Sets, Set Operations, and Venn Diagrams

21 questions

Question 1Question

Given the universal set U={xZ:1x20}U = \{x \in \mathbb{Z} : 1 \le x \le 20\}. Let P={xU:x is a prime number}P = \{x \in U : x \text{ is a prime number}\}, Q={xU:x is an odd integer}Q = \{x \in U : x \text{ is an odd integer}\}, and R={xU:x is a multiple of 3}R = \{x \in U : x \text{ is a multiple of } 3\}. What is the value of n((PQ)R)n((P \cup Q) \cap R')?

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Answer: 8

Answer

8
The correct value is 8. The union PQP \cup Q yields {1,2,3,5,7,9,11,13,15,17,19}\{1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19\}. Intersecting this set with RR' means removing any element that is a multiple of 3. The multiples of 3 in PQP \cup Q are 3, 9, and 15. Removing these 3 elements from the 11 elements of PQP \cup Q leaves exactly 8 elements.

Step-by-Step Solution

1
Identify the elements of sets P, Q, and R within the universal set U.
U={1,2,3,,20}U = \{1, 2, 3, \dots, 20\}, P={2,3,5,7,11,13,17,19}P = \{2, 3, 5, 7, 11, 13, 17, 19\}, Q={1,3,5,7,9,11,13,15,17,19}Q = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}, and R={3,6,9,12,15,18}R = \{3, 6, 9, 12, 15, 18\}.
Listing elements helps accurately compute set unions and intersections.
2
Find the union of sets P and Q, denoted as P ∪ Q.
PQ={1,2,3,5,7,9,11,13,15,17,19}P \cup Q = \{1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19\}, so n(PQ)=11n(P \cup Q) = 11.
Combining elements of both P and Q without repetition gives their union.
3
Find the complement of set R with respect to U, denoted as R'.
R={1,2,4,5,7,8,10,11,13,14,16,17,19,20}R' = \{1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20\}.
The complement set R' contains all elements in U that are not multiples of 3.
4
Determine the intersection of (P ∪ Q) and R'.
(PQ)R={1,2,5,7,11,13,17,19}(P \cup Q) \cap R' = \{1, 2, 5, 7, 11, 13, 17, 19\}. The number of elements is 8.
This removes the multiples of 3 (namely 3, 9, and 15) from the set PQP \cup Q.

Key Concept

Set operations including union, intersection, and set complementation.
Estimated Time:1m 30s
Question 2Question

In a survey of 100100 agricultural market traders in Lagos, 5252 sell cassava, 4545 sell yam, and 6060 sell plantain. Furthermore, 2525 sell both cassava and yam, 2222 sell both yam and plantain, and 2828 sell both cassava and plantain. If the number of traders who sell none of these three crops is twice the number of traders who sell all three crops, how many traders sell exactly two of these crops?

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Answer: 57

Answer

57 traders sell exactly two of these crops.
Using inclusion-exclusion, the total number of traders selling at least one crop is 82+x82 + x. Adding the 2x2x traders selling none gives 82+3x=10082 + 3x = 100, so x=6x = 6. The number of traders selling exactly two crops is (256)+(226)+(286)=19+16+22=57(25 - 6) + (22 - 6) + (28 - 6) = 19 + 16 + 22 = 57.

Step-by-Step Solution

1
Apply the Principle of Inclusion-Exclusion for three set unions.
n(CYP)=82+xn(C \cup Y \cup P) = 82 + x, where x=n(CYP)x = n(C \cap Y \cap P).
Summing single set cardinalities, subtracting pairwise intersections, and adding back the triple intersection accounts for all region overlaps.
2
Set up and solve the universal set cardinality equation.
x=6x = 6
Since total traders U=100|U| = 100 and non-sellers equal 2x2x, the equation 100=(82+x)+2x100 = (82 + x) + 2x simplifies to 3x=183x = 18, giving x=6x = 6.
3
Compute the sum of elements in regions representing exactly two sets.
57
Subtracting x=6x = 6 from each pairwise intersection isolates traders who sell only cassava & yam (19), only yam & plantain (16), and only cassava & plantain (22). Summing these gives 19+16+22=5719 + 16 + 22 = 57.

Key Concept

Principle of Inclusion-Exclusion for Three Sets and Partitioning Venn Diagram Regions
Question 3Question

In a class of 4040 students, 2525 study Mathematics and 1818 study Physics. If 33 students study neither of the two subjects, how many students study both Mathematics and Physics?

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Answer: 6

Answer

6 students study both Mathematics and Physics.
Subtracting the 33 students who study neither subject from the class total of 4040 leaves 3737 students studying at least one subject. Adding those studying Mathematics (2525) and Physics (1818) totals 4343. The excess of 4343 over 3737 represents the 66 students who study both subjects.

Step-by-Step Solution

1
Subtract the number of students studying neither subject from the total number of students in the class.
n(MP)=403=37n(M \cup P) = 40 - 3 = 37
This gives the number of students who belong to at least one of the two sets.
2
Set up the inclusion-exclusion formula n(MP)=n(M)+n(P)n(MP)n(M \cup P) = n(M) + n(P) - n(M \cap P).
37=25+18n(MP)37 = 25 + 18 - n(M \cap P)
Summing n(M)n(M) and n(P)n(P) double-counts the students who study both subjects.
3
Solve for the intersection n(MP)n(M \cap P).
n(MP)=4337=6n(M \cap P) = 43 - 37 = 6
Subtracting the union from the sum of the individual sets isolates the intersection value.

