**Chapter 5 Sets and Relations Miscellaneous Exercise 5**

## Chapter 5 Sets and Relations Miscellaneous Exercise 5

**(I) Select the correct answer from the given alternative.**

**Question 1.For the set A = {a, b, c, d, e} the correct statement is(A) {a, b} ∈ A(B) {a} ∈ A(C) a ∈ A(D) a ∉ AAnswer:**

(C) a ∈ A

**Question 2.If aN = {ax : x ∈ N}, then set 6N ∩ 8N =(A) 8N(B) 48N(C) 12N(D) 24NAnswer:**

(D) 24N

Hint:

6N = {6x : x ∈ N} = {6, 12, 18, 24, 30, ……}

8N = {8x : x ∈ N} = {8, 16, 24, 32, ……}

∴ 6N ∩ 8N = {24, 48, 72, …..}

= {24x : x ∈ N}

= 24N

**Question 3.If set A is empty set then n[P[P[P(A)]]] is(A) 6(B) 16(C) 2(D) 4Answer:**
(D) 4

Hint:

A = Φ

**Question 4.In a city 20% of the population travels by car, 50% travels by bus and 10% travels by both car and bus. Then, persons travelling by car or bus are(A) 80%(B) 40%(C) 60%(D) 70%Answer:**

(C) 60%

Hint:

Let C = Population travels by car

B = Population travels by bus

n(C) = 20%, n(B) = 50%, n(C ∩ B) = 10%

n(C ∪ B) = n(C) + n(B) – n(C ∩ B)

= 20% + 50% – 10%

= 60%

**Question 5.If the two sets A and B are having 43 elements in common, then the number of elements common to each of the sets A × B and B × A is**

**Question 6.Let R be a relation on the set N be defied by {(x, y) / x, y ∈ N, 2x + y = 41} Then R is(A) Reflexive(B) Symmetric(C) Transitive(D) None of theseAnswer:**

(D) None of these

**Question 7.The relation “>” in the set of N (Natural number) is(A) Symmetric(B) Reflexive(C) Transitive(D) Equivalent relationAnswer:**

(C) Transitive

Hint:

For any a ∈ N, a ≯ a

∴ (a, a) ∉ R

∴ > is not reflexive.

For any a, b ∈ N, if a > b, then b ≯ a.

∴ > is not symmetric.

For any a, b, c ∈ N,

if a > b and b > c, then a > c

∴ > is transitive.

**Question 8.A relation between A and B is(A) only A × B(B) An Universal set of A × B(C) An equivalent set of A × B(D) A subset of A × BAnswer:**

(D) A subset of A × B

**Question 9.If (x, y) ∈ N × N, then xy = x ^{2} is a relation that is(A) Symmetric(B) Reflexive(C) Transitive(D) EquivalenceAnswer:**

(D) Equivalence

Hint:

**Question 10.If A = {a, b, c}, The total no. of distinct relations in A × A is(A) 3(B) 9(C) 8(D) 29Answer:**
(D) 29

**(II) Answer the following.**

**Question 1.Write down the following sets in set builder form:(i) {10, 20, 30, 40, 50}(ii) {a, e, i, o, u}(iii) {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}Solution:**
(i) Let A = {10, 20, 30, 40, 50}

∴ A = {x/x = 10n, n ∈ N and n ≤ 5}

(ii) Let B = {a, e, i, o, u}

∴ B = {x/x is a vowel of English alphabets}

(iii) Let C = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}

∴ C = {x/x is a day of a week}

**Question 2.If U = {x/x ∈ N, 1 ≤ x ≤ 12}, A = {1,4, 7,10}, B = {2, 4, 6, 7, 11}, C = {3, 5, 8, 9, 12}. Write down the sets.(i) A ∪ B(ii) B ∩ C(iii) A – B(iv) B ∩ C’(v) A ∪ B ∪ C(vi) A ∩ (B ∪ C)Solution:**
U = {x/x ∈ N, 1 ≤ x ≤ 12} = {1, 2, 3, …., 12}

A = {1, 4, 7, 10}, B = {2, 4, 6, 7, 11}, C = {3, 5, 8, 9, 12}

(i) A ∪ B = {1, 2, 4, 6, 7, 10, 11}

(ii) B ∩ C = {}

(iii) A – B = {1, 10}

(iv) C’ = {1, 2, 4, 6, 7, 10, 11}

∴ B ∩ C’ = {2, 4, 6, 7, 11}

(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(vi) B ∪ C = {2, 3, 4, 5, 6, 7, 8, 9, 11, 12}

∴ A ∩ (B ∪ C) = {4, 7}

**Question 3.In a survey of 425 students in a school, it was found that 115 drink apple juice, 160 drink orange juice, and 80 drink both apple as well as orange juice. How many drinks neither apple juice nor orange juice?Solution:**

Let A = set of students who drink apple juice

B = set of students who drink orange juice

X = set of all students

∴ n(X) = 425, n(A) = 115, n(B) = 160, n(A ∩ B) = 80

No. of students who neither drink apple juice nor orange juice = n(A’ ∩ B’) = n(A ∪ B)’

= n(X) – n(A ∪ B)

= 425 – [n(A) + n(B) – n(A ∩ B)]

= 425 – (115 + 160 – 80)

= 230

**Question 4.In a school, there are 20 teachers who teach Mathematics or Physics. Of these, 12 teach Mathematics and 4 teach both Physics and Mathematics. How many teachers teach Physics?Solution:**

Let A = set of teachers who teach Mathematics

B = set of teachers who teach Physics

∴ n(A ∪ B) = 20, n(A) = 12, n(A ∩ B) = 4

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

20 = 12 + n(B) – 4

∴ n(B) = 12

∴ Number of teachers who teach physics = 12

**Question 5.(i) If A = {1, 2, 3} and B = {2, 4}, state the elements of A × A, A × B, B × A, B × B, (A × B) ∩ (B × A).(ii) If A = {-1, 1}, find A × A × A.Solution:**
(i) A = {1, 2, 3} and B = {2, 4}

A × A = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}

A × B = {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)}

B × A = {(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)}

B × B = {(2, 2), (2, 4), (4, 2), (4, 4)}

∴ (A × B) ∩ (B × A) = {(2, 2)}

(ii) A = {-1, 1}

∴ A × A × A = {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)}

**Question 6.**
If A = {1, 2, 3}, B = {4, 5, 6}, check if the following are relations from A to B. Also, write its domain and range.

