Union:
\(n(A \cup B) = n(A) + n(B) - n(A \cap B)\)
\(n(A \cup B) = n(A) + n(B) - n(A \cap B)\)
Complement:
\(n(\overline{A \cup B}) = n(U) - n(A \cup B)\)
\(n(\overline{A \cup B}) = n(U) - n(A \cup B)\)
Only A:
\(n_0(A) = n(A) - n(A \cap B)\)
\(n_0(A) = n(A) - n(A \cap B)\)
Only B:
\(n_0(B) = n(B) - n(A \cap B)\)
\(n_0(B) = n(B) - n(A \cap B)\)
4. a) A survey was conducted among 125 students of class 10 studying in Gyan Mandir Secondary School regarding the places for educational tour and it was found that 65 students preferred to visit Lumbini (L), 75 preferred to visit Pokhara (P), and 25 preferred both the places.
- Write the cardinality of set of students who preferred to visit both Pokhara and Lumbini in set notational form.
- Show the above information in a Venn-diagram.
- Find the number of students who preferred neither of two places.
- Find the ratio of the number of students who preferred Lumbini only and Pokhara only.
📝 Given Information
Total students, \(n(U)\) = 125
Preferred Lumbini, \(n(L)\) = 65
Preferred Pokhara, \(n(P)\) = 75
Preferred both, \(n(P \cap L)\) = 25
(i) Cardinality notation for both
\(n(P \cap L) = 25\)
(ii) Venn-diagram Representation
(iii) Students who preferred neither of two places
\(n(P \cup L) = n(P) + n(L) - n(P \cap L) = 75 + 65 - 25 = 115\)
\(n(\overline{P \cup L}) = n(U) - n(P \cup L) = 125 - 115 = 10\)
\(n(\overline{P \cup L}) = n(U) - n(P \cup L) = 125 - 115 = 10\)
\(n(\overline{P \cup L}) = 10\)
(iv) Ratio of students who preferred Lumbini only and Pokhara only
\(n_o(L) = n(L) - n(P \cap L) = 65 - 25 = 40\)
\(n_o(P) = n(P) - n(P \cap L) = 75 - 25 = 50\)
\(\text{Ratio} = n_o(L) : n_o(P) = 40 : 50 = 4 : 5\)
\(n_o(P) = n(P) - n(P \cap L) = 75 - 25 = 50\)
\(\text{Ratio} = n_o(L) : n_o(P) = 40 : 50 = 4 : 5\)
Ratio = 4 : 5
4. b) A survey was carried out in a village of Chandrapur municipality of Rautahat district regarding the local languages. Among 1,400 people participated in the survey, it was found that 600 people can speak Bhojpuri, 900 can speak Maithili and 350 people can speak Bhojpuri as well as Maithili languages.
- Write the cardinality of \(n(B \cap M)\).
- Draw a Venn-diagram to show the above data.
- How many people cannot speak both the languages?
- How many more or less people can speak Maithili only than Bhojpuri only?
📝 Given Information
Total surveyed, \(n(U)\) = 1400
Bhojpuri, \(n(B)\) = 600
Maithili, \(n(M)\) = 900
Both, \(n(B \cap M)\) = 350
(i) Cardinality notation for both
\(n(B \cap M) = 350\)
(ii) Venn-diagram Representation
(iii) People who cannot speak both the languages
\(n(B \cup M) = n(B) + n(M) - n(B \cap M) = 600 + 900 - 350 = 1150\)
\(n(\overline{B \cup M}) = n(U) - n(B \cup M) = 1400 - 1150 = 250\)
\(n(\overline{B \cup M}) = n(U) - n(B \cup M) = 1400 - 1150 = 250\)
\(n(\overline{B \cup M}) = 250\)
(iv) Difference between Maithili only and Bhojpuri only
\(n_o(B) = n(B) - n(B \cap M) = 600 - 350 = 250\)
\(n_o(M) = n(M) - n(B \cap M) = 900 - 350 = 550\)
\(\text{Difference} = n_o(M) - n_o(B) = 550 - 250 = 300\)
\(n_o(M) = n(M) - n(B \cap M) = 900 - 350 = 550\)
\(\text{Difference} = n_o(M) - n_o(B) = 550 - 250 = 300\)
300 more people can speak Maithili only than Bhojpuri only.
4. c) During 'Visit Nepal 2020', among 7,500 Chinese tourists who visited Nepal, 60% of them had already visited Bhutan, 50% had visited Sri Lanka, and 30% have visited both the countries.
- Write the number of tourists who visited Bhutan in set notation.
- Illustrate this information in a Venn-diagram.
- How many tourists have visited neither Bhutan nor Srilanka?
- How many tourists have already visited only one of these countries?
📝 Given Information
Total tourists, \(n(U)\) = 7500
Visited Bhutan, \(n(B)\) = 60% of 7500 = 4500
Visited Sri Lanka, \(n(S)\) = 50% of 7500 = 3750
Visited both, \(n(B \cap S)\) = 30% of 7500 = 2250
(i) Set notation for Bhutan
\(n(B) = 4500\)
(ii) Venn-diagram Representation
(iii) Tourists who visited neither Bhutan nor Sri Lanka
\(n(B \cup S) = n(B) + n(S) - n(B \cap S) = 4500 + 3750 - 2250 = 6000\)
\(n(\overline{B \cup S}) = n(U) - n(B \cup S) = 7500 - 6000 = 1500\)
\(n(\overline{B \cup S}) = n(U) - n(B \cup S) = 7500 - 6000 = 1500\)
\(n(\overline{B \cup S}) = 1500\)
(iv) Tourists who visited only one of these countries
\(n_o(B) = n(B) - n(B \cap S) = 4500 - 2250 = 2250\)
\(n_o(S) = n(S) - n(B \cap S) = 3750 - 2250 = 1500\)
\(\text{Only one country} = n_o(B) + n_o(S) = 2250 + 1500 = 3750\)
\(n_o(S) = n(S) - n(B \cap S) = 3750 - 2250 = 1500\)
\(\text{Only one country} = n_o(B) + n_o(S) = 2250 + 1500 = 3750\)
Only One Country = 3750
5. a) In a survey conducted in a community of Gorkha district regarding cultural dances, it was found that out of 500 people, 180 people can perform Sorathi dance (S), 250 can perform Ghatu dance (G) and 120 can perform neither of these two dances.
