# Miscellaneous exercise sets ncert maths solution class 11

# Chapter 1(Miscellaneous) Class 11

**1. Decide, among the following sets, which sets are subsets of one and another.**

**A= {x: x âˆˆ R and x satisfy x ^{2}Â â€“ 8x + 12 = 0}**

**B = {2, 4, 6}**

**C = {2, 4, 6, 8â€¦}**

**D = {6}**

**Solution: **According to the question,

We have,

A = {x: x âˆˆ R and x satisfies x^{2}Â â€“ 8x + 12 =0}

x^{2}Â â€“ 8x + 12 = 0

â‡’ xÂ² – 6x – 2x + 12 = 0

â‡’ x(x – 6) – 2(x – 6) = 0

â‡’ (x – 6)(x – 2) = 0

â‡’ x = 6, 2

Hence, A = {2, 6}

B = {2, 4, 6}, C = {2, 4, 6, 8 â€¦}, D = {6}

Hence, D âŠ‚ A âŠ‚ B âŠ‚ C

**Hence, A âŠ‚ B, A âŠ‚ C, B âŠ‚ C, D âŠ‚ A, D âŠ‚ B, D âŠ‚ C**

**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 âˆˆ A and A âˆˆ B, then x âˆˆ B**

**(ii) If A âŠ‚ B and B âˆˆ C, then A âˆˆ C**

**(iii) If A âŠ‚ B and B âŠ‚ C, then A âŠ‚ C**

**(iv) If A âŠ„ B and B âŠ„ C, then A âŠ„ C**

**(v) If x âˆˆ A and A âŠ„ B, then x âˆˆ B**

**(vi) If A âŠ‚ B and x âˆ‰ B, then x âˆ‰ A**

**Solution: (i) False**

Let A = {1, 2} and B = {1, {1, 2}, {3}}

Now, we have

2 âˆˆ {1, 2} and {1, 2} âˆˆ {1, {1, 2}, {3}}

Hence, A âˆˆ B

We also know

{2} âˆ‰ {1, {1, 2}, {3}}

**(ii) False**

Let us assume that

A = {2}

B = {0, 2}

And, C = {1, {0, 2}, 3}

From the question,

A âŠ‚ B

Hence, B âˆˆ C

But,Â A âˆ‰ C

**(iii) True**

A âŠ‚ B and B âŠ‚ C

LetÂ x âˆˆ A

Then, we have

x âˆˆ B AndÂ x âˆˆ C

Therefore, A âŠ‚ C

**(iv) False**

Since, A âŠ„ B

Also, BÂ âŠ„Â C

LetÂ A = {1, 2}, B = {0, 6, 8}

And, C = {0, 1, 2, 6, 9}

**âˆ´Â A âŠ‚ C**

**(v) False**

Since, x âˆˆ A

Also, AÂ âŠ„Â B

LetÂ A = {3, 5, 7}

Also, B = {3, 4, 6}

We know that,

AÂ âŠ„Â B

âˆ´ 5 âˆ‰ B

**(vi) True**

Since, AÂ âŠ‚Â B

Also, xÂ âˆ‰Â B

LetÂ x âˆˆ A,

We have x âˆˆ B

From the question,

We have, xÂ âˆ‰Â B

â‡’Â x âˆ‰ A

**3. Let A, B and C be the sets such that A âˆª B = A âˆª C and A âˆ© B = A âˆ© C. Show that B = C.**

**Solution: Given**

A âˆª B = A âˆª C And,

A âˆ© B = A âˆ© C

**To show, B = C**

LetÂ x âˆˆ B

â‡’Â x âˆˆ A âˆª BÂ [Given,Â A âˆª B = A âˆª C]

â‡’ x âˆˆ A âˆª C

Hence,

x âˆˆ A or x âˆˆ C

When x âˆˆ A, then

x âˆˆ B

âˆ´ x âˆˆ A âˆ© B [ given A âˆ© B = A âˆ© C ]

â‡’Â x âˆˆ A âˆ© C â‡’Â x âˆˆ A and x âˆˆ C

â‡’ x âˆˆ C

âˆ´ B âŠ‚Â C

Similarly, it can be shown that C âŠ‚ B.

