问题1.如果A和B是两组，使得n(A∪B)= 50，n(A)= 28且n(B)= 32，则求n(A∩B)。
解决方案：
n(A∪B) = 50
n(A) = 28
n(B) = 32
We know the formula, n(A∪B) = n(A) + n(B) – n(A∩B)
Putting the values we get
50 = 28 + 32 – n(A∩B)
50 = 60 – n(A∩B)
–10 = – n(A∩B)
n(A∩B) = 10
问题2.如果P和Q是两个集合，使得P具有40个元素，P∪Q具有60个元素，P∩Q具有10个元素，那么Q有多少个元素?
解决方案：
n(P) = 40
n(P ∪ Q)= 60
n(P ∩ Q) =10
We know, n(P∪Q) = n(P) + n(Q) – n(P∩Q)
Putting the values we get
60 = 40 + n(Q)–10
60 = 30 + n(Q)
N(Q) = 30
Q has 30 elements.
问题3.在学校中，有20位教数学或物理的老师。其中，有12位教数学，有4位教物理和数学。有多少人教授物理?
解决方案：
Teachers teaching physics or math = 20
Teachers teaching physics and math = 4
Teachers teaching maths = 12
Let teachers who teach physics be ‘n(P)’ and for Maths be ‘n(M)’
20 teachers who teach physics or math = n (P ∪ M) = 20
4 teachers who teach physics and math = n (P ∩ M) = 4
12 teachers who teach maths = n (M) = 12
We know the formula
n (P ∪ M) = n (M) + n (P) – n (P ∩ M)
Putting the values we get,
20 = 12 + n (P) – 4
20 = 8 + n (P)
n (P) =12
There are 12 physics teachers.
问题4.在一个70人的小组中，37人喜欢喝咖啡，52人喜欢喝茶，每个人都喜欢至少两种饮料中的一种。有多少人喜欢咖啡和茶?
解决方案：
Total number of people = 70
Number of people who like Coffee = n(C) = 37
Number of people who like Tea = n(T) = 52
Total number = n(C∪T) = 70
Person who likes both would be n(C∩T)
We know as formula
n(C∪T) = n(C) + n(T) – n(C ∩ T)
Putting the values we get
70 = 37 + 52 – n (C ∩ T)
70 = 89 – n (C ∩ T)
n(C∩T)=19
问题5.令A和B为两组，从而：n(A)= 20，n(A∪B)= 42，n(A∩B)= 4。
(i)n(B)
解决方案：
n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
(As we know the formula)
Putting the values we get
42 = 20 + n (B) – 4
42 = 16 + n (B)
n(B) = 26
n(B) = 26
(ii)n(A – B)
解决方案：
We know, the formula
n(A – B) = n (A ∪ B) – n (B)
Putting the values we get
n(A – B) = 42 – 26
= 16
(iii)n(B – A)
n(B – A) = n(B) – n(A ∩ B)
Putting the values we get
n(B – A) = 26 – 4 = 22
n(B–A) = 22
问题6：一项调查显示，有76％的印度人喜欢橙子，而62％的人喜欢香蕉。印度人喜欢桔子和香蕉的百分比是多少?
解决方案：
According to question
People who like oranges = 76%
People who like bananas = 62%
Let assume that people who like oranges be n(O)
Let assume that people who like bananas be n(B)
Total number of people who like oranges or bananas = n (O ∪ B) = 100
People who like both oranges and bananas = n (O ∩ B)
We know the formula
n(O∪B) = n(O) + n(B) – n(O ∩ B)
Substituting the values we get
100 = 76 + 62 – n (O ∩ B)
100 = 138 – n (O ∩ B)
n(O∩B) = 38
问题7.在950人中，有750人会说北印度语，而460人会说英语。找：
(i)有多少人会说北印度语和英语。
(ii)有多少人只会讲北印度语。
(iii)有多少人只会说英语。
解决方案：
Let, total number of people be n (P) = 950
People who can speak English n (E) = 460
People who can speak Hindi n (H) = 750
Given in the question:
(i) People who can speak both Hindi and English = n (H ∩ E)
We know the formula
n(P) = n(E) + n(H) – n(H ∩ E)
Putting the values
950 = 460 + 750 – n (H ∩ E)
950 = 1210 – n(H∩E)
n(H ∩ E) = 260
Number of people who can speak both English and Hindi are 260.
(ii) H is disjoint union of n(H–E) and n(H ∩ E).
(If A and B are disjoint then n (A ∪ B) = n (A) + n (B))
H = n (H–E) ∪ n (H ∩ E)
n (H) = n (H–E) + n (H ∩ E)
750 = n(H – E) + 260
n(H–E) = 490
490 people can speak only Hindi.
(iii) How many can speak English only. (According to Question)
E is disjoint union of n (E–H) and n (H ∩ E)
(If A and B are disjoint then n (A ∪ B) = n (A) + n (B))
E = n(E–H) ∪ n(H ∩ E).
n(E) = n(E–H) + n(H ∩ E).
460 = n(H – E) + 260
n H–E) = 460 – 260 = 200
200 people can speak only English.
问题8：一项调查显示，有76％的印度人喜欢橘子，而62％的香蕉。印度人喜欢桔子和香蕉的百分比是多少?
解决方案：
Let assume that n(p) denote total percentage of Indian.
n(o) denotes the percentage of who like oranges.(let)
n(B) denotes the percentage of who like banana. (let)
n(p) = 100 n(O)=76 n(B)=62
We need to find n(O∩B)
n(p) = n(O)+n(B) – n(O∩B)
100 = 76 + 62 – n(O∩B)
n(O∩B) = 138 – 100
= 38
问题9.在950人中，有750人会说北印度语，而460人会说英语。找
(i)有多少人会说北印度语和英语?
(ii)只会说印地语的人有多少?
(iii)只会说英语的人有多少?
解决方案：
(i) Let assume that n(P) denote that number of person.
n(H) assume that number of people who speak Hindi.
n(E) assume that number of people who speak English.
n(p) = 950 , n(H) = 750 and n(E) = 460
we need to find n(H∩E)
According to formula n(P) = n(H) + n(E) – n(H∩E)
950 = 750 + 460 – n(H∩E)
n(H∩E) = 260
(ii) Here we can say that H is disjoint set union of H-E & H∩E
that implies that H = (H – E) ∪ (H ∩ E)
n(H) = n(H – E) + n(H∩E)
750 = n(H – E) – 260
n(H – E) = 490
(iii) E = (E – H) ∪ ( H ∩ E)
n(E) = n(E-H) + n(H ∩ E)
n(E-H) = 460 – 260
= 200
问题10.在一个50人的小组中，有14人喝茶而不是咖啡，还有30人喝茶。找：
(i)多少人同时喝茶和咖啡?
(ii)有多少人喝咖啡但不喝茶?
(i) Let assume that, n(p) denote the total number of persons
n(T) denote the number of person who drink tea
n(C) denote the number of person who drink coffee
We need to find n(T∩C)
T is clear;y disjoint union of T-C and T∩C
T = (T – C) ∪ (T ∩ C)
30 = 14 + n(T ∩ C)
n(T ∩ C) = 16
(ii) Here we need to find C-T
According to the formula: n(P) = n(C) + n(T) – n(T ∩ C)
50 = n(C) + 14
n(C) = 36
Now , C is the disjoint union of C-T and (T ∩ C)
C = (C-T) ∪ (T ∩ C)
n(C) = n(C-T) + n(C ∩ T)
n(C- T) = 36 – 20
= 20