Key Concept

Principle of Inclusion-Exclusion for Two Sets
Question 4Question

Let the universal set be U={xZ:1x10}U = \{x \in \mathbb{Z} : 1 \le x \le 10\}. If P={x:x is a prime number within U}P = \{x : x \text{ is a prime number within } U\} and Q={x:x is an even number within U}Q = \{x : x \text{ is an even number within } U\}, which of the following represents the set (PQ)(P \cup Q)'?

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Answer: {1,9}\{1, 9\}

Answer

{1,9}\{1, 9\}
The universal set UU contains all integers from 11 to 1010. Prime numbers in this range form P={2,3,5,7}P = \{2, 3, 5, 7\}, and even numbers form Q={2,4,6,8,10}Q = \{2, 4, 6, 8, 10\}. The union PQP \cup Q combines all elements from both sets: {2,3,4,5,6,7,8,10}\{2, 3, 4, 5, 6, 7, 8, 10\}. Taking the complement of this union relative to UU leaves the elements {1,9}\{1, 9\}, making the option stating {1,9}\{1, 9\} correct.

Step-by-Step Solution

1
List all elements of the universal set UU and the subsets PP and QQ
U={1,2,3,4,5,6,7,8,9,10}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}, P={2,3,5,7}P = \{2, 3, 5, 7\}, and Q={2,4,6,8,10}Q = \{2, 4, 6, 8, 10\}
Explicit listing helps identify union and complement elements accurately.
2
Find the union of sets PP and QQ, denoted as PQP \cup Q
PQ={2,3,4,5,6,7,8,10}P \cup Q = \{2, 3, 4, 5, 6, 7, 8, 10\}
The union includes all elements that belong to set PP, set QQ, or both.
3
Determine the complement (PQ)(P \cup Q)' by subtracting PQP \cup Q from the universal set UU
(PQ)=U(PQ)={1,9}(P \cup Q)' = U \setminus (P \cup Q) = \{1, 9\}
The complement contains all elements of UU that are not present in PQP \cup Q.

Key Concept

Set Complement and Union Operations
Question 5Question

In a survey of 150150 subscribers of a digital media platform, 7575 prefer High-Definition Audio (HH), 7070 prefer Offline Downloads (DD), and 6565 prefer Ad-free Listening (AA). It is observed that 3535 subscribers prefer both HH and DD, 3030 prefer both DD and AA, and 2525 prefer both HH and AA. If 1515 subscribers prefer none of these three features, how many subscribers prefer exactly two of these features?

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Answer: 45

Answer

The number of subscribers who prefer exactly two of the features is 45.
To find the number of subscribers who prefer exactly two features, we first calculate the cardinality of the union of all three sets as 15015=135150 - 15 = 135. Applying the 3-set inclusion-exclusion formula gives the number of subscribers preferring all three features as 1515. Subtracting 1515 from each pairwise intersection gives the exclusive regions: 2020 for HH and DD only, 1515 for DD and AA only, and 1010 for HH and AA only. Summing these three exclusive regions gives 20+15+10=4520 + 15 + 10 = 45.

Step-by-Step Solution

1
Determine the total number of subscribers who prefer at least one feature.
HDA=135|H \cup D \cup A| = 135
Subtracting the number of subscribers who prefer none of the features (1515) from the universal set size (150150) gives HDA=15015=135|H \cup D \cup A| = 150 - 15 = 135.
2
Apply the Principle of Inclusion-Exclusion for three sets to find the number of subscribers who prefer all three features.
HDA=15|H \cap D \cap A| = 15
Substitute the known cardinalities into HDA=H+D+AHDDAHA+HDA|H \cup D \cup A| = |H| + |D| + |A| - |H \cap D| - |D \cap A| - |H \cap A| + |H \cap D \cap A| to get 135=75+70+65(35+30+25)+HDA135 = 75 + 70 + 65 - (35 + 30 + 25) + |H \cap D \cap A|, which simplifies to 135=120+HDA135 = 120 + |H \cap D \cap A|, giving HDA=15|H \cap D \cap A| = 15.
3
Calculate the number of subscribers preferring exactly two features by subtracting the triple intersection from each pairwise intersection.
45 subscribers
Subscribers preferring only HH and D=3515=20D = 35 - 15 = 20, only DD and A=3015=15A = 30 - 15 = 15, and only HH and A=2515=10A = 25 - 15 = 10. Summing these exclusive regions yields 20+15+10=4520 + 15 + 10 = 45.

Key Concept

Cardinality of set operations and 3-set inclusion-exclusion principle
Question 6Question

Given the universal set U={xZ:1x15}U = \{x \in \mathbb{Z} : 1 \le x \le 15\}, let A={xU:x is a multiple of 3}A = \{x \in U : x \text{ is a multiple of } 3\} and B={xU:x is an even number}B = \{x \in U : x \text{ is an even number}\}. What is the cardinality of (AB)(A \cup B)'?