Hence, B = C

**4. Show that the following four conditions are equivalent.**

**(i) A âŠ‚ B (ii) A â€“ B = Î¦**

**(iii) A âˆª B = B (iv) A âˆ© B = A**

**Solution: **According to the question,

To prove (i) â¬Œ (ii)

Here, (i) = AÂ âŠ‚Â B and (ii) = A â€“ B â‰ Â Ï•

Let us assume that AÂ âŠ‚Â B

To prove, A â€“ B â‰ Â Ï•

Let A â€“ B â‰ Â Ï•

Hence, there exists X âˆˆ A, X â‰ B, but since AâŠ‚ B, it is not possible.

âˆ´Â A â€“ B =Â Ï•

And AâŠ‚Â B â‡’ A â€“ B â‰ Â Ï•

Let us assume thatÂ A â€“ B â‰ Â Ï•

To prove A âŠ‚ B

Let XâˆˆÂ A

So, XÂ âˆˆÂ B (if XÂ âˆ‰Â B, then A â€“ B â‰ Â Ï•)

Hence, A â€“ B =Â Ï•Â => AÂ âŠ‚Â B

âˆ´ (i) â¬Œ (ii)

Let us assume thatÂ AÂ âŠ‚Â B

To prove, AÂ âˆªÂ B = B

â‡’Â BÂ âŠ‚Â AÂ âˆªÂ B

Let us assume that x âˆˆ Aâˆª B

â‡’Â XÂ âˆˆÂ A or XÂ âˆˆÂ B

Taking Case I, X âˆˆ B

AÂ âˆªÂ B = B

Taking Case II, X âˆˆ A

â‡’Â XÂ âˆˆÂ B (AÂ âŠ‚Â B)

â‡’Â AÂ âˆªÂ BÂ âŠ‚Â B

Let AÂ âˆªÂ B = B

Let us assume that XÂ âˆˆÂ A

â‡’Â XÂ âˆˆÂ AÂ âˆªÂ B (AÂ âŠ‚Â AÂ âˆªÂ B)

â‡’Â XÂ âˆˆÂ B (AÂ âˆªÂ B = B)

âˆ´ AâŠ‚ B

Hence, (i)Â â¬ŒÂ (iii)

To prove (i)Â â¬ŒÂ (iv)

Let us assume that AÂ âŠ‚Â B

AÂ âˆ©Â BÂ âŠ‚Â A

Let XÂ âˆˆÂ A

To prove, XÂ âˆˆÂ Aâˆ©Â B

Since, AÂ âŠ‚Â B and XÂ âˆˆÂ B

Hence, XÂ âˆˆÂ A âˆ©Â B

â‡’Â AÂ âŠ‚Â AÂ âˆ©Â B

â‡’Â A = AÂ âˆ©Â B

Let us assume that AÂ âˆ©Â B = A

Let XÂ âˆˆÂ A

â‡’Â XÂ âˆˆÂ AÂ âˆ©Â B

â‡’Â XÂ âˆˆÂ B and XÂ âˆˆÂ A

â‡’Â AÂ âŠ‚Â B

âˆ´Â (i)Â â¬ŒÂ (iv)

âˆ´ (i) â¬ŒÂ (ii) â¬ŒÂ (iii) â¬ŒÂ (iv)

Hence, proved.

**5. Show that if AÂ âŠ‚Â B, then C â€“ BÂ âŠ‚Â C â€“ A.**

**Solution:**

To show,

C â€“ BÂ âŠ‚Â C â€“ A

According to the question,

Let us assume that x is any element such that XÂ âˆˆÂ C â€“ B

âˆ´ xÂ âˆˆÂ C and x âˆ‰ B

Since, A âŠ‚ B, we have

âˆ´ xÂ âˆˆÂ C and x âˆ‰ A

So, xÂ âˆˆÂ C â€“ A

âˆ´Â C â€“ BÂ âŠ‚Â C â€“ A

Hence, proved.

**6. Assume that P (A) = P (B). Show that A = B.**

**Solution:**

To show,

A = B

According to the question,

P (A) = P (B)

Let x be any element of set A,

xÂ âˆˆÂ A

Since P (A) is the power set of set A, it has all the subsets of set A.

AÂ âˆˆÂ P (A) = P (B)

Let C be an element of set B.

For any C âˆˆÂ P (B),

We have, x âˆˆÂ C

CÂ âŠ‚Â B

âˆ´ xÂ âˆˆÂ B

âˆ´Â AÂ âŠ‚Â B

Similarly, we have

BÂ âŠ‚Â A

So, we get

IfÂ AÂ âŠ‚Â B andÂ B âŠ‚ A

âˆ´Â A = B

**7. Is it true that for any sets A and B, P (A)Â âˆªÂ P (B) = P (AÂ âˆªÂ B)? Justify your answer.**

**Solution: **It is not true that for any sets A and B, P (A)Â âˆªÂ P (B) = P (AÂ âˆªÂ B)

Let us assume

A = {0, 1} And, B = {1, 2}

âˆ´ AÂ âˆªÂ BÂ = {0, 1, 2}

We have,

P (A) = {Ï•, {0}, {1}, {0, 1}}

P (B) = {Ï•, {1}, {2}, {1, 2}}

âˆ´ P (A âˆª B) = {Ï•, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}}

Also,

P (A) âˆª P (B) = {Ï•, {0}, {1}, {2}, {0, 1}, {1, 2}}

âˆ´Â P (A) âˆª P (B â‰ P (A âˆª B)

**Hence, the given statement is false**

**8. Show that for any sets A and B,
A = (AÂ âˆ©Â B)Â âˆªÂ (A â€“ B) and AÂ âˆªÂ (B â€“ A) = (AÂ âˆªÂ B)**

**Solution: To prove, A = (AÂ âˆ©Â B)Â âˆªÂ (A â€“ B)**

Proof:Â Let x âˆˆ A

To show, XÂ âˆˆÂ (AÂ âˆ©Â B)Â âˆªÂ (A â€“ B)

In Case I,

XÂ âˆˆÂ (A âˆ©Â B)

â‡’Â XÂ âˆˆÂ (AÂ âˆ©Â B)Â âŠ‚Â (AÂ âˆªÂ B)Â âˆªÂ (A â€“ B)

In Case II,

XÂ âˆ‰AÂ âˆ©Â B

â‡’Â XÂ âˆ‰Â B or XÂ âˆ‰Â A

â‡’Â XÂ âˆ‰Â B (XÂ âˆ‰Â A)

â‡’Â XÂ âˆ‰Â A â€“ BÂ âŠ‚Â (AÂ âˆªÂ B)Â âˆªÂ (A â€“ B)

âˆ´A âŠ‚ (A âˆ© B) âˆª (A â€“ B) — (i)

It can be concluded that A âˆ© B âŠ‚ A and (A â€“ B) âŠ‚ A

Thus, (A âˆ© B) âˆª (A â€“ B) âŠ‚ A —(ii)

Equating (i) and (ii),

A = (AÂ âˆ©Â B)Â âˆªÂ (A â€“ B)

We also have to show

AÂ âˆªÂ (B â€“ A)Â âŠ‚Â AÂ âˆªÂ B

Let us assume

XÂ âˆˆÂ AÂ âˆªÂ (B â€“ A)

XÂ âˆˆÂ A or XÂ âˆˆÂ (B â€“ A)

â‡’Â XÂ âˆˆÂ A or (XÂ âˆˆÂ B and XÂ âˆ‰A)

