Mathematics Questions and Answers – Practical Problems on Union and Intersection of Two Sets

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This set of Mathematics MCQs focuses on “Practical Problems on Union and Intersection of Two Sets”.

1. Which of the following is correct?
a) n (A ∪ B) = n (A) + n (B) + n (A ∩ B)
b) n (A ∪ B) = n (A) – n (B) – n (A ∩ B)
c) n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
d) n (A ∩ B) = n (A) + n (B) + n (A ∪ B)
View Answer

Answer: c
Explanation: In the Venn diagram,
mathematics-questions-answers-practical-problems-union-intersection-two-sets-q1
n (A ∪ B) = a + b + c
n(A) = a + c, n(B) = b + c and n (A ∩ B) = c
n (A) + n (B) – n (A ∩ B) = (a + c) + (b + c)-c = a + b + c.
Hence, n (A ∪ B) = n (A) + n (B) – n (A ∩ B).
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2. What is A ∪ B for two mutually exclusive sets A and B?
a) n (A) + n (B)
b) n (A) – n (B)
c) n (B) – n (A)
d) n (A) * n (B)
View Answer

Answer: a
Explanation: We know, n (A ∪ B) = n (A) + n (B) – n (A ∩ B).
For mutually exclusive sets A ∩ B=Φ i.e. n (A ∩ B)=0.
n (A ∪ B) = n (A) + n (B).

3. If set A has 10 elements, set B has 20 elements, 5 elements are common to both then find the number of elements in X∪Y.
a) 10
b) 15
c) 20
d) 25
View Answer

Answer: d
Explanation: We know, n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
Given, n (A) = 10, n (B) = 20, n (A ∩ B) = 5.
n (A ∪ B) = 10 + 20 – 5 = 25.
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4. In a family of 10 members, 7 of them like tea or coffee, 4 of them like tea and 5 of them like coffee. How many of them like both tea and coffee?
a) 1
b) 2
c) 7
d) 9
View Answer

Answer: b
Explanation: We know, n (T ∪ C) = n (T) + n (C) – n (T ∩ C)
Given, n (T ∪ C) = 7, n(T)=4, n(C)=5
7 = 4 + 5 – n (T ∩ C)
n (T ∩ C) = 2.
Hence, 2 members like both tea and coffee.

5. In a family of 10 members, 7 of them like tea or coffee, 4 of them like tea and 5 of them like coffee. How many of them like neither tea nor coffee?
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: c
Explanation: Given, n (T ∪ C) = 7.
By de Morgan’s law, n (T’ ∩ C’) = n (T ∪ C)’ = Total – n (T ∪ C) = 10 – 7 = 3.
So, 3 members like neither tea nor coffee.
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6. In a family of 10 members, 7 of them like tea or coffee, 4 of them like tea and 5 of them like coffee. How many of them like only tea?
a) 2
b) 3
c) 4
d) 5
View Answer

Answer: a
Explanation: We know, n (T ∪ C) = n (T) + n (C) – n (T ∩ C)
Given, n (T ∪ C) = 7, n(T)=4, n(C)=5
7=4+5- n (T ∩ C)
n (T ∩ C) = 2.
n (only T) = n (T ∩ C’) = n (T)- n (T ∩ C) = 4-2=2.
2 members like only tea.

7. In a family of 10 members, 7 of them like tea or coffee, 4 of them like tea and 5 of them like coffee. How many of them like only coffee?
a) 2
b) 3
c) 4
d) 5
View Answer

Answer: b
Explanation: We know, n (T ∪ C) = n (T) + n (C) – n (T ∩ C)
Given, n (T ∪ C) = 7, n(T)=4, n(C)=5
7=4+5- n (T ∩ C)
n (T ∩ C) = 2.
n (only C) = n (C ∩ T’) = n (C) – n (T ∩ C) = 5-2=3.
3 members like only coffee.
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8. In a city of 1000 population, all people read newspaper. 600 people read Hindustan Times and 700 people read Times of India. How many people read both newspaper?
a) 100
b) 300
c) 1000
d) 1300
View Answer

Answer: b
Explanation: Since all people read newspaper so, n (T ∪ H) = 1000.
n(H) = 600 and n(T) = 700.
n (T ∪ H) = n (T) + n (H) – n (T ∩ H)
1000 = 700 + 600 – n (T ∩ H)
n (T ∩ H) = 1300-1000 = 300.

