Logical Venn Diagrams Practice Questions — State PSC Exam — Page 2

Question 21Question →

In a state revenue department consisting of $200$ inspectors, assignments are carried out across three specialized sectors: Taxation ( $T$ ), Customs ( $C$ ), and Financial Intelligence ( $F$ ). A departmental survey revealed that $105$ inspectors handle Taxation, $90$ handle Customs, and $85$ handle Financial Intelligence. Furthermore, $45$ inspectors handle both Taxation and Customs, $40$ handle both Customs and Financial Intelligence, and $35$ handle both Taxation and Financial Intelligence. If $15$ inspectors do not work in any of these three sectors, how many inspectors are engaged in EXACTLY TWO of the three sectors?

45

70

95

120

Show answer & explanation

Answer: 45

Answer

45 inspectors are engaged in exactly two of the three sectors.

The option stating 45 is correct. By using the principle of inclusion-exclusion, the total number of inspectors operating in at least one sector is

200 - 15 = 185

. Solving

185 = (105 + 90 + 85) - (45 + 40 + 35) + |T \cap C \cap F|

yields

|T \cap C \cap F| = 25

. To find those in exactly two sectors, we subtract 25 from each pairwise intersection (

45-25=20

,

40-25=15

,

35-25=10

), yielding a total of

20 + 15 + 10 = 45

.

Step-by-Step Solution

1

Determine the total number of inspectors working in at least one sector

|T \cup C \cup F| = 200 - 15 = 185

Inspectors who work in at least one sector equal the total inspectors minus those in none of the sectors.

2

Apply the Principle of Inclusion-Exclusion to find the number of inspectors in all three sectors

|T \cup C \cup F| = |T| + |C| + |F| - (|T \cap C| + |C \cap F| + |T \cap F|) + |T \cap C \cap F| \Rightarrow 185 = 105 + 90 + 85 - (45 + 40 + 35) + |T \cap C \cap F| \Rightarrow 185 = 280 - 120 + |T \cap C \cap F| \Rightarrow |T \cap C \cap F| = 25

The standard set identity relates the union of three sets with their individual sizes and overlapping regions.

3

Calculate the inspectors working in EXACTLY two sectors for each pairwise combination

Taxation and Customs only:

45 - 25 = 20

; Customs and Financial Intelligence only:

40 - 25 = 15

; Taxation and Financial Intelligence only:

35 - 25 = 10

Inspectors in all three sectors are included in every pairwise intersection and must be subtracted to isolate those in exactly two sectors.

4

Sum the counts of inspectors in exactly two sectors

20 + 15 + 10 = 45

Adding these disjoint sets gives the total count of inspectors working in exactly two sectors.

Key Concept

Principle of Inclusion-Exclusion for 3 Sets

Estimated Time:2m 0s

Question 22Question →

A smart grid feasibility study categorized $1000$ commercial buildings based on their adoption of three energy efficiency upgrades: Smart Lighting ( $L$ ), HVAC Optimization ( $H$ ), and Automated Shading ( $A$ ).

The survey revealed the following:
- $150$ buildings have not adopted any of these three upgrades.
- The number of buildings that have adopted exactly one type of upgrade is identical across all three categories.
- The number of buildings with both $L$ and $H$ , but not $A$ , is twice the number of buildings with all three upgrades.
- The number of buildings with both $H$ and $A$ , but not $L$ , is three times the number of buildings with all three upgrades.
- The number of buildings with both $L$ and $A$ , but not $H$ , is equal to the number of buildings with all three upgrades.
- The total number of buildings that have adopted HVAC Optimization ( $H$ ) is $430$ .

Based on this data, what is the total number of buildings that have adopted exactly two types of upgrades?

240

280

330

40

Show answer & explanation

Answer: 240

Answer

240 buildings

By defining the 'exactly one' regions as

x

and the 'all three' region as

y

, we can model the 'exactly two' regions as

2y

,

3y

, and

y

. This gives us two linear equations based on the given totals:

3x + 7y = 850

(from the union of

1000 - 150

) and

x + 6y = 430

(from the components of set

H

). Solving this system yields

y = 40

. The question asks for the total number of buildings with exactly two upgrades, which is

2y + 3y + y = 6y

. Multiplying

6 \times 40

gives

240

.