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Answer: 55

Answer

The cardinality of (AB)(A \cup B)' is 55.
The universal set contains 1515 elements. The set of multiples of 33 within UU contains 55 elements, and the set of even numbers contains 77 elements. Two numbers (66 and 1212) belong to both sets. Subtracting the overlapping count gives 1010 unique elements in the union. Subtracting 1010 from the universal set total of 1515 yields 55 elements in the complement.

Step-by-Step Solution

1
List elements of sets UU, AA, and BB, and determine their cardinalities
U={1,2,3,,15}U = \{1, 2, 3, \dots, 15\}, so n(U)=15n(U) = 15. A={3,6,9,12,15}A = \{3, 6, 9, 12, 15\} (n(A)=5n(A) = 5). B={2,4,6,8,10,12,14}B = \{2, 4, 6, 8, 10, 12, 14\} (n(B)=7n(B) = 7).
Establishing explicit set memberships allows accurate counting of set operations.
2
Find the intersection ABA \cap B and compute the union cardinality n(AB)n(A \cup B)
AB={6,12}A \cap B = \{6, 12\}, so n(AB)=2n(A \cap B) = 2. Using inclusion-exclusion: n(AB)=n(A)+n(B)n(AB)=5+72=10n(A \cup B) = n(A) + n(B) - n(A \cap B) = 5 + 7 - 2 = 10.
The union includes all elements that are either multiples of 3, even, or both, avoiding double counting.
3
Calculate the complement cardinality n((AB))n((A \cup B)') relative to UU
n((AB))=n(U)n(AB)=1510=5n((A \cup B)') = n(U) - n(A \cup B) = 15 - 10 = 5. Explicitly, (AB)={1,5,7,11,13}(A \cup B)' = \{1, 5, 7, 11, 13\}.
The complement set (AB)(A \cup B)' consists of all elements in the universal set UU that are not in ABA \cup B.

Key Concept

Complement of Set Union and Inclusion-Exclusion Principle
Estimated Time:1m 0s
Question 7Question

In a sports academy of 8585 athletes, 5252 participate in track events, 4343 participate in field events, and 1212 participate in neither track nor field events. How many athletes participate in both track and field events?

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Answer: 22

Answer

The number of athletes participating in both track and field events is 22.
To find the number of athletes in both events, first calculate the total number of athletes who take part in at least one event by subtracting the 12 non-participants from 85, giving 73. Adding the 52 track athletes and 43 field athletes yields 95, which double-counts those who participate in both. The difference between 95 and 73 is 22, representing the athletes in the intersection.

Step-by-Step Solution

1
Determine the cardinality of the union of track and field athletes
N(TF)=8512=73N(T \cup F) = 85 - 12 = 73
Athletes participating in neither event are outside the union of track and field sets.
2
Formulate the two-set inclusion-exclusion equation
N(TF)=N(T)+N(F)N(TF)N(T \cup F) = N(T) + N(F) - N(T \cap F)
Adding individual set cardinalities double-counts the intersection.
3
Substitute values and solve for the intersection
N(TF)=52+4373=22N(T \cap F) = 52 + 43 - 73 = 22
Rearranging the equation yields the number of athletes in both sets.

Key Concept

Cardinality of Sets and Principle of Inclusion-Exclusion
Question 8Question

In an agricultural survey of 100100 local farmers in a community, 5555 grow Cassava (CC), 4545 grow Yam (YY), and 4040 grow Maize (MM). It is observed that 2020 farmers grow both Cassava and Yam, 1515 grow both Yam and Maize, and 1616 grow both Cassava and Maize. If 77 farmers grow all three crops, how many farmers grow none of these three crops?

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Answer: 44

Answer

The number of farmers who grow none of the three crops is 4.
The number of farmers who grow at least one crop is found using n(CYM)=55+45+40201516+7=96n(C \cup Y \cup M) = 55 + 45 + 40 - 20 - 15 - 16 + 7 = 96. Subtracting this union from the total number of farmers (100100) gives 10096=4100 - 96 = 4 farmers who grow none of the three crops.

Step-by-Step Solution

1
Apply the Principle of Inclusion-Exclusion for three sets to find the total number of farmers growing at least one crop n(CYM)n(C \cup Y \cup M).
n(CYM)=n(C)+n(Y)+n(M)n(CY)n(YM)n(CM)+n(CYM)n(C \cup Y \cup M) = n(C) + n(Y) + n(M) - n(C \cap Y) - n(Y \cap M) - n(C \cap M) + n(C \cap Y \cap M)
Elements in pairwise intersections are double-counted and elements in all three sets are triple-counted, requiring correction.
2
Substitute the given cardinalities into the formula.
n(CYM)=55+45+40201516+7=96n(C \cup Y \cup M) = 55 + 45 + 40 - 20 - 15 - 16 + 7 = 96
Calculating the total number of elements inside the union of the three sets.
3
Subtract the number of farmers in the union from the universal set size n(U)=100n(U) = 100.
n((CYM))=10096=4n((C \cup Y \cup M)') = 100 - 96 = 4
The complement of the union gives the number of elements belonging to none of the sets.

