Sets Class- 7th RS Aggarwal Exe-6 D Goyal Brothers ICSE Maths Solution . We provide step by step Solutions of lesson-6 Sets for ICSE Class-7 **Foundation RS Aggarwal Mathematics** of Goyal Brothers Prakashan . Our Solutions contain all type Questions of Exe-6 D to develop skill and confidence. Visit official Website **CISCE** for detail information about ICSE Board Class-7 Mathematics. we would learn in this article about Cardinal properties of sets . What is cardinal of a set with an example? What is the cardinal number of set? What are cardinal formula sets?

## Sets Class- 7th RS Aggarwal Exe-6 D Goyal Brothers ICSE Maths Solution

Board | ICSE |

Publications | Goyal brothers Prakashan |

Subject | Maths |

Class | 7th |

Chapter-6 | Sets |

Writer | RS Aggrawal |

Book Name | Foundation |

Topics | Solution of Exe-6 D |

Academic Session | 2023 – 2024 |

**What is Cardinal properties of sets**

The word cardinal numbers or cardinal was coined to express the cardinality (size) of sets, which is the size of a set. The number of elements in a finite set is called cardinality, and it is used to characterize the size of the sets.

**What is cardinal of a set with an example?**

The cardinality of a set is defined as the number of elements in a mathematical set. It can be finite or infinite. For example, the cardinality of the set A = {1, 2, 3, 4, 5, 6} is equal to 6 because set A has six elements.

**What are cardinal formula sets?**

The cardinal number of a set A is denoted as n(A), where A is any set and n(A) is the number of members in set A. Consider a set A consisting of the prime numbers less than 10. Set A ={2, 3, 5, 7}. As the set A consists of 4 elements, therefore, the cardinal number of set A is given as n(A) = 4.

**Exercise – 6 D**

Sets Class- 7th RS Aggarwal Goyal Brothers ICSE Maths Solution

**1. Let A and B be two sets such that n(A) = 52, n(B) = 60 and n(A ∩B) = 16. Draw a venn diagram and find : (i) n(A ∪ B) (ii) n(A – B) (iii) n(B – A).**

**Solution:** (i) n(A ∪ B)

n(A) = 52, n(B) = 60 and n(A ∩B) = 16

But n(A ∪ B) = n(A) + n(B) – n(A ∩B)

n(A ∪ B) = 52 + 60 – 16

n(A ∪ B) = 112 – 16

n(A ∪ B) = 96

**Solution: **(ii) n(A – B)

n(A – B) = n(A) – n(A ∩B)

n(A – B) = 52 – 16

n(A – B) = 36

**Solution: **(iii) n(B – A)

n(B – A) = n(B) – n(A ∩B)

n(B – A) = 60 – 16

n(B – A) = 44

**2. Let P and Q be two sets such that n(P ∪ Q) = 70, n(P) = 45 and n(Q) = 38. Draw a venn diagram and find : (i) n(P ∩ Q) (ii) n(P – Q) (iii) n(Q – P).**

**Solution: **(i) n(P ∩ Q)

n(P ∪ Q) = n(P) + n(Q) – n(P ∩ Q)

70 = 45 + 38 – n(P ∩ Q)

70 = 83 – n(P ∩ Q)

n(P ∩ Q) = 83 – 70

n(P ∩ Q) = 13

**Solution: **(ii) n(P – Q)

n(P – Q) = n(P) – n(P ∩ Q)

n(P – Q) = 45 – 13

n(P – Q) = 32

**Solution: **(iii) n(Q – P)

n(Q – P) = n(Q) – n(P ∩ Q)

n(Q – P) = 38 – 13

n(Q – P) = 25

**3. In a city, there are 25 Hindi medium schools, 18 English medium schools and 7 schools have both the mediums. Find**

**(i) how many school are there in all in the city ;**

**(ii) how many school have Hindi medium only ;**

**(iii) how many school have English medium only.**

**Solution: **Let H be the set of Hindi medium school, E be the set of English medium school.

**(i) There in all in the city ;**

∴ n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

n(H ∪ E) = 25 + 18 – 7

n(H ∪ E) = 43 – 7

n(H ∪ E) = 36

**(ii) Only Hindi medium schools ;**

∴ n(H ∪ E) = n(H) – n(H ∩ E)

n(H ∪ E) = 25 – 7

n(H ∪ E) = 18

**(iii) Only English medium schools**

∴ n(H ∪ E) = n(E) – n(H ∩ E)

n(H ∪ E) = 18 – 7

n(H ∪ E) = 11

**4. There is a group of 50 persons who can speak English or Tamil or both. Out of these persons, 37 can speak English and 30 can speak Tamil.**

**(i) How many can speak both English and Tamil?**

**(ii) How many can speak English only?**

**(iii) How many can speak Tamil only?**

**Solution:** No. of person who can speak English or Tamil or both n(E ∪ T) = 50

No. of person who can speak English n(E) = 37

No. of person who can speak Tamil n(T) = 30

Let E be the set of person speaking English and T be the set of person speaking Tamil

∴ n(E ∪ T) = 50

n(E) = 37

n(T) = 30

**(i) No. of person who can speak English or Tamil**

n(E ∪ T) = n(E) + n(T) – n(E ∩ T)

50 = 37 + 30 – n(E ∩ T)

