Soru 7Soru

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?

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Cevap: 60

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Principle of Inclusion-Exclusion for Three Sets
Soru 8Soru

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?

Cevabı ve açıklamayı göster

Cevap: 160160

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Principle of Inclusion-Exclusion for Three Sets and Mutually Exclusive Region Calculation
Tahmini Süre:1m 30s
Soru 9Soru

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?

Cevabı ve açıklamayı göster

Cevap: 60

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

3-Set Venn Diagram Region Isolation (Inclusion-Exclusion Principle)
Tahmini Süre:1m 30s
Soru 10Soru

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?

Cevabı ve açıklamayı göster

Cevap: 90

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

3-Set Venn Diagram Region Isolation
Soru 11Soru

A survey was conducted among 500500 civil service aspirants in a coaching institute regarding their daily newspaper reading habits among three publications: *The Hindu*, *The Indian Express*, and *Business Standard*.

The survey revealed the following data:
- The total number of aspirants reading *The Hindu*, *The Indian Express*, and *Business Standard* are 240240, 210210, and 180180 respectively.
- Exactly 2020 aspirants read all three newspapers.
- The ratio of the number of aspirants who read ONLY *The Hindu* and *The Indian Express* to those who read ONLY *The Indian Express* and *Business Standard* to those who read ONLY *The Hindu* and *Business Standard* is 3:2:43 : 2 : 4.
- The number of aspirants who read ONLY *Business Standard* is 7070.
- The number of aspirants who read ONLY *The Hindu* is equal to the number of aspirants who read ONLY *The Indian Express*.

Based on the data provided, how many aspirants in total do NOT read any of the three newspapers?

Cevabı ve açıklamayı göster

Cevap: 4545

Cevap

The total number of aspirants who do not read any of the three newspapers is 45.
By resolving the 7 disjoint regions of the Venn diagram using the given ratio 3k:2k:4k3k : 2k : 4k, we find k=15k = 15 from the total for Business Standard (180=70+4k+2k+20180 = 70 + 4k + 2k + 20). This yields exclusive intersection values of 4545, 3030, and 6060. Substituting these into the total for Indian Express (210210) yields Only(E)=115\text{Only}(E) = 115, which also equals Only(H)\text{Only}(H). Summing all seven disjoint regions gives 115+115+70+45+30+60+20=455115 + 115 + 70 + 45 + 30 + 60 + 20 = 455. Subtracting from the total population of 500500 gives 4545 aspirants who do not read any of the three newspapers.

Adım Adım Çözüm

1
Define disjoint regions of the Venn diagram using given variables.
Let HH, EE, and BB denote the sets of readers. Let Only(HE)=3k\text{Only}(H \cap E) = 3k, Only(EB)=2k\text{Only}(E \cap B) = 2k, and Only(HB)=4k\text{Only}(H \cap B) = 4k. The center intersection All Three=20\text{All Three} = 20.
Setting up explicit variables for mutually exclusive regions allows algebraic representation of total set sizes.
2
Calculate the ratio constant kk using the total for set BB (Business Standard).
B=Only(B)+Only(HB)+Only(EB)+All Three    180=70+4k+2k+20    90=6k    k=15|B| = \text{Only}(B) + \text{Only}(H \cap B) + \text{Only}(E \cap B) + \text{All Three} \implies 180 = 70 + 4k + 2k + 20 \implies 90 = 6k \implies k = 15.
All sub-regions comprising set BB are known except kk, making it possible to solve for kk directly.
3
Determine the exact counts of two-set exclusive intersections.
Only(HE)=3(15)=45\text{Only}(H \cap E) = 3(15) = 45, Only(EB)=2(15)=30\text{Only}(E \cap B) = 2(15) = 30, and Only(HB)=4(15)=60\text{Only}(H \cap B) = 4(15) = 60.
Multiplying the ratio multipliers by k=15k = 15 gives exact counts for each overlap.
4
Calculate Only(E)\text{Only}(E) and Only(H)\text{Only}(H).
E=Only(E)+45+30+20=210    Only(E)=115|E| = \text{Only}(E) + 45 + 30 + 20 = 210 \implies \text{Only}(E) = 115. Since Only(H)=Only(E)\text{Only}(H) = \text{Only}(E), Only(H)=115\text{Only}(H) = 115.
Subtracting known intersection counts of set EE from E=210|E| = 210 gives Only(E)\text{Only}(E), which equals Only(H)\text{Only}(H).
5
Calculate the total number of aspirants who read at least one newspaper and find the complement.
HEB=115+115+70+45+30+60+20=455|H \cup E \cup B| = 115 + 115 + 70 + 45 + 30 + 60 + 20 = 455. Aspirants reading none =500455=45= 500 - 455 = 45.
Summing all 7 disjoint regions gives the union. Subtracting the union from the universal set gives the number of aspirants reading no newspaper.

