Class 10 Mathematics • Set
Exercise 1.2 Solution
Exel in Mathematics(Vedanta Publication)
1. a)A survey conducted in a group of students showed that 50 liked singing, 40 liked dancing, 35 liked acting, 20 liked singing as well as dancing, 12 liked dancing as well as acting, 18 liked acting as well as singing, and 7 liked all three activities. If every student liked at least one activity, answer the following questions.
(a)If S, D and A represent the sets of students who liked singing, dancing and acting respectively, write the cardinality notation for the set of students who liked all three activities.
(b)Draw a Venn-diagram to show the above information.
(c)How many students were asked this question?
(d)How many students liked only one activity?
Solution
Given
Here,
n(S) = 50
n(D) = 40
n(A) = 35
n(S ∩ D) = 20
n(D ∩ A) = 12
n(A ∩ S) = 18
n(S ∩ D ∩ A) = 7
n(S ∪ D ∪ A) = 0
n(S ∩ D ∩ A) = 7
Drawing a Venn-diagram to show the above information:
We have,
n (S ∪ D ∪ A) = n (S) + n (D) + n (A) – n (S ∩ D) – n (D ∩ A) – n (A ∩ S) + n (S ∩ D ∩ A)
= 50 + 40 + 35 – 20 – 12 – 18 + 7 = 82
Hence, the question was asked to 82 students.
The number of students who liked only one activity, nₒ(S) + nₒ(D) + nₒ(A)
= 19 + 15 + 12
= 46
1. b)In an examination, out of 150 students, 65 succeeded in English, 75 in Mathematics, 60 in Nepali, 30 in English as well as in Mathematics, 25 in English as well as in Nepali, 40 in Nepali as well as in Mathematics, and 10 succeeded in all three subjects.
(a)If E, M and N denote the sets of students who succeed in English, Mathematics and Nepali respectively, write the cardinality notation to represent the set of students who succeed in all three subjects.
(b)Show the above information in a Venn-diagram.
(c)Find the number of students who didn’t succeed in all three subjects
(d)Find the number of students who succeeded in exactly one subject.
Solution
Given
Here,
n(U) = 150
n(E) = 65
n(M) = 75
n(N) = 60
n(E ∩ M) = 30
n(N ∩ E) = 25
n(M ∩ N) = 40
n(E ∩ M ∩ N) = 10
n(E ∩ M ∩ N) = 10
Showing the above information in a Venn-diagram:
We have,
n (E ∪ M ∪ N) = n (E) + n (M) + n (N) – n (E ∩ M) – n (M ∩ N) – n (N ∩ E) + n (E ∩ M ∩ N)
= 65 + 75 + 60 – 30 – 25 – 40 + 10
= 115
Again, n( E ∪ M ∪ N ) = n(U) – n (E ∪ M ∪ N)
= 150 – 115
= 35
Hence, 35 students didn’t succeed in all three subjects.
The number of students who succeeded in exactly one subject, nₒ(E) + nₒ(M) + nₒ(N)
= 20 + 15 + 5
= 40
1. c)A hotel provides the extra facilities: swimming, gym and fun-park facilities to its guests. Among 300 guests arrived in the hotel, it was found that 60% guests used swimming, 50% used gym and 40% used fun-park. Moreover, 30% used swimming and gym facilities, 20% used gym and fun-park, 15% used fun-park and swimming while 10% used all three facilities. By using the Venn-diagram, find the number of guests who used
(a)none of the facilities
(b)only one facility
(c)only swimming and gym
Solution
Given
Let, S, G and F denote the sets of guests who used swimming, gym and fun-park respectively. Then,
n(U) = 300
n(S) = 60% of 300 = 180
n(G) = 50% of 300 = 150
n(F) =40% of 300 = 120
n(S ∩ G) = 30% of 300 = 90
n(G ∩ F) = 20% of 300 = 60
n(F ∩ S) = 15% of 300 = 45
n(S ∩ G ∩ F) = 10% of 300 = 30
The given data is shown in the Venn-diagram.
We have,
n(S ∪ G ∪ F) = n (S) + n(G) + n(F) – n(S ∩ G) – n(G ∩ F) – n(F ∩ S) + n(S ∩ G ∩ F)
= 180 + 150 + 120 – 90 – 60 – 45 + 30
= 285
Again, n( S ∪ G ∪ F ) = n(U) – n (S ∪ G ∪ F)
= 300 – 285
= 15
Hence, 15 guests used none of the facilities.
From Venn-diagram,
The number of guests who used only one facility, nₒ(S) + nₒ(G) + nₒ(F)
= 75 + 30 + 45
= 150
The number of guests who used only swimming and gym, nₒ(S ∩ F) = 60
2. a)In an examination, 60% students passed in Nepali (N), 45% passed in English (E) and 40% passed in Science (S). Likewise, 20% of them passed in Nepali and English, 15% passed in English and Science, 25% passed in Nepali and Science, and 5% students passed in all three subjects.
(a)What does n(N ∪ E ∪ S) denote?
(b)Represent the above information in a Venn-diagram.
(c)If 50 students failed in all three subjects, find the total number of students.
(d)If those students who failed in all the subjects were passed in Science, what would be the ratio of number of students who passed in Nepali only and Science only?
Solution
Given
Let, n(U) = 100 then
n(N) = 60
n(E)= 45
n(S) = 40
n(N ∩ E) = 20
n(E ∩ S) = 15
n(N ∩ S) = 25
n(N ∩ E ∩ S) = 5
n(N ∪ E ∪ S) denotes the number of students who passed either in Nepali or English or Science.
The given data is shown in the Venn-diagram.
We have,
n(N ∪ E ∪ S) = n (N) + n(E) + n(S) – n(N ∩ E) – n(E ∩ S) – n(S ∩ N) + n(N ∩ E ∩ S)
= 60 + 45 + 40 – 20 – 15 – 25 + 5
= 90
Again, n( N ∪ E ∪ S ) = n(U) – n (N ∪ E ∪ S)
= 100 – 90
= 10
Hence, 10% students failed in all three subjects.
Let, the total number of students, n(U) = x.
Then, 10% of x = 50
or, x = 500
Hence, the total number of students was 500.
When those students who failed in all the subjects were passed in Science then, the number of students who passed in Science only, nₒ(S) = (5 + 10)% of 500 = 75
Also, the number of students who passed in Nepali only, nₒ(N) = 20% of 500 = 100
Ratio of number of students who passed in Nepali only and Science only = 75100 =3: 4
2. b)Out of some people interviewed in a community about the preference of social media, it was recorded that 62% people use Facebook, 60% use Twitter and 53% use Instagram. Moreover, 35% people use Facebook and Twitter, 30% use Twitter and Instagram, 25% use Facebook and Instagram, and 10% use all these three social media.
(a)Write the cardinalities of given sets.
(b)Represent the above information in a Venn-diagram.
(c)If 30 people did not use any of these media, find the total number of people participated in the survey.
(d)If the people who did not use any of the social media used Twitter, find the ratio of number of people who would use Facebook only and Twitter only.
Solution
Given
Let, F, T and I represent the sets of people who use Facebook, Twitter and Instagram respectively.
n(F) = 62
n(T)= 60
n(T) = 53
n(F ∩ T) = 35
n(T ∩ I) = 30
n(I ∩ F) = 25
n(F ∩ T ∩ I) = 10
Suppose, n(U) = 100 then n(F) = 62, n(T)= 60, n(T) = 53, n(F ∩ T) = 35,
n(T ∩ I) = 30, n(I ∩ F) = 25 and n(F ∩ T ∩ I) = 10
The given data is shown in the Venn-diagram.
We have,
n(F ∪ T ∪ I) = n (F) + n(T) + n(I) – n(F ∩ T) – n(T ∩ I) – n(I ∩ F) + n(F ∩ T ∩ I)
= 62 + 60 + 53 – 35 – 30 – 25 + 10
= 95
Again, n( F ∪ T ∪ I ) = n(U) – n (F ∪ T ∪ I)
= 100 – 95
= 5
Hence, 5% people do not use any of these media.
Let, the total number of students, n(U) = x.
Then, 5% of x = 30
or, x = 600
Hence, the total number of students was 500.
When the people who did not use any of the social media used Twitter, number of people who use Twitter only, nₒ(T) = (5 + 5)% of 600 = 60
Also, number of people who use Facebook only, nₒ(F) = 12% of 600 = 72
Ratio of number of people who use Facebook only and Twitter only = 7260 =6: 5
2. c)A survey was conducted on the students studying in grade 10 in a school to determine suitable place for educational tour among Pokhara, Lumbini and Ilam. It was found that 50 students liked Pokhara, 40 students liked Lumbini, 35 students liked Ilam, 15 students liked all three places and 5 students did not like any places.
(a)If P, L and I denote the sets of students who liked Pokhara, Lumbini and Ilam respectively, write the cardinality notation of students who liked all three places.
(b)Show the above information in a Venn diagram.
(c)How many students are studying in grade 10 in the school?
(d)If 5 students, who did like any places in the survey, liked Ilam, what would be the ratio of students who liked only Pokhara and only Ilam?
Solution
Given
Here,
n(P) = 50
n(L)= 40
n(I) = 35
n(P ∩ L ∩ I) = 15
n(P ∪ L ∪ I) = 5
The cardinality notation of students who liked all three places, n(P ∩ L ∩ I) = 15
The given data is shown in the Venn-diagram.
