# KSEEB SSLC Solutions for Class 10 Maths Chapter 2 Sets

In this chapter, we provide KSEEB Solutions for Class 10 Maths Chapter 2 Sets for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB Solutions for Class 10 Maths Chapter 2 Sets pdf, free KSEEB solutions for Class 10 Maths Chapter 2 Sets book pdf download. Now you will get step by step solution to each question.

## KSEEB SSLC Solutions for Class 10 Maths – Sets (English Medium)

#### KSEEB SSLC Solutions for Class 10 Maths Chapter 2 Exercise 2.1

Question 1:

Verify the commutative property of union and intersection of sets for the following.
A = {l, m, n, o, p, q} B = {m, n, o, r, s, t}

Solution :

Commutative property of union of sets,

A ∪ B = B ∪  A
Now A ∪ B = {l, m, n, o, p, q}  ∪  {m, n, o, r, s, t}
A ∪ B = {l, m, n, o, p, q, r, s, t}  …….(i)
Now B ∪ A = {m, n, o, r, s, t} ∪ {l, m, n, o, p, q}

B ∪ A = {l, m, n, o, p, q, r, s, t}  ……(ii)
from (i) and (ii) A ∪ B = B ∪ A
Commutative property of intersection of sets,
A ∩ B = B ∩ A
A ∩ B = {l, m, n, o, p, q} ∩ {m, n, o, r, s, t}
= {m, n, o}  …………….(iii)
B ∩ A = {m, n, o, r, s, t} ∩ {l, m, n, o, p, q}
= {m, n, o} ……………..(iv)
from (iii) and (iv)
A ∩ B = B ∩ A

Question 2:

Given P = {a, b, c, d, e}, Q = {a, e, i, o, u} and R = {a, c, e, g}, verify associative property of union and intersection of sets.

Solution :

Associative property of union
(P ∪ Q) ∪ R = P ∪ (Q ∪ R)
Let P ∪ Q = {a, b, c, d, e}  ∪  {a, e, i, o, u}
= {a, b, c, d, e, i, o, u}
∴ (P ∪ Q) ∪ R = {a, b, c, d, e, i, o, u}  ∪ {a, c, e, g}
(P ∪ Q) ∪ R = {a, b, c, d, e, g, i, o, u}  ………….(i)
Now (Q ∪ R) = {a, c, e, g, i, o, u}
and P ∪ (Q ∪ R) = {a, b, c, d, e} ∪ {a, c, e, g, i, o, u}
P ∪ (Q ∪ R) = {a, b, c, d, e, g, i, o, u} …….(ii)
From (i) and (ii) (P ∪ Q) ∪ R = P ∪ (Q ∪ R)
Associative property of intersection ,
(P ∩ Q) ∩ R = P ∩ (Q ∩ R)
Let P ∩ Q = {a, b, c, d, e} ∩ {a, e, i, o, u} = {a, e}
(P ∩ Q) ∩ R = {a, e} ∩ {a, c, e, g}
(P ∩ Q) R = {a, e}  ……..(iii)
Now (Q ∩ R) = {a, e, i, o, u} ∩ {a, c, e, g}
(Q ∩ R) = {a, e}
∴ P ∩ (Q ∩ R) = {a, b, c, d, e} ∩ {a, e}
P ∩ (Q ∩ R) = {a, e}   ……..(iv)
From (iii) and (iv)
We get (P ∩ Q) ∩ R = P ∩ (Q ∩ R)

Question 3:

If A = {-3, -1,0,4,6,8, 10}, B = {-1, -2, 3, 4, 5, 6} and C = {-6, -4, -2, 2, 4, 6}
show that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Solution :

Let B ∩ C = {-1, -2, 3, 4 ,5, 6} ∩ {-6, -4, -2,2,4, 6}
(B ∩ C) = {-2, 4, 6}
∴ A ∪ (B ∩ C) = {-3, -1,0,4,6,8, 10}  ∪  {-2, 4, 6}
A ∪ (B ∩ C) = {-3, -2, -1,0,4,6,8, 10} ……(1)
Now A ∪ B = {-3, -1,0,4,6,8, 10}  ∪ {-1, -2,3,4,5, 6}
(A ∪ B) = {-3, -2, -1,0,3,4,5,6,8, 10}
(A ∪ C) = {-3, -1,0,4,6,8, 10} ∪ {-6, -4, -2,2,4, 6}
(A ∪  C) = {-6, -4, -3, -2, -1,0,2,4,6,8, 10}
∴ (A ∪  B) ∩ (A ∪ C) = {-3,-2,-1, 0, 3, 4, 5, 6, 8, 10}  ∩ {-6, -4, -3, -2, -1, 0, 2, 4, 6, 8, 10}
(A ∪  B) ∩ (A ∪ C) = {-3, -2, -1,0,4,6,8, 10}  ….. (2)
From (1) and (2)
∴  A ∪ (B ∩ C) = (A ∪ B) ∩ (A  ∪  C)

