The availability of step-by-step OP Malhotra Class 11 Solutions Chapter 1 Sets Ex 1(d) can make challenging problems more manageable.
S Chand Class 11 ICSE Maths Solutions Chapter 1 Sets Ex 1(d)
using part (iv) and part (v) ; we have (A ∪ B)’ = A’ ∩ B’
Question 1.
If n (ξ) = 80, n (A) = 48, n (B) = 40 and n (A ∩ B) = 25, draw a Venn diagram to find :
(i) n (A∪B)
(ii) n (A ∪ B)’
(iii) n (A – B)
(iv) n (B – A)
(v) n (A ∩ B’)
(vi) n (A’ ∩ B)
Solution:
Given n (ξ) = 80 ;
n (A) = 48 ;
n (B) = 40
and n (A ∩ B) = 25
(i) n(A∪B) = n (A) + n (B) – n (A ∩ B)
= 48 + 40 – 25 = 63
(ii) n(A∪B)’ = n(ξ) – n(A∪B)
= 80 – 63 = 17
(iii) n(A – B) = n (A) – n(A∩B)
= 48 – 25 = 23
(iv) n(B – A) = n (B) – n(A∩B)
= 40 – 25 = 15
(v) n (A ∩ B’) = n (A – B) = 23
(vi) n(A’∩B) = n(B ∩ A’) = n (B – A) = 15
Question 2.
If ξ = {x | x ∈ N, x < 10},
A = {x | x is a prime number, x < 10}, B = {x | x is an even number, x < 10} draw a Venn diagram to find :
(i) n (A ∪ B)
(ii) n (A ∩ B)
(iii) n (A B)’
(iv) n (A ∩ B’)
(v) n (A’ ∩ B)
Solution:
Given ξ = {x | x ∈ N, x < 10} = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {x | x is a prime number, x < 10}
= {2, 3, 5, 7}
B = {x | x is an even number}
= {2, 4, 6, 8}
Thus n (ξ) = 9 ; n (A) = 4 ; n (B) = 4
∴ A ∩ B = {2, 3, 5, 7} ∩ {2, 4, 6, 8} = {2}
Thus n (A∩ B) = 1
(i) ∴ n(A∪B) = n(A) + n(B) – n(A∩B)
= 4 + 4 – 1 = 7
(ii) n (A ∩ B) = 1
(iii) n (A ∪ B)’ = n (ξ) – n(A∪B) = 9 – 7 = 2
(iv) n(A∩B’) = n(A) – n(A∩B) = 4 – 1 = 3
(v) n (A’ ∩ B) = n (B) – n(A∩B) = 4 – 1 = 3
Question 3.
Given n (ξ) = 40, n (A’) = 12, n (B) = 15 and B ⊂ A. Draw a Venn diagram to illustrate this information. Hence find n (A – B).
Solution:
Given n (ξ) = 40 ; n (A’) = 12 ; n (B) = 15 Also B ⊂ A
∴ (A – B) = 13
since n (A’) =12 ⇒ n (ξ) – n (A) = 12
n (A) = 40 – 12 = 28
∴ (A – B) = n (A) – n(A∩B)
= n (A) – n (B)
= 28 – 15 = 13
Question 4.
The Venn diagram shows :
ξ = {pupils in class 8}
A = {pupils who play cricket}
B = {pupils who play basketball}
How many pupils:
(i) are in class 8 ?
(ii) play cricket
(iii) play basketball ?
(iv) play both cricket and basketball
(v) play neither cricket nor basketball ?
Solution:
ξ = {pupils in class 8}
(i) number of pupils who are in class 8 = n(ξ) = 15 + 8 + 9 + 13 = 45
(ii) No. of pupils who play cricket = n(A) = 15 + 8 = 23
(iii) No. of pupils who play basketball = n (B) = 8 + 9=17
(iv) No. of pupils who play both cricket and basketball = n(A∩B) = 8
(v) No. of pupils who play neither cricket nor basketball = n (A’ ∩ B’) = n(A ∪ B)’
= n(ξ) – n(A∪B)
= 45 – (15 + 8 + 9) = 13
Question 5.
In the Venn diagram
ξ = {people at a function}
A = {those who asked for tea}
B = {those who watched the ballet}
Write down the number who :
(i) asked for tea
(ii) asked for tea and watched the ballet
(iii) neither asked for tea nor watched the ballet
(iv) attended the function
Solution:
(i) Required no. of peoples who asked for tea = n( A) = 59 + 23 = 82
(ii) Required no. of peoples who asked for tea and watched the ballet = n (A ∩ B) = 23
(iii) Required no. of peoples who neither asked for tea nor watched the ballet
= n( A’ ∩ B’) = n(A∪ B)’
= n(ξ) – n (A ∪ B) = 25
(iv) Required no. of peoples who attended the function = 59 + 23 + 47 + 25 = 154
Question 6.
In a group of 30 people, 18 play squash and 19 play tennis. How many play both games, provided everyone plays at least one game ?
Solution:
Let S = set of all peoples who play squash
T = set of all peoples who play Tennis
∴ n(S) = 18 ; n(T) = 19
Also given n (S ∪ T) = 30
∴ No. of peoples who play both games
= n(S∩T) = n(S) + n(T) – n(S∪T)
= 18 + 19 – 30 = 7
Question 7.