**Question 7.Determine the domain and range of the following relations.(i) R = {(a, b) / a ∈ N, a < 5, b = 4}(ii) R = {(a, b) / b = |a – 1|, a ∈ Z, |a| < 3}Solution:**
(i) R = {(a, b) / a ∈ N, a < 5, b = 4}

∴ Domain (R) = {a / a ∈ N, a < 5} = {1, 2, 3, 4}

Range (R) = {b / b = 4} = {4}

(ii) R = {(a, b) / b = |a – 1|, a ∈ Z, |a| < 3}

Since a ∈ Z and |a| < 3,

a < 3 and a > -3

∴ -3 < a < 3

∴ a = -2, -1, 0, 1, 2

b = |a – 1|

When a = -2, b = 3

When a = -1, b = 2

When a = 0, b = 1

When a = 1, b = 0

When a = 2, b = 1

Domain (R) = {-2, -1, 0, 1, 2}

Range (R) = {0, 1, 2, 3}

**Question 8.Find R : A → A when A = {1, 2, 3, 4} such that(i) R = {(a, b) / a – b = 10}(ii) R = {(a, b) / |a – b| ≥ 0}Solution:**
R : A → A, A = {1, 2, 3,4}

(i) R = {(a, b)/a – b = 10} = { }

(ii) R = {(a, b) / |a – b| ≥ 0}

= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}

A × A = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}

∴ R = A × A

**Question 9.R : {1, 2, 3} → {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}. Check if R is(i) reflexive(ii) symmetric(iii) transitiveSolution:**

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}

(i) Here, (x, x) ∈ R, for x ∈ {1, 2, 3}

∴ R is reflexive.

(ii) Here, (1, 2) ∈ R, but (2, 1) ∉ R.

∴ R is not symmetric.

(iii) Here, (1, 2), (2, 3) ∈ R,

But (1, 3) ∉ R.

∴ R is not transitive.

**Question 10.Check if R : Z → Z, R = {(a, b) | 2 divides a – b} is an equivalence relation.Solution:**

(i) Since 2 divides a – a,

(a, a) ∈ R

∴ R is reflexive. .

(ii) Let (a, b) ∈ R

Then 2 divides a – b

∴ 2 divides b – a

∴ (b, a) ∈ R

∴ R is symmetric.

(iii) Let (a, b) ∈ R, (b, c) ∈ R

Then a – b = 2m, b – c = 2n,

∴ a – c = 2(m + n), where m, n are integers.

∴ 2 divides a – c

∴ (a, c) ∈ R

∴ R is transitive.

Thus, R is an equivalence relation.

**Question 11.Show that the relation R in the set A = {1, 2, 3, 4, 5} Given by R = {(a, b) / |a – b| is even} is an equivalence relation.Solution:**
(i) Since |a – a| is even,

∴ (a, a) ∈ R

∴ R is reflexive.

(ii) Let (a, b) ∈ R

Then |a – b| is even

∴ |b – a| is even

∴ (b, a) ∈ R

∴ R is symmetric.

(iii) Let (a, b), (b, c) ∈ R

Then a – b = ±2m, b – c = ±2n

∴ a – c = ±2(m + n), where m, n are integers.

∴ (a, c) ∈ R

∴ R is transitive

Thus, R is an equivalence relation.

**Question 12.Show that the following are equivalence relations:(i) R in A is set of all books given by R = {(x, y) / x and y have same number of pages}(ii) R in A = {x ∈ Z | 0 ≤ x ≤ 12} given by R = {(a, b) / |a – b| is a multiple of 4}(iii) R in A = (x ∈ N/x ≤ 10} given by R = {(a, b) | a = b}Solution:**
(i) a. Clearly (x, x) ∈ R

∴ R is reflexive.

b. If (x, y) ∈ R then (y, x) ∈ R.

∴ R is symmetric.

c. Let (x, y) ∈ R, (y, x) ∈ R.

Then x, y, and z are 3 books having the same number of pages.

∴ (x, z) ∈ R as x, z has the same number of pages.

∴ R is transitive.

Thus, R is an equivalence relation.

(ii) a. Since |a – a| is a multiple of 4,

(a, a) ∈ R

∴ R is reflexive.

b. Let (a, b) ∈ R

Then a – b = ±4m,

∴ b – a = ±4m, where m is an integer

∴ (b, a) ∈ R

∴ R is symmetric.

c. Let (a, b), (b, c) ∈ R

a – b = ± 4m, b – c = ± 4n,

∴ a – c = ±4(m + n), where m, n are integers

∴ (a, c) ∈ R

∴ R is transitive

Thus, R is an equivalence relation.

(iii) a. Since a = a

∴ (a, a) ∈ R

∴ R is reflexive.

b. Let (a, b) ∈ R Then a = b

∴ b = a

∴ (b, a) ∈ R

∴ R is symmetric.

c. Let (a, b), (b, c) ∈ R

Then, a = b, b = c

∴ a = c

∴ (a, c) ∈ R

∴ R is transitive.

Thus, R is an equivalence relation.