Set Theory Concepts and Venn Diagrams Practice Questions — GRE General Test

Question 1Question →

A museum surveyed $150$ visitors regarding their attendance at two special exhibitions: a Fine Art exhibit and a Natural History exhibit. Among the visitors surveyed, $85$ attended the Fine Art exhibit, $70$ attended the Natural History exhibit, and $20$ attended neither exhibit. How many visitors attended both the Fine Art exhibit and the Natural History exhibit?

5

25

45

65

155

Show answer & explanation

Answer: $25$

Answer

The correct answer is

25

visitors.

Using the inclusion-exclusion formula

\text{Total} = |A| + |B| - |A \cap B| + \text{Neither}

, we substitute the given values:

150 = 85 + 70 - |A \cap B| + 20

. Simplifying gives

150 = 175 - |A \cap B|

, which yields

|A \cap B| = 25

. Thus,

25

visitors attended both exhibits.

Step-by-Step Solution

1

Identify the given set values and formula.

Total visitors

= 150

, Fine Art attendees

|A| = 85

, Natural History attendees

|H| = 70

, Neither

= 20

.

The principle of inclusion-exclusion for two sets states that

\text{Total} = |A| + |H| - |A \cap H| + \text{Neither}

.

2

Calculate the number of visitors who attended at least one exhibit.

|A \cup H| = 150 - 20 = 130

.

Subtracting those who attended neither exhibit from the total population yields the total number of unique visitors who attended at least one of the two exhibits.

3

Solve for the intersection

|A \cap H|

.

130 = 85 + 70 - |A \cap H| \implies 130 = 155 - |A \cap H| \implies |A \cap H| = 25

.

Subtracting the union

|A \cup H|

from the sum of the individual sets

|A| + |H|

eliminates the double-counted intersection.

Key Concept

Two-Set Principle of Inclusion-Exclusion

Estimated Time:1m 0s

Question 2Question →

A survey of $120$ university students evaluated course enrollment in Computer Science and Mathematics. Of the students surveyed, $70$ are enrolled in Computer Science, $65$ are enrolled in Mathematics, and $25$ are enrolled in neither course. How many students are enrolled in both Computer Science and Mathematics?

15

25

40

50

95

Show answer & explanation

Answer: 40

Answer

40 students are enrolled in both Computer Science and Mathematics.

The correct answer is 40. From the total population of 120 students, 25 take neither subject, meaning

120 - 25 = 95

students take at least one subject. According to the Principle of Inclusion-Exclusion,

|C \cup M| = |C| + |M| - |C \cap M|

. Substituting the known values yields

95 = 70 + 65 - |C \cap M|

, which simplifies to

95 = 135 - |C \cap M|

, giving

|C \cap M| = 40

.

Step-by-Step Solution

1

Calculate the total number of students enrolled in at least one of the two courses.

Total in at least one course =

120 - 25 = 95

.

Subtracting the students enrolled in neither course from the total surveyed gives the union of the two sets,

|C \cup M|

.

2

Apply the Principle of Inclusion-Exclusion for two sets.

|C \cup M| = |C| + |M| - |C \cap M|

, so

95 = 70 + 65 - |C \cap M|

.

Adding

|C|

and

|M|

double-counts the students taking both courses, so subtracting the union yields the intersection.

3

Solve for the intersection

|C \cap M|

.

|C \cap M| = 135 - 95 = 40

.

Direct arithmetic evaluation gives the required count of students enrolled in both subjects.

Key Concept

Principle of Inclusion-Exclusion for Two Sets

Estimated Time:1m 0s

Question 3Question →

In a department of $90$ employees, $55$ speak French, $45$ speak Spanish, and $10$ speak neither French nor Spanish. How many employees speak both French and Spanish?

Show answer & explanation

Answer: 20

Answer

20

Subtracting the 10 employees who speak neither language from the total of 90 leaves 80 employees who speak at least one language. By the principle of inclusion-exclusion, Total(At least one) = French + Spanish - Both. Substituting the known values gives 80 = 55 + 45 - Both, which simplifies to 80 = 100 - Both, so Both = 20.

Step-by-Step Solution

1

Find the number of employees who speak at least one of the languages

80 employees

Subtract the 10 employees who speak neither language from the total department size of 90.

2

Sum the total counts for each language group

100

Add the number of French speakers (55) to Spanish speakers (45).

3

Calculate the overlap (intersection) using inclusion-exclusion

20 employees

Subtract the count of employees speaking at least one language (80) from the sum of the individual language groups (100).

Key Concept

Principle of Inclusion-Exclusion for Two Sets

Estimated Time:45s

Question 4Question →

A survey of $200$ financial executives analyzed their personal investment portfolios across three asset classes: Equities ( $E$ ), Real Estate ( $R$ ), and Commodities ( $C$ ). Exactly $20$ executives hold no investments in any of these three asset classes. Among the remaining executives, $115$ hold Equities, $95$ hold Real Estate, and $80$ hold Commodities. Furthermore, $45$ executives hold both Equities and Real Estate, $35$ hold both Real Estate and Commodities, and $40$ hold both Equities and Commodities. How many of the surveyed executives hold investments in exactly two of these three asset classes?

80

90

100

110

120

Show answer & explanation

Answer: 90

Answer

90

To find the number of executives holding investments in exactly two asset classes, we first determine the total number of executives invested in at least one asset class:

200 - 20 = 180

. Applying the three-set inclusion-exclusion formula,

180 = 115 + 95 + 80 - (45 + 35 + 40) + |E \cap R \cap C|

, which simplifies to

180 = 170 + |E \cap R \cap C|

, giving

|E \cap R \cap C| = 10

. The number of executives holding exactly two asset classes is calculated by subtracting the triple intersection (10) from each pairwise overlap:

(45 - 10) + (35 - 10) + (40 - 10) = 35 + 25 + 30 = 90

.

Step-by-Step Solution

1

Calculate the total number of executives holding at least one asset class.

The total number in the union

|E \cup R \cup C| = 200 - 20 = 180

.

Subtracting executives holding no investments gives the cardinality of the union.