In a class of 50 students, 22 like History, 25 like Geography and 10 like both subjects. Draw a Venn diagram and find the number of students who
(i) do not like History
(ii) do not like Geography
(iii) like neither History nor Geography
Solution:
Let H = set of all students who like History
G = set of all students who like geography
and ξ = set of all students
Thus, n (ξ) = 50 ; n (H) = 22 ; n (G) = 25 and n (G ∩ H) = 10
(i) ∴ Required no. of students who donot like History
= n (H’) = n (ξ) – n (H)
= 50 – 22 = 28
(ii) Required no. of students who do not like Geography = 25
(iii) Required no. of students who like neither History nor Geography = n (H’ ∩ G’)
= n(H∪G)’ = n(ξ) – n(H∪G)
= 50 – 37 = 13
Question 8.
2000 candidates appear in a written test in Mathematics and General Awareness for a Government job. 1800 passed in at least one subject. If 1200 passed in Mathematics and 1500 in General Awareness find :
(i) how many passed in both the subjects ?
(ii) how many passed in Mathematics only ?
(iii) how many failed in General Awareness ?
Solution:
Let M = set of all candidates who passed in Mathematics
G = set of all candidates who passed G.A test
Then n (M) = 1200 ; n (G) = 1500 and n(G∪M) = 1800
(i) Required no. of candidates who passed in both the subjects = n(M∩G)
= n (M) + n (G) – n (M ∪ G)
= 1200 + 1500 – 1800 = 900
(ii) ∴ Required no. of candidates who passed in Maths only = n (M ∩ G’)
= n (M) – n (M ∩ G)
= 1200 – 900 = 300
(iii) Required no. of candidates who failed in G.A = n (G’) = n(ξ) – n (G)
= 2000 – 1500 = 500
Question 9.
In a group of 80 people, 40 like Indian food, 36 like Chinese food and 27 do not like any kind of these foods. Draw Venn diagram to find :
(i) how many like both kind of food ?
(ii) how many like only the Indian food ?
(iii) how many like only the Chinese food ?
Solution:
Let A = set of peoples who like Indian food
B = set of peoples who like Chinese food
Then n (A) = 40 ; n (B) = 36 and n (ξ) = 80
(i) Also, no. of peoples who donot like any kind of peoples
= n (A’ ∩ B’) = n(A ∪ B)’
27 = n(ξ) – n(A∪B)
⇒ n (A ∪ B) = 80 – 27 = 53
∴ n (A ∩ B) = n (A) + n(B) – n(A∪B)
= 40 + 36 – 53 = 23
(ii) No. of peoples who like only indian food
= n(A ∩ B’) = 17
(iii) No. of peoples who like only Chinese food
= n(A’ ∩ B) = 13
Question 10.
In a group of people, two-seventh speak Bengali only and three-seventh speak Hindi only. If 20 people speak none of these languages and 80 speak both, find using Venn diagram the total number of people in the group.
Solution:
Let the total number of peoples in the group be x.
∴ n(ξ) = x
Let B = set of all peoples who speak Bengali
and H = set of all peoples who speak Hindi
Then n (B ∩ H’) = \(\frac { 2 }{ 7 }\); n (H ∩ B’) = \(\frac { 3 }{ 7 }\)x
given no. of peoples who speak none of these languages = 20
and Number of peoples who speak both these languages = 80
From Venn diagram, we have
\(\frac { 2 }{ 7 }\)x + 80 + \(\frac { 3x }{ 7 }\) + 20 = x
⇒ \(\frac { 5x }{ 7 }\) + 100 = x
⇒ 100 = x – \(\frac { 5x }{ 7 }\) = \(\frac { 2x }{ 7 }\)
⇒ 2x = 700 ⇒ x = 350
Hence, the required no. of peoples in the group be 350.
Question 11.
In a class of 150 students, the following results were obtained in a certain exami-nation ; 45 students failed in Maths, 50 students failed in Physics, 48 students failed in Chemistry, 30 students failed in both Maths and Physics, 32 failed in Physics and Chemistry. 35 failed in both Maths and Chemistry, 25 failed in all the three subjects. Draw a Venn diagram corresponding to this data and find the number of students who have failed in at least one subject.
Solution:
Let ξ = set of all students in a class
M = set of all students who failed in Maths
P = set of all students who failed in Physics
C = set of all students who failed in Chemistry
Then n (ξ) = 150 ; n (M) = 45 ; n (P) = 50 ; n (C) = 48
n (M ∩ P) = 30 ;
n (P ∩ C) = 32
and n (M ∩ C) = 35
and n (M ∩ P ∩ C) = 25
Hence the required number of students who have failed in atleast one subject
= n(M∪P∪C)
= 5 + 25 + 10 + 6 + 7 + 13
= 71
Question 12.
A firm has 40 workers working in the factory premises, 30 working in its office and 20 working in both the places. How many workers are there in the firm ? How many are working in the (i) factory (ii) office alone ?
Solution:
Let F be the set of all workers working in factory premises and O be the set of all workers working in office
Then n (F) = 40 and n (O) = 30
and n (F ∩ O) = 20
Total no. of workers working in the firm = 20 + 20+ 10 = 50
- Total no. of workers working in factory alone = n(F ∩ O’) = 20
- Total no. of workers working in office alone = n (O ∩ F’) = 10
