The availability of step-by-step ML Aggarwal Maths for Class 11 Solutions Chapter 1 Sets Ex 1.4 can make challenging problems more manageable.

## ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.4

Question 1.

If A and B are two sets such that n (A) = 28, n (B) = 32 and n (A ∪ B) = 50, find n (A ∩ B).

Solution:

Given n (A) = 28,

n (B) = 32

and n (A ∪ B) = 50

we know that

n (A ∪ B) = n (A) + n (B) – n (A ∩ B)

⇒ 50 = 28 + 32 – n (A ∩ B)

⇒ n (A ∩ B) = 60 – 50 = 10

Question 2.

If X and Y are two sets such that X has 21 elements, Y has 32 elements and X ∩ Y has 11 elements, how many elements does X ∪ Y have?

Solution:

Since X be a set containing 21 elements

∴ n (X) = 21 and Y has 32 elements

∴ n (Y) = 32

Also set X ∩ Y contains 11 elements

∴ n (X ∩ Y) = 11

∴ n (X ∪ Y) = n (X) + n (Y) – n (X ∩ Y)

= 21 + 32 – 11 = 42

Hence the set X ∪ Y contains 42 elements.

Question 3.

If n (A) = 20, n (B) = 18 and n (A ∩ B) = 5, calculate

(i) n (A ∪ B)

(ii) n (A – B)

(iii) n (B – A).

Solution:

Given n (A) = 20 ;

n (B) = 18

and n (A ∩ B) = 5

(i) n (A ∪ B) = n(A) + n(B) – n (A ∩ B)

= 20 + 18 – 5 = 33

(ii) n (A – B) = n (A) – n (A ∩ B)

= 20 – 5 = 15

(iii) n (B – A) = n(B) – n (A ∩ B)

= 18 – 5 = 13

Question 4.

If n (A ∪ B) = 18, n (A – B) = 5, n (B – A) = 3 then find the following : find n (A ∩ B).

Solution:

Given n (A ∪ B) = 18,

n (A – B) = 5 ;

n (B – A) = 3

We know that

n (A ∪ B) = n (A – B) + n (A ∩ B) + n (B – A)

18 = 5 + n (A ∩ B) + 3

n (A ∩ B) = 18 – 8 = 10

Question 5.

If A and B are two sets such that n (A – B) = 5, n (B – A) = 3 and n (A ∩ B) = 10, then find the following:

(i) n (A ∪ B)

(ii) n (A)

(iii) n (B).

Solution:

Given n (A – B) = 5 ;

n (B – A) = 3 ;

n (A ∩ B) = 10

(i) ∴ n (A ∪ B) = n (A – B) + n (A ∩ B) + n (B – A)

= 5 + 10 + 3 = 18

(ii) We know that

n (A – B) = n (A) – n (A ∩B)

⇒ 5 = n (A) – 10

⇒ n(A) = 15

(iii) We know that,

n (B – A) = n (B) – n (A ∩ B)

⇒ 3 = n (B) – 10

⇒ n

(B) = 13

Question 6.

If n (ξ) = 50, n (A) = 28 and n (B) = 30, then what is the greatest value of n (A ∪ B) ?

Solution:

Given n (ξ) = 50,

n (A) = 28 ;

n (B) = 30

Since every set is a subset of universal set.

∴ A ∪ B ⊂ ξ

n (A ∪ B) ≤ n(ξ) = 50 ………………(1)

Hence the greatest value of n (A ∪ B) = 50

Question 7.

If n (ξ) = 48 and n (A’ ∪ B’) = 36, then find n (A ∩ B).

Solution:

Given n(ξ) = 48 ;

n (A’ ∪ B’) = 36

⇒ n {(A ∩ B)’} = n (ξ) – n (A ∩ B)

[using Demorgan’s law]

⇒ 36 = 48 – n(A ∩ B)

⇒ n (A ∩ B) = 12

Question 8.

If n (ξ) = 40, n (A) = 22, n (A ∩ B) = 8 and n ((A ∪ B)’) = 6, determine n (B).

