Utilizing OP Malhotra Class 11 Solutions Chapter 1 Sets Chapter Test as a study aid can enhance exam preparation.
S Chand Class 11 ICSE Maths Solutions Chapter 1 Sets Chapter Test
Question 1.
Answer true or false :
(i) If A = {1, 2, 3, 4, 5, 6} and B = {whole numbers less than 6}, then A = B.
(ii) The empty set has no subsets.
(iii) The empty set may be represented by (φ)
(iv) {1, 2} = {2, 1}
(v) Φ = {0}
(vi) Φ ⊂ {1, 2}
(vii) Given that AS = {x | x is a square} and B = {x | x is a rectangle}.
State which of the following is true : A = B or A ≠ B.
(viii) The total number of subsets of a finite set which contains n elements is 2n.
(ix) If A ⊂ B and C ⊂ B, then A ⊂ C.
(x) The cardinal number of the set of the letters in the word ‘INDIA’ is 4.
(xi) There are as many numbers in the set of natural numbers as in the set of natural numbers divisible by 17.
(xii) If n (A) = n (B), then A ↔ B i.e., A and B are equivalent sets.
(xiii) If x ∈ A’, then x ∉ (A’)’
(xiv) If A = {0, 2, 4}, and B = {0, 3, 5, 7}, then A ∩ B = φ
(xv) ξ ∩ φ = ξ
(xvi) The union of two overlapping (inter-secting) sets is either of the two sets.
(xvii) A ∩ B = φ then A ∩ φ = B.
(xviii) For any sets X and Y, (X ∪ Y)’ = X’ ∩ Y’.
(xix) For any sets A and B, A∪B = B∪A.
Solution:
(i) Given A = {1, 2, 3, 4, 5, 6}
and B = {whole numbers less than 6} = {0, 1, 2, 3, 4, 5}
Clearly 6 ∈ A but 6 ∉ B
Thus A ≠ B
∴ given statement is false.
(ii) False, since every set is a subset of itself. i.e. Φ ⊆ Φ
(iii) False, empty set is represented by Φ or { }.
(iv) True, {1, 2} = {2, 1} since in both sets distincts elements are 1 and 2.
(v) False, Φ = { } i.e. empty set contains no element while {0} contains an element 0.
(vi) True ; clearly Φ be a proper subset of {1,2}.
(vii) Given A = {x | x is square}
and B = {x | x is a rectangle}
Every square is a rectangle but every rectangle is not a square.
∴ A ≠ B
(viii) True, since the total no. of subset of set containing n elements be 2n.
Total no. of subsets containing no element of a given set = nC0
Total no. of subsets containing one element of given set = nC1
Total no. of subsets containing 2 elements of given set = nC2
and so on
Total no. of subsets containing n elements of given set = nCn
Hence Total no. of subsets of given set
=nC0 + nC1 + nC2 + …. + nCn
= 2n
(ix) False, e.g. : Given A = {1, 2, 3} ;
B = {1,2, 3,4} ; C = {1,2}
Hence A ⊂ B, C ⊂ B but A ⊄ C
since 3 ∈ A but 3 ∉ C
(x) True, clearly cardinal number of the set of the letters in the word INDIA
= no. of distinct letters in INDIA = 4
(xi) True, N = {1, 2, 3, 4, ….}
A = {17, 34, 51, 68 }
Clearly both sets are infinite sets as the counting of elements of both sets never comes to an end.
(xii) True, By definition, two sets A and B are equivalent if the no. of elements in both sets are equal i.e. n (A) = n (B)
(xiii) True, if x ∈ A ⇒ x ∉ A’ ⇒ x ∈ (A’)’
if x ∉ A ⇒ x ∈ A’ ⇒ x ∉ (A’)’
(xiv) Given A = {0, 2, 4} and B = {0, 3, 5, 7}
∴ A ∩ B = {0} ≠ Φ
∴ given statement is false.
