Exercise-1.1
Q.1: Which of the
following are sets. Explain your answer.
(a). The collection
of all day in a week which have the first letter S.
Answer:
In a year we can easily identify all the days in a week which
starts with the letter S. So
it will form a clearly defined collection of objects.
Hence, the given
collection can be a set.
(b). The collection
of ten most famous singers of India.
Answer:
If we talk about the most famous
singers then it is not a well-defined collection, because there
are different parameters to be famous. So, it does not fall in this category.
Hence, the given
collection cannot be a set.
(c). A group of
best football strikers of the world.
Answer:
A group of best football
strikers cannot be determined because each individual have
different point of view to identify the best football strikers. So it does not
form a well-defined collection.
Hence, this group is not
a set.
(d). The collection
of all girls in your school.
Answer:
The collection of all girls in
your school can easily be identified as this category is clearly defined.
Hence, the given
collection can be a set.
(e). The collection
of all odd numbers below 50.
Answer:
The collection of all odd
numbers below 50 can be identified by calculating, so it forms a
well-defined collection. .
Hence, the given
collection can be a set.
(f). A collection
of poems written by the poet Shakespeare.
Answer:
A collection of all the poems
written by Shakespeare can be identified easily as he has the copyright on all
his poems. So it forms a well-defined collection.
Hence the given
collection can be a set.
(g). The collection
of all prime numbers.
Answer:
All prime
numbers can be identified after doing some calculations and thus
it forms a well-defined collection.
Hence, the given
collection can be a set.
(h). The collection
of questions in science book.
Answer:
We can find out the given question
in a science book, so it will form a well-defined collection of objects.
Hence, the given
collection can be a set.
(i). A collection
of most dangerous reptiles in India.
Answer:
If we are talking about the most dangerous reptiles in India, and then it would be
difficult to tell. This category will vary from person to person. So it does
not fall in the category of well-defined collection.
Hence, the given
collection cannot be a set.
Q.2:
Let P = {2, 3, 4, 5, 6, 7}. Insert the correct symbol ∈or∉ inside the given
blank spaces below:
(a). 2 . . . . . .
. . . . P
(b). 9 . . . . . .
. . . P
(c). 11 . . . . . .
. . P
(d). 4 . . . . . .
. . P
(e). 0 . . . . . .
. . P
(f). 7 . . . . . .
. . P
Answer:
(a). 2∈P
(b). 9∉P
(c). 11∉P
(d). 4∈P
(e). 0∉P
(f). 7∈P
Q.3: Write the given sets
in roster form:
(a). P = {y: y is
an integer and -4 < y < 6}.
(b). Q = {y: y is a
natural number which is <8}
(c). R = {y: y is a
2 digit natural number in which the sum of its digits is 9}
(d). S = {y: y is a
prime number which is a divisor of 70}
(e). T = The set of
all letters in the word ELEPHANT
(f). U = The set of
all letters in the word DIVISION
Answer:
(a). P = {y: y is an integer and -4 < y < 6}
The elements from this given set are –3, –2,-1, 0, 1, 2, 3, 4, and 5 only.
Hence, we can write the following set in the roaster form as given below:
P = {–3, –2,-1, 0, 1, 2,
3, 4, 5}
(b). Q = {y: y is a natural number which is <6}
The elements from this given set are 1, 2, 3, 4, 5, 6, and 7 only.
Hence, we can write the following set in the roaster form as given below:
Q = {1, 2, 3, 4, 5, 6, 7}
(c). R = {y: y is a 2 digit natural number in which the sum of its
digits is 9}
The elements from this given set are 18, 27, 36, 45, 54, 63, 72, 81 and 90 only.
Hence, we can write the following set in the roaster form as given below:
R = {18, 27, 36, 45, 54,
63, 72, 81, 90}
(d). S = {y: y is a prime number which is a divisor of 140}
70 = 2 x 2 x 5 x 7
The elements from this given set are 2, 5, and 7 only.
Hence, we can write the following set in the roaster form as given below:
S = {2, 5, 7}.
(e). T = The set of all letters in the word ELEPHANT
There are 8 letters in the given word ELEPHANT, out of which L is repeated.
Hence, we can write the following set in the roaster form as given below:
T = {E, L, P, H, A, N, T}
(f). U = The set of all letters in the word DIVISION
There are 8 letters in the given word DIVISION, out of which I is repeated.
Hence, we can write the following set in the roaster form as given below:
U = {D, I, V, S, O, N}
Q.4: Write the given sets
in set-builder form:
(a). {4, 8, 12, 16,
20}
(b). {3, 9, 27, 81}
(c). {4, 16, 64,
256, 1024}
(d). {1, 3, 5, 7…}
(e). {1, 8, 27,
64….1000}
Answer:
(a).
{4, 8, 12, 16, 20} = {y: y = 4n, n∈Pand1≤n≤5}
(b). {3, 9, 27, 81}
We can see here
that 3=31, 9=32, 27=33, 81=34.
Hence, {3, 9, 27, 81} =
{y: y = 3n, n∈Pand1≤n≤4
(c). {4, 16, 64,
256, 1024}
We can see here
that 4=41,16=42,64=43,256=44,1024=45
Hence, {4, 16, 64, 256,
1024} = {y: y = 4n, n∈Pand1≤n≤5
(d). {1, 3, 5, 7…}
Above mentioned that the
numbers are the set of odd natural
numbers.
Hence, {1, 3, 5, 7…} = {y: y is an odd natural numbers}
(e). {1, 8, 27,
64….1000}
We can see here
that 1=13,2=23,3=33,4=43,…1000=103
Hence, {1, 8, 27,
64….1000} = {y: y = n2, n∈Pand1≤n≤10
Q.5: List all the
elements from the given sets:
(a). P = {y: y is
even natural number}
(b).
Q = {y: y is an integer, −12<y<92}
(c).
R = {y: y is an integer; y2≤4}
(d). S = {y: y is
letter in the word “TIFFIN”}
(e). T = {y: y is a
month of a year having 31 days}
(f). U = {y: y is a
consonant in the English alphabet which precedes m}
Answer:
(a). P = {y: y is
even natural number} = {2, 4, 6, 8, 10 …..}
(b).
Q = {y: y is an integer, −12<y<92}
We can see that −12=0.5and92=4.5
Hence, Q = {0, 1, 2, 3,
4}
(c).
R = {y: y is an integer; y2≤4}
We can see that (−1)2=1≤4;(−2)2=4≤4;(−3)2=9>4
⇒
02=0≤4
⇒
12=1≤4
⇒ 22=4≤4
⇒ 32=9>4
Hence, R = {-2, -1, 0, 1,
2}
(d). S = {y: y is
letter in the word “TIFFIN”}
= {T, I, F, N}
(e). T = {y: y is a
month of a year having 31 days}
= {January, March, May, July, September, November}
(f). U = {y: y is a
non-vowel alphabet in English which comes before m}
= {b, c, d, f, g, h, j, k, l}
Q.6: Match the following:
(A)
|
{1, 2, 3, 6}
|
(i)
|
{y: y is a divisor of 6 and also a prime
number}
|
(B)
|
{T, R, I, G, O, N, M, E, Y}
|
(ii)
|
{y: y is less than 10 and also an odd
number}
|
(C)
|
{2, 3}
|
(iii)
|
{y: y is natural number divisor of 6}
|
(D)
|
{1, 3, 5, 7, 9}
|
(iv)
|
{y: y is a letter of the word TRIGONOMETRY}
|
Answer:
(A)
|
{1, 2, 3, 6}
|
(iii)
|
{y: y is natural number divisor of 6}
|
(B)
|
{T, R, I, G, O, N, M, E, Y}
|
(iv)
|
{y: y is a letter of the word TRIGONOMETRY}
|
(C)
|
{2, 3}
|
(i)
|
{y: y is a divisor of 6 and also a prime number}
|
(D)
|
{1, 3, 5, 7, 9}
|
(ii)
|
{y: y is less than 10 and also an odd number}
|
Exercise 1.2
Q.1: Which of the
following given below is null set?
(i). Set of odd
natural numbers which is divisible by 2.
(ii). Set of even
numbers which are prime
(iii).
{x: x is a natural number, x<5 and x>7}
(iv). {y: y is a
point common to any two parallel lines}
Answer:
(i). A
set of odd natural numbers which
are divisible by 2 is a null set as
none of the odd numbers is divisible
by 2.
(ii). A
set of even prime numbers is
not null set as
there is number 2 which is prime and also divisible by 2.
(iii). {x:
x is a natural number, x<5 and x>7} is a null set as
any number cannot be less than 5 and also
greater than 7.
(iv). {y:
y is a point common to any two parallel lines} is a null set as parallel lines do not intersect.
Therefore, there is no common point.
Q.2: State whether
the following sets are infinite or finite:
(i). A set of
months of a year.
(ii). {1, 2, 3 ….}
(iii). {1, 2, 3…99,
100}
(iv). The set of
positive integers which are greater than 100.
(v). The set of
prime numbers which are less than 99
Answer:
(i). The
set of months of a year has 12 elements. Therefore,
it is a finite set.
(ii). {1,
2, 3 ….} has infinite numbers in the set. Therefore, it is infinite set.
(iii). {1,
2, 3…99, 100} has elements from 1 to 100. Therefore, it is finite set.
(iv). The
set of positive integers which are greater than 100 has infinite elements as
there are infinite such elements. Therefore,
it I infinite set.
(v). The
set of prime numbers which are less than 99 has finite numbers in this set
which is less than 99. Therefore, it is finite
set.
Q.3: State whether the
following sets are infinite or finite:
(i). The set of
lines parallel to the x – axis.
(ii). The set of
letters in the vowels.
(iii). The set of
numbers multiple of 10.
(iv). The set of
humans living on Earth.
(v). The set of
circles passing through the origin (0, 0).
Answer:
(i). The
set of lines which are parallel to the x-axis has infinite elements. Therefore, it is an infinite set.
(ii). The
set of letters in the vowels has a finite element that is 5 elements. Therefore, it is a finite set.
(iii). The
set of numbers which are multiple of 5 has infinite elements. Therefore, it is an infinite set.
(iv). The
set of animals living on the earth has a finite number of elements. Therefore, it is finite set.
(v). The
set of circles passing through the origin (0, 0) has infinite elements as
number of circles can pass through the origin. Therefore, it is an infinite set.
Q.4: In the following set
given below, state whether A = B or not:
(i). A = {w, x, y, z}
B = {z, y, x, w}
(ii). A = {5, 9,
13, 17}
B = {9, 5, 17, 19}
(iii). A = {4, 2,
6, 10, 8}
B = {x: x is positive even integer and x≤10 }
(iv). A = {x:
x is a multiple of 10}
B = {10, 15, 20, 25, 30 …}
Answer:
(i). A
= {w, x, y, z}
B = {z, y, x, w}
Both the sets have same elements but the order is
different. Therefore, A = B
(ii). A
= {5, 9, 13, 17}
B = {9, 5, 17, 19}
It can be seen that
13 ∈ A
but 13 ∉ B. Therefore A ≠ B
(iii). A
= {4, 2, 6, 10, 8}
B = {x: x is positive
even integer and x≤10 }
= {2, 4, 6, 8, 10}
Therefore, A = B
(iv).