Question 4:

If U = {4, 8, 12, 16, 20, 24, 28}, A = {8, 16, 24},
B = {4, 16, 20, 28}. Verify that
(i) (A ∪ B)’ = A’ ∩ B’ and
(ii) (A ∩ B)’ = A’ ∪ B’

Solution :

Now A’ = U A = {4, 12,20, 28}
B’ = U B = {8, 12, 24}
(i) Let A ∪ B = {8, 16, 24} ∪ {4, 16,20, 28}
= {4, 8, 16,20,24, 28}
(A ∪ B)’ = U (A ∪  B) = {12}  ….. (1)
A’ ∩ B’ = {4, 12,20, 28} ∩ {8, 12, 24}
A’ ∩ B’ = {12}   ..…. (2)
From (1) and (2), we have,
∴ (A ∪  B)’ = A’ ∩ B’
(ii) A ∩ B = {8, 16, 24} ∩ {4, 16,20, 28} = {16}
(A ∩ B)’ = U (A ∩ B) = {4, 8, 12, 20, 24, 28} ….. (3)
Now A’ ∪ B’ = {4, 12, 20, 28}  ∪ {8, 12, 24}
A’ ∪ B’ = {4, 8, 12, 20, 24, 28} ….. (4)
From (3) and (4)
(A ∩ B)’ = A’ ∪  B’

Question 5:

If A = {1, 2, 3}, B = {2, 3, 4, 5}, C = {2, 4, 5, 6} are the subsets of U = {1, 2, 3, 4, 5, 6, 7, 8}, verify (A ∪ B)’ = A’∩ B’ and (A ∩ B)’ = A’ ∪ B’.

Solution :

Now A’ = U A = {4, 5, 6, 7, 8}
B’ = U B = {1, 6, 7, 8}
Consider
(i) (A ∪ B)’ = A’ ∩ B’
A ∪ B = {1, 2, 3} ∪ {2, 3, 4, 5}
(A ∪ B) = {1, 2, 3, 4, 5}
(A ∪ B)’ = U (A ∪ B) = {6, 7, 8} …..(1)
A’ ∩ B’ = {4, 5, 6, 7, 8} ∩ {1, 6, 7, 8}
A’ ∩ B’ = {6, 7, 8}   ……(2)
From (1) and (2)
(A  ∪ B)’ = A’ ∩ B’
(ii) (A ∩ B)’ = A’ ∪  B’
(A ∩ B) = {1, 2, 3} ∩ {2, 3, 4, 5}
(A ∩ B) = {2, 3}
∴ (A ∩ B)’ = U (A ∩ B) = {1, 4, 5, 6, 7, 8} …….(3)
Now A’ ∪ B’ = {4, 5, 6, 7, 8} ∪ {1, 6, 7, 8}
A’ ∪   B’ = {1, 4, 5, 6, 7, 8}   .…….(4)
From (3) and (4)
(A ∩ B)’ = A’ ∪ B’

Question 6:

If A = {2, 3, 5, 7, 11, 13}, B = {5, 7,9, 11, 15} are the subsets of  U = {2, 3, 5, 7, 9, 11, 13, 15}, verify DeMorgan’s laws.