9. In a city of 1000 population, all people read newspaper. 600 people read Hindustan Times and 700 people read Times of India. How many people read only Hindustan Times?
a) 100
b) 200
c) 300
d) 600
View Answer

Answer: c
Explanation: Since all people read newspaper so, n (T ∪ H) = 1000.
n(H) = 600 and n(T) = 700.
n (T ∪ H) = n (T) + n (H) – n (T ∩ H)
1000 = 700 + 600 – n (T ∩ H)
n (T ∩ H) = 1300-1000 = 300.
n (T’ ∩ H) = n(H)- n (T ∩ H) = 600-300 = 300.
300 persons read only Hindustan Times.
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10. In a city of 1000 population, all people read newspaper. 600 people read Hindustan Times and 700 people read Times of India. How many people read only Times of India?
a) 200
b) 400
c) 600
d) 700
View Answer

Answer: b
Explanation: Since all people read newspaper so, n (T ∪ H) = 1000.
n(H) = 600 and n(T) = 700.
n (T ∪ H) = n (T) + n (H) – n (T ∩ H)
1000 = 700 + 600 – n (T ∩ H)
n (T ∩ H) = 1300-1000 = 300.
n (T ∩ H’) = n (T) – n (T ∩ H) = 700-300 = 400.
400 persons read only Hindustan Times.

11. In a committee, if 50 people speak French, 20 people speak Spanish, 10 people speak both the languages, then how many people speak at least one of the two languages?
a) 10
b) 20
c) 50
d) 60
View Answer

Answer: d
Explanation: We know, n (F∪S) = n (F) + n (S) – n(F∩S)
Given, n (F)=50, n (S)=20, n (F∩S) = 10
n (F∪S) = 50 + 20 – 10 = 60.
60 people speak at least one of the two languages.

12. A group conducted a survey of 100 consumers and reported that 72 consumers like apples and 45 consumers like oranges, what is the least number that must have liked both products?
a) 14
b) 15
c) 16
d) 17
View Answer

Answer: d
Explanation: n(U)=100, n(A)=72, n(O)=45
n (A∪O) = n (A) + n (O) – n(A∩O)
n (A∪O) ≤ n(U) => n (A∪O) ≤ 100
Greatest value of n (A∪O) is 100 which gives least value of n(A∩O)
100=72+45 – n(A∩O)
n(A∩O) = 17 least value of n(A∩O).

13. In a population of 1000 people, 700 people play football, 600 people play cricket, and 100 people play both. Is the data correct?
a) True
b) False
View Answer

Answer: b
Explanation: Given, n(U)=1000, n(F)=700, n(C)=600, n(F∩C) =100
n (F∪C) = n (F) + n (C) – n(F∩C)
n (F∪C) = 700+600-100=1200
We know, n (A∪O) ≤ n(U). Here this is not valid so, data is incorrect.

14. In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the student knows either Hindi or English. How many students are there in the group?
a) 75
b) 100
c) 125
d) 175
View Answer

Answer: c
Explanation: Given, n(H)=100, n(E)=50, n(H∩E) = 25
We know, n (H∪E) = n (H) + n (E) – n(H∩E)
n (H∪E) = 100+50-25=125
Since each of the student knows either Hindi or English, so n(U) = n (H∪E) = 125.

15. If 200 people like tea, 300 people like coffee and 250 people like both. Is the data correct?
a) True
b) False
View Answer

Answer: b
Explanation: Given, n(T)=200, n(C)=300, n(T∩C) = 250
We know, n(T∩C) ≤ n(T) and n(T∩C) ≤ n(C)
But here n(T∩C) ≤ n(T) is not valid so, data given is incorrect.

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter