LESSON: SET
Exercise 1.1
1. For the set A= { x : x Є N and x < 10} and ф, find the followings –
(a) n(A) and n(ф) (b) n(Aꓴф) and n(A ꓵ ф)
Solution:
Given, A= { x : x Є N and x < 10}
(a) n(A)=10 and n(ф)=0
(b) n(Aꓴф)= n(A) + n(ф)
=10 + 0
= 10
And n(A ꓵ ф)= n(A) + n(B) – n(A ꓴ ф)
= 10 + 0 – 10
2. Let A and B be two sets and U be their Universal set. If n(U)= 120, n(A)= 42, n(B)= 50 and n(AꓵB)= 21 then find,
(i) n (AUB), n(A – B), n(B – A) and n(AˊꓵBˊ)
(ii) n(Bˊ), n(Aˊ), n(AꓴB)ˊ
(iii) n(PUQ) and n(PꓵQ) where P= A – B and Q= AꓵB
(iv) How many elements are there in the set U – (AUB)
Solution:
Given, n(U)=120, n(A)= 42, n(B)= 50, n(AꓵB)= 21
i) We know that
n (AUB)= n(A)+n(B) – n(AꓵB)
=42+50 – 21
=71
Again, n (A ̶ B)= n(A) ̶ n(AꓵB)
= 42 – 21
= 21
n (B ̶ A)= n(B) ̶ n(AꓵB)
= 50 – 21
= 29
And n(AˊꓵBˊ)= n(AꓴB)ˊ
= n(U) – n(AUB)
= 120 – 71
= 49
(ii) n(Bˊ)= n(U) – n(B)
= 120 – 50
= 70
n (Aˊ)= n(U) – n(A)
= 120 – 42
= 78
And n(AUB)ˊ= n(U) – n(AUB)
= 120 – 71
= 49
(iii) n(PUQ)= n [(A ̶ B) U (AꓵB)]
= n[(AꓵBˊ) U (AꓵB)]
= n[Aꓵ(BˊUB)] (distribution law)
= n [AꓵU]
= n(A)
= 42
And n(PꓵQ)= n [(A ̶ B) ꓵ (AꓵB)]
= n[(AꓵBˊ) ꓵ (AꓵB)]
= n[ A ꓵ (BꓵBˊ)]
= n [A ꓵ ф]
= n(ф)
= 0
(iv) n[U – (AUB)] = n(U) – n(AUB)
= 120 – 71
= 49
3. If n(AꓵB)= 36, n(A – B)= 25, n(B – A)= 20, Then find n(AUB), n(A) and n(B).
Solution:
We know that,
n(AUB)= n(A – B) + n(B – A) + n(AꓵB)
= 25 + 20 + 36
= 81
n(A)= n(A – B) + n(AꓵB)
= 25 + 36
= 61
n(B)= n(B – A) + n(AꓵB)
= 20 + 36
= 56
5. In a class test in mathematics and English, it was found that 55 students have passed in Mathematics, 46 have passed in English and 35 passed in both the subjects. If the number of students who appeared in the test is 100, the find
(i) the percentage of unsuccessful students in both subjects.
(ii) the percentage of students who have passed in Mathematics only.
(iii) the percentage of students who have passed in English only.
Solution:
Let, set of students who have passed in Mathematics be M
Set of students who have passed in English be E
Given, n(U)=100, n(M)= 55, n(E)= 46, and n(MꓵE)= 35
We know that,
n(MUE)= n(M) + n(E) – n(MꓵE)
= 55 + 46 – 35
= 66
(i) Number of students unsuccessful both subjects= n(MꓵE)ˊ
= n(U) - n(MUE)
= 100 – 66
= 34
Therefore, the percentage of unsuccessful students in both subjects= 34%
(ii) Number of students who have passed in Mathematics= n(M – E)
= n(M) - n(MꓵE)
= 55 – 35
= 20
Therefore, the percentage of students who have passed in Mathematics= 20%
(iii) Number of students who have passed in English = n(E – M)
= n(E) - n(MꓵE)
= 46 – 35
= 11
Therefore, the percentage of students who have passed in English= 11%
6. In a survey of 550 students in a school, it was found that 175 students drink milk, 300 students drink tea and 110 students drink both milk and tea. Find the number of students who drinks neither milk nor tea.
Solution:
Let, the set of students drink milk be M
The set of students drink tea be T
Given, n(M)= 175, n(T)= 300 and n(MꓵT)= 110
We know that,
n(MUT)= n(M) + n(T) - n(MꓵT)
= 175 + 300 – 110
= 365
Therefore, the number of students who drink neither milk nor tea= 550 – 365
= 185
7. In a survey among a group of employees of a central Govt. office in Assam, it was found that 80 of them can speak the Assamese, 70 of them can speak English and 50 of them can speak both Assamese and English. If each of the employees who took part in the survey can speak either Assamese or English or both of the two languages, then find
(i) the total number of employees who participated in the survey?
(ii) how many of them can speak Assamese only?
(iii) how many of them can speak English only?
Solution:
Let the set of employees can speak Assamese be A and the set of employees can speak English be E
Given, n(A)= 80, n(E)= 70 and n(AꓵE)= 50
(i) We know that, n(AUE)= n(A) + n(E) – n(AꓵE)
= 80 + 70 – 50
= 100
Therefore, the total number of employees= 100
(ii) the number of employees who can speak Assamese only = n(AUE) – n(E)
= 100 – 70
= 30
(iii) the number of employees who can speak English only = n(AUE) – n(A)
= 100 – 80
= 20
8. In a club of 250 members, it is found that 130 of them drink tea and 85 of them drink tea but not coffee. If each of the members of the club drinks at least one of the items between tea and coffee, the find-
(i) how many members drink coffee
(ii) how many members drink coffee, but not tea?