(xv) True, since Φ = ξ – Φ = ξ
∴ ξ ∩ Φ’ = ξ ∩ ξ = ξ
(xvi) False, e.g. : Let A = {1, 2, 3} and B = {3, 4, 5}
Here A ∩ B = {3} ≠ Φ
∴ A ∪ B = {1, 2, 3, 4, 5} ≠ A or B
(xvii) Given A ∩ B = Φ
∴ A ∩ Φ = A ∩ (A ∩ B) = (A ∩A) ∩ B
= A ∩ B = Φ
∴ given statement is false.
(xviii) True, Let x ∈ (X ∪ Y)’ be any arbitrary element
⇒ x ∉ X ∪ Y ⇒ x ∉ X and x ∉ Y
⇒ x ∈ X’ and x ∈ Y’
⇒ x ∈ X’ ∩ Y’
∴ (X∪Y)’ ⊆ X’ ∩ Y’ …(1)
Let x ∈ (X’ ∩ Y’) be any arbitrary element
⇒ x ∈ X’ and x ∈ Y’
⇒ x ∉ X and x ∉ Y
⇒ x ∉ (X∪Y) ⇒ x ∈ (X ∪ Y)’
∴ X’ ∩ Y’ ⊆ (X ∪ Y)’ … (2)
∴ from (1) and (2);
we have (X ∪ Y)’ = X’ ∩ Y’
(xix) True, Let x ∈ A ∪ B be any arbitrary element
⇔ x ∈ A or x ∈ B
⇔ x ∈ B or x ∈ A
⇔ x ∈ B ∪ A
Thus A ∪ B = B ∪ A
[∵ A ∪ B ⊆ B ∪ A and B ∪ A ⊆ A ∪ B]
Question 2.
A is the set of all integers from 20 to 70, both inclusive.
B = {x: x ∈ A, x is a perfect square},
C = {x : x ∈ A, x is a prime number}
D = {x: x ∈ A, is the first digit of x > its second digit}.
List the following sets.
(i) B ∩ C
(ii) B ∩ D
(iii) B ∩ C D
Solution:
Given A = {20, 21, 22, , 68, 69, 70}
B = (x ; x ∈ A, x is a perfect square} = {35, 36, 49, 64}
C = {x : x ∈ A, x is a prime number = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67}
D = {x: x ∈ A, x is the first digit of x > its second digit}
= {20, 21, 30, 31, 32, 40, 41, 42, 43, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 65}
(ii) ∴ B ∩ C = {x : x ∈ B and x∈C} = { } or Φ
(iii) B ∩ D = {x;x ∈ B and x ∈ D} = {64}
(iii) B ∩ C ∩ D = Φ ∩ D = Φ
Question 3.
If X = {a, b, c, d) and Y = (f, b, d, g} find (i) X – Y (ii) Y – X (iii) X ∩ Y (iv) X ∪ Y
Solution:
Given X = (a, b, c, d} and Y = {f, b, d, g)
(i) X – Y = set of all elements which are in X but not in Y
= {a, b, c, d} – {f, b, d, g} = {a, c}
(ii) Y – X = set of all elements which are in Y but not in X
= {f, b, d, g) – (a, b, c, d} = {f, g}
(iii) X ∩ Y = {x : x ∈ X and x ∈ Y} = {b, d}
(iv) X ∪ Y = {x : x ∈ X or x ∈ Y}
= {a, b, c, d, f g}
Question 4.