A = {x: x is a multiple of 10}
B = {10, 15, 20, 25, 30 …}
It can be seen that
15 ∈ B
but 15 ∉ A.
Therefore A ≠ B
Q.5 In the following set
given below, is the pair of sets equal?
(i). A = {3, 4}
B = {y: y is solution of y2+5y+6=0}
(ii). A = {a: a is
a letter in the word FOLLOW}
B = {b: b is a letter in the word WOLF}
Answer:
(i) A
= {3, 4}
B = {y: y is solution
of y2+5y+6=0}
The equation given y2+5y+6=0 can be solved as:
y (y + 3) + 2(y + 3) = 0
(y + 2) (y + 3) = 0
y = –2 or y = –3
Therefore, A = {2, 3} and B = {–2, –3}
Therefore,
A ≠ B
(ii) A
= {x: x is a letter in the word FOLLOW}
= {F, O, L, W}
B = {y: y is a letter in the word WOLF}
= {W, O, L, F}
Both the sets have same
elements but the order is different.
Therefore, A = B
Q.6: From the following
sets, select equal sets:
A = {2, 4, 8, 12}
B = {1, 2, 3, 4}
C = {4, 8, 12, 14}
D = {3, 1, 4, 2}
E = {–1, 1}
F = {0, a}
G = {1, –1}
H = {0, 1}
Answer:
A = {2, 4, 8, 12}
B = {1, 2, 3, 4}
C = {4, 8, 12, 14}
D = {3, 1, 4, 2}
E = {–1, 1}
F = {0, a}
G = {1, –1}
H = {0, 1}
We can see that:
8 ∈ A, 8 ∉ B,
8 ∉ D,
8 ∉ E,
8 ∉ F,
8 ∉ G
and 8 ∉ H
Therefore,
A ≠ B, A ≠ D, A ≠ E, A ≠ F, A ≠ G and A ≠ H
Also,
2 ∈ A and 2 ∉ C
Therefore,
A ≠ B
Also, 3 ∈ B, 3 ∉ C,
3 ∉ E,
3 ∉ F,
3 ∉ G
and 3 ∉ H
Therefore,
B ≠ C, B ≠ E, B ≠ F, B ≠ G, B ≠ H
Also,
12 ∈ C, 12 ∉ D,
12 ∉ E,
12 ∉ F,
12 ∉ G,
12 ∉ H
Therefore, C ≠ D, C ≠ E, C ≠ F, C ≠ G and C ≠ H
Also,
4 ∈ D, 4 ∉ E,
4 ∉ F,
4 ∉ G,
4 ∉ H
Therefore, D ≠ E, D ≠ F, D ≠ G, D ≠ H
Similarly, E ≠ F, E ≠ G, E ≠ H, F ≠ G, F ≠ H and G ≠ H
The order in which
elements of the set are listed is not significant.
Therefore, B = D and E =
G
Therefore, they are
equal.
Exercise 1.3
Q.1:
Fill in the blanks properly using ⊂ and ⊄.
(i). {3, 4, 5} ____
{2, 3, 4, 5, 6}
(ii). {a, b, c}
____ {d, c, d}
(iii). {y: y is a
pupil of Class 11 of the school} ____ {y: y is students of the school}
(iv). {y: y is a
circle in the plane} ____ {y: y is a circle in the same plane with radius 2
unit}
(v). {y: y is an
equilateral triangle in a plane} ____ {y: y is a rectangle in the same plane}
(vi). {y: y is an
equilateral triangle in a plane} ____ {y: y is a triangle in the plane}
(vii). {y: y is an
odd natural number} ____ {x: x is an integers}
Answer:
(i). {3,
4, 5} ⊂ {2,
3, 4, 5, 6}
(ii). {a,
b, c} ⊄ {d,
c, d}
(iii). {y:
y is a pupil of Class 11 of the school} ⊂ {y:
y is students of the school}
(iv). {y:
y is a circle in the plane} ⊄ {y:
y is a circle in the same plane with radius 2 unit}
(v). {y:
y is an equilateral triangle in a plane} ⊄ {y:
y is a rectangle in the same plane}
(vi). {y:
y is an equilateral triangle in a plane} ⊄ {y:
y is a triangle in the plane}
(vii). {y:
y is an odd natural number} ⊂ {x:
x is an integers}
Q.2: State whether the
given statements are true or false:
(i).
{b, c} ⊄ {c, d, e}
(ii).
{a, e, i} ⊂ {x: x is a vowel in the English alphabets}
(iii).
{1, 2, 3} ⊂{1, 2, 4, 5}
(iv).
{c} ⊂ {b, c, d}
(v).
{b} ∈ {a, b, c, d}
(vi).
{y: y is an even natural no. less than 6} ⊂ {y: y is a natural
no. which can divide 36}
Answer:
(i). False
Since, Each element of {d, c} is
present in {c, d, e}
(ii). True
Since, a, e and i are
the three vowels of the English
alphabet.
(iii). False
Since, 3 ∈ {1, 2, 3} but 3 ∉ {1, 2, 4, 5}
(iv). True
Since, Element c of {c} is also present in {b, c,
d}.