2

Use the Principle of Inclusion-Exclusion for three sets to find the triple intersection

|E \cap R \cap C|

.

|E \cup R \cup C| = |E| + |R| + |C| - (|E \cap R| + |R \cap C| + |E \cap C|) + |E \cap R \cap C|

, so

180 = 115 + 95 + 80 - (45 + 35 + 40) + |E \cap R \cap C|

, yielding

180 = 290 - 120 + |E \cap R \cap C| = 170 + |E \cap R \cap C|

, hence

|E \cap R \cap C| = 10

.

The inclusion-exclusion formula relates the union, individual set sizes, pairwise overlaps, and the triple overlap.

3

Determine the number of executives in each pairwise intersection that belong ONLY to those two sets.

Equities and Real Estate only:

45 - 10 = 35

; Real Estate and Commodities only:

35 - 10 = 25

; Equities and Commodities only:

40 - 10 = 30

.

Subtracting the triple intersection count from each pairwise intersection isolates those holding exactly two asset classes.

4

Sum the counts for executives holding exactly two asset classes.

35 + 25 + 30 = 90

.

Adding these disjoint sets yields the total number of executives in exactly two categories.

Key Concept

Three-Set Inclusion-Exclusion Principle and Venn Diagram Region Decomposition

Estimated Time:2m 0s

Question 5Question →

An agricultural research station evaluated $300$ soil plots for the presence of three specific mineral deficiencies: Nitrogen ( $N$ ), Phosphorus ( $P$ ), and Potassium ( $K$ ). The survey revealed the following data:
- Exactly $60$ plots exhibited none of the three deficiencies.
- $140$ plots exhibited Nitrogen deficiency ( $N$ ).
- $130$ plots exhibited Phosphorus deficiency ( $P$ ).
- $120$ plots exhibited Potassium deficiency ( $K$ ).
- Exactly $70$ plots exhibited exactly two of the three deficiencies.

Which of the following statements MUST be true? Select all such statements.

Select all that apply

Exactly

40

plots exhibited all three mineral deficiencies.

Exactly

130

plots exhibited exactly one mineral deficiency.

The number of plots exhibiting at least two mineral deficiencies is

110

.

Exactly

170

plots exhibited at most two mineral deficiencies.

The number of plots with Nitrogen deficiency only is

70

.

Show answer & explanation

Answer: Exactly $40$ plots exhibited all three mineral deficiencies.; Exactly $130$ plots exhibited exactly one mineral deficiency.; The number of plots exhibiting at least two mineral deficiencies is $110$ .

Answer

The statements confirming that exactly 40 plots exhibited all three deficiencies, exactly 130 plots exhibited exactly one deficiency, and 110 plots exhibited at least two deficiencies are all correct.

The system of set equations shows that

n_3 = 40

plots have all three deficiencies,

n_1 = 130

plots have exactly one deficiency, and

n_2 + n_3 = 70 + 40 = 110

plots have at least two deficiencies. Therefore, the statements asserting

40

plots for all three deficiencies,

130

plots for exactly one deficiency, and

110

plots for at least two deficiencies are all guaranteed to be true.

Step-by-Step Solution

1

Determine the total number of plots exhibiting at least one deficiency.

The total number of plots with at least one deficiency is

300 - 60 = 240

.

Plots with no deficiencies are excluded from the set union

|N \cup P \cup K|

.

2

Set up the inclusion-exclusion equations for set membership.

Let

n_1

be the number of plots with exactly one deficiency,

n_2 = 70

be the number of plots with exactly two deficiencies, and

n_3

be the number of plots with all three deficiencies.
Equation 1 (Total elements in union):

n_1 + n_2 + n_3 = 240 \implies n_1 + 70 + n_3 = 240 \implies n_1 + n_3 = 170

.
Equation 2 (Sum of individual set cardinalities):

|N| + |P| + |K| = n_1 + 2n_2 + 3n_3 \implies 140 + 130 + 120 = n_1 + 2(70) + 3n_3 \implies 390 = n_1 + 140 + 3n_3 \implies n_1 + 3n_3 = 250

.

Each element in an individual set sum is counted once for single-set membership, twice for double-set membership, and three times for triple-set membership.

3

Solve the system of linear equations for

n_1

and

n_3

.

Subtract Equation 1 from Equation 2:

(n_1 + 3n_3) - (n_1 + n_3) = 250 - 170 \implies 2n_3 = 80 \implies n_3 = 40

.
Substitute

n_3 = 40

into Equation 1:

n_1 + 40 = 170 \implies n_1 = 130

.

This yields

40

plots with all three deficiencies and

130

plots with exactly one deficiency.

4

Evaluate each option statement against the computed set cardinalities.

1) All three deficiencies (

n_3

) =

40

(True).
2) Exactly one deficiency (

n_1

) =

130

(True).
3) At least two deficiencies (

n_2 + n_3

) =

70 + 40 = 110

(True).
4) At most two deficiencies = Total -

n_3 = 300 - 40 = 260 \neq 170

(False).
5) Nitrogen deficiency only depends on how the

70

dual-deficiency plots are split between

N \cap P

,

N \cap K

, and

P \cap K

, which is not uniquely determined by the given data (False).

Only statements meeting the exact numerical bounds and logical guarantees must be true.

Key Concept

Three-Set Principle of Inclusion-Exclusion and Partitioning into Disjoint Regions

Question 6Question →

An agricultural research station evaluated a sample of $120$ organic farm plots for contamination by three types of crop fungi: *Fusarium* ( $F$ ), *Rhizoctonia* ( $R$ ), and *Sclerotinia* ( $S$ ). The audit revealed the following data:

- $55$ plots were contaminated with *Fusarium*.
- $50$ plots were contaminated with *Rhizoctonia*.
- $45$ plots were contaminated with *Sclerotinia*.
- $20$ plots were contaminated with both *Fusarium* and *Rhizoctonia*.
- $15$ plots were contaminated with both *Rhizoctonia* and *Sclerotinia*.
- $18$ plots were contaminated with both *Fusarium* and *Sclerotinia*.
- $8$ plots were contaminated with all three fungi.

Which of the following statements must be true? Select all such statements.

Select all that apply

Exactly 15 farm plots were free from all three types of crop fungi.

Exactly 29 farm plots were contaminated by exactly two of the three crop fungi.

Exactly 37 farm plots were contaminated by *Fusarium* only.

Exactly 53 farm plots were contaminated by at least two of the three crop fungi.

Exactly 50 percent of all plots contaminated by *Rhizoctonia* were also contaminated by *Fusarium*.