Solution:
Let the set of members drink tea be T and
The set of members drink coffee be C
Given, n(T)= 130, n(TUC)= 250, n(T – C)= 85
(i) The number of members who drink coffee, n(C)= n(TUC) – n(T – C)
= 250 – 85
= 165
(ii) n(TꓵC)= n(T) + n(C) – n(TUC)
= 130 + 165 – 250
=45
Therefore, number of members drink coffee, but not tea, n(C – T)= n(C) – n(TꓵC)
= 165 – 45
= 120
9. In a class of 90 students, 60 students play volleyball, 53 students play badminton and 35 students play both of the games. Find the number of students
(i) who do not play any one of the two games
(ii) who play badminton only, but not volleyball
(iii) who play volleyball only, but not badminton
(iv) who play atleast one of two games?
Solution:
Let the set students play volleyball be V and
The set of students play badminton be B and set of universal be U
Given, n(U)= 90, n(V)= 60, n(B)= 53 and n(VꓵB)= 35
Therefore, n(VUB)= n(V) + n(B) - n(VꓵB)
= 60 + 53 – 35
= 78
(i) number of students who do not play any one of the two games,=n(U)-n(VUB)
= 90 – 78
= 12
(ii) the number of students who play badminton only, but not volleyball,
n(B – V)= n(B) - n(VꓵB)
= 53 – 35
= 18
(iii) the number of students who play volleyball only, but not badminton,
n(V – B)= n(V) - n(VꓵB)
= 60 – 35
= 25
(iv) the number of students who play atleast one of two games,
n(VUB)= n(V) + n(B) - n(VꓵB)
= 60 + 53 – 35
= 78
10. In a survey among 1500 families of a town it was found that 1263 families have TV, 639 families have radio and 197 families have neither TV nor Radio.
(i) how many families in that town have both radio and TV?
(ii) how many families have TV only, but no radio?
( iii) how many families have radio only, but no TV.
Solution:
Let total set of families be U
The set of families have TV be T
The set of families have radio be R
Given, n(U)= 1500, n(T)= 1263, n(R)=639 and n(TˊꓵRˊ)= 197
Now, n(TˊꓵRˊ)= n(U) – n(TUR)
ð 197 = 1500 - n(TUR)
ð n(TUR)= 1500 – 197
ð n(TUR)= 1306
(i) the number of families in that town have both radio and TV,
n(TꓵR) = n(T) + n(R) – n(TUR)
= 1263 + 639 – 1303
= 599
(ii) the number of families have TV only, but no radio,
n(T – R)= n(T) - n(TꓵR)
= 1263 – 599
= 664
(iii) the number of families have radio only, but no TV,
n(R – T)= n(R) - n(TꓵR)
= 639 – 599
= 40
11. Out of 180 students of a class, 76 of them study Mathematics, 81 of them study Physics and 80 of them study Chemistry. Also, 34 of them study Mathematics and Physics, 30 of them study Mathematics and Chemistry and 33 of them study Physics and Chemistry. If 18 students study all the three subjects then find –
(i) the number of students who study only Physics
(ii) the number of students who study only Chemistry
(iii )the number of students who study only Mathematics
(iv ) the number of students who study mathematics and Physics, but not Chemistry
( v) the number of students who study Physics and Chemistry, but not Mathematics
(vi ) the number of students who study chemistry and Mathematics, but not Physics
(vii ) the number of students who study none of the three subjects.
Solution:
Let the set of total students are U
Set of students study Mathematics be M
Set of students study Physics be P
Set of students study Chemistry be C
Given, n(U)= 180, n(M)= 76, n(P)= 81, n(C)= 80,
n(MꓵP)= 34, n(MꓵC)= 30, n(PꓵC)= 33 and n(MꓵPꓵC)= 18
(i) the number of students who study only Physics= n(P) – n(MꓵP)–n(PꓵC)+ n(MꓵPꓵC)
= 81 – 34 – 33 + 18
= 32
(ii) the number of students who study only Chemistry=n(C)–n(MꓵC)- n(PꓵC)+n(MꓵPꓵC)
= 80 – 30 – 33 + 18
= 35
(iii) the number of students who study only Mathematics
= n(M)–n(MꓵP)–n(MꓵC)+n(MꓵPꓵC)
= 76 – 34 – 30 + 18
= 30
(iv) the number of students who study mathematics and Physics, but not Chemistry
= n(MꓵP) - n(MꓵPꓵC)
= 34 – 18
= 16
(v) the number of students who study Physics and Chemistry, but not Mathematics
= n(PꓵC) - n(MꓵPꓵC)
= 33 – 18
= 15
(vi) the number of students who study chemistry and Mathematics, but not Physics
= n(CꓵM) - n(MꓵPꓵC)
= 30 – 18
= 12
(vii) the number of students who study none of the three subjects
= n(U) – n(MUPUC)
= n(U) – [n(M) + n(P) + n(C) – n(MꓵP) – n(PꓵC) – n(MꓵC) + n(MꓵPꓵC)]
= 180 – [ 76 + 81 + 80 – 34 – 33 – 30 + 18]
= 180 – 158
= 22
No comments:
Post a Comment