(v). False
Since, Element b of {b} is also present in {a, b,
c, d}.
(vi). True
Since, {y: y is an even natural number less than 6} = {2, 4}
{y: y is a natural no. which can divide 36} = {1, 2, 3, 4, 6, 9,
12, 18, 36}
Each elements of {2, 4}
are present in {1, 2, 3, 4, 6, 9, 12, 18, 36}
Q.3: Let X = {11, 12,
{13, 14}, 15}. According to the given set which of the given statements are
false? Explain why.
(i).
{13, 14} ⊂ X
(ii).
{13, 14} ∈ X
(iii).
{{13, 14}} ⊂ X
(iv).
11 ∈ X
(v).
11 ⊂ X
(vi).
{11, 12, 15} ⊂ X
(vii).
{11, 12, 15} ∈ X
(viii).
{11, 12, 13} ⊂ X
(ix).
Ø ∈ X
(x).
Ø ⊂ X
(xi).
{Ø} ⊂ X
Answer:
Given:
X = {11, 12, {13, 14},
15}
(i).
The Statement {13, 14} ⊂ X is False
Since, 13 ∈ {13, 14}
But, 13 ∉ X
(ii).
The Statement {13, 14} ∈ X is True
Since, {13, 14} is an element of X
(iii).
The Statement {{13, 14}} ⊂ X is True
Since, {13, 14} ∈ {{13, 14}} and {13, 14} ∈ X
(iv).
The Statement 11 ∈ X is True
Since, 11 is an element of X
(v).
The Statement 11 ⊂ X is False
Since, an element of a set can never be subset of itself.
(vi).
The Statement {11, 12, 15} ⊂ X is True
Since, each element of {11, 12, 15} is present in X
(vii).
The Statement {11, 12, 15} ∈ X is False
Since, {11, 12, 13} is not an element of X.
(viii).
The Statement {11, 12, 13} ⊂ X is False
Since, 13 ∈ {11, 12, 13}
But, 13 ∉ X
(ix).
The Statement Ø∈ X is False
Since, X does not contain element Ø
(x).
The Statement Ø⊂ X is True
Since, Ø is subset of every set.
(xi).
The Statement {Ø} ⊂ X is False
Since, Ø∈ {Ø}
Ø∈ X.
Q.4: Write all the
subsets of the given sets:
(i). {b}
(ii). {b, c}
(iii). {2, 3, 4}
(iv). Ø
Answer:
(i). {b}:
Subsets are
as given: Ø and {b}
(ii). {b, c}:
Subsets are
as given: Ø, {b}, {c} and {b, c}
(iii). {2, 3, 4}:
Subsets are
as given: Ø, {2}, {3}, {4}, {2, 3},
{3, 4}, {4, 2} and {2, 3, 4}
(iv). Ø:
Subsets are
as given: Ø
Q.5: How many elements
has P(X), if X = Ø ?
Answer:
If X has m elements that is n(X) =
m, then n[P(X)] = 2m
If X = Ø,
then n(X) = 0
Therefore, n[P(X)]
= 20 = 1
Therefore, P(X) has one
element.
Q.6: Write the given in the form of intervals:
(i).
{y: y ∈ R, –5 < y ≤ 7}
(ii).
{y: y ∈ R, –13 < y < –11}
(iii).
{y: y ∈ R, 1 ≤ y < 8}
(iv).
{y: y ∈ R, 4 ≤ y ≤ 5}
Answer:
(i). {y:
y ∈ R,
–5 < y ≤ 7} = (-5, 7]
(ii). {y:
y ∈ R,
–13 < y < –11} = (-13, -11)
(iii). {y:
y ∈ R,
1 ≤ y < 8} = [1, 8)
(iv). {y:
y ∈ R,
4 ≤ y ≤ 5} = [4, 5]
Q.7: Write the
given intervals in the form of set – builder:
(i). (–4, 1)
(ii). [7, 13]
(iii). (7, 13]
(iv). [–24, 6)
Answer:
(i).
(–4, 1) = {y: y ∈ R,
-4 < y < 1}
(ii).
[7, 13] = {y: y ∈ R,
7 ≤ y ≤ 13}
(iii).
(7, 13] = {y: y ∈ R,
7 < y ≤ 13}
(iv).
[–24, 6) = {y: y ∈ R,
–24 ≤ x
< 6}
Q.8: What universal set/
sets would you propose for the given sets?
(i). The set of
right triangles
(ii). The set of
isosceles triangles
Answer:
(i). The set of
right triangles
Proposed universal sets are:
·
The set of triangles
·
The set of polygons
(ii). The set of
isosceles triangles
Proposed universal sets
are:
·
The set of triangles
·
The set of polygons
·
The set of two –
dimensional figures
Q.9: X = {1, 3, 5}, Y =
{2, 4, 6} and Z = {0, 2, 4, 6, 8}
Which of the given sets
can be considered as the universal set for the given sets X, Y and Z?
(i). {0, 1, 2, 3,
4, 5, 6}
(ii). Ø
(iii). {0, 1, 2, 3,
4, 5, 6, 7, 8, 9, 10}
(iv). {1, 2, 3, 4,
5, 6, 7, 8}
Answer:
(i). {0, 1, 2, 3,
4, 5, 6}
X ⊂ {0, 1, 2, 3, 4, 5, 6}
Y ⊂ {0, 1, 2, 3, 4, 5, 6}
Z ⊄ {0, 1, 2, 3, 4, 5, 6}
Therefore, the set {0, 1,
2, 3, 4, 5, 6} cannot be the universal set for the sets X, Y and Z.