We have,
n(P ∪ L ∪ I) = n (P) + n(L) + n(I) – n(P ∩ L) – n(L ∩ I) – n(I ∩ P) + n(P ∩ L ∩ I)
= 50 + 40 + 35 – 15 – 15 – 15 + 15
= 95
Again,
Total number of students studying in grade 10, n(U) = n (P ∪ L ∪ I) +n( P ∪ L ∪ I )
= 95 + 5 = 100
When 5 students, who did like any places in the survey, liked Ilam then the number of students who liked only Ilam, nₒ(I) = 20 + 5 = 25
Also, number of students who liked only Pokhara, nₒ(P) = 35
Ratio of students who liked only Pokhara and only Ilam = 3525 =7: 5
3. a)In a survey of 900 tourists who arrived in Nepal during ‘Visit Nepal 2020’, 450 preferred to go trekking, 300 preferred rafting, 400 preferred forest safari, and 100 preferred none of these activities. Likewise, 200 preferred trekking and rafting, 110 preferred trekking and safari, 100 preferred rafting and safari.
(a)How many of them preferred either of these activities?
(b)How many of them preferred all of these activities?
(c)Illustrate the information in a Venn-diagram.
(d)How many of them preferred only two activities?
Solution
Given
Let, T, R and S denote the sets of tourists who preferred trekking, rafting and forest safari respectively. Then,
n(U) = 900
n(T) = 450
n(R) = 300
n(S) = 400
n(T ∩ R) = 200
n(R ∩ S) = 100
n(T ∩ S) = 110
n(T ∪ R ∪ S) = 100
We have, n(T ∪ R ∪ S) = n(U) – n( T ∪ R ∪ S ) = 900 – 100 = 800
So, 800 people preferred either trekking or rafting or forest safari.
We have,
n(T ∪ R ∪ S) = n (T) + n(R) + n(S) – n(T ∩ R) – n(R ∩ S) – n(S ∩ T) + n(T ∩ R ∩ S)
or, 800 = 450 + 300 + 400 – 200 – 100 – 110 + n(T ∩ R ∩ S)
or, 800 = 740 + n(T ∩ R ∩ S)
or, 60 = n(T ∩ R ∩ S)
Hence, 60 people preferred all three activities.
The given data is shown in the Venn-diagram.
From Venn-diagram, nₒ(T ∩ R) + nₒ(R ∩ S) + nₒ(S ∩ T)
= 140 + 40 + 50
= 230
Hence, 230 tourists preferred only two activities.
3. b)In a group of 80 people, 50 prefer modern songs, 45 prefer classic songs and 40 prefer folk songs. Likewise, 25 prefer modern and classic songs, 20 prefer classic and folk songs, while 15 prefer modern and folk songs. Each person prefers at least one type of song. By using a Venn-diagram, find the number of people who prefer:
(a)at least one type of songs
(b)all three types songs
(c)only two types of songs
Solution
Given
Let M, C and F denote the sets of people who preferred modern songs, classic songs and folk songs respectively. Then,
n(U) = 80
n(M) = 50
n(C) = 45
n(F) = 40
n(M ∩ C) = 25
n(C ∩ F) = 20
n(M ∩ F) = 15
n(M ∪ C ∪ F) = 0
We have, n(M ∪ C ∪ F) = n(U) – n( M ∪ C ∪ F ) = 80 – 0 = 80
So, 80 people preferred at least one type of songs.
We have,
n(M ∪ C ∪ F) = n (M) + n(C) + n(F) – n(M ∩ C) – n(C ∩ F) – n(F ∩ M) + n(M ∩ C ∩ F)
or, 80 = 50 + 45 + 40 – 25 – 20 – 15 + n(M ∩ C ∩ F)
or, 80 = 75 + n(M ∩ C ∩ F)
or, 5 = n(M ∩ C ∩ F)
Hence, 5 people prefer all three types songs.
nₒ(M ∩ C) = n(M ∩ C) – n(M ∩ C ∩ F) = 25 – 5 = 20
nₒ(C ∩ F) = n(C ∩ F) – n(M ∩ C ∩ F) = 20 – 5 = 15
nₒ(F ∩ M) = n(F ∩ M) – n(M ∩ C ∩ F) = 15 – 5 = 10
Also, nₒ(M ∩ C) + nₒ(C ∩ F) + nₒ(F ∩ M) = 20 + 15 + 10 = 45
Hence, 45 people refer only two types of songs.
4. a)In a group of students, 20 study Economics, 18 study History, 21 study Science, 7 study Economics only, 10 study Science only, 6 study Economics and Science only and 3 study Science and History only.
(a)Represent the above information in Venn-diagram
(b)How many students study all the subjects?
(c)How many students are there altogether?
Solution
Given
Let E, H, and S be the sets of students who study Economics, History and Science respectively. Here,
n(E) = 20
n(H) = 18
n(S) = 21
nₒ (E) = 7
nₒ (S) = 10
nₒ (E ∩ S) = 6
nₒ (S ∩ H) = 3
The above information is presented in Venn-diagram:
Hint:
n(E ∩ H ∩ S) = n(S) – nₒ(S) – nₒ(E ∩ S) – nₒ(H ∩ S)
= 21 – 10 – 6 – 3
= 2
From Venn-diagram, n(E ∩ H ∩ S) = 2
So, 2 students study all the subjects.
n(M ∪ C ∪ F) = 7 + 8 + 10 + 5 + 3 + 6 + 2 = 41
So, there are 41 students altogether.
4. b)In a survey of a group of people, it was found that 30 of them have business, 35 have services, 25 have farming, 12 have business only, 15 have services only, 10 have business and services only and 6 have services and farming only.
(a)Draw a Venn-diagram to illustrate the above information
(b)Find how many people have all three occupations?
(c)How many people were there in the survey?
Solution
Given
Let B, S, and F be the sets of people who have business, services and farming respectively. Then,
n(B) = 30
n(S) = 35
n(F) = 25
nₒ (B) = 12
nₒ (S) = 15
nₒ (B ∩ S) = 10
nₒ (S ∩ F) = 6
The above information is presented in Venn-diagram:
Hint:
n(E ∩ H ∩ S) = n(S) – nₒ(S) – nₒ(B ∩ S) – nₒ(S ∩ F)
= 35 – 15 – 10 – 6
= 4
From Venn-diagram, n(B ∩ S ∩ F) = 4
So, 4 people have all three occupations.
n(B ∪ S ∪ F) = 12 + 15 + 11 + 10 + 6 + 4 + 4 = 62
So, 62 people were there in the survey.
5. a)Each student in a class of 32 plays at least one game: cricket, football, or basketball. 20 play cricket, 18 play basketball and 25 play football. 9 play cricket and basketball, 13 play football and basketball, and 5 play all three.
(a)Find the number of students who play cricket and football
(b)Find the number of students who play cricket and football but not basketball
(c)Show the information in a Venn-diagram.
Solution
Given
Let C, F, and B be the sets of student who play cricket, football and basketball respectively. Then,
n(U) = n(C ∪ F ∪ B) = 32
n(C) = 20
n(B) = 18
n(F) =25
n(C ∩ B) = 9
n(F ∩ B) = 13
n(C ∩ F ∩ B) = 5
We have,
n(C ∪ F ∪ B) = n (C) + n(F) + n(B) – n(C ∩ F) – n(F ∩ B) – n(B ∩ C) + n(C ∩ F ∩ B)
or, 32 = 20 + 25 + 18 – n(C ∩ F) – 9 – 13 + 5
or, 32 = 46 – n(C ∩ F)
or, n(C ∩ F) = 14
So, 14 students play cricket and football.
nₒ(C ∩ F) = n(C ∩ F) – n(C ∩ F ∩ B) = 14 – 5 = 9
So, 9 students play cricket and football but not basketball.
The above information is shown in Venn-diagram:
5. b)In a group of 150 people, the number of people who like Coke, Pepsi and Sprite are equal. There are 35 people who like Coke and Pepsi, 40 people like Pepsi and Sprite, 44 like Sprite and Coke, 25 people like all three drinks and 34 people don’t like any of these drinks.
(a)Show the information in a Venn-diagram.
(b)How many people like Coke?
(c)How many people like Pepsi but not Coke or Sprite?
Solution
Given
Let C, P, and S be the sets of people who like Coke, Pepsi and Sprite respectively. Then,
n(U) = 150
n(C ∩ P) = 35
n(P ∩ S) = 40
n(S ∩ C) = 44
n(C ∩ P ∩ S) = 25
n( C ∪ P ∪ S ) = 34
The above information is shown in Venn-diagram:
We have, n(C ∪ P ∪ S) = n (U) – n( C ∪ P ∪ S )
= 150 – 34 = 116
Let, n(C) = n(P) = n(S) = x
We have,
n(C ∪ P ∪ S) = n(C) + n(P) + n(S) – n(C ∩ P) – n(P ∩ S) – n(S ∩ C) + n(C ∩ P ∩ S)
or, 116 = x + x + x – 35 – 40 – 44 + 25
or, 116 = 3x – 94
or, 210 = 3x
or, x = 70
∴ n(C) = n(P) = n(S) = 70
No. of people who like Coke, n(C) = 70
No. of people who like Pepsi but not Coke or Sprite, nₒ(P) = 70 – (10 + 25 + 15) = 20