- Write the set notation that represents the number of people who can perform neither Sorathi nor Ghatu dance.
- Represent the above information in a Venn-diagram.
- How many people can perform both the dances?
- Compare the number of people who can perform Sorathi only and Ghatu only.
📝 Given Information
Total surveyed, \(n(U)\) = 500
Sorathi, \(n(S)\) = 180
Ghatu, \(n(G)\) = 250
Neither, \(n(\overline{S \cup G})\) = 120
(i) Set notation for neither
\(n(\overline{S \cup G}) = 120\)
(ii) Venn-diagram Representation
(iii) People who can perform both the dances
Let \(n(S \cap G) = x\)
From Venn-diagram,
\(n(U) = n_o(S) + n_o(G) + n(S \cap G) + n(\overline{S \cup G})\)
\(500 = (180-x) + (250-x) + x + 120\)
\(500 = 550 - x \implies x = 50\)
From Venn-diagram,
\(n(U) = n_o(S) + n_o(G) + n(S \cap G) + n(\overline{S \cup G})\)
\(500 = (180-x) + (250-x) + x + 120\)
\(500 = 550 - x \implies x = 50\)
\(n(S \cap G) = 50\)
(iv) Comparison of Sorathi only and Ghatu only
\(n_o(S) = 180 - 50 = 130\)
\(n_o(G) = 250 - 50 = 200\)
\(\text{Difference} = 200 - 130 = 70\)
\(n_o(G) = 250 - 50 = 200\)
\(\text{Difference} = 200 - 130 = 70\)
The number of people who can perform Sorathi only is 70 less than Ghatu only. (Ratio = 13 : 20)
5. b) Last Wednesday, a School successfully accomplished its school's day. For this program, every student participated in at least one of the activities, athletics or music. In a class of 45 students, 21 participated in athletics (A) and 29 participated in music (M).
- Write the cardinality of \(n(\overline{A \cup M})\).
- Draw a Venn-diagram to represent the above information.
- How many students participated in both the activities?
- Compare the number of students who participated in athletics only and music only.
📝 Given Information
Total students, \(n(U)\) = 45
Athletics, \(n(A)\) = 21
Music, \(n(M)\) = 29
Neither, \(n(\overline{A \cup M})\) = 0
(i) Cardinality of complement
\(n(\overline{A \cup M}) = 0\)
(ii) Venn-diagram Representation
(iii) Students who participated in both
Let \(n(A \cap M) = x\)
\(45 = (21-x) + (29-x) + x + 0\)
\(45 = 50 - x \implies x = 5\)
\(45 = (21-x) + (29-x) + x + 0\)
\(45 = 50 - x \implies x = 5\)
\(n(A \cap M) = 5\)
(iv) Comparison of Athletics only and Music only
\(n_o(A) = 21 - 5 = 16\)
\(n_o(M) = 29 - 5 = 24\)
\(\text{Ratio} = n_o(A) : n_o(M) = 16 : 24 = 2 : 3\)
\(n_o(M) = 29 - 5 = 24\)
\(\text{Ratio} = n_o(A) : n_o(M) = 16 : 24 = 2 : 3\)
The number of students in music only is 8 more than athletics only. (Ratio = 2 : 3)
5. c) In a survey conducted among 150 students of a school, it was found that 70 students liked cricket, 62 liked basketball and 30 students did not like any of these two games.
- Write the cardinality of \(n(\overline{B \cup C})\).
- Present the information in a Venn-diagram.
- Find the number of students who liked cricket only.
- Compare the number of students who liked both games and who liked except these two games.
📝 Given Information
Total students, \(n(U)\) = 150
Cricket, \(n(C)\) = 70
Basketball, \(n(B)\) = 62
Neither, \(n(\overline{B \cup C})\) = 30
(i) Cardinality of neither
\(n(\overline{B \cup C}) = 30\)
(ii) Venn-diagram Representation
(iii) Students who liked Cricket only
Let \(n(B \cap C) = x\)
\(150 = (62-x) + (70-x) + x + 30 \implies 150 = 162-x \implies x = 12\)
\(n_o(C) = n(C) - n(B \cap C) = 70 - 12 = 58\)
\(150 = (62-x) + (70-x) + x + 30 \implies 150 = 162-x \implies x = 12\)
\(n_o(C) = n(C) - n(B \cap C) = 70 - 12 = 58\)
\(n_o(C) = 58\)
(iv) Comparison of both games and neither
\(n(B \cap C) = 12\)
\(n(\overline{B \cup C}) = 30\)
\(\text{Difference} = 30 - 12 = 18\)
\(n(\overline{B \cup C}) = 30\)
\(\text{Difference} = 30 - 12 = 18\)
The number of students who liked both games is 18 less than who liked neither. (Ratio = 2 : 5)
6. a) Among 54 SEE appeared students from a school, 18 students got ‘A+’ grade in Mathematics only, 25 got ‘A+’ grade in English only and 7 students did not get ‘A+’ grade in these two subjects.
- Write the cardinalities of the given sets in set notational forms.
- Show the above information in Venn-diagram.
- How many students got ‘A+’ grade in Mathematics?
- How many more or less students got ‘A+’ grade not in English than not in Mathematics?
📝 Given Information
Total students, \(n(U)\) = 54
Math only, \(n_o(M)\) = 18
English only, \(n_o(E)\) = 25
Neither, \(n(\overline{M \cup E})\) = 7
(i) Cardinality notations
\(n(U)=54, n_o(M)=18, n_o(E)=25, n(\overline{M \cup E})=7\)
(ii) Venn-diagram Representation
(iii) Students who got 'A+' in Mathematics
Let \(n(M \cap E) = x\)
\(n(U) = n_o(M) + n_o(E) + n(M \cap E) + n(\overline{M \cup E})\)
\(54 = 18 + 25 + x + 7 \implies 54 = 50 + x \implies x = 4\)
\(n(M) = n_o(M) + n(M \cap E) = 18 + 4 = 22\)
\(n(U) = n_o(M) + n_o(E) + n(M \cap E) + n(\overline{M \cup E})\)
\(54 = 18 + 25 + x + 7 \implies 54 = 50 + x \implies x = 4\)
\(n(M) = n_o(M) + n(M \cap E) = 18 + 4 = 22\)
\(n(M) = 22\)
(iv) Difference between not in English and not in Mathematics
\(n(E) = n_o(E) + n(M \cap E) = 25 + 4 = 29\)
\(n(\overline{E}) = n(U) - n(E) = 54 - 29 = 25\)
\(n(\overline{M}) = n(U) - n(M) = 54 - 22 = 32\)
\(\text{Difference} = n(\overline{M}) - n(\overline{E}) = 32 - 25 = 7\)
\(n(\overline{E}) = n(U) - n(E) = 54 - 29 = 25\)
\(n(\overline{M}) = n(U) - n(M) = 54 - 22 = 32\)
\(\text{Difference} = n(\overline{M}) - n(\overline{E}) = 32 - 25 = 7\)
The students who got ‘A+’ grade not in English is 7 less than not in Mathematics.