Anahtar Kavram

3-Set Principle of Inclusion-Exclusion and Disjoint Region Analysis
Soru 12Soru

In a regional administrative office of 8080 officers, 4545 officers completed a training course in Data Analytics, 4040 officers completed a course in Public Policy, and 1515 officers completed both training courses. How many officers completed neither of the two training courses?

Cevabı ve açıklamayı göster

Cevap: 1010

Cevap

10 officers completed neither of the two training courses.
To find the number of officers who completed neither course, calculate the union of the two sets using inclusion-exclusion: n(Data AnalyticsPublic Policy)=45+4015=70n(\text{Data Analytics} \cup \text{Public Policy}) = 45 + 40 - 15 = 70. The number of officers who completed neither course is the complement of this union relative to the total group: 8070=1080 - 70 = 10.

Adım Adım Çözüm

1
Identify the given set values and total universe
Total officers n(U)=80n(U) = 80, Data Analytics n(A)=45n(A) = 45, Public Policy n(B)=40n(B) = 40, Both n(AB)=15n(A \cap B) = 15.
Establish baseline data to apply the set inclusion-exclusion principle.
2
Calculate the number of officers who completed at least one training course using the principle of inclusion-exclusion
n(AB)=n(A)+n(B)n(AB)=45+4015=70n(A \cup B) = n(A) + n(B) - n(A \cap B) = 45 + 40 - 15 = 70.
Avoid double-counting the officers who completed both courses.
3
Calculate the number of officers who completed neither course
n(Neither)=n(U)n(AB)=8070=10n(\text{Neither}) = n(U) - n(A \cup B) = 80 - 70 = 10.
Subtract the union of the two sets from the total population to find the complement set.

Anahtar Kavram

Two-set Principle of Inclusion-Exclusion and Complementary Sets
Tahmini Süre:45s
Soru 13Soru

In a survey of 100100 residents in a locality, 6060 residents read Newspaper X, 5050 read Newspaper Y, and 2020 read both Newspaper X and Newspaper Y. How many residents read neither Newspaper X nor Newspaper Y?

Cevabı ve açıklamayı göster

Cevap: 1010

Cevap

The number of residents who read neither Newspaper X nor Newspaper Y is 10.
The total number of residents who read at least one newspaper is given by N(XY)=N(X)+N(Y)N(XY)=60+5020=90N(X \cup Y) = N(X) + N(Y) - N(X \cap Y) = 60 + 50 - 20 = 90. Thus, the number of residents who read neither newspaper is 10090=10100 - 90 = 10.

Adım Adım Çözüm

1
Calculate the number of residents who read at least one newspaper using the Inclusion-Exclusion principle.
XY=60+5020=90|X \cup Y| = 60 + 50 - 20 = 90
Simply adding total readers of X and Y double-counts the residents who read both newspapers.
2
Subtract the number of residents reading at least one newspaper from the total surveyed population.
Neither = 10090=10100 - 90 = 10
The universe of residents consists of those reading at least one newspaper and those reading neither.

Anahtar Kavram

Principle of Inclusion-Exclusion for two sets
Tahmini Süre:1m 0s
Soru 14Soru

In a survey of 120120 civil service aspirants, 6565 read Newspaper A, 5555 read Newspaper B, and 4545 read Newspaper C. Additionally, 2525 read both A and B, 2020 read both B and C, 1515 read both A and C, and 88 read all three newspapers. How many aspirants read exactly two of these newspapers?

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Cevap: 36

Cevap

36 aspirants read exactly two newspapers.
To find the number of aspirants who read exactly two newspapers, we must subtract the number of aspirants who read all three newspapers (8) from each of the two-newspaper intersection groups. The number of aspirants reading only A and B is 258=1725 - 8 = 17, only B and C is 208=1220 - 8 = 12, and only A and C is 158=715 - 8 = 7. Summing these exclusive regions gives 17+12+7=3617 + 12 + 7 = 36.

Adım Adım Çözüm

1
Identify the given set values and intersections
Total aspirants N=120N = 120; n(AB)=25n(A \cap B) = 25; n(BC)=20n(B \cap C) = 20; n(AC)=15n(A \cap C) = 15; n(ABC)=8n(A \cap B \cap C) = 8.
We need to extract overlapping region counts to isolate the 'exactly two' regions.
2
Calculate aspirants reading ONLY two newspapers for each pair
Only A and B = 258=1725 - 8 = 17; Only B and C = 208=1220 - 8 = 12; Only A and C = 158=715 - 8 = 7.
The given pairwise intersections n(AB)n(A \cap B) include those who read all three newspapers, so n(ABC)n(A \cap B \cap C) must be removed from each pair.
3
Sum the exclusive two-set regions
17+12+7=3617 + 12 + 7 = 36.
Adding these three mutually exclusive regions yields the total number of aspirants reading exactly two newspapers.