Key Concept

Principle of Inclusion-Exclusion for Three Sets and Universal Complement
Question 9Question

In a medical study of 120120 hospital outpatients, 6868 are prescribed Green tea extract (GG), 5252 are prescribed Herbal infusion (HH), and 4444 are prescribed Chamomile syrup (CC). Furthermore, 2424 patients receive both Green tea extract and Herbal infusion, 1818 receive both Green tea extract and Chamomile syrup, 1616 receive both Herbal infusion and Chamomile syrup, while 88 receive all three prescriptions. How many patients receive at most one of these three prescriptions?

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Answer: 7878

Answer

The number of patients who receive at most one of the three prescriptions is 7878.
To find the number of patients who receive at most one prescription, we must calculate the number of patients who receive exactly one prescription as well as those who receive none (the complement of the union relative to the universal set of 120 outpatients). The number receiving only Green tea is 3434, only Herbal infusion is 2020, and only Chamomile syrup is 1818, giving 7272 patients taking exactly one prescription. The total taking at least one prescription is 114114, leaving 120114=6120 - 114 = 6 patients receiving none. Adding these together yields 72+6=7872 + 6 = 78.

Step-by-Step Solution

1
Calculate the number of patients in each exclusive intersection region.
Patients receiving all three n(GHC)=8n(G \cap H \cap C) = 8. Patients receiving only GH=248=16G \cap H = 24 - 8 = 16. Patients receiving only GC=188=10G \cap C = 18 - 8 = 10. Patients receiving only HC=168=8H \cap C = 16 - 8 = 8.
Subtracting the triple intersection isolates the regions representing patients taking exactly two prescriptions.
2
Calculate the number of patients receiving exactly one prescription.
n(G only)=68(16+10+8)=34n(G \text{ only}) = 68 - (16 + 10 + 8) = 34. n(H only)=52(16+8+8)=20n(H \text{ only}) = 52 - (16 + 8 + 8) = 20. n(C only)=44(10+8+8)=18n(C \text{ only}) = 44 - (10 + 8 + 8) = 18. Total receiving exactly one =34+20+18=72= 34 + 20 + 18 = 72.
Each set total is reduced by its overlapping intersection regions to yield the single-prescription regions.
3
Determine the number of patients receiving at least one prescription using the inclusion-exclusion principle.
n(GHC)=(68+52+44)(24+18+16)+8=16458+8=114n(G \cup H \cup C) = (68 + 52 + 44) - (24 + 18 + 16) + 8 = 164 - 58 + 8 = 114.
The principle of inclusion-exclusion accounts for double-counted pairwise intersections and triple-counted intersections.
4
Calculate the number of patients receiving none of the prescriptions.
n((GHC))=120114=6n((G \cup H \cup C)') = 120 - 114 = 6.
Subtracting the total taking at least one prescription from the universal set gives the complement count.
5
Sum the patients receiving exactly one prescription and those receiving zero prescriptions.
Total taking at most one =72+6=78= 72 + 6 = 78.
'At most one' encompasses both 'zero prescriptions' and 'exactly one prescription'.

Key Concept

Three-set principle of inclusion-exclusion and complementary sets within a universal set
Question 10Question

In a technology conference of 6060 software developers, 3535 write code in Python, 2828 write in Java, and 1212 write in both Python and Java. How many developers write in neither of these two languages?

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Answer: 9

Answer

The number of developers who write in neither Python nor Java is 99.
The total number of developers writing in at least one language is given by 35+2812=5135 + 28 - 12 = 51. Subtracting this from the universal set size of 6060 gives 6051=960 - 51 = 9 developers who write in neither language.

Step-by-Step Solution

1
Calculate the number of developers who write in at least one of the languages.
n(PJ)=35+2812=51n(P \cup J) = 35 + 28 - 12 = 51
By the principle of inclusion-exclusion for two sets, n(PJ)=n(P)+n(J)n(PJ)n(P \cup J) = n(P) + n(J) - n(P \cap J).
2
Subtract from the universal set total to get the complement.
n((PJ))=6051=9n((P \cup J)') = 60 - 51 = 9
The number of elements outside the union is the universal set total minus the union cardinality.

Key Concept

Two-set inclusion-exclusion and complement cardinality
Estimated Time:1m 0s
Question 11Question

In a department of 7070 university lecturers, 4040 publish research in Journal AA, 3030 publish in Journal BB, and 2525 publish in Journal CC. It is known that 1515 publish in both Journals AA and BB, 1212 publish in both Journals BB and CC, 1010 publish in both Journals AA and CC, and 55 publish in all three journals. How many lecturers do not publish in any of these three journals?

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Answer: 77

Answer

The number of lecturers who do not publish in any of the three journals is 77.
Using the inclusion-exclusion principle for three sets, n(ABC)=40+30+25151210+5=63n(A \cup B \cup C) = 40 + 30 + 25 - 15 - 12 - 10 + 5 = 63. The number of lecturers publishing in none of the journals is the complement of this union relative to the universal set of 7070, which is 7063=770 - 63 = 7.