Solution:

Given n(ξ) = 40,

n (A) = 22,

n (A ∩ B) = 8

and n {(A ∪ B)’ = 6

⇒ 6 = n {(A ∪ B)’)

= n(ξ) – n (A ∪ B)

⇒ 6 = 40 – [n (A) + n (B) – n (A ∩ B)]

⇒ 6 = 40 – [22 + n (B) – 8]

⇒ 6 = 26 – n (B)

⇒ n (B) = 26 – 6 = 20

Question 9.

If n (ξ) = 50, n (A) = 20, n (B) =26 and n ((A ∪ B)’) = 6, find

(i) n (A ∩ B)

(ii) (A – B).

Solution:

Given n (ξ) = 50,

n (A) = 20,

n (B) = 26,

n ((A ∪ B)’) = 6

(i) Now 6 = n ((A ∪ B)’)

= n (ξ) – n (A ∪ B)

⇒ 6 = 50 – {n (A) + n (B) – n (A ∩ B)}

⇒ 6 = 50 – {20 + 26 – n (A ∩ B)

⇒ 6 = 50 – 46 + n (A ∩ B)

n (A ∩ B) =6 – 4 = 2

(ii) n (A – B) = n (A) – n (A ∩ B)

= 20 – 2 = 18

Question 10.

In a class of 35 students, 24 like to play cricket and 16 like to play football. Also, each student likes to play atleast one of the two games. How many students like to play both cricket and football?

Solution:

Let us define the sets are as follows:

A : set of students who like to play cricket.

B : set of students who like to play football.

Then n (A) = 24 ;

n (B) = 26

and n (A ∪ B) = 35

We know that,

n (A ∪ B) = n (A) + n (B) – n (A ∩ B)

⇒ n (A ∩ B) = 24 + 16 – 35 = 5

Thus required no. of students who like to play both cricket and football be 5.

Question 11.

In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many can speak both Hindi and English?

Solution:

Let H be the set of people who can speak Hindi

and E be the set of people who can speak English.

Here we have given that

n (H ∪ E) = 400 ;

n (H) = 250 ;

n (E) = 200

We know that,

n (H ∪ E) = n (H) + n (E) – n (H ∩E)

400 = 250 + 200 – n (H ∩ E)

n (H ∩ E) = 450 – 400 = 50

Hence required number of people who can speak Hindi and English both be 50.

Question 12.

If n (E) = 50, n (A) = 30 and n (B) = 28, find

(i) the greatest value of n (A ∪ B)

(ii) the least value of n (A ∩ B).

Solution:

Given n (ξ) = 50 ;

n (A) = 30

and n (B) = 28

(i) Since every set be a subset of universal set ξ.

∴ A ∩ B ⊂ ξ

⇒ n (A ∪ B) ≤ n(ξ) = 50 ……………………(1)

Thus, the greatest value of n (A ∪ B) = 50

(ii) We know that,

n (A ∪ B) = n (A) + n (B) – n (A ∩ B)

⇒ n (A) + n (B) – n (A ∩ B) ≤ 50

⇒ 30 + 28 – n (A ∩ B) ≤ 50

⇒ n (A ∩ B) ≥ 58 – 50 = 8

∴ least value of n (A ∩ B) = 8.

Question 13.

In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Find the number of students who play neither.

Solution:

Let C and T be the set of students who play cricket and tennis respectively and be the set of all students.

Then n (ξ) = 60 ;

n (C) = 25 ;

n (T) = 20

and n (C ∩ F) = 10

Thus the required no. of students who play neither of two games = n (C’ ∩ T’)

= n (C ∪ T)’

n (ξ) – n (C ∪ T)

= n (ξ) – {n (C) + n (T) – n (C ∩ T)}

= 60 – {25 + 20 – 10}

= 60 – 35 = 25

Question 14.

In a group of people, 50 people read newspaper A, 20 read newspaper B and 10 read both newspapers. How many people read atleast one of the two newspapers?

Solution:

Let A and B be the set of people who read newspaper A and newspaper B respectively.