Step-by-Step Solution

1

Determine the total number of buildings with at least one upgrade.

Total Union

= 1000 - 150 = 850

buildings.

Subtracting the buildings with none of the upgrades from the total surveyed gives the union of the three sets.

2

Define algebraic variables for the distinct regions of the Venn diagram.

Let the number of buildings with exactly one upgrade in any category be

x

. Let the number of buildings with all three upgrades be

y

.

Parameterizing the unknown regions allows us to express the given relationships algebraically.

3

Express the regions representing exactly two upgrades in terms of

y

.

n(L \cap H \text{ only}) = 2y

,

n(H \cap A \text{ only}) = 3y

, and

n(L \cap A \text{ only}) = y

. The total for exactly two upgrades is

2y + 3y + y = 6y

.

Translating the comparative statements in the prompt into exact algebraic expressions.

4

Formulate a system of linear equations using the Total Union and Total H.

Total Union equation:

3x + (2y + 3y + y) + y = 850 \Rightarrow 3x + 7y = 850

. Total H equation:

x + 2y + 3y + y = 430 \Rightarrow x + 6y = 430

.

The union includes all distinct regions. The Total H set includes H only, all intersections involving H, and the central intersection.

5

Solve the system of equations to find

y

and calculate the final target (

6y

).

From the second equation,

x = 430 - 6y

. Substituting into the first:

3(430 - 6y) + 7y = 850 \Rightarrow 1290 - 18y + 7y = 850 \Rightarrow 11y = 440 \Rightarrow y = 40

. The total for exactly two is

6(40) = 240

.

This isolates

y

, giving the base unit needed to find the number of buildings with exactly two upgrades.

Key Concept

Algebraic Formulation of Multi-Set Logical Venn Diagrams

Question 23Question →

At an annual fitness convention, $220$ individuals registered for various sessions. The convention offered three main workshop categories: Yoga ( $Y$ ), Pilates ( $P$ ), and Aerobics ( $A$ ).

According to the attendance records:
- $100$ individuals participated in Yoga.
- $110$ individuals participated in Pilates.
- $90$ individuals participated in Aerobics.
- $50$ individuals participated in exactly two of these workshop categories.
- $20$ individuals did not participate in any of the three workshops.

Based on this information, how many individuals participated in all three workshop categories?

15

25

50

75

Show answer & explanation

Answer: 25

Answer

25

To find the number of people in all three categories, we analyze the structure of the sets. The sum of people in the individual categories is

100 + 110 + 90 = 300

. The actual number of active participants (the union) is the total population minus those who did not participate:

220 - 20 = 200

. When we subtract the union from the sum of the individual sets (

300 - 200 = 100

), this difference represents the overcounted overlaps. Specifically, people in exactly two categories (

E_2

) were counted twice, leaving one extra count. People in all three categories (

E_3

) were counted three times, leaving two extra counts. Therefore,

100 = E_2 + 2E_3

. Since we are given

E_2 = 50

, we substitute to get

100 = 50 + 2E_3

, which simplifies to

2E_3 = 50

, meaning

E_3 = 25

.

Step-by-Step Solution

1

Determine the number of individuals who participated in at least one workshop (the union of all three sets).

220 - 20 = 200

individuals.

The universal set contains

220

people, but

20

participated in none. The union (

N(Y \cup P \cup A)

) is the total minus the 'none' group.

2

Calculate the total sum of individuals across the three individual sets.

100 + 110 + 90 = 300

.

This sum counts individuals in exactly one category once, individuals in exactly two categories twice, and individuals in all three categories three times.

3

Set up an equation relating the sum of individual sets, the union, the 'exactly two' group (

E_2

), and the 'all three' group (

E_3

).

\text{Sum of Sets} - \text{Union} = E_2 + 2E_3

. Substituting values:

300 - 200 = 50 + 2E_3

.

Subtracting the union (which counts every participant exactly once) from the sum of sets leaves behind the extra counts: one extra count for everyone in exactly two, and two extra counts for everyone in all three.