Show answer & explanation

Answer: Exactly 15 farm plots were free from all three types of crop fungi.; Exactly 29 farm plots were contaminated by exactly two of the three crop fungi.

Answer

The correct statements are that exactly 15 farm plots were free from all three types of crop fungi, and exactly 29 farm plots were contaminated by exactly two of the three crop fungi.

The statement regarding 15 farm plots being free from all three fungi is correct because the total number of plots contaminated by at least one fungus is calculated using inclusion-exclusion: 55 + 50 + 45 - 20 - 15 - 18 + 8 = 105, leaving 120 - 105 = 15 plots clean. The statement regarding 29 farm plots contaminated by exactly two fungi is correct because summing the mutually exclusive regions containing exactly two fungi yields (20 - 8) + (15 - 8) + (18 - 8) = 12 + 7 + 10 = 29.

Step-by-Step Solution

1

Calculate the non-overlapping regions for each intersection.

Determining exact two-set regions requires removing elements that belong to all three sets.

2

Calculate single-fungus-only regions.

|F only| = 55 - (12 + 10 + 8) = 25.
|R only| = 50 - (12 + 7 + 8) = 23.
|S only| = 45 - (10 + 7 + 8) = 20.

Subtract all double and triple overlaps from total individual set counts.

3

Determine union of all three sets and the neither region.

Total contaminated = 25 + 23 + 20 + 12 + 7 + 10 + 8 = 105.
Plots free from all fungi = 120 - 105 = 15.

Subtracting the total contaminated plots from the overall sample size gives the plots outside all sets.

4

Verify statement validity.

Plots free of all fungi = 15 (Valid).
Plots with exactly two fungi = 12 + 7 + 10 = 29 (Valid).

Match calculated regional counts against the presented options.

Key Concept

Three-Set Principle of Inclusion-Exclusion and Venn Diagram Partitioning

Question 7Question →

An urban planning department surveyed a total of $360$ commercial buildings regarding three sustainability certifications: LEED ( $L$ ), BREEAM ( $B$ ), and WELL ( $W$ ). Exactly $50$ of the surveyed buildings held none of the three certifications. The survey revealed the following data:

• $180$ buildings held the LEED certification.
• $140$ buildings held the BREEAM certification.
• $130$ buildings held the WELL certification.
• $45$ buildings held both the LEED and BREEAM certifications.
• $40$ buildings held both the BREEAM and WELL certifications.
• $25$ buildings held all three certifications.

How many of the surveyed buildings held EXACTLY TWO of the three certifications?

Show answer & explanation

Answer: 90

Answer

90

To find the number of buildings with exactly two certifications, we first determine the total number of buildings holding at least one certification:

360 - 50 = 310

. Applying the three-set inclusion-exclusion formula,

310 = 180 + 140 + 130 - (45 + 40 + |L \cap W|) + 25

, which simplifies to

310 = 390 - |L \cap W|

, giving

|L \cap W| = 80

. Next, we isolate the regions holding exactly two certifications by subtracting the 25 triple-certified buildings from each pairwise intersection: LEED & BREEAM only =

45 - 25 = 20

, BREEAM & WELL only =

40 - 25 = 15

, and LEED & WELL only =

80 - 25 = 55

. Summing these three exclusive regions gives

20 + 15 + 55 = 90

.

Step-by-Step Solution

1

Determine the total size of the union of all three sets

|L \cup B \cup W| = 360 - 50 = 310

Buildings holding at least one certification represent the entire surveyed population minus those holding no certifications.

2

Solve for the unknown overlap of LEED and WELL certifications using the inclusion-exclusion formula

|L \cap W| = 80

Using

|L \cup B \cup W| = |L| + |B| + |W| - |L \cap B| - |B \cap W| - |L \cap W| + |L \cap B \cap W|

, we have

310 = 180 + 140 + 130 - 45 - 40 - |L \cap W| + 25 = 390 - |L \cap W|

.

3

Calculate the number of buildings in each region corresponding to exactly two certifications

LEED and BREEAM only:

45 - 25 = 20

; BREEAM and WELL only:

40 - 25 = 15

; LEED and WELL only:

80 - 25 = 55

Each total pairwise intersection includes the 25 buildings that hold all three certifications, so subtracting 25 isolates those holding exclusively two certifications.

4

Sum the three isolated regions

20 + 15 + 55 = 90

The total number of buildings holding exactly two certifications is the sum of the three non-overlapping regions representing two certifications.

Key Concept

Three-Set Inclusion-Exclusion Principle and Partitioning Venn Diagrams

Estimated Time:2m 0s

Question 8Question →

A total of 120 software engineers at a firm were surveyed about their attendance at three professional development workshops: Machine Learning ( $M$ ), Cloud Computing ( $C$ ), and Cybersecurity ( $S$ ). The survey results revealed the following:

- 65 engineers attended Machine Learning
- 55 engineers attended Cloud Computing
- 48 engineers attended Cybersecurity
- 25 engineers attended both Machine Learning and Cloud Computing
- 18 engineers attended both Cloud Computing and Cybersecurity
- 20 engineers attended both Machine Learning and Cybersecurity
- 10 engineers attended all three workshops

How many of the surveyed software engineers attended exactly one of the three workshops?

42

48

72

105

115

Show answer & explanation

Answer: 72

Answer

72 engineers attended exactly one of the three workshops.

To find the number of engineers attending exactly one workshop, calculate the exclusive single-set counts for each category using a 3-set Venn diagram model. Subtracting the overlapping regions gives 30 for Machine Learning only, 22 for Cloud Computing only, and 20 for Cybersecurity only. Summing these exclusive regions yields

30 + 22 + 20 = 72

. Alternatively, using the algebraic formula

|\text{Exactly 1}| = (|M| + |C| + |S|) - 2(|M \cap C| + |C \cap S| + |M \cap S|) + 3|M \cap C \cap S| = (65 + 55 + 48) - 2(25 + 18 + 20) + 3(10) = 168 - 126 + 30 = 72

.

Step-by-Step Solution

1

Find the number of engineers in the regions corresponding to exactly two workshops.

Machine Learning and Cloud Computing only =

25 - 10 = 15

; Cloud Computing and Cybersecurity only =

18 - 10 = 8

; Machine Learning and Cybersecurity only =

20 - 10 = 10

.

The given pairwise intersection values include engineers who attended all three workshops.