(ii). Ø
X ⊄ Ø
Y ⊄ Ø
Z ⊄ Ø
Therefore, the set Ø cannot be the universal set for the sets X, Y and Z.
(iii). {0, 1, 2, 3,
4, 5, 6, 7, 8, 9, 10}
X ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Y ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Z ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Therefore, the set {0, 1,
2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the sets X, Y and Z.
(iv). {1, 2, 3, 4,
5, 6, 7, 8}
X ⊂ {1, 2, 3, 4, 5, 6, 7, 8}
Y ⊂ {1, 2, 3, 4, 5, 6, 7, 8}
Z ⊄ {1, 2, 3, 4, 5, 6, 7, 8}
Therefore, the
set {1, 2, 3, 4, 5, 6, 7, 8} cannot be the universal set for the sets
X, Y and Z.
Exercise 1.4
Q.1: Find the union of
each of the following pairs of sets:
(i). P = {1, 4, 6} and Q
= {1, 3, 4}
(ii). X = {a, e, i, o, u}
and Y = {x, y, z}
(iii). X = {x: x is a
natural number and multiple of 3} and Y = {x: x is a natural number less than
6}
(iv). X = {x: x is a
natural number and 1 < x ≤ 5} and Y = {x: x is a natural number and 5 < x
< 10}
(v). X = {4, 5, 6} and Y
= Φ
Answer:
(i).
Given:
P = {1,
4, 6} and Q = {1,
3, 4}
Therefore,
P ∪ Q = {1, 3, 4, 6}
(ii).
Given:
X = {a, e,
i, o, u} and Y = {x,
y, z}
Therefore,
X ∪ Y = {x, y, z, a, e, i, o, u}
(iii).
Given:
X = {x:
x is a natural number and multiple of 3} and Y = {x: x is a natural number less than 5}
X = {3, 6, 9 ….}
Y = {1, 2, 3, 4}
X ∪ Y = {1, 2, 3, 4, 6, 9 …. }
Therefore,
X ∪ Y = {a:a = 1, 2, 3, 4 or multiple of 3}
(iv).
Given:
X = {x:
x is a natural number and 1 < x ≤ 5} = {2,
3, 4, 5}
Y = {x:
x is a natural number and 5 < x < 10} = {6, 7, 8, 9}
X ∪ Y = {2, 3, 4, 5, 6, 7, 8, 9}
Therefore, X ∪ Y = {a:a ∈ N and 1<a<10}
(v).
Given:
X = {4,
5, 6} and Y = Φ
Therefore,
X ∪ Y = {4, 5, 6}
Q.2: Let A = {x, y}
and B = {x, y, z}
Is A ⊂ B? What is A ∪ B?
Answer:
Given:
A = {x, y} and B = {a, b,
c}
Yes, A ⊂ B that is A is subset of B.
Therefore,
A ∪ B = {a, b, c} = B
Q.3: If
X and Y are two sets such that X ⊂ Y, then what is X ∪ Y?
Answer:
If X and Y are two sets such that X ⊂ Y, then X ∪ Y
= Y as all the elements of set X are
present in set Y.
Q.4: If P = {1, 2, 3, 4},
Q = {3, 4, 5, 6}, R = {5, 6, 7, 8} and S = {7, 8, 9, 10}
Find the following:
(i).
P ∪ Q
(ii).
P ∪ R
(iii).
Q ∪ R
(iv).
Q ∪ S
(v).
P ∪ Q ∪ R
(vi).
P ∪ Q ∪ S
(vii).
Q ∪ R ∪ S
Answer:
Given:
P = {1,
2, 3, 4]
Q = {3,
4, 5, 6}
R= {5,
6, 7, 8} and,
S = {7,
8, 9, 10}
Now,
(i). P ∪ Q = {1, 2, 3, 4, 5, 6}
(ii). P ∪ R = {1, 2, 3, 4, 5, 6, 7, 8}
(iii). Q ∪ R = {3, 4, 5, 6, 7, 8}
(iv).
Q ∪ S
= {3, 4, 5, 6, 7, 8, 9, 10}
(v). P ∪ Q ∪ R
= {1, 2, 3, 4, 5, 6, 7, 8}
(vi).
P ∪ Q ∪ S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(vii).
Q ∪ R ∪ S = {3, 4, 5, 6, 7, 8, 9, 10}
Q.5: Find the
intersection of each of the following pairs of sets:
(i). P = {1, 4, 6}
and Q = {1, 3, 4}
(ii). X = {a, e, i,
o, u} and Y = {x, y, z}
(iii). X = {x: x is
a natural number and multiple of 3} and Y = {x: x is a natural number less than
6}
(iv). X = {x: x is
a natural number and 1 < x ≤ 5} and Y = {x: x is a natural number and 5 <
x < 10}
(v). X = {4, 5, 6}
and Y = Φ
Answer:
(i).
Given:
P = {1,
4, 6} and Q = {1,
3, 4}
Therefore,
P ∩ Q = {1, 3}
(ii).
Given:
X = {a,
e, i, o, u} and Y = {x,
y, z}
Therefore,
X ∩ Y = Φ
(iii).
Given:
X = {x:
x is a natural number and multiple of 3} and Y = {x: x is a natural number less than 5}
X = {3, 6, 9 ….}
Y = {1, 2, 3, 4}
Therefore,
X ∩ Y = {3}
(iv).