Step-by-Step Solution

1
Apply the Principle of Inclusion-Exclusion for three sets to find the total number of lecturers who publish in at least one journal, n(ABC)n(A \cup B \cup C).
n(ABC)=n(A)+n(B)+n(C)[n(AB)+n(BC)+n(AC)]+n(ABC)n(A \cup B \cup C) = n(A) + n(B) + n(C) - [n(A \cap B) + n(B \cap C) + n(A \cap C)] + n(A \cap B \cap C)
Elements counted multiple times in pairwise intersections must be subtracted, and the central triple intersection must be added back.
2
Substitute the given numerical values into the formula.
n(ABC)=40+30+25(15+12+10)+5=9537+5=63n(A \cup B \cup C) = 40 + 30 + 25 - (15 + 12 + 10) + 5 = 95 - 37 + 5 = 63
To evaluate the total cardinality of the union.
3
Subtract n(ABC)n(A \cup B \cup C) from the universal set size n(U)n(U).
n((ABC))=n(U)n(ABC)=7063=7n((A \cup B \cup C)') = n(U) - n(A \cup B \cup C) = 70 - 63 = 7
The number of lecturers publishing in none of the journals corresponds to the complement of the union of all three sets.

Key Concept

Principle of Inclusion-Exclusion for Three Sets and Set Complement
Question 12Question

In a survey of 120120 town residents regarding three newspapers—*The Herald* (HH), *The Express* (EE), and *The Nation* (NN)—it was found that 5252 read *The Herald*, 4545 read *The Express*, and 6060 read *The Nation*. Furthermore, 1515 read both *The Herald* and *The Express*, 2222 read both *The Express* and *The Nation*, and 1818 read both *The Herald* and *The Nation*. If the number of residents who read none of these three newspapers is twice the number of those who read all three newspapers, how many residents read exactly two of the newspapers?

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Answer: 3737

Answer

The number of residents who read exactly two newspapers is 3737.
Using the principle of inclusion-exclusion, n(HEN)=52+45+60(15+22+18)+x=102+xn(H \cup E \cup N) = 52 + 45 + 60 - (15 + 22 + 18) + x = 102 + x, where xx is the number of residents reading all three newspapers. Since the number of residents reading none is 2x2x, the total universe equation is 102+x+2x=120102 + x + 2x = 120, giving 3x=183x = 18 and x=6x = 6. The number of residents reading exactly two newspapers is (156)+(226)+(186)=9+16+12=37(15 - 6) + (22 - 6) + (18 - 6) = 9 + 16 + 12 = 37.

Step-by-Step Solution

1
Formulate the principle of inclusion-exclusion for three sets.
n(HEN)=n(H)+n(E)+n(N)n(HE)n(EN)n(HN)+n(HEN)n(H \cup E \cup N) = n(H) + n(E) + n(N) - n(H \cap E) - n(E \cap N) - n(H \cap N) + n(H \cap E \cap N)
To express the total number of residents reading at least one newspaper in terms of the unknown number of residents who read all three.
2
Substitute the given values and set up the equation for the universal set.
n(HEN)=52+45+60152218+x=102+xn(H \cup E \cup N) = 52 + 45 + 60 - 15 - 22 - 18 + x = 102 + x. Let n(HEN)=2xn(H \cup E \cup N)' = 2x. Then 102+x+2x=120    102+3x=120    3x=18    x=6102 + x + 2x = 120 \implies 102 + 3x = 120 \implies 3x = 18 \implies x = 6.
The sum of elements in the union and its complement must equal the universal set size of 120120.
3
Calculate the number of residents reading exactly two newspapers.
(n(HE)x)+(n(EN)x)+(n(HN)x)=(156)+(226)+(186)=9+16+12=37 (n(H \cap E) - x) + (n(E \cap N) - x) + (n(H \cap N) - x) = (15 - 6) + (22 - 6) + (18 - 6) = 9 + 16 + 12 = 37
Subtracting the triple intersection xx from each pairwise intersection isolates the regions corresponding to exactly two newspapers.

Key Concept

Three-set inclusion-exclusion principle and Venn diagram cardinal region decomposition
Estimated Time:1m 30s
Question 13Question

On an international flight carrying 150150 passengers, each passenger was offered three meal options: Chicken (CC), Fish (FF), and Vegetarian (VV). A survey of their choices showed that 7575 passengers chose Chicken, 6060 chose Fish, and 5050 chose Vegetarian. Additionally, 1515 passengers chose both Chicken and Fish, 1212 chose both Fish and Vegetarian, 1818 chose both Chicken and Vegetarian, while 88 passengers chose all three meals. How many passengers chose none of the three meal options?

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Answer: 2

Answer

2 passengers chose none of the meal options.
Using the inclusion-exclusion principle for three overlapping sets, the total number of passengers taking at least one meal is calculated as CFV=75+60+50(15+12+18)+8=148|C \cup F \cup V| = 75 + 60 + 50 - (15 + 12 + 18) + 8 = 148. Subtracting this value from the total count of 150150 passengers yields 150148=2150 - 148 = 2 passengers who selected none of the meal options.