2

Calculate the number of engineers who attended only one specific workshop.

Machine Learning only =

65 - (15 + 10 + 10) = 30

; Cloud Computing only =

55 - (15 + 8 + 10) = 22

; Cybersecurity only =

48 - (10 + 8 + 10) = 20

.

Subtract all overlapping regions (two-workshop-only and three-workshop) from each set's total.

3

Sum the single-workshop regions to find the total attending exactly one workshop.

30 + 22 + 20 = 72

.

The subsets representing single workshop attendance are disjoint.

Key Concept

Principle of Inclusion-Exclusion for Three Sets and Venn Diagram Region Partitioning

Estimated Time:2m 0s

Question 9Question →

A health club surveyed $100$ members regarding their use of cardio machines and weight equipment. Of the members surveyed, $60$ use cardio machines, $50$ use weight equipment, and $15$ use neither cardio machines nor weight equipment. Which of the following statements must be true? Indicate all such statements.

Select all that apply

Exactly

25

members use both cardio machines and weight equipment.

Exactly

35

members use cardio machines but do not use weight equipment.

Exactly

10

members use both cardio machines and weight equipment.

Exactly

110

members use at least one of the two types of equipment.

Exactly

85

members use at least one of the two types of equipment.

Show answer & explanation

Answer: Exactly $25$ members use both cardio machines and weight equipment.; Exactly $35$ members use cardio machines but do not use weight equipment.; Exactly $85$ members use at least one of the two types of equipment.

Answer

The true statements are those stating that exactly 25 members use both types of equipment, exactly 35 members use cardio machines only, and exactly 85 members use at least one type of equipment.

The total surveyed population of

100

consists of members using at least one type of equipment and members using neither (

15

). Thus,

100 - 15 = 85

members use at least one type of equipment. Using the inclusion-exclusion formula

\text{Union} = \text{Cardio} + \text{Weights} - \text{Intersection}

, we have

85 = 60 + 50 - \text{Intersection}

, giving an intersection of

25

members using both types of equipment. Subtracting

25

from the

60

total cardio users shows that

35

members use cardio machines exclusively.

Step-by-Step Solution

1

Calculate the total number of members using at least one type of equipment.

Total using at least one type

= 100 - 15 = 85

.

Members using at least one type of equipment represent the total surveyed population minus those who use neither.

2

Calculate the number of members using both cardio machines and weight equipment using the principle of inclusion-exclusion.

\text{Both} = \text{Cardio} + \text{Weights} - \text{At least one} = 60 + 50 - 85 = 25

.

Adding the individual set counts double-counts the overlapping region, so subtracting the total union gives the intersection size.

3

Calculate the number of members using cardio machines only.

\text{Cardio Only} = \text{Cardio} - \text{Both} = 60 - 25 = 35

.

Subtracting the intersection from the total cardio group yields the exclusive cardio users.

Key Concept

Principle of Inclusion-Exclusion for Two Sets

Estimated Time:1m 15s

Question 10Question →

A survey of $250$ registered voters was conducted to analyze their primary news sources: Television ( $T$ ), the Internet ( $I$ ), and Print newspapers ( $P$ ). The survey revealed the following results:

- $140$ voters get news from Television.
- $150$ voters get news from the Internet.
- $80$ voters get news from Print newspapers.
- $45$ voters get news from both Television and Print newspapers.
- $60$ voters get news from both the Internet and Print newspapers.
- $30$ voters get news from all three sources.
- $20$ voters do not get news from any of these three sources.

How many of the surveyed voters get news from Television and the Internet, but NOT from Print newspapers?

Show answer & explanation

Answer: 35

Answer

35 voters get news from Television and the Internet, but not from Print newspapers.

Using the principle of inclusion-exclusion for three sets, the total union size is 230 voters (250 total minus 20 who use none). Setting up the formula 230 = 140 + 150 + 80 - 45 - 60 - |T ∩ I| + 30 allows us to solve for |T ∩ I| = 65. To find those who use Television and Internet but NOT Print newspapers, we subtract the 30 voters who use all three sources from 65, resulting in 35 voters.

Step-by-Step Solution

1

Find the size of the union of all three sets

|T ∪ I ∪ P| = 250 - 20 = 230

Subtracting the 20 voters who use none of the three news sources from the total sample of 250 gives the total number of voters in at least one category.

2

Set up the Principle of Inclusion-Exclusion for three sets

230 = 140 + 150 + 80 - 45 - 60 - |T ∩ I| + 30

The formula sums individual set sizes, subtracts pairwise intersections, and adds back the triple intersection.

3

Solve for the total intersection of Television and Internet

|T ∩ I| = 65

Simplifying the equation gives 230 = 295 - |T ∩ I|, which yields |T ∩ I| = 65.

4

Exclude those who also read Print newspapers

|(T ∩ I) \ P| = 65 - 30 = 35

Subtracting the 30 voters who use all three sources leaves only those who use Television and Internet without Print newspapers.

Key Concept

Three-Set Principle of Inclusion-Exclusion

Question 11Question →

A market research firm surveyed $150$ venture capital investors regarding their investments in three emerging sectors: Artificial Intelligence ( $A$ ), Biotechnology ( $B$ ), and Clean Energy ( $C$ ). The survey revealed the following data:

- $15$ investors did not invest in any of the three sectors.
- $78$ investors invested in Artificial Intelligence.
- $65$ investors invested in Biotechnology.
- $62$ investors invested in Clean Energy.
- $10$ investors invested in all three sectors.

If $50$ investors invested in exactly two of the three sectors, how many investors invested in exactly one of the three sectors?

65

70

75

85

90

Show answer & explanation

Answer: 75

Answer

75 investors invested in exactly one of the three sectors.

The total number of investors who invested in at least one of the three sectors is

150 - 15 = 135

. In a 3-set Venn diagram, the total union is the sum of those who belong to exactly one sector (

E_1

), exactly two sectors (

E_2

), and all three sectors (

E_3

). We are given

E_2 = 50

and

E_3 = 10

. Setting up the region equation

E_1 + E_2 + E_3 = 135

gives

E_1 + 50 + 10 = 135

, which yields

E_1 = 75

.

Step-by-Step Solution

1

Calculate the total number of investors who invested in at least one sector (the union of sets A, B, and C).

|A \cup B \cup C| = 150 - 15 = 135

Subtracting the investors who invested in none of the sectors from the total population yields the size of the union.