Given:
X = {x:
x is a natural number and 1 < x ≤ 5} = {2, 3, 4, 5}
Y = {x:
x is a natural number and 5 < x < 10} = {6, 7, 8, 9}
Therefore,
X ∩ Y = Φ
(v). X = {4, 5, 6}
and Y = Φ
Therefore,
X ∩ Y = Φ
Q.6: If A = {3, 5, 7, 9,
11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}
Find the following:
(i).
A ∩ B
(ii).
B ∩ C
(iii).
A ∩ C ∩ D
(iv).
A ∩ C
(v).
B ∩ D
(vi).
A ∩ (B ∪ C)
(vii).
A ∩ D
(viii).
A ∩ (B ∪ D)
(ix).
(A ∩ B) ∩ (B ∪ C)
(x).
(A∪ D) ∩ (B ∪ C)
Answer:
(i).
A ∩ B = {7, 9, 11}
(ii).
B ∩ C = {11, 13}
(iii). A ∩ C ∩ D
= {A ∩ C} ∩ D
= {11} ∩ {15, 17} = Φ
(iv). A ∩ C = {11}
(v). B ∩ D = Φ
(vi). A ∩ (B ∪ C)
= (A ∩ B) ∪ (A ∩ C)
= {7, 9, 11} ∪ {11}
= {7, 9, 11}
(vii).
A∩ D = Φ
(viii). A∩ (B ∪ D)
= (A ∩ B) ∪ (A ∩ D)
= (7, 9, 11} ∪ Φ = {7, 9, 11}
(ix).
(A ∩ B) ∩ (B ∪ C)
= {7, 9, 11} ) ∩ {7, 9, 11, 13, 15} = {7,
9, 11}
(x). (A∪ D) ∩ (B ∪ C)
= {3, 5, 7, 9, 11, 15,
17} ) ∩ {7,
9, 11, 13, 15} = {7, 9, 11, 15}
Q.7: If A = {y: y is a
natural number}, B ={y: y is an even natural number}, C = {y: y is an odd
natural number} and D = {y: y is a prime number}
Find the following:
(i).
A ∩ B
(ii).
A ∩ C
(iii).
A ∩ D
(iv).
B ∩ C
(v).
B ∩ D
(vi).
C ∩ D
Answer:
A = {y:
y is a natural number} = {1, 2, 3, 4, 5….}
B = {y:
y is an even natural number} = {2,
4, 6, 8 ….}
C = {y:
y is an odd natural number} = {1,
3, 5, 7, 9 ….}
D = {y:
y is a prime number} = {2, 3, 5, 7….}
Now,
(i).
A ∩ B
= {y: y is a natural number} = B
(ii). A ∩ C = {y: y is an odd natural number} = C
(iii). A ∩ D = {y: y is a prime number} = D
(iv).
B ∩ C = Φ
(v).
B ∩ D = {2}
(vi). C ∩ D = {y: y is odd prime
number}
Q.8: Which of the given
pairs of sets are disjoint?
(i). A = {1, 2, 3,
4} and B = {x: x is a natural number and 4 ≤ x ≤ 6}
(ii). A = {a, e, i,
o, u} and B = {c, d, e, f}
(iii). A = {x: x is
an even integer} and B = {x: x is an odd integer}
Answer:
(i). A = {1,
2, 3, 4} and B = {x:
x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}
A ∩ B = {1, 2, 3, 4} ∩ {4,
5 6} = {4}
Therefore, this pair of
set is not disjoint.
(ii). A = {a,
e, i, o, u} and B = {c,
d, e, f}
A ∩ B = {a, e, i, o, u} ∩ {c,
d, e, f} = {e}
Therefore, this pair of
set is not disjoint.
(iii). A = {x: x is an even integer} and B = {x: x is an odd integer}
A ∩ B = {x: x is an even integer} ∩ {x: x is an odd integer} = Φ
Therefore, this pair of
set is disjoint.
Q.9: If A = {3, 6, 9, 12,
15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16} and D =
{5, 10, 15, 20}
Find the following:
(i). A – B
(ii). A – C
(iii). A – D
(iv). B – A
(v). C – A
(vi). D – A
(vii). B – C
(viii). B – D
(ix). C – B
(x). D – B
(xi). C – D
(xii). D – C
Answer:
(i). A
– B = {3, 6, 9, 15, 18, 21}
(ii). A
– C = {3, 9, 15, 18, 21}
(iii).
A – D = {3, 6, 9, 12, 18,
21}
(iv).
B – A = {4, 8, 16, 20}
(v).
C – A = {2, 4, 8, 10, 14,
16}
(vi). D
– A = {5, 10, 20}
(vii). B
– C = {20}
(viii). B
– D = {4, 8, 12, 16}
(ix). C
– B = {2, 6, 10, 14}
(x).
D – B = {5, 10, 15}
(xi).
C – D = {2, 4, 6, 8, 12, 14, 16}
(xii). D
– C = {5, 15, 20}
Q.10: If X = {a, b, c, d}
and Y = {f, b, d, g}
Find the following:
(i). X – Y
(ii). Y – X
(iii).
X ∩ Y
Answer:
(i). X
– Y = {a, c}
(ii). Y
– X = {f, g}
(iii).
X ∩ Y = {b, d}
Q.11: What is R – Q, if R
is the set of real numbers and Q is the set of rational?
Answer:
R: Set
of real numbers
Q: Set
of rational numbers
Hence, R – Q is a set of
irrational numbers.
Q.12: State whether the
following statements are true or false. Give reason.
(i). A = {2, 3, 4,
5} and B = {3, 6} are disjoint sets.