Step-by-Step Solution

1
Identify the cardinalities of the individual sets, pairwise intersections, triple intersection, and the universal set.
N(U)=150N(U) = 150, C=75|C| = 75, F=60|F| = 60, V=50|V| = 50, CF=15|C \cap F| = 15, FV=12|F \cap V| = 12, CV=18|C \cap V| = 18, and CFV=8|C \cap F \cap V| = 8.
Organizing the given information allows for direct application of set cardinality formulas.
2
Calculate the total number of passengers who selected at least one meal using the Principle of Inclusion-Exclusion for three sets.
CFV=75+60+50(15+12+18)+8=18545+8=148|C \cup F \cup V| = 75 + 60 + 50 - (15 + 12 + 18) + 8 = 185 - 45 + 8 = 148.
Adding individual set totals overcounts elements in pairwise intersections, and subtracting pairwise intersections subtracts the triple intersection one too many times, so it must be added back.
3
Find the number of passengers who selected none of the meal choices by taking the complement of the union with respect to the universal set.
N(None)=N(U)CFV=150148=2N(\text{None}) = N(U) - |C \cup F \cup V| = 150 - 148 = 2.
Passengers choosing none of the meal options correspond to the region outside all three sets within the universal set.

Key Concept

Principle of Inclusion-Exclusion for Three Sets and Complement of Union
Estimated Time:1m 30s
Question 14Question

Given the universal set U={1,2,3,4,5,6,7,8,9,10}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} and a subset P={xU:x is a prime number}P = \{x \in U : x \text{ is a prime number}\}, which of the following sets represents the complement of PP, denoted as PP'?

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Answer: {1,4,6,8,9,10}\{1, 4, 6, 8, 9, 10\}

Answer

The set {1,4,6,8,9,10}\{1, 4, 6, 8, 9, 10\}
The prime numbers in UU are 2,3,5,2, 3, 5, and 77, making P={2,3,5,7}P = \{2, 3, 5, 7\}. The complement PP' consists of all elements in UU that do not belong to PP, which yields {1,4,6,8,9,10}\{1, 4, 6, 8, 9, 10\}.

Step-by-Step Solution

1
Identify the elements of the universal set UU
U={1,2,3,4,5,6,7,8,9,10}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}
The universal set defines the boundary of all possible elements under consideration.
2
Identify the elements belonging to set PP (prime numbers in UU)
P={2,3,5,7}P = \{2, 3, 5, 7\}
Prime numbers have exactly two distinct positive divisors: 1 and themselves. Note that 1 is not a prime number.
3
Compute the complement P=UPP' = U \setminus P
P={1,4,6,8,9,10}P' = \{1, 4, 6, 8, 9, 10\}
The complement of set PP consists of all elements present in UU that are not present in PP.

Key Concept

Set Complement and Universal Set Boundaries
Estimated Time:45s
Question 15Question

In a survey of 200200 agricultural exporters regarding three major commodities—Cocoa (CC), Palm Oil (PP), and Rubber (RR)—it was found that 110110 export Cocoa, 9090 export Palm Oil, and 7575 export Rubber. Exactly 2020 exporters export none of the three commodities, and 4545 export Cocoa only. Furthermore, the number of exporters who export Cocoa and Palm Oil only is twice the number of exporters who export all three commodities; the number who export Palm Oil and Rubber only is equal to the number who export all three; and the number who export Cocoa and Rubber only is 55 more than the number who export all three. Find the total number of exporters who export at least two of the three commodities.

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Answer: 80

Answer

The total number of exporters who export at least two of the three commodities is 80.
The correct answer of 80 is obtained by solving for the number of exporters trading in all three commodities (x=15x = 15) using the Cocoa set equation 45+2x+(x+5)+x=11045 + 2x + (x + 5) + x = 110, and then evaluating the region sum for at least two commodities (2x+x+(x+5)+x=5x+5=802x + x + (x + 5) + x = 5x + 5 = 80).

Step-by-Step Solution

1
Assign a variable to the triple intersection
Let x=n(CPR)x = n(C \cap P \cap R) be the number of exporters of all three commodities.
The intersection of all three sets serves as the common parameter for all double-intersection regions.
2
Write algebraic expressions for the three pairwise-only intersections
n(CP only)=2xn(C \cap P \text{ only}) = 2x, n(PR only)=xn(P \cap R \text{ only}) = x, and n(CR only)=x+5n(C \cap R \text{ only}) = x + 5.
These expressions are derived directly from the relationships given in the problem statement.
3
Formulate and solve an equation using the set of Cocoa exporters
45+2x+(x+5)+x=110    50+4x=110    x=1545 + 2x + (x + 5) + x = 110 \implies 50 + 4x = 110 \implies x = 15.
The set of Cocoa exporters consists of four mutually exclusive regions whose cardinalities sum to 110.
4
Sum the regions corresponding to 'at least two commodities'
(2x)+(x)+(x+5)+x=5x+5=5(15)+5=80(2x) + (x) + (x + 5) + x = 5x + 5 = 5(15) + 5 = 80.
'At least two' encompasses everyone who exports exactly two commodities plus those who export all three.

Key Concept

Three-set principle of inclusion-exclusion and cardinal region decomposition
Question 16Question

In a survey of 180180 cloud computing engineers regarding their proficiency in three major platforms—AWS (AA), Azure (BB), and Google Cloud (CC)—it was found that 9595 are proficient in AWS, 8080 in Azure, and 7575 in Google Cloud. Furthermore, 4040 are proficient in both AWS and Azure, 3535 in both Azure and Google Cloud, 3030 in both AWS and Google Cloud, and 1515 are not proficient in any of the three platforms. How many engineers are proficient in exactly one of these platforms?