6. b) In a survey of a group of people, 20% are using cellular data but not Wi-Fi, 65% are using Wi-Fi but not cellular data and 5% of them use neither cellular data nor Wi-Fi.
- Write the cardinality of \(n(\overline{C \cup W})\).
- Represent the above information in a Venn-diagram.
- Find the percent of people who are using Wi-Fi.
- By what percent is the number of people who are not using cellular data more or less than the people who are not using Wi-Fi?
📝 Given Information
Total people, \(n(U)\) = 100%
Cellular only, \(n_o(C)\) = 20%
Wi-Fi only, \(n_o(W)\) = 65%
Neither, \(n(\overline{C \cup W})\) = 5%
(i) Cardinality of neither
\(n(\overline{C \cup W}) = 5\%\)
(ii) Venn-diagram Representation
(iii) Percentage of people using Wi-Fi
Let \(n(C \cap W) = x\)
\(100\% = 20\% + 65\% + x + 5\% \implies 100\% = 90\% + x \implies x = 10\%\)
\(n(W) = n_o(W) + n(C \cap W) = 65\% + 10\% = 75\%\)
\(100\% = 20\% + 65\% + x + 5\% \implies 100\% = 90\% + x \implies x = 10\%\)
\(n(W) = n_o(W) + n(C \cap W) = 65\% + 10\% = 75\%\)
\(n(W) = 75\%\)
(iv) Difference between not using cellular data and not using Wi-Fi
\(n(C) = 20\% + 10\% = 30\%\)
Not using cellular data, \(n(\overline{C}) = 100\% - 30\% = 70\%\)
Not using Wi-Fi, \(n(\overline{W}) = 100\% - 75\% = 25\%\)
\(\text{Difference} = 70\% - 25\% = 45\%\)
Not using cellular data, \(n(\overline{C}) = 100\% - 30\% = 70\%\)
Not using Wi-Fi, \(n(\overline{W}) = 100\% - 75\% = 25\%\)
\(\text{Difference} = 70\% - 25\% = 45\%\)
Not using cellular data is 45% more than not using Wi-Fi.
7. a) In a survey of 15,000 students of different schools, 6,000 of them were found to have tuition classes before the SEE examination. Among them, 3,000 studied only Mathematics, 1,800 only Science and 600 studied other subjects but not these two subjects.
- Write the cardinality of \(n_o(M)\), \(n_o(S)\) and \(n(\overline{M \cup S})\).
- Represent the above information in a Venn-diagram.
- Find the number of students who studied Science.
- What percent of the students of the survey studied either Mathematics or Science?
📝 Given Information
Total survey, \(n(U)\) = 15000
Tuition set total, \(n(U')\) = 6000
Math only, \(n_o(M)\) = 3000
Science only, \(n_o(S)\) = 1800
Other subjects, \(n(\overline{M \cup S})\) = 600
(i) Cardinality values
\(n_o(M) = 3000,\; n_o(S) = 1800,\; n(\overline{M \cup S}) = 600\)
(ii) Venn-diagram Representation
(iii) Students who studied Science
Let \(n(M \cap S) = x\)
\(n(U') = n_o(M) + n_o(S) + n(M \cap S) + n(\overline{M \cup S})\)
\(6000 = 3000 + 1800 + x + 600 \implies 6000 = 5400 + x \implies x = 600\)
\(n(S) = n_o(S) + n(M \cap S) = 1800 + 600 = 2400\)
\(n(U') = n_o(M) + n_o(S) + n(M \cap S) + n(\overline{M \cup S})\)
\(6000 = 3000 + 1800 + x + 600 \implies 6000 = 5400 + x \implies x = 600\)
\(n(S) = n_o(S) + n(M \cap S) = 1800 + 600 = 2400\)
\(n(S) = 2400\)
(iv) Percent of students studying either Math or Science
\(n(M \cup S) = n_o(M) + n_o(S) + n(M \cap S) = 3000 + 1800 + 600 = 5400\)
\(\text{Percent} = \frac{n(M \cup S)}{n(U)} \times 100\% = \frac{5400}{15000} \times 100\% = 36\%\)
\(\text{Percent} = \frac{n(M \cup S)}{n(U)} \times 100\% = \frac{5400}{15000} \times 100\% = 36\%\)
Required Percent = 36%
7. b) Of 200 students who appeared the SEE from Dhaulagiri Secondary School, 25% students got ‘A+’ grade in different subjects. Out of the students who got ‘A+’ grade, 40% students got in English only, 20% got in Mathematics only and 30% in other subjects.
- Write the set notation representing the number of students who got A+ grades.
- Show the above information in a Venn-diagram.
- How many students got ‘A+’ grade in English?
- Find the ratio of number of students who got ‘A+’ grade in English to those students who got in Mathematics.
📝 Given Information
Total, \(n(U)\) = 200
A+ total, \(n(U')\) = 25% of 200 = 50
English only, \(n_o(E)\) = 40% of 50 = 20
Math only, \(n_o(M)\) = 20% of 50 = 10
Other, \(n(\overline{E \cup M})\) = 30% of 50 = 15
(i) Set notation representing got A+
\(n(U') = 50\)
(ii) Venn-diagram Representation
(iii) Students who got 'A+' in English
Let \(n(E \cap M) = x\)
\(n(U') = n_o(E) + n_o(M) + n(E \cap M) + n(\overline{E \cup M})\)
\(50 = 20 + 10 + x + 15 \implies 50 = 45 + x \implies x = 5\)
\(n(E) = n_o(E) + n(E \cap M) = 20 + 5 = 25\)
\(n(U') = n_o(E) + n_o(M) + n(E \cap M) + n(\overline{E \cup M})\)
\(50 = 20 + 10 + x + 15 \implies 50 = 45 + x \implies x = 5\)
\(n(E) = n_o(E) + n(E \cap M) = 20 + 5 = 25\)
\(n(E) = 25\)
(iv) Ratio of English to Math
\(n(M) = n_o(M) + n(E \cap M) = 10 + 5 = 15\)
\(\text{Ratio} = n(E) : n(M) = 25 : 15 = 5 : 3\)
\(\text{Ratio} = n(E) : n(M) = 25 : 15 = 5 : 3\)
Ratio = 5 : 3
8. a) In a survey of a group of people, it was found that 65% of them liked comedy movies, 63% liked action movies, 33% liked both types of movies, and 150 people did not like both types of movies.