(ii). A = {a, e, i,
o, u } and B = {a, b, c, d} are disjoint sets.
(iii). A = {2, 6,
10, 14} and B = {3, 7, 11, 15} are disjoint sets.
(iv). A = {2, 6,
10} and B = {3, 7, 11} are disjoint sets.
Answer:
(i). False
Because, 3 ∈ {2, 3, 4, 5} and 3 ∈ {3,
6}
A ∩ B = {2, 3, 4, 5} ∩ {3,
6} = {3}
(ii). False
Because, a ∈ {a, e, i, o, u} and a ∈ {a, b, c, d}
A ∩ B = {a, e, i, o, u } ∩ {a,
b, c, d} = {a}
(iii). True
Because, A ∩ B= {2, 6, 10, 14} ∩ {3,
7, 11, 15} = Φ
(iv). True
Because, A ∩ B = {2, 6, 10} ∩ {3,
7, 11} = Φ
Exercise 1.5
Q.1: Let U = {1, 2,
3, 4, 5, 6, 7, 8, 9}
X = {1, 2, 3, 4, 5}
Y = {2, 4, 6} and
Z = {2, 4, 5, 6}.
Find the following sets:
(i). X’
(ii). Y’
(iii).
(X∪Y)′
(iv).
(X∪Z)′
(v).
(X′)‘
(vi).
(Y–Z)′
Answer:
Given:
U = {1,
2, 3, 4, 5, 6, 7, 8, 9}
X = {1,
2, 3, 4, 5}
Y = {2,
4, 6, 9} and
Z = {2,
4, 5, 6}
(i). X’ = {6, 7, 8,
9}
(ii). Y’ = {1, 3,
5, 7, 8, 9}
(iii). (X∪Y)′
And, (X∪Y)= {1, 2, 3, 4, 5, 6, 9}
Therefore, (X∪Y)′= {7, 8}
(iv). (X∪Z)′
And, (X∪Z) = {1, 2, 3, 4, 5, 6}
Therefore, (X∪Z)′ = {7, 8, 9}
(v).
(X′)‘ = X = {1, 2, 3, 4, 5}
(vi). (Y–Z)′
Since, (Y – Z) = {9}
Therefore, (Y – Z)’ = {1,
2, 3, 4, 5, 6, 7, 8}
Q.2: If U = {a, b, c, d,
e, f, g, h}, find the complements of the given sets:
(i). W = {a, b, c}
(ii). X = {d, e, f,
g}
(iii). Y = {a, c,
e, g}
(iv). Z = {f, g, h,
a}
Answer:
Given:
U = {a,
b, c, d, e, f, g, h}
(i). W
= {a, b, c}
W’ = {d, e, f, g, h}
(ii). X
= {d, e, f, g}
X’ = {a, b, c}
(iii). Y
= {a, c, e, g}
Y’ = {b, d, f}
(iv). Z
= {f, g, h, a}
Z’ = {b, c, d, e}
Q.3: Take natural numbers
as the universal set. Write the complements of the given sets:
(i). A = {y: y is
an even natural number}
(ii). B = {y: y is
an odd natural number}
(iii). C = {y: y is
a positive multiple of 3}
(iv). D = {y: y is
a prime number}
(v). E = {y: y is a
natural number divisible by 3 and 5}
(vi). F = {y: y is
a perfect square}
(vii). G = {y: y is
perfect cube}
(viii). H = {y: y +
5 = 8}
(ix). I = {y: 2y +
5 = 9}
(x).
J = {y: y ≥ 7}
(xi).
K = {y: y ∈ N and 2y + 1 > 10}
Answer:
Given:
U = set
of natural numbers = N
(i). A’
= {y: y is an even natural number}’
= {y: y is an odd natural
number}
(ii). B’
= {y: y is an odd natural number}
= {y: y is an even
natural number}
(iii). C’
= {y: y is a positive multiple of 3}’
= {y:
y ∈ N and y is not a multiple of 3}
(iv). D’
= {y: y is a prime number}’
= {y: y is a positive
composite number and y = 1}
(v). E’
= {y: y is a natural number divisible by 3 and 5}’
= {y: y is a natural
number that is not divisible by 3 or 5}
(vi). F’
= {y: y is a perfect square}’
= {y:
y ∈ N and y is not a perfect square}
(vii). G’
= {y: y is perfect cube}’
= {y:
y ∈ N and y is not a perfect cube}
(viii). H’
= {y: y + 5 = 8}’
= {y:
y ∈ N and y ≠ 3}
(ix). I’
= {y: 2y + 5 = 9}’
= {y:
y ∈ N and y ≠ 2}
(x). J’
= {y: y ≥ 7}’
= {y: y ∈ N and x < 7}
(xi). K’
= {y: y ∈ N
and 2y + 1 > 10}
= {y:
y ∈ N and y ≤92}
Q.4: If U = {1, 2, 3, 4,
5,6,7,8, 9}
A = {2, 4, 6, 8} and, B =
{2, 3, 5, 7}.
Verify that:
(i).
(A∪B)′ = A′∩B′
(ii).
(A∩B)′ = A′∪B′
Answer:
Given:
U = {1,
2, 3, 4, 5,6,7,8, 9}
A = {2,
4, 6, 8} and B = {2, 3, 5, 7}
(i). (A∪B)′
= {2, 3, 4, 5, 6, 7, 8}’ = {1, 9}
Now, A′∩B′ = {1, 3, 5, 7, 9} ∩ {1,
4, 6, 8, 9} = {1, 9}
Therefore, (A∪B)′ = A′∩B′
(ii). (A∩B)′
= {2}’ = {1, 3, 4, 5, 6, 7, 8, 9}
Now, A′∪B′ = {1, 3, 5, 7, 9} ∪ {1,
4, 6, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9}
Therefore, (A∩B)′ = A′∪B′
Q.5: Draw the Venn
diagrams for the following:
(i).