4

Solve the algebraic equation for

E_3

.

100 = 50 + 2E_3 \implies 50 = 2E_3 \implies E_3 = 25

.

Isolating

E_3

provides the exact number of individuals who participated in all three workshops.

Key Concept

Logical Venn Diagrams and the application of categorical region sums (Inclusion-Exclusion Principle).

Question 24Question →

A municipal transport authority conducted a survey of $500$ daily commuters to analyze their usage of three public transport modes: Bus, Metro, and Train. The survey revealed the following data:
- $210$ commuters use the Bus.
- $190$ commuters use the Metro.
- $160$ commuters use the Train.
- $70$ commuters use exactly two of these transport modes.
- $20$ commuters use all three transport modes.

How many commuters in the surveyed group use none of these three transport modes?

30

50

70

140

Show answer & explanation

Answer: 50

Answer

The number of commuters using none of the three transport modes is 50.

The correct answer is found by determining the total number of commuters using at least one transport mode (the union). The sum of the individual sets (

210 + 190 + 160 = 560

) counts people using exactly one mode once, exactly two modes twice, and exactly three modes three times. Therefore,

560 = (\text{Exactly One}) + 2 \times (\text{Exactly Two}) + 3 \times (\text{Exactly Three})

. Substituting the given values gives

560 = (\text{Exactly One}) + 2(70) + 3(20)

, which simplifies to

560 = (\text{Exactly One}) + 140 + 60

. Thus, Exactly One equals

360

. The total union is Exactly One

+

Exactly Two

+

Exactly Three, which is

360 + 70 + 20 = 450

. Finally, the number of commuters using none of the modes is the total surveyed minus the union:

500 - 450 = 50

.

Step-by-Step Solution

1

Calculate the sum of commuters in the individual transport mode categories.

210 + 190 + 160 = 560

.

This establishes the gross total before adjusting for overlaps, where multi-mode commuters are counted multiple times.

2

Set up the inclusion-exclusion relationship for disjoint regions.

560 = n(\text{Exactly 1}) + 2 \times n(\text{Exactly 2}) + 3 \times n(\text{Exactly 3})

.

When summing the individual sets, commuters using exactly two modes are counted twice, and those using all three are counted three times.

3

Solve for the number of commuters using exactly one mode.

560 = n(\text{Exactly 1}) + 2(70) + 3(20) \Rightarrow 560 = n(\text{Exactly 1}) + 200 \Rightarrow n(\text{Exactly 1}) = 360

.

We need the 'exactly one' count to piece together the total union of commuters using at least one mode.

4

Calculate the total number of commuters using at least one mode (the union).

\text{Union} = 360 + 70 + 20 = 450

commuters.

The union is simply the sum of the disjoint regions: exactly one, exactly two, and exactly three.

5

Subtract the union from the total surveyed commuters to find those using none.

500 - 450 = 50

commuters.

Commuters outside the union represent those who use zero of the surveyed transport modes.

Key Concept

Solving 3-set Venn diagrams using disjoint region equations rather than standard intersection formulas.

Question 25Question →

A cybersecurity agency audited $250$ government web portals and found vulnerabilities across three categories: Data Breach ( $D$ ), Injection Attacks ( $I$ ), and Authentication Flaws ( $A$ ). The audit report stated:
- $110$ portals were vulnerable to Data Breach
- $130$ portals were vulnerable to Injection Attacks
- $100$ portals were vulnerable to Authentication Flaws
- $45$ portals had both $D$ and $I$ vulnerabilities
- $55$ portals had both $I$ and $A$ vulnerabilities
- $40$ portals had both $D$ and $A$ vulnerabilities
- $20$ portals had all three types of vulnerabilities

Based on this data, how many portals had exactly one type of vulnerability?

120

60

220

80

Show answer & explanation

Answer: $120$

Answer

There are

120

portals with exactly one type of vulnerability.

To find the number of portals with exactly one vulnerability, we isolate the non-overlapping portion of each set. For set

D

, this is

110 - 45 - 40 + 20 = 45

. For set

I

, it is

130 - 45 - 55 + 20 = 50

. For set

A

, it is

100 - 40 - 55 + 20 = 25

. Adding these exclusive regions gives

45 + 50 + 25 = 120

.