â‡’Â (XÂ âˆˆÂ A or XÂ âˆˆÂ B) and (XÂ âˆˆÂ A and XÂ âˆ‰A)

â‡’Â XÂ âˆˆÂ (BÂ âˆªA)

âˆ´ A âˆª (B â€“ A) âŠ‚ (A âˆª B) — (iii)

According to the question,

**To prove: **(AÂ âˆªÂ B)Â âŠ‚Â AÂ âˆª (B â€“ A)

Let yÂ âˆˆÂ AâˆªB

y âˆˆ A or y âˆˆ B

(y âˆˆ A or y âˆˆ B) and (x âˆˆ A and x âˆ‰A)

â‡’Â yÂ âˆˆÂ A or (yÂ âˆˆÂ B and yÂ âˆ‰A)

â‡’Â yÂ âˆˆÂ AÂ âˆªÂ (B â€“ A)

Thus, A âˆª B âŠ‚ A âˆª (B â€“ A) —- (iv)

âˆ´ From equations (iii) and (iv), we get

**AÂ âˆªÂ (B â€“ A) = AÂ âˆªÂ B**

**9. Using properties of sets, show that
(i) AÂ âˆªÂ (AÂ âˆ©Â B) = A
(ii) A âˆ© (A âˆª B) = A**

**Solution: **(i) To show: A âˆª (A âˆ© B) = A

We know that,

A âŠ‚Â A

â‡’ A âˆ© B âŠ‚ A

âˆ´ A âˆª (A âˆ© B) âŠ‚ A —Â (i)

Also, according to the question,

We have

AâŠ‚ A âˆª (A âˆ© B) — (ii)

Hence, from equations (i) and (ii),

We have

A âˆª (A âˆ© B)Â = A

(ii) To show,

A âˆ© (A âˆª B) = A

A âˆ© (A âˆª B) = (A âˆ© A) âˆª (A âˆ© B)

= A âˆª (A âˆ© B)

= A

**10. Show that AÂ âˆ©Â B = AÂ âˆ©Â C need not imply B = C.**

**Solution: **Let

A = {0, 1}, B = {0, 2, 3}

And, C = {0, 4, 5}

According to the question,

A âˆ© B = {0} And,

A âˆ© C = {0}

âˆ´ A âˆ© B = A âˆ© C = {0}

But,

2Â âˆˆ B and 2 âˆ‰ C

**Therefore, B â‰ C**

**11. Let A and B be sets. If AÂ âˆ©Â X = BÂ âˆ©Â X = Ï• and AÂ âˆªÂ X = BÂ âˆªÂ X for some set X, show that A = B.
(Hints: A = A âˆ© (A âˆª X) , B = B âˆ© (B âˆª X), and use Distributive law.)**

**Solution: **Let A and B be two sets such that A âˆ© X = B âˆ© X = Ï• and A âˆª X = B âˆª X for some set X.

To show, A = B

Proof: A = A âˆ© (A âˆª X) = A âˆ© (B âˆª X) [A âˆª X = B âˆª X]

= (A âˆ© B) âˆª (A âˆ© X) [Distributive law]

= (A âˆ© B) âˆª Î¦ [A âˆ© X = Î¦]

= A âˆ© B — (i)

Now, B = B âˆ© (B âˆª X)

= B âˆ© (A âˆª X) [A âˆª X = B âˆª X]

= (B âˆ© A) âˆª (B âˆ© X) â€¦ [Distributive law]

= (B âˆ© A) âˆª Î¦ [B âˆ© X = Î¦]

= A âˆ© B — (ii)

Hence, from equations (i) and (ii), **we obtain A = B**

**12. Find sets A, B and C such that A âˆ© B, B âˆ© C and A âˆ© C are non-empty sets and A âˆ© B âˆ© C = Î¦.**

**Solution: **Let , A {0, 1}

B = {1, 2}

And, C = {2, 0}

A âˆ© B = {1}, B âˆ© C = {2}

And, A âˆ© C = {0}

âˆ´ A âˆ© B, B âˆ© C and A âˆ© C are not empty sets.