Let ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A ={1,2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}, find
(i) A’
(ii) B’
(iii) (A ∪ C)’
(iv) (A ∪ B)’
(v) (A ∩ C)’
(vi) (A’)’
(vii) (B – C)’
Solution:
Given ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {1,2, 3, 4} ;
B = {2, 4, 6, 8}
and C = {3, 4, 5, 6}
(i) A’ = ξ – A = {5, 6, 7, 8, 9}
(ii) B’ = ξ – B = {1, 2, 3, 4, 5, 6, 7, 8,9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}
(iii) A∪C = {1, 2, 3, 4} ∪ {3, 4, 5, 6}
= {1,2,3, 4, 5, 6}
∴ (A∪C)’ = ξ – (A∪C) = {7, 8, 9}
(iv) A∪B = {1, 2, 3, 4} ∪ {2, 4, 6,8}
= {1, 2, 3, 4, 6, 8}
∴ (A∪B)’ = ξ – (A∪B) = {5, 7, 9}
(v) A∩C = {1, 2, 3, 4}∩{3, 4, 5,6}
= {3, 4}
∴ (A∩C)’ = ξ – (A∩C) = {1, 2, 5, 6, 7, 8, 9}
(vi) A’ = ξ – A = {5, 6, 7, 8, 9}
∴ (A’)’ = ξ – A’ = {1,2, 3, 4, 5, 6, 7, 8, 9} – {5, 6, 7, 8, 9}
= {1, 2, 3, 4}
(vii) B – C = set of all elements which are in B but not in C = {2, 8}
∴ (B – C)’ = ξ – (B – C)
= {1, 3, 4, 5, 6, 7, 9}
Question 5.
If ξ = {1, 2, 3, 10}, A ={1, 2, 5}, and B = {6, 7}, verify that A – B = A ∩ B’ = B’ – A’.
Solution:
Given ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {1, 2, 5}
and B = {6, 7}
∴ B’ = ξ – B = {1, 2, 3, 4, 5, 8, 9}
L.H.S = A – B = set of all elements which are in A but not in B = {1, 2, 5} – {6, 7}
= {1,2, 5}
R.H.S = A ∩ B’
= {1,2, 5} ∩ {1,2,3, 4, 5, 8,9}
= {1,2,5}
Also, A’ = ξ – A = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 5}
= {3, 4, 6, 7, 8, 9}
∴ B’ – A’= {1,2, 3, 4, 5, 8, 9} – {3, 4, 6, 7, 8, 9}
= {1, 2, 5}
Thus A – B = A ∩ B’ = B’ – A’
Question 6.
(i) In the Venn diagram, shade the set A∪(B ∩ C).
(ii) Express in set notation the subset shaded in the Venn diagram.
(iii) In a class of 36 students, 25 study History, 20 study Geography and 4 study neither History nor Geography. Find how many students study both History and Geography ?
Solution:
(ii) Here the shaded portion consists of all elements which are in P but not in Q.
∴ required equivalent subset be P ∩ Q’
(iii) Let H = set of all students who study history
and G = set of all students who study geography
Then n (H) = 25 ; n (G) = 20
and n (ξ) = 36
where ξ = set of all students in a class
4 = n(H’∩G’) = n(H∪ G)’ = n(ξ) – n(H∪G)
⇒ n (H ∪ G) = n (ξ) – 4 = 36 – 4 = 32
∴ n(H∩G) = n(H) + n(G) – n(H∪G)
= 25 + 20 – 32 = 13
Hence, the required number of students who study both History and Geography
= n(H∩G) = 13
Question 7.
ξ = {x : x is an integer and 1 ≤ x ≤ 8} P = {x : x > 5}, Q = {x : x ≤ 3}
(i) (a) Find the value of n (P ∪ Q)
(b) List the elements of P’ ∩ Q’
(ii) Express, in set notation, as simply as possible, the subset shaded in the Venn diagram.
(iii) It is given that n (ξ) = 40, n (P) = 18, n (Q) = 20 and n(P∩Q) = 7. Find (a) n (P ∪ Q), (b) n (P’ ∪ Q’).
(iv) There are 27 children in a class, of these children, 19 own a bicycle, 15 own a scooter and 3 own neither a bicycle nor a scooter. Using a Venn diagram, or otherwise, find the number of children who own a bicycle but not a scooter.