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Answer: 100

Answer

100 engineers are proficient in exactly one platform.
Using the 3-set inclusion-exclusion formula, the number of engineers proficient in all three platforms is solved as 20. Subtracting the relevant overlap regions from each set gives 45 for AWS only, 25 for Azure only, and 30 for Google Cloud only. Adding these single-set values yields a total of 100 engineers proficient in exactly one platform.

Step-by-Step Solution

1
Determine the cardinality of the union of all three sets.
n(ABC)=18015=165n(A \cup B \cup C) = 180 - 15 = 165.
Subtracting the engineers who are not proficient in any of the three platforms from the universal set.
2
Apply the Principle of Inclusion-Exclusion to calculate the triple intersection n(ABC)n(A \cap B \cap C).
n(ABC)=20n(A \cap B \cap C) = 20.
Substituting known values gives 165=250105+n(ABC)165 = 250 - 105 + n(A \cap B \cap C), which simplifies to n(ABC)=20n(A \cap B \cap C) = 20.
3
Calculate the counts for regions representing exactly two platforms.
AWS & Azure only = 20, Azure & GCP only = 15, AWS & GCP only = 10.
Subtracting the triple intersection count (20) from each pairwise intersection.
4
Calculate the single-set exclusive regions.
AWS only = 45, Azure only = 25, GCP only = 30.
Subtracting all multi-platform overlap regions from each total set size.
5
Sum the single-set exclusive regions.
Total = 45+25+30=10045 + 25 + 30 = 100.
Combining the counts of engineers proficient in AWS only, Azure only, and Google Cloud only.

Key Concept

Principle of Inclusion-Exclusion for Three Sets and Venn Diagram Region Partitioning
Question 17Question

In a survey of 120120 tech entrepreneurs at an innovation hub, 6565 secured Angel Investment (AA), 5555 received Venture Capital (VV), and 5050 obtained Government Grants (GG). Furthermore, 2525 received both AA and VV, 2020 received both VV and GG, 2222 received both AA and GG, while 1212 received no funding from any of these three sources. How many entrepreneurs secured funding from exactly two of these sources?

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Answer: 52

Answer

The number of entrepreneurs who secured funding from exactly two sources is 52.
Subtracting the 1212 unfunded entrepreneurs from the total 120120 yields 108108 funded entrepreneurs. Applying inclusion-exclusion gives 108=65+55+50(25+20+22)+n(AVG)108 = 65 + 55 + 50 - (25 + 20 + 22) + n(A \cap V \cap G), which gives n(AVG)=5n(A \cap V \cap G) = 5. Subtracting 55 from each pairwise intersection yields 2020, 1515, and 1717 for entrepreneurs receiving funding from exactly two sources. Summing these gives 5252.

Step-by-Step Solution

1
Determine the cardinality of the union of all three funding sets
n(AVG)=108n(A \cup V \cup G) = 108
Subtracting the 12 unfunded entrepreneurs from the universal set of 120 gives the total number of entrepreneurs who received at least one form of funding.
2
Solve for the number of entrepreneurs who received funding from all three sources using inclusion-exclusion
n(AVG)=5n(A \cap V \cap G) = 5
Using n(AVG)=n(A)+n(V)+n(G)n(AV)n(VG)n(AG)+n(AVG)n(A \cup V \cup G) = n(A) + n(V) + n(G) - n(A \cap V) - n(V \cap G) - n(A \cap G) + n(A \cap V \cap G), we get 108=65+55+50252022+x108 = 65 + 55 + 50 - 25 - 20 - 22 + x, which simplifies to 108=103+x108 = 103 + x, giving x=5x = 5.
3
Calculate the count for each region representing exactly two funding sources
Only AV=20A \cap V = 20, Only VG=15V \cap G = 15, Only AG=17A \cap G = 17
Subtracting the 3-set intersection (x=5x = 5) from each pairwise intersection isolates the elements belonging to strictly two sets.
4
Sum the three strictly two-set regions
20+15+17=5220 + 15 + 17 = 52
Adding the individual counts for the three disjoint regions gives the total number of entrepreneurs who received funding from exactly two sources.

Key Concept

Principle of Inclusion-Exclusion for three sets and cardinal partitioning of Venn diagrams
Question 18Question

In a survey of 100100 book club members regarding their reading preferences among Science Fiction (SS), Mystery (MM), and Historical Fiction (HH), it was found that 4848 read Science Fiction, 4242 read Mystery, and 3838 read Historical Fiction. Furthermore, 1818 read both Science Fiction and Mystery, 1515 read both Science Fiction and Historical Fiction, 1414 read both Mystery and Historical Fiction, and 88 read all three genres. How many of the members read exactly one of these three genres?

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Answer: 58

Answer

The number of members who read exactly one of the three genres is 5858.
By isolating the disjoint regions of the 3-set Venn diagram, the number of members reading only Science Fiction is 2323, only Mystery is 1818, and only Historical Fiction is 1717. Adding these disjoint sets yields 5858.