2

Express the union as the sum of non-overlapping region counts.

|A \cup B \cup C| = E_1 + E_2 + E_3

, where

E_1

is the number of investors in exactly 1 sector,

E_2

is the number in exactly 2 sectors, and

E_3

is the number in all 3 sectors.

Any element in the union belongs to exactly one, exactly two, or all three sets.

3

Substitute the known values into the partition equation and solve for

E_1

.

135 = E_1 + 50 + 10 \implies E_1 = 135 - 60 = 75

We are given

E_2 = 50

and

E_3 = 10

, leaving

E_1

as the remaining unknown value.

Key Concept

Three-Set Inclusion-Exclusion and Region Partitioning

Estimated Time:2m 0s

Question 12Question →

A specialty coffee roastery surveyed $200$ customers to determine their preferences among three origin beans: Ethiopian ( $E$ ), Colombian ( $C$ ), and Guatemalan ( $G$ ). Every customer surveyed preferred at least one of the three origins. The survey revealed that $105$ customers preferred Ethiopian, $90$ preferred Colombian, and $85$ preferred Guatemalan. Additionally, $35$ customers preferred both Ethiopian and Colombian, $30$ preferred both Colombian and Guatemalan, and $40$ preferred both Ethiopian and Guatemalan. How many customers preferred all three origin beans?

15

20

25

35

50

Show answer & explanation

Answer: $25$

Answer

25 customers preferred all three origin beans.

The correct answer is obtained by setting up the three-set inclusion-exclusion formula

|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |B \cap C| + |A \cap C|) + |A \cap B \cap C|

. Substituting the given values gives

200 = 105 + 90 + 85 - (35 + 30 + 40) + x

. Solving

200 = 175 + x

yields

x = 25

.

Step-by-Step Solution

1

State the 3-set Inclusion-Exclusion Principle formula.

|E \cup C \cup G| = |E| + |C| + |G| - (|E \cap C| + |C \cap G| + |E \cap G|) + |E \cap C \cap G|

This fundamental formula relates the total union of three overlapping sets to their individual sizes, pairwise intersections, and three-set intersection.

2

Substitute the given values into the formula.

200 = 105 + 90 + 85 - (35 + 30 + 40) + x

, where

x = |E \cap C \cap G|

Since every customer preferred at least one origin, the union of all three sets equals the total number of customers surveyed (

200

).

3

Simplify the sum of individual sets and pairwise intersections.

Sum of individual sets =

105 + 90 + 85 = 280

. Sum of pairwise intersections =

35 + 30 + 40 = 105

.

Grouping terms makes evaluating the equation easier and reduces calculation errors.

4

Solve for the unknown three-set intersection

x

.

200 = 280 - 105 + x \implies 200 = 175 + x \implies x = 25

Subtracting

175

from

200

yields the number of customers who prefer all three origin beans.

Key Concept

Three-Set Principle of Inclusion-Exclusion

Question 13Question →

An aviation safety committee audited $240$ international flights for operational delays caused by Weather ( $W$ ), Mechanical issues ( $M$ ), and Air Traffic Control ( $T$ ). The audit revealed the following data:

- $112$ flights experienced Weather delays.
- $98$ flights experienced Mechanical delays.
- $104$ flights experienced Air Traffic Control delays.
- $28$ flights experienced both Weather and Mechanical delays.
- $34$ flights experienced both Mechanical and Air Traffic Control delays.
- $42$ flights experienced both Weather and Air Traffic Control delays.
- The number of flights that experienced none of these three types of delays was equal to twice the number of flights that experienced all three types of delays.

How many flights experienced exactly one of these three types of delays?

Show answer & explanation

Answer: 136

Answer

The number of flights that experienced exactly one of the three types of delays is 136.

The total number of audited flights is partitioned into those experiencing at least one delay and those experiencing none. By applying the Inclusion-Exclusion formula

|W \cup M \cup T| = 112 + 98 + 104 - (28 + 34 + 42) + x = 210 + x

, and setting the total audited flights to

240 = (210 + x) + 2x

, we solve for the triple intersection

x = 10

. Isolating the exclusive single-delay regions yields

52

flights for Weather only,

46

for Mechanical only, and

38

for ATC only, giving a total of

136

flights.

Step-by-Step Solution

1

Define the unknown intersection and external set quantities in terms of a single variable.

Let

x = |W \cap M \cap T|

represent the number of flights experiencing all three types of delays. Consequently, the number of flights experiencing none of the delays is

2x

.

Relating the unknown triple intersection to the 'neither' category allows all regions to be expressed in terms of

x

.

2

Apply the Principle of Inclusion-Exclusion for three overlapping sets to find the total number of flights with at least one delay.

|W \cup M \cup T| = |W| + |M| + |T| - (|W \cap M| + |M \cap T| + |W \cap T|) + |W \cap M \cap T| = 112 + 98 + 104 - (28 + 34 + 42) + x = 210 + x

.

Subtracting pairwise intersections corrects for double-counting, and adding back the triple intersection corrects for over-subtraction.

3

Formulate and solve the linear equation for the total sample population of 240 flights.

\text{Total Flights} = |W \cup M \cup T| + |\text{Neither}| \implies 240 = (210 + x) + 2x \implies 240 = 210 + 3x \implies 3x = 30 \implies x = 10

.

The sum of flights in the union of the sets and flights outside all three sets must equal the total number of flights audited.

4

Determine the number of flights in the two-set intersection regions excluding the triple intersection.

Weather and Mechanical only:

28 - 10 = 18

; Mechanical and ATC only:

34 - 10 = 24

; Weather and ATC only:

42 - 10 = 32

.

Each pairwise intersection given in the stem includes the triple intersection of 10 flights.

5

Compute the number of flights belonging exclusively to single categories and sum them.

Weather only:

112 - (18 + 32 + 10) = 52

; Mechanical only:

98 - (18 + 24 + 10) = 46

; ATC only:

104 - (32 + 24 + 10) = 38

. Sum of single categories:

52 + 46 + 38 = 136

.

Subtracting all overlapping components from each set total isolates the exclusive single-category elements.