- Draw a Venn-diagram to illustrate the above information.
- Find the number of people participated in the survey.
- Find the number of people who liked both types of movies.
- Justify if the number of people who did not like comedy movies is 150 more than action only.
📝 Given Information
Let total people, \(n(U)\) = \(x\)
Comedy, \(n(C)\) = 65% of \(x\) = \(0.65x\)
Action, \(n(A)\) = 63% of \(x\) = \(0.63x\)
Both, \(n(C \cap A)\) = 33% of \(x\) = \(0.33x\)
Neither, \(n(\overline{C \cup A})\) = 150
(i) Venn-diagram Representation
(ii) Number of participants
\(n(U) = n_o(C) + n_o(A) + n(C \cap A) + n(\overline{C \cup A})\)
\(x = 0.32x + 0.30x + 0.33x + 150\)
\(x = 0.95x + 150 \implies 0.05x = 150 \implies x = 3000\)
\(x = 0.32x + 0.30x + 0.33x + 150\)
\(x = 0.95x + 150 \implies 0.05x = 150 \implies x = 3000\)
Total Participants, \(n(U) = 3000\)
(iii) People who liked both
\(n(C \cap A) = 0.33 \times 3000 = 990\)
\(n(C \cap A) = 990\)
(iv) Justification of Amrita's calculation
Did not like Comedy, \(n(\overline{C}) = n(U) - n(C) = 3000 - 0.65(3000) = 1050\)
Action only, \(n_o(A) = 0.30 \times 3000 = 900\)
\(\text{Difference} = 1050 - 900 = 150\)
Action only, \(n_o(A) = 0.30 \times 3000 = 900\)
\(\text{Difference} = 1050 - 900 = 150\)
Yes, \(1050 - 900 = 150\) (Calculation is correct and justified).
8. b) In an examination, 80% examinees passed in English, 70% in Mathematics, 60% passed in both the subjects, and 45 examinees failed in both subjects.
- Draw a Venn-diagram to represent the above information.
- Find the number of examinees participated in the survey.
- Find the number of examinees who passed in both subjects.
- Determine if failed in English is double the number of passed Mathematics only.
📝 Given Information
Let total examinees, \(n(U)\) = \(x\)
English, \(n(E)\) = 80% of \(x\) = \(0.8x\)
Mathematics, \(n(M)\) = 70% of \(x\) = \(0.7x\)
Both, \(n(E \cap M)\) = 60% of \(x\) = \(0.6x\)
Failed both, \(n(\overline{E \cup M})\) = 45
(i) Venn-diagram Representation
(ii) Number of examinees
\(n(U) = n_o(E) + n_o(M) + n(E \cap M) + n(\overline{E \cup M})\)
\(x = 0.2x + 0.1x + 0.6x + 45\)
\(x = 0.9x + 45 \implies 0.1x = 45 \implies x = 450\)
\(x = 0.2x + 0.1x + 0.6x + 45\)
\(x = 0.9x + 45 \implies 0.1x = 45 \implies x = 450\)
Total Examinees, \(n(U) = 450\)
(iii) Examinees who passed both
\(n(E \cap M) = 0.6 \times 450 = 270\)
\(n(E \cap M) = 270\)
(iv) Verification of Ram's calculation
Failed in English, \(n(\overline{E}) = n(U) - n(E) = 450 - 0.8(450) = 90\)
Passed Math only, \(n_o(M) = 0.1 \times 450 = 45\)
\(\frac{n(\overline{E})}{n_o(M)} = \frac{90}{45} = 2 \implies n(\overline{E}) = 2 \times n_o(M)\)
Passed Math only, \(n_o(M) = 0.1 \times 450 = 45\)
\(\frac{n(\overline{E})}{n_o(M)} = \frac{90}{45} = 2 \implies n(\overline{E}) = 2 \times n_o(M)\)
Ram's calculation is correct.
9. a) In a survey of a community, it was found that 65% of people liked folk songs, 55% liked modern songs and 10% of people did not like both types of songs.
- Illustrate the above information in a Venn-diagram.
- What percentage of people liked either folk songs or modern songs?
- If 360 people liked both types of songs, how many people were surveyed?
- How many times is the number of people who liked both type of songs of the number of people who did not like both type of songs?
📝 Given Information
Universal set, \(n(U)\) = 100%
Folk, \(n(F)\) = 65%
Modern, \(n(M)\) = 55%
Neither, \(n(\overline{F \cup M})\) = 10%
(i) Venn-diagram Representation
(ii) Percent of people who liked either folk or modern
\(n(F \cup M) = n(U) - n(\overline{F \cup M}) = 100\% - 10\% = 90\%\)
Liked either = 90%
(iii) Total number of people surveyed
Let \(n(F \cap M) = x\%\)
\(100\% = (65\% - x) + (55\% - x) + x + 10\% \implies 100\% = 130\% - x \implies x = 30\%\)
Let total surveyed $= y$
\(30\% \text{ of } y = 360 \implies y = 1200\)
\(100\% = (65\% - x) + (55\% - x) + x + 10\% \implies 100\% = 130\% - x \implies x = 30\%\)
Let total surveyed $= y$
\(30\% \text{ of } y = 360 \implies y = 1200\)
Total Surveyed, \(n(U) = 1200\)
(iv) Comparison of both with neither
Both $= 360$
Neither $= 10\% \text{ of } 1200 = 120$
\(\frac{\text{Both}}{\text{Neither}} = \frac{360}{120} = 3 \implies \text{Both} = 3 \times \text{Neither}\)
Neither $= 10\% \text{ of } 1200 = 120$
\(\frac{\text{Both}}{\text{Neither}} = \frac{360}{120} = 3 \implies \text{Both} = 3 \times \text{Neither}\)
The number of people who liked both is 3 times of neither.