Anahtar Kavram

Venn Diagram set decomposition and region isolation
Soru 15Soru

In a survey of 500500 civil service aspirants preparing for State PSC examinations regarding their daily newspaper reading habits:
- 260260 aspirants read Newspaper A
- 220220 aspirants read Newspaper B
- 180180 aspirants read Newspaper C
- 9090 aspirants read both Newspaper A and Newspaper B
- 7070 aspirants read both Newspaper B and Newspaper C
- 8080 aspirants read both Newspaper A and Newspaper C
- 3030 aspirants read all three newspapers

How many aspirants read exactly one of these three newspapers?

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Cevap: 270270

Cevap

The number of aspirants who read exactly one newspaper is 270270.
The correct answer is derived by determining the number of readers exclusive to each single newspaper set. Subtracting all overlapping regions (both the exclusive two-paper readers and three-paper readers) from each newspaper's total gives 120120 for Newspaper A only, 9090 for Newspaper B only, and 6060 for Newspaper C only. Adding these exclusive values yields 270270.

Adım Adım Çözüm

1
Calculate the number of aspirants who read ONLY Newspaper A, ONLY Newspaper B, and ONLY Newspaper C by isolating two-set and three-set intersections.
Disjoint 2-set intersection counts (excluding all 3 newspapers):
- Reading A and B only: 9030=6090 - 30 = 60
- Reading B and C only: 7030=4070 - 30 = 40
- Reading A and C only: 8030=5080 - 30 = 50
The given two-newspaper intersection counts include the 3030 aspirants who read all three newspapers.
2
Subtract the exclusive two-set and three-set intersection counts from each total newspaper count to find single-newspaper readers.
- Only Newspaper A: 260(60+50+30)=120260 - (60 + 50 + 30) = 120
- Only Newspaper B: 220(60+40+30)=90220 - (60 + 40 + 30) = 90
- Only Newspaper C: 180(50+40+30)=60180 - (50 + 40 + 30) = 60
To find readers of 'only' one paper, all overlaps must be subtracted from the total set count.
3
Sum the exclusive counts for Newspaper A, Newspaper B, and Newspaper C.
Total reading exactly one newspaper = 120+90+60=270120 + 90 + 60 = 270
These three categories are mutually exclusive, so their sum gives the total count for 'exactly one'.

Anahtar Kavram

Principle of Inclusion-Exclusion for Three Sets
Soru 16Soru

A survey was conducted among 300300 State PSC aspirants regarding their preparation for three subjects: General Studies (GSGS), General Aptitude Test (CC), and Optional Subject (OO). The data collected is as follows:
- 180180 candidates prepare for GSGS
- 140140 candidates prepare for CC
- 120120 candidates prepare for OO
- 8080 candidates prepare for both GSGS and CC
- 5050 candidates prepare for both CC and OO
- 6060 candidates prepare for both GSGS and OO
- 3030 candidates prepare for all three subjects

Which of the following statements are correct? (Select all correct statements)

Geçerli olan tümünü seçin

Cevabı ve açıklamayı göster

Cevap: The number of candidates preparing for exactly two subjects is 100100.; The number of candidates preparing for at least two subjects is 130130.; The number of candidates who do not prepare for any of the three subjects is 2020.

Cevap

The correct statements are those asserting that the number of candidates preparing for exactly two subjects is 100, the number of candidates preparing for at least two subjects is 130, and the number of candidates preparing for none of the three subjects is 20.
The statements asserting that exactly two subjects equal 100, at least two subjects equal 130, and none of the subjects equal 20 are all mathematically accurate based on set region decomposition: exactly two subjects count is 50+20+30=10050 + 20 + 30 = 100; at least two subjects is 100+30=130100 + 30 = 130; and outside all sets is 300280=20300 - 280 = 20.

Adım Adım Çözüm

1
Identify the central region (all three subjects).
The number of candidates preparing for all three subjects n(GSCO)=30n(GS \cap C \cap O) = 30.
This value serves as the base subtraction term for all pairwise intersections.
2
Calculate the counts for candidates preparing for exactly two subjects.
GS and C only = 8030=5080 - 30 = 50; C and O only = 5030=2050 - 30 = 20; GS and O only = 6030=3060 - 30 = 30. Total exactly two subjects = 50+20+30=10050 + 20 + 30 = 100.
Subtracting the triple intersection from each dual intersection isolates regions with exactly two subjects.
3
Calculate candidates preparing for only one subject.
Only GS = 180(50+30+30)=70180 - (50 + 30 + 30) = 70; Only C = 140(50+30+20)=40140 - (50 + 30 + 20) = 40; Only O = 120(30+30+20)=40120 - (30 + 30 + 20) = 40.
Subtracting all overlapping regions from total set cardinalities yields single-subject counts.
4
Calculate the total union and the remainder outside all sets.
Total in at least one subject = 70+40+40+50+20+30+30=28070 + 40 + 40 + 50 + 20 + 30 + 30 = 280. Neither subject = 300280=20300 - 280 = 20.
Applying inclusion-exclusion principle determines the complete universe coverage.

Anahtar Kavram

Three-Set Principle of Inclusion-Exclusion
Tahmini Süre:2m 0s
Venn Diagrams and Set-Based Data Alıştırma Soruları — State PSC Exam | Examkin