Key Concept

Three-Set Inclusion-Exclusion Principle and Venn Diagram Region Partitioning

Estimated Time:2m 30s

Question 14Question →

A technology firm surveyed $100$ software developers regarding their proficiency in three programming languages: Python ( $P$ ), Java ( $J$ ), and C++ ( $C$ ). The survey revealed the following results:
- $55$ developers are proficient in Python.
- $50$ developers are proficient in Java.
- $35$ developers are proficient in C++.
- $25$ developers are proficient in both Python and Java.
- $20$ developers are proficient in both Java and C++.
- $15$ developers are proficient in both Python and C++.
- $10$ developers are proficient in none of these three languages.

Which of the following statements must be true? Select all such statements.

Select all that apply

Exactly

10

developers are proficient in all three programming languages.

Exactly

50

developers are proficient in at least two of these programming languages.

The number of developers proficient in Python only is

25

.

The total number of developers proficient in exactly one of these languages is

50

.

The percentage of Python-proficient developers who are proficient in Python only is

25\%

.

Show answer & explanation

Answer: Exactly $10$ developers are proficient in all three programming languages.; The number of developers proficient in Python only is $25$ .; The total number of developers proficient in exactly one of these languages is $50$ .

Answer

The correct statements are those asserting that exactly 10 developers know all three languages, that 25 developers know Python only, and that 50 developers know exactly one of these languages.

The statements confirming that 10 developers know all three languages, 25 know Python only, and 50 know exactly one language are correct based on standard 3-set Venn diagram calculations. Using Inclusion-Exclusion,

|P \cup J \cup C| = 90

, which gives

|P \cap J \cap C| = 10

. Subtracting overlapping regions yields

25

for Python only,

15

for Java only,

10

for C++ only, summing to

50

for exactly one language.

Step-by-Step Solution

1

Calculate the total number of developers proficient in at least one language.

N_{\text{at least one}} = 100 - 10 = 90

.

Subtracting the developers proficient in none of the languages from the total survey population yields the union

|P \cup J \cup C|

.

2

Apply the Principle of Inclusion-Exclusion for 3 sets to find the triple intersection

|P \cap J \cap C|

.

90 = 55 + 50 + 35 - (25 + 20 + 15) + |P \cap J \cap C| \implies 90 = 80 + |P \cap J \cap C| \implies |P \cap J \cap C| = 10

.

The formula

|P \cup J \cup C| = |P| + |J| + |C| - (|P \cap J| + |J \cap C| + |P \cap C|) + |P \cap J \cap C|

links all given set quantities.

3

Calculate the count of developers proficient in exactly two languages for each pair.

Python and Java only =

25 - 10 = 15

; Java and C++ only =

20 - 10 = 10

; Python and C++ only =

15 - 10 = 5

.

Subtracting the triple intersection count (

10

) from each pairwise intersection yields the exclusive two-set regions.

4

Calculate the single-language proficiency counts (exactly one language).

Python only =

55 - (15 + 5 + 10) = 25

; Java only =

50 - (15 + 10 + 10) = 15

; C++ only =

35 - (5 + 10 + 10) = 10

. Total exactly one =

25 + 15 + 10 = 50

.

Subtracting all overlapping regions containing each language from its total count yields the single-language region size.

5

Evaluate the statement options based on computed regional values.

Statements stating 10 all three, 25 Python only, and 50 exactly one language are true.

Comparing calculated values (

10

for all three,

25

for Python only,

50

for exactly one language) against each option validates the true choices.

Key Concept

Three-set inclusion-exclusion principle and Venn diagram region decomposition

Question 15Question →

A technology institute surveyed a cohort of $300$ software engineers regarding their proficiency in three programming paradigms: Functional ( $F$ ), Object-Oriented ( $O$ ), and Concurrent ( $C$ ). Every surveyed engineer is proficient in at least one of these three paradigms. The ratio of the total number of engineers proficient in $F$ , $O$ , and $C$ is $5 : 6 : 4$ , respectively. Furthermore, exactly $20\%$ of the engineers proficient in $F$ are proficient in all three paradigms. If exactly $54$ engineers are proficient in both $F$ and $O$ , $48$ are proficient in both $O$ and $C$ , and $30$ are proficient in both $F$ and $C$ , how many engineers in the cohort are proficient in exactly one programming paradigm?

96

141

222

273

78

Show answer & explanation

Answer: 222

Answer

222 engineers are proficient in exactly one programming paradigm.

Using the Principle of Inclusion-Exclusion for three sets,

|F \cup O \cup C| = |F| + |O| + |C| - (|F \cap O| + |O \cap C| + |F \cap C|) + |F \cap O \cap C|

. Substituting

|F|=5k

,

|O|=6k

,

|C|=4k

,

|F \cap O \cap C|=k

, and the total cohort

300

gives

300 = 16k - 132

, yielding

k = 27

. Consequently, the triple intersection is

27

. Decomposing into disjoint regions:

F

only

= 78

,

O

only

= 87

, and

C

only

= 57

. Summing these gives

78 + 87 + 57 = 222

.

Step-by-Step Solution

1

Set up algebraic representations for the set sizes using the given ratio.

Let

|F| = 5k

,

|O| = 6k

, and

|C| = 4k

for some positive constant

k

.

The total proficiencies follow the ratio

5:6:4

.

2

Express the triple intersection

|F \cap O \cap C|

in terms of

k

.

|F \cap O \cap C| = 0.20 \times |F| = 0.20 \times 5k = k

.

Exactly 20% of engineers proficient in

F

are proficient in all three paradigms.

3

Apply the Principle of Inclusion-Exclusion for three sets to solve for

k

.

|F \cup O \cup C| = |F| + |O| + |C| - (|F \cap O| + |O \cap C| + |F \cap C|) + |F \cap O \cap C| \implies 300 = 5k + 6k + 4k - (54 + 48 + 30) + k \implies 300 = 16k - 132 \implies 16k = 432 \implies k = 27

.

Every engineer is proficient in at least one paradigm, so

|F \cup O \cap C| = 300

.

4

Calculate the total size of each set and each exclusive intersection region.

|F| = 135

,

|O| = 162

,

|C| = 108

, and

|F \cap O \cap C| = 27

.
Exclusively

F \cap O = 54 - 27 = 27

.
Exclusively

O \cap C = 48 - 27 = 21

.
Exclusively

F \cap C = 30 - 27 = 3

.

Subtracting the triple intersection from pairwise intersections yields the two-set-only regions.

5

Determine the number of engineers proficient in exactly one paradigm.

