11th NCERT Sets Miscellaneous Exercise Questions 16

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Question (1)

Decide, among the following sets, which sets are subsets of one and another: A = {x: x ∈ R and x satisfy xB = {2, 4, 6}, C = {2, 4, 6, 8…}, D = {6}.

Solution

xx = 2 or x = 6

A = { 2, 6}

B = {2, 4, 6}, C = {2, 4, 6, 8…}, D = {6}.

D ⊂ A ⊂ B ⊂ C.

Hence, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C

Question (2)

In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.(i) If x ∈ A and A ∈ B, then x ∈ B

Solution

(i) It is FalseLet A = {1, 2} and B = {1, {1, 2}, {3}} Now, 2 ∈ A, { 1, 2} ∈ B.

But 2 ∉ B.

x ∈ A and A ∈ B, then x ∈ B

(ii) If A ⊂ B and B ∈ C, then A ∈ C

Solution

It is False statement.Let A = {3}, B = {2 , 3 } , and C = { 1, {2,3},5} .

Here A ⊂ B, B ∈ C, but {3} ∉ C.

(iii) If A ⊂ B and B ⊂ C, then A ⊂ C

Solution

It is true statement.Let x ∈ A,

then x ∈ B, as A ⊂ B

x ∈ C , as B ⊂ C

So all x ∈ A , x ∈ C.

∴ A ⊂ C.

(iv) If A ⊄ B and B ⊄ C, then A ⊄ C

Solution

The statement is false.Let A = {a,b }, B = {b, c. d } C = { a, b, c, e}

Here A B, B ⊄ C, but A ⊂ C.

(v) If x ∈ A and A ⊄ B, then x ∈ B

Solution

It is false statement.Let A = {3, 5, 7} and B = {3, 4, 6}

Now, 5 ∈ A and A ⊄ B

However, 5 ∉ B

(vi) If A ⊂ B and x ∉ B, then x ∉ A

Solution

It is true.Let A ⊂ B and x ∉ B.

To show: x∉ A

If possible, suppose x ∈ A.

Then, x ∈ B, which is a contradiction as x ∉ B

∴ x ∉ A

Question (3)

Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. show that B = C.Solution

Let x ∈ B⇒ X ∈ A ∪ B [ B ⊂ A ∪ B ]

⇒ X ∈ A ∪ C [ A ∪ B = A ∪ C ]

⇒ x ∈ A, or x ∈ C

Case I : x ∈ A and also, x ∈ B

∴ x ∈ A ∩ B

⇒ X ∈ A ∩ C [ A ∩ B = A ∩ C ]

∴ x ∈ A and x ∈ C

∴ For all x ∈ B, x ∈ C

∴ B ⊂ C

Similarly, we can show that C ⊂ B.

∴ B = C

Question (4)

Show that the following four conditions are equivalent:(i) A ⊂ B (ii) A – B = Φ

(iii) A ∪ B = B (iv) A ∩ B = A

Solution

Let us consider A ⊂ BSo every x ∈ A, x ∈ B.

So A ∩ B = A

A - B = { x: x ∈ A, x ∉ B }

= A - ( A ∩ B)

= A - A

= φ

So (ii) is ture.

A ∪ B = { x : x ∈ A or x ∈ B}

Since all x ∈ A, x ∈ B.

∴ A ∪ B = B

so (iii) is true.

A ∩ B = { x: x ∈ A and x∈ B}

Since all x ∈ A, x ∈ B.

∴ A ∩ B = A

so (iv) is true.

So all are equivalent statements.

Question (5)

Show that if A ⊂ B, then C – B ⊂ C – A.Solution

A ⊂ Ball x ∈ A, x ∈ B

Let y ∈ (C - B)

⇒ y ∈ C , but y ∉ B

Since y ∉ B , y ∉ A, [ A ⊂ B]

But y ∈ C, and y ∉ A

So y ∈ ( C - A )

So every y ∈ (C - B ) ,y ∈ (C - A)

So C - B ⊂ C - A.

Question (6)

Assume that P (A) = P (B). Show that A = B.Solution

Let x ∈ A, then A ∈ P(A)As P(A) = P(B) , A ∈ P(B)

⇒ x ∈ A, x ∈ B

So A ⊂ B.

Similiarly y ∈ B, then B ∈ P(B)

As P(A) = P(B) , B ∈ P(A)

⇒ y ∈ B, y ∈ A

So B ⊂ A.

Since A ⊂ B and B ⊂ A

⇒ A = B.

Question (7)

Is it true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)? Justify your answer.Solution

It is False statement.Let A = {0, 1} and B = {1, 2}

∴ A ∪ B = {0, 1, 2}

P(A) = {Φ, {0}, {1}, {0, 1}}

P(B) = {Φ, {1}, {2}, {1, 2}}

P(A ∪ B) = {Φ, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}}

P(A) ∪ P(B) = {Φ, {0}, {1}, {0, 1}, {2}, {1, 2}}

∴ P(A) ∪ P(B) ≠ P(A ∪ B)

Question (8)

Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)Solution

(i) A = (A ∩ B) ∪ (A – B)RHS = (A ∩ B) ∪ (A – B)

= ( A ∩ B) ∪ ( A ∩ B')

= [ A ∪ ( A ∩ B')] ∩ [ B ∪ ( A ∩ B')]

= [( A ∪A) ∩ ( A ∪ B')] ∩ [( B ∪ A) ∩ ( B ∪ B')]

= [ A ∩ ( A ∪ B')] ∩ [( B ∪ A) ∩ U]

= [ (A ∩ A) ∪ ( A ∩ B')] ∩ [B ∪ A]

= [ A ∪( A ∩ B'] ∩ [B ∪ A]

= A ∩ [B ∪ A]

= A

= LHS.