50 = 67 – n(E ∩ T)

n(E ∩ T) = 67 – 50

n(E ∩ T) = 17

**(ii) No. of person who can speak English only**

n(E ∪ T) = n(E) – n(E ∩ T)

n(E ∪ T) = 37 – 17

n(E ∪ T) = 20

**(iii) No. of person who can speak Tamil only**

n(E ∪ T) = n(T) – n(E ∩ T)

n(E ∪ T) = 30 – 17

n(E ∪ T) = 13

**5. In a class of 40 students, each one plays either Tennis or Badminton or both. If 28 play Tennis and 26 play Badminton, find**

**(i) how many play both the games;**

**(ii) how many play Tennis only;**

**(iii) how many play Badminton only.**

**Solution:** No. of student who can play either Tennis or Badminton or both n(T ∪ B) = 40

No. of student who can play Tennis n(T) = 18

No. of student who can play Badminton n(B) = 26

T be the set of student who can play Badminton

∴ n(T ∪ B) = 40

n(T) = 18

n(B) = 26

**(i) No. of student who can play either Tennis or Badminton or both**

n(T ∪ B) = n(T) + n(B) – n(T ∩ B)

40 = 28 + 26 – n(T ∩ B)

40 = 54 – n(T ∩ B)

n(T ∩ B) = 54 – 40

n(T ∩ B) = 14

**(ii) No. of student who can play Tennis only**

n(T ∪ B) = n(T) – n(T ∩ B)

n(T ∪ B) = 28 – 14

n(T ∪ B) = 14

**(iii) No. of student who can play Badminton only**

n(T ∪ B) = n(B) – n(T ∩ B)

n(T ∪ B) = 26 – 14

n(T ∪ B) = 12

**6. In a class of 45 pupils, 21 play chess, 23 play cards and 5 play both the games. Find**

**(i) how many do not play any of the games;**

**(ii) how many play chess only;**

**(iii) how many play cards only.**

**Solution:** Let C be the set of pupils playing chess and K playing cards

Total number of pupils = 45

No. of pupils who can play chess n(C) = 21

No. of pupils who can play cards n(K) = 23

**(i) No. of pupils who can play chess or cards or both**

∴ n(C ∪ K) = n(C) + n(K) – n(C ∩ K)

n(C ∪ K) = 21 + 23 – 45

n(C ∪ K) = 44 – 45

n(C ∪ K) = 1

5 play both the games

5 + 1 = 6

**(ii) No. of pupils who can play chess**

n(C ∪ K) = n(C) – n(C ∩ K)

n(C ∪ K) = 21 – 5

n(C ∪ K) = 16

**(iii) No. of pupils who can play cards**

n(C ∪ K) = n(K) – n(C ∩ K)

n(C ∪ K) = 23 – 5

n(C ∪ K) = 18

**7. In a group of 36 girls, each one can either stitch or weave or can do both. If 25 girls can stitch and 17 can stitch only, how many can weave only?**

**Solution:** No. of girls who can either stitch or weave or both = 36

No. of girls who can stitch = 25

No. of girls who can stitch only = 17

Let S be the set of girls who can stitch and W be the set of girls who can weave

∴ n(S ∪ W) = 36

n(S) = 25

No. of girls who can stitch only = 17

No. of girls who can weave only = 36 – 25

No. of girls who can weave only = 11

**8. In a group of 24 children, each on plays cricket or hockey or both. If 16 play cricket and 12 play cricket only, find how many play hockey only.**

[**Hint.** n(C ∪ H) = 24, n(C) = 16, n(C – H) = 12]

**Solution:** Total number of children = 24

Let C be the set of children playing cricket and H be the set of children playing hockey

∴ n(C ∪ H) = 24

n(C) = 16

n(C – H) = 12

n(C) – n(C ∩ H) = n(C – H)

16 – n(C ∩ H) = 12

n(C ∩ H) = 16 -12

n(C ∩ H) = 4

∴ No. of children who can play hockey only = 24 – 16

No. of children who can play hockey only = 8

**9. In a group of 40 persons, 10 drink tea but not coffee and 26 drink tea. How many drink coffee but not tea?**

**Solution:** Total number of persons = 40

people who can drink only tea but not coffee = 10

people who can drink only tea = 26

Let T be the set of drinking tea and C be the set of drinking coffee

∴ n(T ∪ C) = 40

n(T) = 26

n(T ∩ C) = 10

∴ No. of person drinking coffee not tea = 40 – 26

No. of person drinking coffee not tea = 14

**10. All the people in a locality read the daily newspaper Indian Express or Hindustan Times or both. If 120 read Indian Express and 150 read Hindustan Times and 36 read both, find :**

**(i) How many people are there in the locality;**

**(ii) How many people read only Indian Express.**

**Solution:** No. of people who read Indian Express = 120

No. of people who read Hindustan Times = 150

No. of people who read both = 36

Let I be the set of people reading Indian Express and H be the set of people reading Hindustan Times

Then, n(I) = 120

n(I ∩ H) = 36

**(i) No. of people in the locality**

n(I ∪ H) = n(I) + n(H) – n(I ∩ H)

n(I ∪ H) = 120 + 150 – 36

n(I ∪ H) = 270 – 36

n(I ∪ H) = 234

∴ No. of people in the locality = 234

**(ii) No. of people who read only Indian Express**

n(I ∪ H) = n(I) – n(I ∩ H)

n(I ∪ H) = 120 – 36

n(I ∪ H) = 84

**— : end of Sets Class- 7th RS Aggarwal Exe-6 D Goyal Brothers ICSE Maths Solution:–**

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