9. b) In a survey of some farmers in a community, 70% of them are found cultivating rice, 60% cultivating wheat, 20% are not cultivating both the crops, and 450 farmers are found cultivating both the crops.
- Draw a Venn-diagram to illustrate the above information.
- Find the percentage of farmers who are cultivating either rice or wheat.
- Find the total number of farmers participated in the survey.
- By how many times is the number of farmers who are cultivating both the crops more than the number of farmers who are cultivating none of these crops?
📝 Given Information
Universal set, \(n(U)\) = 100%
Rice, \(n(R)\) = 70%
Wheat, \(n(W)\) = 60%
Neither, \(n(\overline{R \cup W})\) = 20%
Both, \(n(R \cap W)\) = 450
(i) Venn-diagram Representation
(ii) Percentage of farmers cultivating either
\(n(R \cup W) = n(U) - n(\overline{R \cup W}) = 100\% - 20\% = 80\%\)
Cultivating either = 80%
(iii) Total number of farmers
Let \(n(R \cap W) = x\%\)
\(100\% = (70\% - x) + (60\% - x) + x + 20\% \implies 100\% = 150\% - x \implies x = 50\%\)
Let total farmers $= y$
\(50\% \text{ of } y = 450 \implies y = 900\)
\(100\% = (70\% - x) + (60\% - x) + x + 20\% \implies 100\% = 150\% - x \implies x = 50\%\)
Let total farmers $= y$
\(50\% \text{ of } y = 450 \implies y = 900\)
Total Farmers = 900
(iv) Comparison of both with none
Both $= 450$
None $= 20\% \text{ of } 900 = 180$
\(\frac{\text{Both}}{\text{None}} = \frac{450}{180} = 2.5 \implies \text{Both} = 2.5 \times \text{None}\)
None $= 20\% \text{ of } 900 = 180$
\(\frac{\text{Both}}{\text{None}} = \frac{450}{180} = 2.5 \implies \text{Both} = 2.5 \times \text{None}\)
The number of farmers cultivating both is 2.5 times of none.
10. a) 75 students in a class like picnic (P) or hiking (H) or both. Out of them, 10 like both the activities. The ratio of the number of students who like picnic to those who like hiking is 2 : 3.
- If \(n(P) = 2x\), what is the value of \(n(H)\)?
- Represent the above information in a Venn-diagram.
- Find the number of students who like picnic.
- Find the percentage of students who like picnic only.
📝 Given Information
Total, \(n(U)\) = 75
Both, \(n(P \cap H)\) = 10
Neither, \(n(\overline{P \cup H})\) = 0
Ratio \(n(P) : n(H)\) = 2 : 3
(i) Value of \(n(H)\)
If \(n(P) = 2x\), then:
\(n(H) = 3x\)
(ii) Venn-diagram Representation
(iii) Students who like picnic
\(n(U) = n_o(P) + n_o(H) + n(P \cap H) + n(\overline{P \cup H})\)
\(75 = (2x - 10) + (3x - 10) + 10 + 0\)
\(75 = 5x - 10 \implies 5x = 85 \implies x = 17\)
\(n(P) = 2x = 2 \times 17 = 34\)
\(75 = (2x - 10) + (3x - 10) + 10 + 0\)
\(75 = 5x - 10 \implies 5x = 85 \implies x = 17\)
\(n(P) = 2x = 2 \times 17 = 34\)
\(n(P) = 34\)
(iv) Percentage of students who like picnic only
\(n_o(P) = n(P) - n(P \cap H) = 34 - 10 = 24\)
\(\text{Percentage} = \frac{24}{75} \times 100\% = 32\%\)
\(\text{Percentage} = \frac{24}{75} \times 100\% = 32\%\)
Required Percent = 32%
10. b) A survey is conducted in a community regarding the activities that are beneficial for healthy life. Out of 100 people participating in the survey, it is found that the ratio of number of people who are doing yoga (Y) and going for jogging (J) regularly in the early morning is 4 : 5. If 25 people are following both the activities and 35 people are not following both of these activities, find:
- If \(n(Y) = 4x\), what is the value of \(n(J)\)?
- Draw a Venn-diagram to represent the above information.
- Find the number of people who are going for jogging.
- Find the percent of people who are going for jogging only.
📝 Given Information
Total, \(n(U)\) = 100
Both, \(n(Y \cap J)\) = 25
Neither, \(n(\overline{Y \cup J})\) = 35
Ratio \(n(Y) : n(J)\) = 4 : 5
(i) Value of \(n(J)\)
If \(n(Y) = 4x\), then:
\(n(J) = 5x\)
(ii) Venn-diagram Representation
(iii) People who jog regularly
\(n(U) = n_o(Y) + n_o(J) + n(Y \cap J) + n(\overline{Y \cup J})\)
\(100 = (4x - 25) + (5x - 25) + 25 + 35\)
\(100 = 9x + 10 \implies 9x = 90 \implies x = 10\)
\(n(J) = 5x = 5 \times 10 = 50\)
\(100 = (4x - 25) + (5x - 25) + 25 + 35\)
\(100 = 9x + 10 \implies 9x = 90 \implies x = 10\)
\(n(J) = 5x = 5 \times 10 = 50\)
\(n(J) = 50\)
(iv) Percentage of jogging only
\(n_o(J) = n(J) - n(Y \cap J) = 50 - 25 = 25\)
\(\text{Percentage} = \frac{25}{100} \times 100\% = 25\%\)
\(\text{Percentage} = \frac{25}{100} \times 100\% = 25\%\)
Jogging Only = 25%
10. c) In a group of students, the ratio of the number of students who liked music (M) and sports (S) is 9 : 7. Out of which 25 liked both the activities, 20 liked music only, and 15 liked none of the activities.
- If \(n(M) = 9x\), what is the value of \(n(S)\)?
- Represent the above information in a Venn-diagram.
- Find the total number of students in the group.
- Find the number of students who liked at most one of these activities.