Solution:
Given ξ = (x : x is an integer and 1 ≤ x ≤ 8} = {1,2, 3, 4, 5, 6, 7,8}
P = {x : x > 5} = {6, 7, 8}
Q = {x : x ≤ 3} = {1, 2, 3}
(i) (a) ∴ P∪Q = {6, 7, 8} ∪ {1,2, 3}
= {1,2, 3, 6, 7,8}
Thus n(P∪Q) = no. of distinct elements in P ∪ Q = 8
(ii) (b) P’ = ξ – P= {1, 2, 3, 4, 5}
Q’ = ξ – Q = {4, 5, 6, 7,8}
∴ P’ ∩ Q’ = {1, 2, 3, 4, 5} ∩ {4, 5, 6, 7, 8} = {4, 5}
Clearly the shaded portion consist of all common elements which are in B and C and A’
∴ required subset be B∩C∩A’.
(iii) Given n (ξ) = 40 ; n (P) = 18 ;
n (Q) = 20 and n (P ∩ Q) = 7
(a) We know that
n(P∪Q) = n (P) + n (Q) – n (P ∩ Q)
∴ n(P∪Q) = 18 + 20 – 7 = 31
(b) n (P’ ∪ Q’) = n(P∩ Q)’
= n (ξ ) – n (P ∩ Q) = 40 – 7 = 33
(iv) Let B = set of all children who have own bicycle
S = set of all children who have own scooter
ξ = set of all children in a class
Then n (B) = 19 ; n (S) = 15 ; n (ξ) = 27
3 = n(B’∩S’) = n(B∪ S)’
⇒ 3 = n(ξ ) – n(B∪S)
⇒ 3 = 27 – n (B ∪ S)
⇒ n (B ∪ S) = 27 – 3 = 24
∴ n(B∩S) = n(B) + n(S) – n (B ∪ S)
= 19 + 15 – 24 = 10
Thus required number of children who own a bicycle but not a scooter = n(B ∩ S’) = 9
Question 8.
(i) In the Venn diagram, shade P ∪ Q’.
(ii) A group of 60 children attend an often school club, of these, 35 children play football and 29 play hockey. 3 children do not play either football or hockey. By drawing a Venn diagram or other-wise, find the number of children who play only hockey.
Solution:
(i)
(ii) Let ξ = set of all children in a school who attend an often school club F = set of all children who play football H = set of all children who play Hockey Then n (ξ) = 60 ; n (F) = 35 ; n (H) = 29
given no. of children who do not play either football or Hockey = 3
⇒ 3 = K(F’∩H’) = n(F ∪ H)’
⇒ 3 = n(ξ) – n(F∪H)
⇒ n (F ∪ H) = 60 – 3= 57
∴ (F∩H) = n(F) + n (H) – n(F ∪ H)
= 35 + 29 – 57 = 7
Hence the required number of children
who play only hockey = n(H∩F’) = 22
Question 9.
(i) A, B and C are subsets of universal set ξ, if A = {2, 4, 6, 8, 12, 20} ;
B = {3, 6, 9, 12, 15},
C = {5, 10, 15, 20}
and ξ is the set of whole numbers, draw a Venn diagram showing the relation of A, B and C.
(ii) If a set A has n elements, then the number of elements in power set of A is
(a) n
(b) 2n
(c) n²
(d) none of these
(Roorkee)
(iii) If a finite set S contains n elements, then the number of non-empty proper subsets of S is
(i) 2.2n-1
(ii) 2 (2n – 1)
(iii) (2n-1 – 1)
(iv) 2 (2n-1 – 1)
Solution:
(i) Given ξ = {0, 1, 2, ……. }
A = {2, 4, 6, 8, 12, 20}
B = {3, 6, 9, 12, 15}
and C = {5, 10, 15, 20}
(ii) We know that power set of A be the set of all subsets of A
Total no. of subsets containing no element = nC0
Total no. of subsets containing one element of given set = nC1
Total no. of subsets containing 2 elements of given set = nC2
and so on
Total no. of subsets containing n elements of given set = nCn
Hence Total no. of subsets of given set = nC0 + nC1 + nC2 + …. + nCn = 2n
(iii) True, since the total no. of subset of set containing n elements be 2n.