Step-by-Step Solution

1
Determine exclusive two-set intersection regions
n(SM only)=10n(S \cap M \text{ only}) = 10, n(SH only)=7n(S \cap H \text{ only}) = 7, n(MH only)=6n(M \cap H \text{ only}) = 6
The given pairwise totals include members who read all three genres, so subtracting n(SMH)=8n(S \cap M \cap H) = 8 isolates those in exactly two sets.
2
Determine exclusive single-set regions
n(S only)=23n(S \text{ only}) = 23, n(M only)=18n(M \text{ only}) = 18, n(H only)=17n(H \text{ only}) = 17
Subtracting all overlapping regions within each set's boundary gives the number of members reading only that specific genre.
3
Sum the single-set regions
23+18+17=5823 + 18 + 17 = 58
The set of members reading exactly one genre is the disjoint union of the three exclusive single-set regions.

Key Concept

3-Set Venn Diagram Cardinality and Disjoint Region Analysis
Question 19Question

In a survey of 9090 high school students regarding their participation in sports clubs, 4040 play Badminton (BB), 3535 play Volleyball (VV), and 4242 engage in Swimming (SS). It was found that 1414 play both Badminton and Volleyball, 1212 play both Volleyball and Swimming, and 1515 play both Badminton and Swimming. If 88 students participate in none of these three sports, how many students participate in all three sports?

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Answer: 66

Answer

The number of students participating in all three sports is 66.
First, determine the number of students in the union of the three sports sets by subtracting the 88 non-participating students from the universal set of 9090, giving 8282. Then apply the Principle of Inclusion-Exclusion: 82=40+35+42(14+12+15)+x82 = 40 + 35 + 42 - (14 + 12 + 15) + x, where xx is the number of students participating in all three sports. Simplifying gives 82=76+x82 = 76 + x, which yields x=6x = 6.

Step-by-Step Solution

1
Calculate the cardinality of the union of the three sets.
n(BVS)=n(U)n((BVS))=908=82n(B \cup V \cup S) = n(U) - n((B \cup V \cup S)') = 90 - 8 = 82
Subtracting the students who do not participate in any sport from the total universal set gives the number of students participating in at least one sport.
2
Set up the Principle of Inclusion-Exclusion formula for three sets.
n(BVS)=n(B)+n(V)+n(S)[n(BV)+n(VS)+n(BS)]+n(BVS)n(B \cup V \cup S) = n(B) + n(V) + n(S) - [n(B \cap V) + n(V \cap S) + n(B \cap S)] + n(B \cap V \cap S)
This fundamental relation accounts for overlapping subsets in a three-set system.
3
Substitute the known numerical values into the formula and solve for n(BVS)n(B \cap V \cap S).
82=40+35+42(14+12+15)+n(BVS)    82=11741+n(BVS)    82=76+n(BVS)    n(BVS)=8276=682 = 40 + 35 + 42 - (14 + 12 + 15) + n(B \cap V \cap S) \implies 82 = 117 - 41 + n(B \cap V \cap S) \implies 82 = 76 + n(B \cap V \cap S) \implies n(B \cap V \cap S) = 82 - 76 = 6
Simplifying the arithmetic expression directly isolates the unknown intersection value.

Key Concept

Principle of Inclusion-Exclusion for Three Sets
Estimated Time:1m 15s
Question 20Question

Given the universal set U={xZ:1x20}U = \{x \in \mathbb{Z} : 1 \le x \le 20\}, with subsets P={xU:x is a prime number}P = \{x \in U : x \text{ is a prime number}\} and Q={xU:x is an odd integer}Q = \{x \in U : x \text{ is an odd integer}\}, what is the cardinal number of (PQ)(P \cup Q)'?

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Answer: 99

Answer

99
The universal set contains 2020 elements. The union PQP \cup Q consists of all odd numbers and prime numbers between 11 and 2020, giving 1111 unique elements: {1,2,3,5,7,9,11,13,15,17,19}\{1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19\}. Subtracting these 1111 elements from the total 2020 elements in UU gives n((PQ))=2011=9n((P \cup Q)') = 20 - 11 = 9.

Step-by-Step Solution

1
Identify the elements of the universal set UU and subsets PP and QQ.
U={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}, so n(U)=20n(U) = 20.
P={2,3,5,7,11,13,17,19}P = \{2, 3, 5, 7, 11, 13, 17, 19\}
Q={1,3,5,7,9,11,13,15,17,19}Q = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}
Explicitly listing the set elements helps correctly calculate the union.
2
Determine the union PQP \cup Q.
PQ={1,2,3,5,7,9,11,13,15,17,19}P \cup Q = \{1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19\}, which contains 1111 elements.
The union combines all unique elements present in set PP, set QQ, or both.
3
Calculate the complement set (PQ)(P \cup Q)' and its cardinality.
(PQ)=U(PQ)={4,6,8,10,12,14,16,18,20}(P \cup Q)' = U \setminus (P \cup Q) = \{4, 6, 8, 10, 12, 14, 16, 18, 20\}, so n((PQ))=2011=9n((P \cup Q)') = 20 - 11 = 9.
The complement of a set contains all elements in the universal set UU that are not in the given set.

Key Concept

Set Operations and Complement of Sets
Estimated Time:1m 0s
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