Only

F = 135 - (27 + 3 + 27) = 78

.
Only

O = 162 - (27 + 21 + 27) = 87

.
Only

C = 108 - (3 + 21 + 27) = 57

.
Total exactly one =

78 + 87 + 57 = 222

.

Subtracting all overlapping regions from each total set size gives the single-category populations.

Key Concept

Three-set Principle of Inclusion-Exclusion and Venn Diagram region decomposition.

Question 16Question →

A survey of $300$ urban commuters evaluated their usage of three transit services: the Bus ( $B$ ), the Commuter Rail ( $R$ ), and the Express Ferry ( $F$ ). The survey revealed the following information:

- $160$ commuters use the Bus.
- $140$ commuters use the Commuter Rail.
- $110$ commuters use the Express Ferry.
- $60$ commuters use both the Bus and the Commuter Rail.
- $45$ commuters use both the Commuter Rail and the Express Ferry.
- $50$ commuters use both the Bus and the Express Ferry.
- $25$ commuters use all three transit services.

How many of the surveyed commuters use none of these three transit services?

20

45

70

280

145

Show answer & explanation

Answer: 20

Answer

20 commuters use none of the three transit services.

Using the inclusion-exclusion principle for three overlapping sets, the total number of commuters using at least one of the transit services is

160 + 140 + 110 - (60 + 45 + 50) + 25 = 280

. Subtracting this value from the total surveyed group of

300

commuters gives

300 - 280 = 20

commuters who use none of the three services.

Step-by-Step Solution

1

Apply the Principle of Inclusion-Exclusion for three sets to find the total number of commuters who use at least one transit service,

|B \cup R \cup F|

.

|B \cup R \cup F| = |B| + |R| + |F| - (|B \cap R| + |R \cap F| + |F \cap B|) + |B \cap R \cap F|

Simply adding set sizes double-counts elements in pairwise intersections and triple-counts elements in all three sets.

2

Substitute the given numerical values into the inclusion-exclusion formula.

|B \cup R \cup F| = 160 + 140 + 110 - (60 + 45 + 50) + 25 = 410 - 155 + 25 = 280

Combining the sums and differences yields the exact count of commuters using at least one mode of transit.

3

Subtract the number of commuters using at least one service from the total surveyed population to find those using none.

\text{None} = 300 - 280 = 20

The universe of surveyed commuters consists of those using at least one service plus those using none.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Estimated Time:1m 30s

Question 17Question →

An environmental auditing agency surveyed $250$ manufacturing plants regarding their compliance with three environmental standards: Air Quality ( $A$ ), Water Discharge ( $W$ ), and Waste Management ( $M$ ). The survey yielded the following data:

- $130$ plants meet Air Quality standards ( $A$ ).
- $140$ plants meet Water Discharge standards ( $W$ ).
- $120$ plants meet Waste Management standards ( $M$ ).
- $40$ plants meet all three standards.
- $20$ plants meet none of the three standards.
- The number of plants meeting both Air Quality and Water Discharge standards is equal to the number of plants meeting both Water Discharge and Waste Management standards.
- The number of plants meeting both Air Quality and Waste Management standards is $10$ fewer than the number meeting both Air Quality and Water Discharge standards.

How many of the surveyed plants meet exactly one of the three environmental standards?

Show answer & explanation

Answer: 110

Answer

110

By setting up the 3-set inclusion-exclusion equation, the unknown pairwise intersections are found to be 70, 70, and 60. Subtracting the 40 plants that meet all three standards gives the exclusive double-overlap regions (30, 30, and 20). Subtracting these along with the central intersection from each single set yields 40 plants meeting only Air Quality, 40 meeting only Water Discharge, and 30 meeting only Waste Management, totaling 110 plants.

Step-by-Step Solution

1

Find total number of plants meeting at least one standard

|A ∪ W ∪ M| = 250 - 20 = 230

Subtracting plants that meet no standards from the total surveyed gives the union of all three sets.

2

Set up algebraic expressions for pairwise intersections

|A ∩ M| = k, |A ∩ W| = k + 10, |W ∩ M| = k + 10

Define the smallest pairwise intersection as k and express the other two based on the given relationships.

3

Apply the Principle of Inclusion-Exclusion (PIE) for three sets to solve for k

230 = 130 + 140 + 120 - (k + 10 + k + 10 + k) + 40 => k = 60

Substitute set sizes and the triple intersection into the 3-set inclusion-exclusion formula.

4

Calculate the number of plants in each exclusive region

Only (A ∩ W) = 30, Only (W ∩ M) = 30, Only (A ∩ M) = 20

Subtract the triple intersection (40) from each pairwise intersection.

5

Calculate plants meeting exactly one standard and sum them

Only A = 40, Only W = 40, Only M = 30; Total = 40 + 40 + 30 = 110

Subtract all double-overlap and triple-overlap regions from each individual set total.

Key Concept

Three-Set Principle of Inclusion-Exclusion and Venn Diagram Region Decomposition

Question 18Question →

A university surveyed a cohort of $150$ freshmen regarding their membership in three student organizations: the Art Club ( $A$ ), the Music Society ( $M$ ), and the Theater Guild ( $T$ ). The survey revealed the following data:

- $68$ students belong to the Art Club.
- $62$ students belong to the Music Society.
- $54$ students belong to the Theater Guild.
- $22$ students belong to both the Art Club and the Music Society.
- $18$ students belong to both the Music Society and the Theater Guild.
- $15$ students belong to both the Art Club and the Theater Guild.
- $8$ students belong to all three organizations.

How many of the surveyed students belong to exactly one of these three organizations?

Show answer & explanation

Answer: 98

Answer

98 students belong to exactly one of the three organizations.

To find the number of students belonging to exactly one organization, analyze the regions of a 3-set Venn diagram starting from the innermost region (all three clubs =

8

). Subtracting

8

from each pairwise intersection gives the students in exactly two clubs: Art & Music only (

14

), Music & Theater only (

10

), and Art & Theater only (

7

). Next, subtract the overlapping regions from each club total: Art only is

68 - (14 + 7 + 8) = 39

; Music only is

62 - (14 + 10 + 8) = 30

; Theater only is

54 - (7 + 10 + 8) = 29

. Summing these single-club regions gives

39 + 30 + 29 = 98

.

