(i) \(\ n(A) \) and \(\ n(\emptyset) \)
(ii) \(\ n(A \cup \emptyset) \) and \(\ n(A \cap \emptyset) \)
Solution:
Here, \(\ A= \) { \(\ x: x \in N \)} = {1,2,3,4,5,6,7,8,9,10}
\(\ n(A) = 10 \) and \(\ n(\emptyset) = 0 \)
(i) \(\ n(A \cup \emptyset) = n(A) + n(\emptyset) =10 + 0 = 10 \) and
(ii) \(\ n(A \cap \emptyset)= n(A) + n(B)-n(A \cup \emptyset) =10 + 0-10 = 0 \)
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 \(\ n(A\ \cap B)= 21 \) then find,
(i) \(\ n (A\cup B), n(A-B), n(B-A) \) and \(\ n(Aˊ\cap Bˊ) \)
(ii) \(\ n(Bˊ), n(Aˊ), n(A \cup B)ˊ \)
(iii) \(\ n(P\cup Q) \) and \(\ n(P\cap Q) \) where P = A-B and \(\ Q= A\cap B \)
(iv) how many elements are there in the set \(\ U-(A\cup B) \)
Solution:
Here, \(\ n( \cup )=120, n(A)= 42, n(B)= 50, n(A\cap B) = 21 \)
(i) We know, \(\ n (A \cup B)= n(A)+n(B)-n(A\cap B) \)
\(\ = 42+50-21 = 71 \)
Also, \(\ n (A-B) = n(A)- n(A\cap B) = 42-21 = 21 \)
And, \(\ n (B-A) = n(B)- n(A\cap B) = 50-21 = 29 \)
Now, \(\ n(Aˊ\cap Bˊ) = n(A \cup B)ˊ \)
\(\ = n( U )-n(A\cup B) = 120-71 = 49 \)
(ii) \(\ n(Bˊ)= n( \cup )-n(B) = 120-50 = 70 \)
\(\ n (Aˊ)= n( \cup )-n(A) = 120-42 = 78 \)
Now, \(\ n(A \cup B)ˊ= n( \cup )-n(A \cup B) = 120-71 = 49 \)
(iii) \(\ n(P \cup Q)= n [(A-B) \cup (A\cap B)] = n[(A\cap Bˊ) U (A\cap B)] \)
\(\ = n[A\cap (Bˊ\cup B)] \) (distribution law)
\(\ = n [A\cap U] = n(A) = 42 \)
Now, \(\ n(P\cap Q)= n [(A-B) \cap (A\cap B)] = n[(A\cap Bˊ) \cap (A\cap B)] \)
\(\ = n[ A \cap (B\cap Bˊ)] = n [A \cap \emptyset]= n(\emptyset) = 0 \)
(iv) \(\ n[U-(A \cup B)] = n( \cup )-n(A \cup B) = 120-71 = 49 \)
3. If n(\(A\cap B\)) = 36, n(A-B) = 25, n(B-A) = 20 then find n(\(A\cup B)\), n(A) and n(B).
Solution:
Here, \(\ n(A\cap B) = 36, n(A-B) = 25, n(B-A) = 20 \)
We know, \(\ n(A\cup B) = n(A-B) + n(B-A) + n(A\cap B) \)
\(\ = 25 + 20 + 36 = 81 \)
Also, \(\ n(A)= n(A-B) + n(A\cap B) = 25 + 36 = 61 \)
And, \(\ n(B)= n(B-A) + n(A\cap B) = 20 + 36 = 56 \)
4. Draw a Venn diagram to demarcate the regions \(\ A\cap B, A-B \) and \(\ B-A \) with respect to the question (3) above and hence verify the results obtained already.
Solution:
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, then 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, U = the set of students who have passed in Mathematics
And, E = the set of students who have passed in English
Then, \(\ n(U)=100, n(M) = 55, n(E) = 46, n(M\cap E) = 35 \)
We know, \(\ n(M\cup E)= n(M) + n(E)- n(M\cap E) \)
\(\ = 55 + 46-35 = 66 \)
(i) The number of unsuccessful students in both subjects,
\(\ n(M\cap E)ˊ = n(U)-n(MUE) = 100-66 = 34 \)
Therefore, the percentage of unsuccessful students in both subjects is 34%
(ii) Number of students who have passed in Mathematics,
\(\ n(M-E) = n(M)-n(M\cap E) = 55-35 = 20 \)
Therefore, the percentage of students who have passed in Mathematics is 20%
(iii) Number of students who have passed in English,
\(\ n(E-M) = n(E)-n(M\cap E) = 46-35 = 11 \)
Therefore, the percentage of students who have passed in English is 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, M = the set of students drink milk And, T = the set of students drink tea Then, \(\ n(M) = 175, n(T) = 300, n(M\cap T) = 110 \)
also U = 550
We know, \(\ n(M\cup T) = n(M) + n(T)-n(M\cap T) \)
\(\ = 175 + 300-110 = 365 \)
So, the required number of students who drinks neither milk nor tea,
\(\ n(M\cup T)’ = U-n(M\cup T) = 550-365 = 185 \)
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, A = the set employees who can speak Assamese
And, E = the set of employees who can speak English
Then, \(\ n(A) = 80, n(E) = 70, n(A\cap E) = 50 \)
(i) The total number of employees who participated in the survey,
\(\ n(A\cup E) = n(A) + n(E)-n(A\cap E) \)
\(\ = 80 + 70-50 = 100 \)
(ii) The number of employees who can speak Assamese only,
\(\ n(A-E) = n(A)-n(A\cap E) = 80-50 = 30 \)
(iii) The number of employees who can speak English only,
\(\ n(E-A) = n(E)-n(A\cap E) = 70-50 = 20 \)
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, then find-
(i) How many members drink coffee?