📝 Given Information
Both, \(n(M \cap S)\) = 25
Music only, \(n_o(M)\) = 20
Neither, \(n(\overline{M \cup S})\) = 15
Ratio \(n(M) : n(S)\) = 9 : 7
(i) Value of \(n(S)\)
If \(n(M) = 9x\), then:
\(n(S) = 7x\)
(ii) Venn-diagram Representation
(iii) Total number of students
We have, \(n_o(M) = n(M) - n(M \cap S) = 9x - 25\)
Since \(n_o(M) = 20 \implies 9x - 25 = 20 \implies 9x = 45 \implies x = 5\)
\(n_o(S) = 7x - 25 = 7(5) - 25 = 10\)
\(n(U) = n_o(M) + n_o(S) + n(M \cap S) + n(\overline{M \cup S})\)
\(n(U) = 20 + 10 + 25 + 15 = 70\)
Since \(n_o(M) = 20 \implies 9x - 25 = 20 \implies 9x = 45 \implies x = 5\)
\(n_o(S) = 7x - 25 = 7(5) - 25 = 10\)
\(n(U) = n_o(M) + n_o(S) + n(M \cap S) + n(\overline{M \cup S})\)
\(n(U) = 20 + 10 + 25 + 15 = 70\)
Total Students = 70
(iv) Students who liked at most one activity
\(\text{At most one} = n(U) - n(M \cap S) = 70 - 25 = 45\)
At most one = 45
10. d) Out of 120 students appeared in an examination, the number of students who passed in Mathematics only is twice the number of students who passed in Science only. Also, 50 students passed in both subjects and one-third students failed in both subjects.
- Write the cardinality of \(n(\overline{M \cup S})\).
- Present the given information in a Venn-diagram.
- Find the number of students who passed in Mathematics.
- If the failed students were successful in Science, find the ratio of students who passed in only Mathematics and only Science.
📝 Given Information
Total examinees, \(n(U)\) = 120
Both, \(n(M \cap S)\) = 50
Failed both, \(n(\overline{M \cup S})\) = \(\frac{1}{3}\) of 120 = 40
Math only to Science only ratio = 2 : 1
(i) Cardinality of complement
\(n(\overline{M \cup S}) = 40\)
(ii) Venn-diagram Representation
(iii) Students who passed in Mathematics
Let \(n_o(S) = x \implies n_o(M) = 2x\)
\(n(U) = n_o(M) + n_o(S) + n(M \cap S) + n(\overline{M \cup S})\)
\(120 = 2x + x + 50 + 40 \implies 120 = 3x + 90 \implies 3x = 30 \implies x = 10\)
\(n(M) = n_o(M) + n(M \cap S) = 2(10) + 50 = 70\)
\(n(U) = n_o(M) + n_o(S) + n(M \cap S) + n(\overline{M \cup S})\)
\(120 = 2x + x + 50 + 40 \implies 120 = 3x + 90 \implies 3x = 30 \implies x = 10\)
\(n(M) = n_o(M) + n(M \cap S) = 2(10) + 50 = 70\)
\(n(M) = 70\)
(iv) Conditional ratio of Math only to Science only
\(n_o(M) = 2x = 20\)
New \(n_o(S) = x + 40 = 10 + 40 = 50\)
\(\text{Ratio} = 20 : 50 = 2 : 5\)
New \(n_o(S) = x + 40 = 10 + 40 = 50\)
\(\text{Ratio} = 20 : 50 = 2 : 5\)
Ratio = 2 : 5
11. a) In the local level election, Mr. Harka and Mrs. Sunita were two candidates for the post of the mayor in a municipality and 25,000 voters were in the voter list. Voters were supposed to cast the vote for a single candidate. 12,000 people cast vote for Harka only, 10,000 people cast vote for Sunita only and 1,000 people cast vote even for both the candidates.
- State the given sets in cardinality notations.
- Show the above information in a Venn-diagram.
- How many people didn’t cast the vote?
- Find the percentage of valid votes.
📝 Given Information
Total Voters, \(n(U)\) = 25000
Harka only, \(n_o(H)\) = 12000
Sunita only, \(n_o(S)\) = 10000
Both, \(n(H \cap S)\) = 1000
(i) Cardinality notations
\(n(U) = 25000,\; n_o(H) = 12000,\; n_o(S) = 10000,\; n(H \cap S) = 1000\)
(ii) Venn-diagram Representation
(iii) People who didn’t cast vote
\(n(H \cup S) = n_o(H) + n_o(S) + n(H \cap S) = 12000 + 10000 + 1000 = 23000\)
\(n(\overline{H \cup S}) = n(U) - n(H \cup S) = 25000 - 23000 = 2000\)
\(n(\overline{H \cup S}) = n(U) - n(H \cup S) = 25000 - 23000 = 2000\)
Didn't cast vote = 2000
(iv) Percentage of valid votes
Valid votes $= n_o(H) + n_o(S) = 12000 + 10000 = 22000$
Total cast $= n(H \cup S) = 23000$
\(\text{Percentage} = \frac{22000}{23000} \times 100\% \approx 95.65\%\)
Total cast $= n(H \cup S) = 23000$
\(\text{Percentage} = \frac{22000}{23000} \times 100\% \approx 95.65\%\)
Valid votes % = 95.65%
11. b) There are 400 students in Sarada Secondary School. The students are allowed to cast vote either only for Ajay or for Binita as their school prefect. 200 students cast vote for Ajay only, 175 cast vote for Binita only and 15 of them cast vote even for both.
- What does \(n(A \Delta B)\) represent?
- Represent the above data in a Venn-diagram.
- How many students did not cast vote?
- Calculate the percentage of valid votes.
📝 Given Information
Total students, \(n(U)\) = 400
Ajay only, \(n_o(A)\) = 200
Binita only, \(n_o(B)\) = 175
Both, \(n(A \cap B)\) = 15
(i) Representation of \(n(A \Delta B)\)
\(n(A \Delta B) = n_o(A) + n_o(B)\)
\(n(A \Delta B)\) represents the total number of valid votes.
(ii) Venn-diagram Representation
(iii) Students who did not cast vote
\(n(A \cup B) = n_o(A) + n_o(B) + n(A \cap B) = 200 + 175 + 15 = 390\)
\(n(\overline{A \cup B}) = n(U) - n(A \cup B) = 400 - 390 = 10\)
\(n(\overline{A \cup B}) = n(U) - n(A \cup B) = 400 - 390 = 10\)
Did not cast vote = 10
(iv) Percentage of valid votes
Valid votes $= n_o(A) + n_o(B) = 200 + 175 = 375$
Total cast $= n(A \cup B) = 390$
\(\text{Percentage} = \frac{375}{390} \times 100\% \approx 96.15\%\)
Total cast $= n(A \cup B) = 390$
\(\text{Percentage} = \frac{375}{390} \times 100\% \approx 96.15\%\)
Valid votes % = 96.15%
12. a) In a survey conducted among the participants in a picnic programme, 60 people liked meat, 55 people didn’t like meat, 48 didn’t like fish and 25 liked meat but not fish.
- Write the relation among \(n(U)\), \(n(M)\) and \(n(\overline{M})\).
- How many people were surveyed?
- How many people like fish but not meat?
- Find the ratio of vegetarians to non-vegetarians.
📝 Given Information
Liked meat, \(n(M)\) = 60
Disliked meat, \(n(\overline{M})\) = 55
Disliked fish, \(n(\overline{F})\) = 48
Meat only, \(n_o(M)\) = 25
(i) Relation among universal, main, and complement sets
\(n(U) = n(M) + n(\overline{M})\)
(ii) People surveyed
\(n(U) = n(M) + n(\overline{M}) = 60 + 55 = 115\)
Surveyed, \(n(U) = 115\)
(iii) People who like fish but not meat
\(n(F) = n(U) - n(\overline{F}) = 115 - 48 = 67\)
\(n(M \cap F) = n(M) - n_o(M) = 60 - 25 = 35\)
\(n_o(F) = n(F) - n(M \cap F) = 67 - 35 = 32\)
\(n(M \cap F) = n(M) - n_o(M) = 60 - 25 = 35\)
\(n_o(F) = n(F) - n(M \cap F) = 67 - 35 = 32\)
\(n_o(F) = 32\)
(iv) Ratio of vegetarians to non-vegetarians
Non-vegetarians, \(n(M \cup F) = n_o(M) + n_o(F) + n(M \cap F) = 25 + 32 + 35 = 92\)
Vegetarians (Neither), \(n(\overline{M \cup F}) = n(U) - n(M \cup F) = 115 - 92 = 23\)
\(\text{Ratio} = n(\overline{M \cup F}) : n(M \cup F) = 23 : 92 = 1 : 4\)
Vegetarians (Neither), \(n(\overline{M \cup F}) = n(U) - n(M \cup F) = 115 - 92 = 23\)
\(\text{Ratio} = n(\overline{M \cup F}) : n(M \cup F) = 23 : 92 = 1 : 4\)
Ratio = 1 : 4
12. b) In a group of 1,150 youths, each uses smart phones, either I-phone or Samsung or both or neither, 150 of them use only I-phone, 770 of them use Samsung and 900 use only one of these two smart phones.
- Write the cardinality of \(n(I \Delta S)\).
- Show the above information in a Venn-diagram.
- Find the number of youths who use both of these smart phones.
- Compare the number of youths who use either I-phone or Samsung, and none of these smart phones.
📝 Given Information
Total youth, \(n(U)\) = 1150
I-phone only, \(n_o(I)\) = 150
Samsung, \(n(S)\) = 770
Symmetric difference, \(n(I \Delta S)\) = 900
(i) Cardinality of symmetric difference
\(n(I \Delta S) = 900\)
(ii) Venn-diagram Representation
(iii) Youth who use both of these smart phones
\(n(I \Delta S) = n_o(I) + n_o(S) \implies 900 = 150 + n_o(S) \implies n_o(S) = 750\)
Let \(n(I \cap S) = x\)
\(n(S) = n_o(S) + x \implies 770 = 750 + x \implies x = 20\)
Let \(n(I \cap S) = x\)
\(n(S) = n_o(S) + x \implies 770 = 750 + x \implies x = 20\)
\(n(I \cap S) = 20\)
(iv) Comparison of either and none
Either, \(n(I \cup S) = n_o(I) + n(S) = 150 + 770 = 920\)
None, \(n(\overline{I \cup S}) = n(U) - n(I \cup S) = 1150 - 920 = 230\)
\(\text{Ratio} = 920 : 230 = 4 : 1\)
None, \(n(\overline{I \cup S}) = n(U) - n(I \cup S) = 1150 - 920 = 230\)
\(\text{Ratio} = 920 : 230 = 4 : 1\)
The number of youths who use either phone is 4 times those who use none.
13. a) A marketing company found that, of 200 households surveyed, 80 used neither brand A nor B soaps, 60 used only brand A soap and for every household that used both brands of soap, 3 used only brand B soap.
- If the number of households that used both brands of soaps is x, write the number of households that used only brand B soap.
- Draw a Venn-diagram to show the above information.
- How many households used only one brand of soap?
- What percentage of the total number of households were found to use brand B soap?
📝 Given Information
Total surveyed, \(n(U)\) = 200
Neither, \(n(\overline{A \cup B})\) = 80
Soap A only, \(n_o(A)\) = 60
Both, \(n(A \cap B)\) = \(x\)
(i) Soap B only representation
\(n_o(B) = 3x\)
(ii) Venn-diagram Representation
(iii) Households using only one brand of soap
\(n(U) = n_o(A) + n_o(B) + n(A \cap B) + n(\overline{A \cup B})\)
\(200 = 60 + 3x + x + 80 \implies 200 = 140 + 4x \implies 4x = 60 \implies x = 15\)
\(\text{Only one brand} = n_o(A) + n_o(B) = 60 + 3(15) = 105\)
\(200 = 60 + 3x + x + 80 \implies 200 = 140 + 4x \implies 4x = 60 \implies x = 15\)
\(\text{Only one brand} = n_o(A) + n_o(B) = 60 + 3(15) = 105\)
Only one brand = 105
(iv) Percentage of households using Brand B
\(n(B) = n_o(B) + n(A \cap B) = 3(15) + 15 = 60\)
\(\text{Percentage} = \frac{60}{200} \times 100\% = 30\%\)
\(\text{Percentage} = \frac{60}{200} \times 100\% = 30\%\)
Brand B Percent = 30%
13. b) 100 employees in an office were asked about their preference for tea and coffee. It was observed that for every person who preferred both the drinks, there were 2 people who preferred coffee and 3 people who preferred tea. The number of people who drink neither of the two drinks is same as those who drink both.
- How many people preferred both the drinks?
- How many people preferred only one drink?
- How many people preferred at most one drink?
📝 Given Information
Total employees, \(n(U)\) = 100
Let both, \(n(T \cap C)\) = \(x\)
Tea, \(n(T)\) = \(3x\)
Coffee, \(n(C)\) = \(2x\)
Neither, \(n(\overline{T \cup C})\) = \(x\)
(i) Employees who preferred both
\(n(U) = n(T) + n(C) - n(T \cap C) + n(\overline{T \cup C})\)
\(100 = 3x + 2x - x + x \implies 100 = 5x \implies x = 20\)
\(100 = 3x + 2x - x + x \implies 100 = 5x \implies x = 20\)
\(n(T \cap C) = 20\)
(ii) Employees who preferred only one drink
\(n_o(T) = n(T) - n(T \cap C) = 3x - x = 2x = 40\)
\(n_o(C) = n(C) - n(T \cap C) = 2x - x = x = 20\)
\(\text{Only one drink} = n_o(T) + n_o(C) = 40 + 20 = 60\)
\(n_o(C) = n(C) - n(T \cap C) = 2x - x = x = 20\)
\(\text{Only one drink} = n_o(T) + n_o(C) = 40 + 20 = 60\)
Only one drink = 60
(iii) Employees who preferred at most one drink
\(\text{At most one} = n(U) - n(T \cap C) = 100 - 20 = 80\)
At most one = 80
13. c) Due to the heavy rainfall during a few monsoon days, 140 households were victimized throughout the country. In the first phase, Nepal government has decided to provide the support of either food, shelter or both to a few victimized households. 80 households got food support, 70 got shelter and 50 households got the support of food and shelter both. The government has managed the budget of Rs 10,000 per household for food, Rs 25,000 per household for shelter and Rs 35,000 per household for food and shelter.
- Calculate the amount of budget for food only.
- Calculate the amount of budget for shelter only.
- Calculate the total amount of budget allocated.
- How many households were remained to get support in the first phase?
📝 Given Information
Total victimized, \(n(U)\) = 140
Food, \(n(F)\) = 80
Shelter, \(n(S)\) = 70
Both, \(n(F \cap S)\) = 50
(i) Budget for food only
\(n_o(F) = n(F) - n(F \cap S) = 80 - 50 = 30\)
\(\text{Budget} = 30 \times \text{Rs. } 10,000 = \text{Rs. } 3,00,000\)
\(\text{Budget} = 30 \times \text{Rs. } 10,000 = \text{Rs. } 3,00,000\)
Food only Budget = Rs. 3,00,000
(ii) Budget for shelter only
\(n_o(S) = n(S) - n(F \cap S) = 70 - 50 = 20\)
\(\text{Budget} = 20 \times \text{Rs. } 25,000 = \text{Rs. } 5,00,000\)
\(\text{Budget} = 20 \times \text{Rs. } 25,000 = \text{Rs. } 5,00,000\)
Shelter only Budget = Rs. 5,00,000
(iii) Total budget allocated
Food only Budget $= \text{Rs. } 3,00,000$
Shelter only Budget $= \text{Rs. } 5,00,000$
Both Budget $= 50 \times \text{Rs. } 35,000 = \text{Rs. } 17,50,000$
\(\text{Total Budget} = 3,00,000 + 5,00,000 + 17,50,000 = \text{Rs. } 25,50,000\)
Shelter only Budget $= \text{Rs. } 5,00,000$
Both Budget $= 50 \times \text{Rs. } 35,000 = \text{Rs. } 17,50,000$
\(\text{Total Budget} = 3,00,000 + 5,00,000 + 17,50,000 = \text{Rs. } 25,50,000\)
Total Budget = Rs. 25,50,000
(iv) Households remained to get support
Received support, \(n(F \cup S) = n_o(F) + n_o(S) + n(F \cap S) = 30 + 20 + 50 = 100\)
Remained, \(n(\overline{F \cup S}) = n(U) - n(F \cup S) = 140 - 100 = 40\)
Remained, \(n(\overline{F \cup S}) = n(U) - n(F \cup S) = 140 - 100 = 40\)
Remained households = 40
13. d) In a survey, one-third of the number of children like only mango and 22 do not like mango at all. Also, two-fifth of the number of children like orange but 12 like none of them.
- Draw a Venn-diagram to show the above information.
- Find the total number of children in the survey.
- Find the number of children who like only one fruit.
- Find the difference between those who do not like orange at all and those who do not like mango at all.
📝 Given Information
Let total children, \(n(U)\) = \(x\)
Mango only, \(n_o(M)\) = \(\frac{x}{3}\)
Disliked Mango, \(n(\overline{M})\) = 22
Orange, \(n(O)\) = \(\frac{2x}{5}\)
Neither, \(n(\overline{M \cup O})\) = 12
(i) Venn-diagram Representation
(ii) Total number of children
From Venn-diagram,
\(n(U) = n_o(M) + n(O) + n(\overline{M \cup O})\)
\(x = \frac{x}{3} + \frac{2x}{5} + 12 \implies x = \frac{5x + 6x + 180}{15}\)
\(15x = 11x + 180 \implies 4x = 180 \implies x = 45\)
\(n(U) = n_o(M) + n(O) + n(\overline{M \cup O})\)
\(x = \frac{x}{3} + \frac{2x}{5} + 12 \implies x = \frac{5x + 6x + 180}{15}\)
\(15x = 11x + 180 \implies 4x = 180 \implies x = 45\)
Total Children, \(n(U) = 45\)
(iii) Children who like only one fruit
\(n_o(M) = \frac{x}{3} = \frac{45}{3} = 15\)
Since \(n(\overline{M}) = n_o(O) + 12 = 22 \implies n_o(O) = 10\)
\(\text{Only one fruit} = n_o(M) + n_o(O) = 15 + 10 = 25\)
Since \(n(\overline{M}) = n_o(O) + 12 = 22 \implies n_o(O) = 10\)
\(\text{Only one fruit} = n_o(M) + n_o(O) = 15 + 10 = 25\)
Only one fruit = 25
(iv) Difference between those who do not like orange and mango at all
Do not like orange, \(n(\overline{O}) = n_o(M) + n(\overline{M \cup O}) = 15 + 12 = 27\)
Do not like mango, \(n(\overline{M}) = 22\)
\(\text{Difference} = 27 - 22 = 5\)
Do not like mango, \(n(\overline{M}) = 22\)
\(\text{Difference} = 27 - 22 = 5\)
The number of children who do not like orange at all is 5 more than those who do not like mango at all.