Total no. of subsets containing no element of a given set = nC0
Total no. of subsets containing one element of given set = nC1
Total no. of subsets containing 2 elements of givne set = nC2
and so on
Total no. of subsets containing n elements of given set = nCn
Hence Total no. of subsets of given set = nC0 + nC1 + nC2 + …. + nCn = 2n
∴ required no. of non-empty proper subsets of A = 2n – 2 = 2 (2n-1 – 1)
[excluding Φ and given set]
Question 10.
If A = {3, 6, 8, 15, 19} and B = {1, 2, 6, 8, 14, 15}, then verify that A ∆ B = (A ∪ B) – (A ∩ B).
Solution:
Given A ={3,6, 8, 15, 19} and B = {1,2, 6, 8, 14, 15}
A ∆ B = (A – B) ∪ (B – A)
∴ A – B = set of all elements which are in A but not in B
= {3, 6, 8, 15, 19} – {1,2, 6, 8, 14, 15}
= {3, 19}
B – A = set of all elements which are in B but not in A
= {1, 2, 6, 8, 14, 15} – {3, 6, 8, 15, 19} = {1, 2, 14}
Thus A ∆ B = (A – B) ∪ (B – A)
= {3, 19} ∪ {1,2, 14}
= {1, 2, 3, 14, 19} …(1)
A ∩ B = {3, 6, 8, 15, 19} ∩{1,2, 6, 8, 14, 15}
A ∪ B = {3,6, 8, 15, 19} ∪ {1,2, 6, 8, 14, 15}
= {1,2, 3, 6, 8, 14, 15, 19}
∴ (A∪B) – (A∩B)= {1,2, 3, 14, 19} …(2)
∴ from eqn. (1) and eqn. (2) ; we have
A ∆ B = (A – B) ∪ (B – A)
= (A∪B) – (A∩B)
Question 11.
(i) If A, B, C are three non-zero sets, then (A∩B)∩(B∩C)∩(C∩ A) is equal to
(a) A∩B∩C
(b) A∪B∪C
(c) Φ
(d) none of these
(ii) Consider the given Venn diagram if n (ξ) = 42, n (A) = 15, n (B) = 12 and n (A ∪ B) = 22, then the area represented by the shaded portion in the Venn diagram is
(a) 25
(b) 27
(c) 32
(d) 37
(iii) In a class of 100 students, 55 students have passed in mathematics and 67 students have passed in physics. Then, the number of students who have passed in physics only is
(a) 22
(b) 33
(c) 10
(d) 45
Solution:
(i) (A∩B)∩(B∩C)∩(C∩A) = [{(A ∩ B)∩B}∩ {(A ∩ B) ∩ C}] ∩ (C∩A)
[using distributive law]
= [{(A∩B)∩(A∩B)}∩C]∩(C∩A)
= {(A∩B)∩C) ∩(C∩A)
[using associative law]
= [(A ∩ C)∩( B∩ C)] ∩(C∩A)
= (A∩C)∩(B∩C) [∵ A∩A = A]
= [(A∩ C) ∩ B] ∩ [(A ∩C)∩C]
= (A∩C∩B)∩(A∩C)
= A∩C∩B = A∩B∩C
Aliter :
(ii) Given n (ξ) = 42 ; n (A) =15;
n (B) = 12 and n (A ∪ B) = 22
∴ n (A ∩ B) = n (A) + n (B) – n(A∪B)
= 15 + 12 – 22 = 5
∴ area of shaded region = n (ξ) – n (A ∩ B) = 42 – 5 = 37
(iii) Let ξ = set of all students in a class = P ∪ M
M = set of all students who have passed in mathematics
P = set of all students who have passed in physics
Then n (ξ ) = 100 ; n (M) = 55 ; n (P)= 67
i.e. n(P∪M) = 100
∴ n(P∩M) = n (P) + n (M) – n(P∪M)
= 67 + 55 – 100 = 22
Therefore, the no. of students who have passed in physics only = n(P ∩ M’)
= n (P) – n(P∩M)
= 67 – 22 = 45