Solution :

DeMorgon’s laws are as follows:
(i) (A ∪ B)’= A’ ∩ B’
(ii) (A ∩ B)’ = A’ ∪  B’
Now  A’ = U A = {9,15}
B’ = U B = {2, 3,13}
(i) A ∪ B= {2, 3, 5, 7, 11, 13}  ∪  {5, 7, 9, 11, 15}
A ∪ B = {2,3, 5,7, 9, 11, 13, 15}
(A ∪ B)’ = U (A ∪ B) = { }  …….(1)
A’ ∩ B’ = {9, 15} ∩ {2, 3, 13}
A’ ∩ B’ = {}  …….(2)
∴ From (1) and (2) (A ∪ B)’ = A’ ∩ B’
(ii) A ∩ B = {2, 3, 5, 7, 11, 13} ∩ {5, 7, 9, 11, 15}
A ∩ B = {5, 7, 11}
∴(A ∩ B)’ = U (A ∩ B) = {2, 3, 9, 13, 15} ……(3)
Now A’ ∪ B’ = {9, 15}  ∪  {2, 3, 13}
A’ ∪ B’ = {2, 3,9, 13, 15}  ……(4)
∴ From (3) and (4) (A ∩ B)’ = A’ ∪  B’

Question 7(i):

Draw Venn diagrams to illustrate the following:
(A ∪ B)

Solution :

Question 7(ii):

Draw Venn diagrams to illustrate the following:
(A ∪ B)’

Solution :

Question 7(iii):

Draw Venn diagrams to illustrate the following:
A’ ∩ B

Solution :

Question 7(iv):

Draw Venn diagrams to illustrate the following:
A ∩ B’

Solution :

Question 7(v):

Draw Venn diagrams to illustrate the following:
A B

Solution :

Question 7(vi):

Draw Venn diagrams to illustrate the following:
A ∩ (B C)

Solution :

Question 7(vii):

Draw Venn diagrams to illustrate the following:
A ∪ (B ∩ C)

Solution :

Question 7(viii):

Draw Venn diagrams to illustrate the following:
C ∩ (B ∪ A)

Solution :

Question 7(ix):

Draw Venn diagrams to illustrate the following:
C ∩ (B A)

Solution :

Question 7(x):

Draw Venn diagrams to illustrate the following:
A (B ∩ C)

Solution :

Question 7(xi):

Draw Venn diagrams to illustrate the following:
(A B) ∪ (A C)

Solution :

Question 7(xii):

Draw Venn diagrams to illustrate the following:
(A ∪ B) (A ∪ C)

Solution :

Question 8(i):

Using Venn diagram, verify whether the following are true:
A ∪ B = B ∪ A

Solution :

Question 8(ii):

Using Venn diagram, verify whether the following are true:
A B ≠ B A

Solution :

Question 8(iii):

Using Venn diagram, verify whether the following are true:
(A ∪ B)’ = A’ ∩ B’

Solution :

Question 8(iv):

Using Venn diagram, verify whether the following are true:
A (B ∪ C) = (A B) ∩ (A C)

Solution :

Question 8(v):

Using Venn diagram, verify whether the following are true:
A ∪ (B ∩ C)= (A ∪ B) ∩ (A ∪ C)

Solution :

Question 8(vi):

Using Venn diagram, verify whether the following are true:
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Solution :

Question 9:

Using Venn diagram, show that (A ∩ B) ∪ (A B) = A

Solution :

Exercise 2.2:

Question (I)1:

If A and B are the sets such that n (A) = 37, n(B) = 26 and n (A ∪ B) = 51, find n (A ∩ B).

Solution :

n(A ∩ B) = n(A) +n(B) – n(A ∪ B)
= 37 + 26 – 51= 63 – 51
n(A ∩ B) = 12 Question (I)2:

In a group of 50 persons, 30 like tea, 25 like coffee and 16 like both. How many like?

1. either tea or coffee
2. neither tea nor coffee

Solution :

No. of persons who like tea, n(T) = 30
No. of persons who like coffee, n(C) = 25
No. of persons who like both = n(T ∩ C) = 16

• No. of persons who like either tea or coffee
= n(T ∪ C) = n(T) + n(C) –  n(T ∩ C)
= 30 + 25 – 16 = 39
• No. of persons who like neither tea nor coffee
= n(T ∪ C)’
= U – n( T ∪ C)
= 50 – 39
= 11

Question (I)3:

In a group of passengers, 100 know Kannada, 50 know English and 25 know both. If passengers know either Kannada or English, how many passengers are in the group?