Hence, we get

**A âˆ© B âˆ© C = Î¦**

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

**Solution: **Let us assume that,

U = The set of all students who took part in the survey

T = The set of students taking tea

C = The set of students taking coffee

Total number of students in a school, n (U) = 600

Number of students taking tea, n (T) = 150

Number of students taking coffee, n (C) = 225

Also,Â n (T âˆ© C) = 100

Now, we have to find the number of students taking neither coffee nor tea =Â n (T ‘ âˆ© Câ€™)

Since,Â n ( T ‘ âˆ© Câ€™ )= n( T âˆ© C )â€™

= n (U) â€“ n (T âˆ© C)

= n (U) â€“ [n (T) + n(C) â€“ n (T âˆ© C)]

= 600 â€“ [150 + 225 â€“ 100]

= 600 â€“ 275

= 325

**âˆ´Â Number of students taking neither coffee nor tea = 325 students**

**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 = The set of all students in the group

E = The set of students who know English

H = The set of students who know Hindi

âˆ´ H âˆª E = U

Number of students who know Hindi n (H) = 100

Number of students who know English, n (E) = 50

Number of students who know both, n (H âˆ© E)Â = 25

We have to find the total number of students in the group =Â n (U)

We know that

n (U) = n(H) + n(E) â€“ n(H âˆ© E)

= 100 + 50 â€“ 25

= 125

**âˆ´Â Total number of students in the group = 125 students**

**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:** **(i)Â **Let,

A = The set of people who read newspaper H

B = The set of people who read newspaper T

C = The set of people who read newspaper I

Given,

Number of people who read newspaper H= n (A) = 25

Number of people who read newspaper T = n (B) = 26

Number of people who read the newspaper I = n (C) = 26

Number of people who read both newspaper H and I = n (A âˆ© C) = 9

Number of people who read both newspaper H and T = n (A âˆ© B) = 11

Number of people who read both newspaper T and I = n (B âˆ© C) = 8

And, the number of people who read all three newspapers H, T and I = n (A âˆ© B âˆ© C) = 3

Now, we have to find the number of people who read at least one of the newspapers

â‡’n(AâˆªBâˆªC)Â = 25 + 26 + 26 â€“ 11 â€“ 8 â€“ 9 + 3

= 80 â€“ 28

= 52

**âˆ´Â There are a total of 52 students who read at least one newspaper.**

(ii) Let,

a = The number of people who read newspapers H and T only

b = The number of people who read newspapers I and H only

c = The number of people who read newspapers T and I only

d = The number of people who read all three newspapers

D = n(A âˆ© B âˆ© C) = 3

Now, we have

n(A âˆ© B) = a + d

n(B âˆ© C) = c + d

And,Â n(C âˆ© A) = b + d

âˆ´Â a + d + c +d + b + d = 11 + 8 + 9

a + b + c + d = 28 â€“ 2d

= 28 â€“ 6

= 22

âˆ´ The number of people who read exactly one newspaper = 52 â€“ 22

**= 30 people**

**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 three products. Find how many liked product C only.**

**Solution: **Let A, B and C = the set of people who like product A, product B and product C, respectively.

Now, according to the question,

Number of students who like product A, n (A) = 21

Number of students who like product B, n (B) = 26

Number of students who like product C, n (C) = 29

Number of students who like both products A and B, n (A âˆ© B) = 14

Number of students who like both products A and C, n (C âˆ© A) = 12

Number of students who like both products C and B, n (B âˆ© C) = 14

Number of students who like all three products, n (A âˆ© B âˆ© C) = 8

From the Venn diagram, we get

Number of students who only like product C = {29 â€“ (4 + 8 + 6)}

= {29 â€“ 18}

**= 11 students**