Step-by-Step Solution

1

Calculate the number of portals vulnerable ONLY to Data Breach (

D

).

|D_{only}| = |D| - |D \cap I| - |D \cap A| + |D \cap I \cap A| = 110 - 45 - 40 + 20 = 45

Subtracting the pairwise intersections removes the portals with multiple vulnerabilities, but it subtracts the triple intersection twice, so it must be added back once.

2

Calculate the number of portals vulnerable ONLY to Injection Attacks (

I

).

|I_{only}| = |I| - |D \cap I| - |I \cap A| + |D \cap I \cap A| = 130 - 45 - 55 + 20 = 50

Applying the same region isolation principle to set

I

.

3

Calculate the number of portals vulnerable ONLY to Authentication Flaws (

A

).

|A_{only}| = |A| - |D \cap A| - |I \cap A| + |D \cap I \cap A| = 100 - 40 - 55 + 20 = 25

Applying the same region isolation principle to set

A

.

4

Sum the three isolated regions to find the total number of portals with exactly one vulnerability.

45 + 50 + 25 = 120

These regions represent portals that have one and only one type of vulnerability.

Key Concept

Applying set theory and the inclusion-exclusion principle to isolate specific regions within a three-set Venn diagram.

Question 26Question →

A language institute conducted a survey among $400$ civil service aspirants to assess their proficiency in three foreign languages: French, German, and Spanish. The survey revealed the following data:
- $180$ aspirants are proficient in French.
- $150$ aspirants are proficient in German.
- $160$ aspirants are proficient in Spanish.
- $40$ aspirants are proficient in exactly French and German, but not Spanish.
- $30$ aspirants are proficient in exactly German and Spanish, but not French.
- $50$ aspirants are proficient in exactly French and Spanish, but not German.
- $70$ aspirants are not proficient in any of these three languages.

What is the number of aspirants who are proficient in all three languages?

Show answer & explanation

Answer: 20

Answer

20

By determining the union of the three sets (

400 - 70 = 330

) and applying the inclusion-exclusion principle while correctly distinguishing between 'exactly two' and the full intersection of two sets, we find that

20

aspirants are proficient in all three languages.

Step-by-Step Solution

1

Determine the number of aspirants proficient in at least one of the three languages.

n(F \cup G \cup S) = 400 - 70 = 330

The total population consists of those who speak at least one language and those who speak none.

2

Set up an equation using the Principle of Inclusion-Exclusion for three sets. Let

x

be the number of aspirants proficient in all three languages.

n(F \cap G) = 40 + x

,

n(G \cap S) = 30 + x

, and

n(F \cap S) = 50 + x

The total intersection of any two sets includes those in exactly those two sets plus those in all three sets.

3

Substitute all values into the union formula.

330 = 180 + 150 + 160 - (40 + x) - (30 + x) - (50 + x) + x

The formula

n(F \cup G \cup S) = n(F) + n(G) + n(S) - n(F \cap G) - n(G \cap S) - n(F \cap S) + n(F \cap G \cap S)

accounts for all overlapping regions.

4

Simplify the equation and solve for

x

.

330 = 490 - 120 - 2x \implies 330 = 370 - 2x \implies 2x = 40 \implies x = 20

Basic algebraic simplification yields the final value for the intersection of all three sets.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Alternative Method

Instead of using the union formula, use a region-based approach in a Venn diagram. Let the central 'all three' region be

x

. Calculate the 'only one' regions in terms of

x

: Only French =

180 - (40 + 50 + x) = 90 - x

. Only German =

150 - (40 + 30 + x) = 80 - x

. Only Spanish =

160 - (50 + 30 + x) = 80 - x

. The sum of all disjoint regions inside the union is

(90 - x) + (80 - x) + (80 - x) + 40 + 30 + 50 + x = 370 - 2x

. Since the union is

400 - 70 = 330

, we have

370 - 2x = 330

, which gives

x = 20

.

Estimated Time:2m 0s

Logical Venn Diagrams

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