Step-by-Step Solution

1

Find the number of students belonging strictly to each pair of organizations (two-set intersections only).

Art and Music only =

14

; Music and Theater only =

10

; Art and Theater only =

7

.

The given pairwise totals include students who belong to all three organizations (

8

), so subtracting

8

isolates those in exactly two groups.

2

Determine the number of students belonging to each individual organization exclusively.

Art only =

39

; Music only =

30

; Theater only =

29

.

Subtract all shared membership regions (both two-group only and three-group) from each total organization membership.

3

Add the counts of students belonging to exactly one group.

39 + 30 + 29 = 98

.

The question requests the sum of all students in the non-overlapping single-set regions.

Key Concept

Three-Set Venn Diagram Region Partitioning

Question 19Question →

At a technology conference attended by $80$ software engineers, $50$ engineers write code in Python, $40$ write code in Java, and $15$ write code in neither Python nor Java. How many of the software engineers write code in both Python and Java?

10

15

25

30

35

Show answer & explanation

Answer: 25

Answer

The number of software engineers who write code in both Python and Java is 25.

Out of

80

engineers,

15

write neither language, which means

80 - 15 = 65

engineers write Python, Java, or both. Using the inclusion-exclusion principle,

|P \cup J| = |P| + |J| - |P \cap J|

, substituting the values gives

65 = 50 + 40 - |P \cap J|

. Solving for the overlap gives

|P \cap J| = 90 - 65 = 25

. Thus, 25 engineers write code in both languages.

Step-by-Step Solution

1

Find the number of engineers who write code in at least one of the two languages (Python or Java).

|P \cup J| = 80 - 15 = 65

Subtracting the engineers who write neither language from the total gives the union of the two sets.

2

Apply the Principle of Inclusion-Exclusion formula for two sets.

|P \cup J| = |P| + |J| - |P \cap J|

The total number of engineers in the union is equal to the sum of the individual sets minus their intersection.

3

Substitute the known values into the equation to solve for the intersection

|P \cap J|

.

65 = 50 + 40 - |P \cap J| \implies 65 = 90 - |P \cap J| \implies |P \cap J| = 25

Solving the linear equation yields the number of engineers writing both languages.

Key Concept

Principle of Inclusion-Exclusion for Two Sets

Estimated Time:1m 0s

Question 20Question →

An environmental protection agency audited $240$ coastal wetland sites to evaluate contamination by three specific pollutants: Microplastics ( $M$ ), Heavy metals ( $H$ ), and Agricultural runoff ( $A$ ). The audit revealed the following findings:

- Exactly $30$ of the audited wetlands showed no contamination from any of the three pollutants.
- The total number of wetlands containing Microplastics, Heavy metals, and Agricultural runoff were $130$ , $110$ , and $100$ , respectively.
- The number of wetlands containing Microplastics and Heavy metals but NOT Agricultural runoff was $25$ .
- The number of wetlands containing Heavy metals and Agricultural runoff but NOT Microplastics was $35$ .
- The number of wetlands containing Microplastics and Agricultural runoff but NOT Heavy metals was $20$ .

Based on the audit data, which of the following statements must be true? Select all such statements.

Select all that apply

Exactly

25

wetlands contain all three pollutants.

The number of wetlands containing only Microplastics is

60

.

The total number of wetlands containing exactly one pollutant is

85

.

The percentage of all surveyed wetlands that contain at least two pollutants is

50\%

.

The ratio of wetlands containing only Heavy metals to wetlands containing only Agricultural runoff is

5

to

4

.

Show answer & explanation

Answer: Exactly $25$ wetlands contain all three pollutants.; The number of wetlands containing only Microplastics is $60$ .; The ratio of wetlands containing only Heavy metals to wetlands containing only Agricultural runoff is $5$ to $4$ .

Answer

The statements confirming that exactly 25 wetlands contain all three pollutants, that 60 wetlands contain only Microplastics, and that the ratio of wetlands containing only Heavy metals to only Agricultural runoff is 5 to 4 are all correct.

The correct options are those stating that 25 wetlands contain all three pollutants, that 60 wetlands contain only Microplastics, and that the ratio of Heavy metals only to Agricultural runoff only is 5 to 4. Each of these follows directly from setting up the inclusion-exclusion equation for three sets and determining all eight mutually exclusive regions of the Venn diagram.

Step-by-Step Solution

1

Determine the number of wetlands containing at least one pollutant.

Total in union

|M \cup H \cup A| = 240 - 30 = 210

.

The total audited set is 240, and 30 sites have no pollutants.

2

Apply the Principle of Inclusion-Exclusion for three sets.

|M \cup H \cup A| = |M| + |H| + |A| - (|M \cap H| + |H \cap A| + |A \cap M|) + |M \cap H \cap A|

, which simplifies to

210 = 130 + 110 + 100 - S_2 + x \implies S_2 - x = 130

, where

S_2

is the sum of pairwise intersections and

x

is the triple intersection.

This relates the total union to the individual set sizes and intersection regions.

3

Express

S_2

in terms of the given 'exactly two' regions and solve for

x

.

The number of wetlands with exactly two pollutants is

25 + 35 + 20 = 80

. Since

S_2 - 3x = 80

, substituting

S_2 = 130 + x

gives

(130 + x) - 3x = 80 \implies 2x = 50 \implies x = 25

.

The sum of pairwise overlaps counts the triple intersection three times, so subtracting

3x

yields the 'exactly two' region.

4

Calculate single-category region sizes.

Microplastics only =

130 - (25 + 20 + 25) = 60

; Heavy metals only =

110 - (25 + 35 + 25) = 25

; Agricultural runoff only =

100 - (20 + 35 + 25) = 20

. Total single-category =

60 + 25 + 20 = 105

.

Subtracting all dual and triple overlap counts from each set total isolates the exclusive membership.

5

Verify each choice statement against the calculated region counts.

Triple intersection is 25 (True). Microplastics only is 60 (True). Exactly one pollutant total is 105, not 85 (False). Percentage of total sample with at least two pollutants is

105 / 240 = 43.75\%

, not

50\%

(False). Ratio of Heavy metals only to Agricultural runoff only is

25:20 = 5:4

(True).

Validates exact values against option assertions.

Key Concept

Three-set Principle of Inclusion-Exclusion and Venn diagram region decomposition

Set Theory Concepts and Venn Diagrams

Answer

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