12 questions · self-marked practice — reveal the answer and mark yourself.
R in A = (x ∈ N/x ≤ 10} given by R = {(a, b) | a = b}
R in A = {x ∈ Z | 0 ≤ x ≤ 12} given by R = {(a, b) / |a – b| 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.
R in A is set of all books given by R = {(x, y) / x and y have same number of pages}
∴ 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.
∴ (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.
(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.
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
(i) No. of persons exposed to chemical A but not to chemical B = n(A ∩ B’)
= n(A) – n(A ∩ B)
= 150 – 36
= 114
2. No. of persons exposed to chemical B but not to chemical A = n(A’ ∩ B) = n(B) – n(A ∩ B) = 74 – 36 = 38
3. No. of persons exposed to chemical A or chemical B = n(A ∪ B)
= n(A) + n(B) – n(A ∩ B)
= 150 + 74 – 36
= 188
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) = 35 + 40 + 40 – 20 – 17 – 15 + 5 = 68 ∴ No. of students who did not fail in any exam = n(X) – n(A ∪ B ∪ C) = 200 – 68 = 132
2. No. of students who failed in NEET or JEE entrance = n(B ∪ C)
= n(B) + n(C) – n(B ∩ C)
= 40 + 40 – 17
= 63
(i)(A ∪ B ∪ C) (ii) (A ∩ B ∩ C)