∴ A = (A ∩ B) ∪ (A – B)

(ii) A ∪ (B – A) = (A ∪ B)

LHS = A ∪ (B – A)

= A ∪ ( B ∩ A')

= ( A ∪ B) ∩ ( A ∪ A')

= ( A ∪ B) ∩ U

= (A ∪ B)

= RHS

∴ A ∪ (B – A) = (A ∪ B)

Question (9)

Using properties of sets show that(i) A ∪ (A ∩ B) = A (ii) A ∩ (A ∪ B) = A

Solution

(i) LHS = A ∪ (A ∩ B)= (A ∪ A) ∩ ( A ∪ B )

= A ∩ ( A ∪ B )

= A

= RHS

(ii) A ∩ (A ∪ B) = A

LHS = A ∩ (A ∪ B)

= (A ∩ A ) ∪ ( A ∩ B)

= A ∪ ( A ∩ B)

= A

= RHS

Question (10)

Show that A ∩ B = A ∩ C need not imply B = CSolution

Let x ∈ B, and x ∉ A.x ∉ A ∩ B.

x ∉ A ∩ C. [ A ∩ B = A ∩ C ]

x ∉ C

All x ∈ B, x ∉ C.

⇒ B ≠ C.

Question (11)

Let A and B be sets. If A ∩ X = B ∩ X = Φ and A ∪ X = B ∪ X for some set X, show that A = B.(Hints,A = A ∩ (A ∪ X) B = B ∩ (B ∪ X) and use distributive law)

Solution

We can writeA = A ∩ (A ∪ X)

= A ∩ ( B ∪ X) [A ∪ X = B ∪ X]

= ( A ∩ B) ∪ (A ∩ X)

= ( A ∩ B) ∪ φ [ A ∩ X = Φ ]

= ( A ∩ B) -------(1)

B = B ∩ (B∪ X)

= B ∩ ( A ∪ X) [A ∪ X = B ∪ X]

= ( B ∩ A) ∪ (B ∩ X)

= ( A ∩ B) ∪ φ [ B ∩ X = Φ ]

= ( A ∩ B) -------(2)

From (1) and (2) we get,

A = B.

Question (12)

Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = Φ.Solution

Let A = {0, 1}, B = {1, 2}, and C = {2, 0}.Accordingly, A ∩ B = {1}, B ∩ C = {2}, and A ∩ C = {0}.

∴ A ∩ B, B ∩ C, and A ∩ C are non-empty.

However, A ∩ B ∩ C = Φ

Question (13)

In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?Solution

Let U be the set of all students who took part in the survey.Let T be the set of students taking tea.

Let C be the set of students taking coffee.

Accordingly, n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100

To find: Number of student taking neither tea nor coffee i.e., we have to find n(T’ ∩ C’).

n(T’ ∩ C’) = n(T ∪ C)’

= n(U) – n(T ∪ C)

= n(U) – [n(T) + n(C) – n(T ∩ C)]

= 600 – [150 + 225 – 100]

= 600 – 275

= 325

Hence, 325 students were taking neither tea nor coffee.

Question (14)

In a group of students 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?Solution

Let U be the set of all students in the group.Let E be the set of all students who know English.

Let H be the set of all students who know Hindi.

∴ H ∪ E = U

Accordingly, n(H) = 100 and n(E) = 50

n( H ∩ E) = 25

n(U) = n(H) + n(E) – n(H ∩ E)

= 100 + 50 – 25

= 125

Hence, there are 125 students in the group.

Question (15)

In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I,11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:(i) the number of people who read at least one of the newspapers.

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

Solution

Let A be the set of people who read newspaper H.Let B be the set of people who read newspaper T.

Let C be the set of people who read newspaper I.

Accordingly, n(A) = 25, n(B) = 26, and n(C) = 26

n(A ∩ C) = 9, n(A ∩ B) = 11, and n(B ∩ C) = 8

n(A ∩ B ∩ C) = 3

Let U be the set of people who took part in the survey.n (U) = 60

We will draw the venn diagram for the above information as follows.

From the venn diagram,

(i) n( reads atleaast one of news paper )

= 8 + 8 + 3 + 6 + 10 + 5 + 12

= 52

52people read atleast one news paper.

(ii) n( people reads exactly one newspaper )

= 8 + 12 + 10

= 30

30 people reads exactly one news paper.

Question (16)

In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.Solution

Let A, B, and C be the set of people who like product A, product B, and product C respectively.Accordingly, n(A) = 21, n(B) = 26, n(C) = 29, n(A ∩ B) = 14, n(C ∩ A) = 12,

n(B ∩ C) = 14, n(A ∩ B ∩ C) = 8

The Venn diagram for the given problem can be drawn as

It can be seenfrom venn diagram that

number of people who like product C only

= 11