Maths 5 Mark Question

iii. X = {x | x is a prime number between 80 and 100}
∴ X = {83, 89, 97}
Y = {y | y is an odd number between 90 and 100}
∴ Y = {91, 93, 95, 97, 99}

ii. n (E ∪ F) = n (E) + n (F) – n(E ∩ F)
100 = 72 + 43 – n (E ∩ F)
n (E ∩ F) = 72 + 43 – 100
∴ n(E ∩ F) = 15
Number of people who speak English and French = 15

iii. Number of people who speak only English = n(E) – n(E ∩ F)
= 72 – 15 = 57

iv. Number of
people who speak only French = n(F) – n(E ∩ F)
= 43 – 15 = 28

Alternate Method:
Let U be the set of all the persons,
E be the set of persons who speak English,
F be the set of persons who speak French and x people speak both the languages.

Since, each one out of 100 persons speak at least one language.
∴ n(U) = n(E ∪ F) = 100
∴ 72 – x + x + 43 – x = 100
∴ 115 – x= 100
∴ x = 115 – 100= 15.
Number of people who speak English and French = 15
Number of people who speak only English = 72 – x = 72 – 15 = 57
Number of people who speak only French = 43 – x = 43 – 15 = 28

ii. Let A = {3, 5}, B = {3, 5, 8, 9, 10}
Since, all elements of set A are present in set B.
∴ A ⊆ B
Also, A ∩ B = {3, 5} = A
∴ If A ⊆ B, then A ∩B = A.

iii. Let A = {2, 3, 8, 10}, B = {3,8}
A ∩ B = {3, 8} = B
Also, all the elements of set B are present in set A
∴ B ⊆ A
∴ If A ∩ B = B, then B ⊆ A.

iv. Let A = {2, 3, 8, 10}, B = {3, 8}, A ∩B = {3, 8}
Since, all the elements of set A n B are present in set A and B
A ∩ B ⊆ A and A ∩B ⊆B

v. Let U= {3, 4, 6, 8}, A = {6, 4}
∴ A’ = {3, 8}
∴ A ∩ A’= { } = φ

vi. A ∩ φ = { } = φ

vii. Let A = {6, 4}
∴ A ∩ A = {6, 4}
∴ A ∩ A = A

ii. n(R ∪ S)=n (R) + n (S) – n (R ∩ S)
= 130 + 180 – 110
∴n (R ∪ S) = 200
∴ 200 students follow the hobby of rock climbing or sky watching.

iii. Total number of students = 220.
Number of students who do not follow the hobby of rock climbing or sky watching
= n (U) – n (R ∪ S)
= 220 – 200
= 20

iv. Number of students who follow the hobby of rock climbing only
= n (R) – n(R ∩ S)
= 130 – 110
= 20

v. Number of students who follow the hobby of sky watching only
= n (S) – n (R ∩ S)
= 180 – 110
= 70

Alternate Method:
Let U be the set of students of class IX,
R be the set of students who follow the hobby of rock climbing and
S be the set of students who follow the hobby of sky watching.

From the Venn diagram
i. Students who follow the hobby of rock climbing or sky watching
= n(R ∪ S)
= 20 + 110 + 70
= 200

ii. Number of students who do not follow the hobby of rock climbing or sky watching
= n (U) – n(R ∪S)
= 220 – 200
= 20

iii. Number of students who follow the hobby of rock climbing only
= n (R) – n(R ∩S)
= 130 – 110
= 20

iv. Number of students who follow the hobby of sky watching only
= n (S) – n (R ∩ S)
= 180 – 110
= 70

Alternate Method:
Let U be the set of students who appeared for the exam,
E be the set of students who passed in English and M be the set of students who passed in Maths.

Since, none of the students failed in both subjects
∴ Total student = n(E ∪M)
= 10 + 40 + 20
= 70
∴ The number of students who passed at least in one of the subjects is 70.

ii. n(T ∪ C) = n(T) + n(C) – n(T ∩ C)
= 80 + 60 – 20
∴ n(T ∪ C) = 120
∴ 120 students drink tea or coffee
Also, there are 125 students in the hostel.

iii. Number of students who do not drink tea or coffee = n(U) – n(T ∪ C)
= 125 – 120
= 5
∴ 5 students do not drink tea or coffee.

Alternate Method:
Let U be the set of students in the hostel, T be the set of students who drink tea and C be the set of students who drink coffee.

From Venn diagram,
Student who drinks tea or coffee = n(T ∪ C) = 60 + 20 + 40 = 120
∴ The number of students who do not drink tea or coffee = n(U) – n(T ∪ C)
= 125 – 120 = 5
∴ 5 students do not drink tea or coffee.

ii. The residents of Pune, Madhya Pradesh, Indore and Maharashtra are all residents of India.
∴ B can be the Universal set for the above sets.

U = {1, 2, 3, 4, …… ,18, 19, 20}
Y = { y | y ∈ N, y is a prime number from 1 to 20}
∴ Y = {2, 3, 5, 7, 11, 13, 17, 19}

ii. A = {a, b, c, d, e}, D = {a, e}
A ⊆ D
False
Since, all the elements of A are not present in D.

iii. D = {a, e}, B = {c, d, e, f}
D ⊆ B
False
Since, all the elements of D are not present in B.

iv. D = {a, e}, A = {a, b, c, d, e}
D ⊆ A
True
Since, all the elements of D are present in A.

v. B = {c, d, e, f}, A = {a, b, c, d, e}
B ⊆ A
False
Since, all the elements of B are not present in A.

vi. C = {b, d}, A= {a, b, c, d, e}
C ⊆A
True
Since, all the elements of C are present in A.

B = {x | x is a natural number but x is neither prime nor composite}
1 is the only number which is neither prime nor composite,
∴ x = 1
∴ B = {1} …(ii)

C = {x | x G N, x < 2}
1 is the only natural number less than 2.
∴ x = 1
∴ C = {1} …(iii)
∴ The element in sets A, B and C is identical. … [From (i), (ii) and (iii)]
∴ A, B and C are equal sets.