Also we have given

n(A) = 50 ;

n(B) = 20

and n (A ∩ B) = 10

Thus, the required no. of people who read atleast one of the two newspaper = n (A ∪ B)

= n (A) + n (B) – n (A ∩ B)

= 50 + 20 – 10 = 60

Question 15.

In a group of 65 students, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?

Solution:

Let C and T be the set of students who like cricket and tennis respectively

given n (C ∩ T) = 10

and n (C) = 40

n (C ∪ T) = 65

∴ n(C ∪ T) = n (C) + n (T) – n (C ∩ T)

65 = 40 + n (T) – 10

n (T) = 75 – 40 = 35

hence the required no. of students who like tennis = 35.

Thus required no. of students who like tennis only but not cricket

= n (T ∩ C^{c})

= n (T) – n (T ∩ C)

= 35 – 10 = 25.

Question 16.

In class XI of a certain school, 50 students cat burger and 42 students eat noodles in lunch time. If 24 students eat both burger and noodles, find the number of students who eat burger only noodles only any of the two food items?

Solution:

Let B and N be the set of students who eat burger and noodles in lunch time respectively.

Then we have given that n (B) = 50;

n (N) = 42

and n (B ∩ N) = 24

(i) ∴ required no. of students who eat burger only = n (B ∩ N^{c})

= n (B) – n (B ∩ N)

= 50 – 24 = 26

(ii) ∴ required no. of students who eat noodles only = n (N ∩ B^{c})

= n (N) – (N ∩ B)

= 42 – 24 = 18

(iii) required no. of students who eat any of the two food items = n (B ∪ N)

= n (B) + n (N) – n (B ∩ N)

= 50 + 42 – 24 = 68

Question 17.

In a survey of 600 students in a school, 150 students were found to be drinking tea and 225 drinking milk and 100 students were drinking both tea and milk. How many students were drinking neither tea nor milk? What do you think which drink should a student prefer and why? (Value Based)

Solution:

Let be the set of all students in a school.

Let T and M be the set of all students who were found to be drinking tea and milk.

Then n (ξ) = 600 ;

n (T) = 150 ;

n (M) = 225

and n (T ∪ M) = 100

∴ required number of students who were drinking neither tea nor milk = n (T^{c} ∩ M^{c})

= n [(T ∪ M)^{c}]

= n (ξ) – n (T ∪ M)

= n (ξ) – [n (T) + n (M) – n (T ∩ M)]

= 600 – [150 + 225 – 100]

= 600 – 275 = 325

So students should prefer milk since milk be a complete food.

It makes body fit and healthy since it gives all nutrients to the body.

Question 18.

In a survey of 60 people, it was found that 25 people read Newspaper H, 26 read Newspaper T, 26 read Newspaper 1,9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all the three Newspapers. Find

(i) the number of people who read atleast one of the three Newspapers.

(ii) the number of people who read exactly one Newspaper.

Solution:

Let be the set of all peoples.

Let H, T and I be the set of all people who read newspaper H, newspaper T and news paper I.

Then n (ξ) = 60 ;

n (H) = 25 ;

n (T) = 26 ;

n (I) = 26 ;

n (H ∩ I) = 9 ;

n (H ∩ T) = 11;

n (T ∩ I) = 8

and n (H ∩ I ∩ T) = 3

(i) ∴ required no. of people who read atleast one of the three newspaper = n (H ∪ T ∪ I)

= n (H) + n (T) + n (l) – n (H ∩ T) – n (T ∩ I) – n (H ∩ I) + n (H ∩ T ∩ I)

= 25 + 26 + 26 – 9 – 11 – 8 + 3 = 52

(ii) ∴ required no. of people who read exactly one Newspaper

= n (H ∩ T^{c} ∩ I^{c}) + n (H^{c} ∩T ∩ I^{c}) + n (H^{c} ∩ T^{c} ∩ I)

= n [H ∩ (T ∪ I)^{c}] + n [T ∩ (H ∪ I)^{c}] + n [I ∩ (H ∪ T)^{c}] ………………..(1)

n [H ∩ (T ∪ I)^{c}] = n (H) – n (H ∩ (T ∪ I))

= n (H) – n [(H ∩ T) ∪ (H ∩ I)]