(ii) How many members drink coffee, but not tea?
Solution:
Let, T = the set of members who drink tea
And, C = the set of members who drink coffee
Then, \(\ n(T) = 130, n(T-C) = 85 \)
also, \(\ n(T\cup C) = 250 \)
We know, \(\ n(T) = n(T-C) + n(T\cap C) \)
\(\ ⇨ 130-85 = n(T\cap C) \)
\(\ ⇨ n(T\cap C) = 45 \)
(i) The number of members who drink coffee,
\(\ n(C) = n(TUC)-n(T-C) = 250-85 = 165 \)
(ii) The number of members who drink coffee but not tea,
\(\ n(C-T) = n(C)-n (T\cap C) = 165-45 = 120 \)
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 at least one of two games?
Solution:
Let, V = the set of students who play Volleyball
and, B = the set of students who play badminton
So, \(\ n(V) = 60, n(B) = 53, n(V\cap B) = 35 \)
also \(\ n(U) = 90 \)
Therefore, \(\ n(VUB) = n(V) + n(B)-n(V\cap B) = 60 + 53-35 = 78 \)
The number of students who do not play any one of the two games,
\(\ n(V’ \cap B’) = n(VUB)’ = n(U)-n(VUB) = 90-78 = 12 \)
The number of students who play badminton only, but not volleyball,
\(\ n(B-V) = n(B)-n(V\cap B) = 53-35 = 18 \)
The number of students who play volleyball only, but not badminton,
\(\ n(V-B) = n(V)-n(V\cap B) = 60-35 = 25 \)
The number of students who play at least one of two games,
\(\ n(V\cup B) = n(V) + n(B)-n(V\cap B) = 60 + 53-35 = 78 \)
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, T = the set of families having TV
And, R = the set of families having radio
Them, \(\ n(U) = 1500, n(T) = 1263, n(R) = 639, n(T’\cap R’) = 197 \)
We know, \(\ n(T’\cap R’) = n(U)-n(T\cup R) \)
[by de-Morgan’s law, \(\ n(T’\cap R’) = n(T\cup R)’ \) ]
\(\ ⇨ 197 = 1500-n(T\cup R) \)\(\ ⇨ n(T\cup R) = 1500-197 = 1303 \)
(i) the number of families in that town have both radio and TV,
\(\ n(T\cap R) = n(T) + n(R) - n(T\cup R) \)
\(\ = 263 + 639-1303 = 599 \)
(ii) the number of families have TV only, but no radio,
\(\ n(T-R) = n(T)-n(T\cap R) = 1263-599 = 664 \)
(iii) the number of families have radio only, but no TV,
\(\ n(R-T) = N(R)-n(T\cap R) = 639-599 = 40 \)
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
Simplified formula to be used:
(i) \(\ n(A\cap B’\cap C’) = n(A)-n(A\cap B)-n(A\cap C) + n(A\cap B\cap C) \); for finding the number of elements in A only but not in B and C.
(ii) \(\ n[(A\cap B)\cap C’] = n(A\cap B)-n(M\cap P\cap C \); for finding the number of elements in A and B but not in C.
Solution:
Let, M = the set of students who study Mathematics P = the set of students who study Physics C = the set of students who study Chemistry So, \(\ n( \cup ) = 180, n(M) = 76, n(P) = 81, n(C) = 80, \)
\(\ n(M \cap P) = 34, n(M \cap C) = 30, n(P \cap C) = 33 \)
Also, \(\ n(M \cap P \cap C) = 18 \)
then,
(i) The number of students who study only Physics,
\(\ n(M’ \cap P \cap C’) = n(P)-n(M \cap P)-n(P \cap C) + n(M \cap P \cap C) \)
\(\ = 81-34-33 + 18 = 32 \)
The number of students who study only Chemistry,
\(\ n(M’ \cap P’ \cap C) = n(C)-n(M \cap C)-n(P \cap C) + n(M \cap P \cap C) \)
\(\ = 80-30-33 + 18 = 35 \)
The number of students who study only Mathematics,
\(\ n(M \cap P’ \cap C’) = n(M)-n(M \cap P)-n(M \cap C) + n(M \cap P \cap C) \)
\(\ = 76-34-30 + 18 = 30 \)
The number of students who study Mathematics and Physics, but not Chemistry,
\(\ n[(M \cap P) \cap C’] = n(M \cap P)-n(M \cap P \cap C) \)
\(\ = 34-18 = 16 \)
The number of students who study Physics and Chemistry, but not Mathematics,
\(\ n[M’ \cap (P \cap C)] = n(P \cap C)-n(M \cap P \cap C) \)
\(\ = 33-18 = 15 \)
The number of students who study Chemistry and Mathematics, but not Physics,
\(\ n[(M\cap C)\cap P’] = n(M\cap C)-n(M \cap P \cap C) \)
\(\ = 30-18 = 12 \)
The number of students who study none of the three subjects,
\(\ n(M’\cap P’\cap C’) = n(M\cup P\cup C)’ \) [de-Morgan’s law]
\(\ = n(U)-n(M\cup P\cup C) \)
\(\ = 180-[n(M) + n(P) + n(C)-n(M\cap P)-n(P\cap C)-n(M\cap C) + n(M\cap P\cap C)] \)
\(\ = 180-[ 76 + 81 + 80-34-33-30 + 18] = 180-158 = 22 \)