Solution :

n(K) = 100
n(E) = 50
n(K ∩ E) = 25
n(K ∪ E) =?
n(K ∪ E) = n(K) + n(E) – n(K ∩ E) = 100 + 50 – 25
n(K ∪ E) = 125
∴ 125 Passengers are there in the group. Question (I)4:

In a class, 50 students offered Mathematics, 42 offered Biology and 24 offered both the subjects. Find the number of students, who offer

1. Mathematics only
2. Biology only. Also find the total number of students in the class.

Solution :

n(M) = 50
n(B) = 42
n(M ∩ B) = 24
n(M B) =?
n(B M) =?
n(M ∪ B) = ?
Number of students who offered Mathematics only
= n(M B) = n(M) – n(M ∩ B)
= 50- 24 = 26 Students
Number of students who offered Biology only
= n(B M)
= n(B) – n(B ∩ M)
= 42 – 24 = 18
Now n(M ∪ B) = n(M) + n(B) – n(M ∩ B)
= 50 + 42 – 24 = 92 – 24
n(M ∪ B) = 68
∴ Total number of students in the class = 68 Question (I)5:

In a medical examination of 150 people, it was found that 90 had eye problem, 50 had heart problem and 30 had both complaints. What percentage of people had either eye trouble or heart trouble?

Solution :

Question (II)1:

A radio station surveyed 190 audiences to determine the types of music they liked. The survey revealed that 114 liked rock music, 50 liked folk music, 41 liked classical music, 14 liked rock music and folk music, 15 liked rock music and classical music, and 11 liked classical music and folk music. 5 liked all the three types of music.

Find

1. how many did not like any of the 3 types?
2. how many liked any two types only?
3. how many liked folk music but not rock music?

Solution :

Number of audiences who liked rock music = n(R)
Number of audiences who liked folk music = n(F)
Number of audiences who liked classical music = n(C)

• Number of audiences who did not like any of the three
= 190 – (114 + 50 + 41 – 14 – 15 – 11 + 5) = 20
• Number of audiences who liked any two types only
= (14 – 5) + (15 – 5) + (11 – 5)
= 9 + 10 + 6 = 25
•  Number of audiences who like folk but not rock music = (50 – 14) = 36

Question (II)2:

An advertising agency finds that, of its 170 people, 115 use television, 110 use radio and 130 use magazines. Also, 85 use television and magazines, 75 use television and radio, 95 use radio and magazines, 70 use all the three.

Find,

1. how many use only television?
2. how many use only radio?
3. how many use television and magazine but not radio?

Solution :

Number of people = 170
Number of people who used Television = n(T)
Number of people who use Magazine = n(M) • Number of people who use only television
= 115 – 85 – 75 + 70
= 25
• Number of people who use only radio
= 110 – 75 – 95 + 70
= 10
• Number of people who use television and magazine but not radio = 25 + 15 + 20 = 60

Question (II)3:

In a village, out of 120 farmers, 93 farmers have grown vegetables, 63 have grown flowers, 45 have grown sugarcane, 45 farmers have grown vegetables and flowers, 24 farmers have grown flowers and sugarcane, 27 farmers have grown vegetables and sugarcane. Find how many farmers have grown vegetables, flowers and sugarcane?

Solution :

Let the number of farmers growing vegetable be n(V),
the number of farmers growing flowers be n(F) and
the number of farmers growing sugarcane be n(S).

15 farmers have grown vegetables, flowers and sugarcane.
We know that n(V ∩ F ∩ S) = n(V ∪ F ∪ S) – n(V) – n(F) – n(S) + n(V ∩ F) + n(F ∩ S) + n(V ∩ S)
Therefore,
n(V ∩ F ∩ S) = 120 – 93 – 63 – 45 + 45 + 24 + 27
= 15

Question (II)4:

In a town 85% use bicycle, 40% use motorbike and 20% use car. Also, 32% use bicycle and motorbike, 13% use bicycle and car and 10% use motorbike and car. Find the percentage of people who use all the three types of vehicles.

Solution :

Let the percentage of bicycle users be n(B), the percentage of motorbike users be n(M) and the percentage of car users be n(C).
n(B ∪ M ∪ C) = 100%
We know that n(B ∩ M ∩ C) = n(B ∪ M ∪ C) – n(B) – n(M) – n(C) + n(B ∩ M) + n(M ∩ C) + n(B ∩ C)
Thus, we have,
n(B ∩ M ∩ C)
= 100% – 85% – 40% – 20% + 32% + 13% + 10%
= 10% ∴ Percentage of people who use all the three types of vehicles is 10%.

All Chapter KSEEB Solutions For Class 10 Maths

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All Subject KSEEB Solutions For Class 10

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