= n (H) – [n (H ∩ T) + n (H ∩ J ) – n (H ∩ T ∩ I)]

= 25 – [11 + 9 – 3]

= 25 – 17 = 8

n [T ∩ (H ∩ I)^{c}] = n (T) – [n (T ∩ (H ∪ I)]

= n (T) – [n (T ∩ H) ∪ (T ∩ I)]

= n (T) – [n (T ∩ H) + n (T ∩ I) – n (H ∩T ∩ I)]

= 26 – [11 + 8 – 3]

= 26 – 16 = 10

n [I ∩ (T ∪ H)^{c}] = n (I) – n [I ∩ (T ∪ H)]

= n (I) – n [(T ∩ I) ∪ (I ∩ H)]

= n (I) – [n (T ∩ I) + n (T ∩ H) – n (T ∩ I ∩ H)]

= 26 – (8 + 9 – 3]

= 26 – 14 = 12

∴ from (1) ; we have

reqd. no. of students = 8 + 10 + 12 = 30

Aliter :

From Venn diagram, we have

(i) Thus, required no. of people who read atleast one of the three newspaper

= n (H ∪ T ∪ I)

= 8 + 8 + 3 + 6 + 10 + 5 + 1 = 252

(ii) ∴ required no. of people who read exactly one newspaper = 8 + 10 + 12 = 30.

Question 19.

In a survey of 100 students regarding watching T.V., it was found that 28 watch action movies, 30 watch comedy serials, 42 watch news channels, 8 watch action movies and comedy serials, 10 watch action movies and news channels, 5 watch comedy serials and news channels and 3 watch all the three programs. Draw a Venn diagram to illustrate this information and find

(i) how many watch news channels only?

(ii) how many do not watch any of the three programs?

Solution:

Let E) be the total no. of students.

Let A, C and N be the set of students who watch action movies comedy serials and news channels.

∴ n (A) = 28 ;

n (C) = 30 ;

n (N) = 42;

n (A ∩ C) = 8 ;

n (A ∩ N) = 10 ;

n (C ∩ N) = 5 ;

n (A ∩ N ∩ C) = 3

and n (ξ) = 100

From Venn diagram, we have

(i) Reqd. no. of students who watch news channel only = 30

(ii) Required no. of students who do not watch any of the three programmes

= 100 – [13 + 5 + 3 + 7 + 20 + 2 + 30]

= 100 – 80 = 20

Question 20.

In a survey of 25 students, it was found that 15 had taken Mathematics, 12 had taken Physics and 11 had taken Chemistry. 5 had taken Mathematics and Chemistry, 9 had taken Mathematics and Physics, 4 had taken Physics and Chemistry and 3 had taken all the three subjects. Find the number of students that had taken

(i) only Chemistry

(i) only Mathematics

(iii) only Physics

(iv) Physics and Chemistry but not Mathematics

(v) Mathematics and Physics but not Chemistry

(vi) only one of the subjects

(vii) atleast one of the three subjects

(viii) none of the three subjects.

Solution:

Let denotes the total number of survey students.

Let M, P and C denotes the set of students who had taken mathematics, physics and chemistiy respectively.

We have given that

n (M) = 15 ;

n (P) = 12 ;

n (C) = 11 ;

n (ξ) = 25 ;

n (M ∩ C) = 5 ;

n (M ∩ P) = 5 ;

n (P ∩ C) = 4 ;

n (P ∩ M ∩ C) = 3

From Venn diagram ; we have

(i) required no. of students who had taken only chemistry = 5

(ii) required number of students had taken only mathematics = 4

(iii) required no. of students had taken only physics = 2

(iv) reqd. no. of students that had taken Physics and Chemistry but not Mathematics = 1

(v) reqd. no. of students that had taken mathematics and physics but not chemistry = 6

(vi) reqd. no. of students that had taken only one of the subjects = 2 + 5 + 4 = 11

(vii) reqd. no. of students that had taken atleast one of the three subjects = 4 + 6 + 3 + 2 + 2 + 1 + 5 = 23

(viii) reqd. no. of students that had taken none of the three subjects = 25 – 23 = 2.