(A∪B)′
(ii).
A′∩B′
(iii).
(A∩B)′
(iv).
A′∪B′
Answer:
(i).
(A∪B)′:
(ii).
A′∩B′:
(iii).
(A∩B)′:
(iv).
A′∪B′:
Q.6:
Let U be the universal set that is the set of all triangles in a plane. If X is
the set of all triangles with at least one angle different from 60∘, what is X’ ?
Answer:
X’ is the set of all
equilateral triangles.
Q.7: Complete the given
statements using proper symbols:
(i).
A∪A′
(ii).
Ø ∩A
(iii).
A∩A′
(iv). U′∩A
Answer:
(i). A∪A′ = U
(ii). Ø ∩A = U∩A = A
(iii). A∩A′ = Ø
(iv). U′∩A =
Ø ∩A = Ø
Exercise 1.6
Q.1:
If A and B are two sets such that n(A) = 16, n(B) = 24 and n(A∪B) = 39. Find n(A∩B).
Answer:
Given:
n(A) = 16, n(B) = 23 and n(A∪B) = 39
As we know that n(A∪B)=n(A)+n(B)–n(A∩B)
39 = 16 + 24 – n(A∩B)
n(A∩B) = 40 – 39 = 1
Therefore, n(A∩B) = 1
Q.2: If
A and B are two sets such that A∪B has 17 elements, A has 9 elements and B has 14 elements.
How many elements does A∩B have?
Answer:
Given:
n(A) = 9, n(B) = 14 and, n(A∪B) = 17
As we know that n(A∪B)=n(A)+n(B)–n(A∩B)
17 = 9 + 14 – n(A∩B)
n(A∩B) = 23 – 17 = 6
Therefore, A∩B = 6
Q.3: In a group there are
450 people, 200 speaks Hindi and 270 can speak English. How many people can
speak both English and Hindi?
Answer:
Let us assume that:
H =
the set of people who speaks Hindi
And, E =
the set of people who speaks English
Given:
n(H∪E) = 450, n(H)
= 200 and
n(E) = 270
n(H∩E) = ?
As we know that, n(H∪E)=n(H)+n(E)–n(H∩E)
450 = 200 + 270 – n(H∩E)
450 = 470 – n(H∩E)
n(H∩E) = 470 – 450
n(H∩E) = 20
Therefore, 20 people can
speak both English and Hindi.
Q.4: If
X and Y are two sets such that X has 22 elements, Y has 34 elements, and n(X∩Y) has 10 elements, how many elements does n(X∪Y) have?
Answer:
Given:
n(X∩Y) = 10, n(X)
= 22 and,
n(Y) = 34
n(X∪Y) = ?
As we know that, n(X∪Y)=n(X)+n(Y)–n(X∩Y)
n(X∪Y) = 22 + 34 – 10
n(X∪Y) = 56 – 10
n(X∪Y) = 46
Therefore,
the set n(X∪Y) has 46 elements.
Q.5: If A and B are
two sets such that A has 45 elements, n (A ∩ B) has 15 elements and n
(A ∪ B) has 70 elements,
how many elements does B have?
Answer:
Given:
n(A) = 45, n(A∩B) = 15 and, n(A∪B) = 70
n(B)= ?
As we know that, n(A∪B)=n(A)+n(B)–n(A∩B)
70 = 45 + n(B) – 15
70 = 30 + n(B)
n(B) = 40
Therefore, the set n(B)
has 40 elements.
Q.6:There are 70 people,
out of which 35 like tea, 55 like coffee, and each person likes at least one of
the two beverages. How many people like both tea and coffee?
Answer:
Let us assume that:
T =
the set of people who like tea
And, C =
the set of people who like coffee
Given:
n(T) = 35, n(C) = 55 and n(C∪T) = 70
n(C∩T) = ?
As we know that n(C∪T)=n(C)+n(T)–n(C∩T)
70 = 55 + 35 – n(C∩T)
70 = 90 – n(C∩T)
n(C∩T) = 20
Therefore, there are 20
people who like both tea and coffee.
Q.7: There are 70
students in a group, 35 like cricket, 15 like both tennis and cricket. How many
like tennis only and not cricket? How many like tennis?
Answer:
Let us assume that:
C =
the set of students who like cricket
And, T =
the set of students who like tennis
Given:
n(C) = 35, n(C∪T) = 70, n(C∩T) = 15
n(T) = ?
As we know that n(C∪T)=n(C)+n(T)–n(C∩T)
70 = 35 + n(T) – 15
70 = 20 + n(T)
n(T) = 50
Therefore, there are 50
students who like to play tennis.
Now, (T–C)∪(T∩C)=T
Also, (T–C)∩(T∩C) = Ø
Therefore, n(T) = n(T –
C) + n(T∩C)
50 = n(T – C) + 15
n(T – C) = 50 – 15
n(T – C) = 35
Therefore, there are 35
students who play only tennis.
Q.8: In a committee, 60
people speak French, 30 speak Spanish and 20 speak both Spanish and French. How
many speak at least one of these two languages?
Answer:
Let us assume that:
F =
the set of people who speaks French
And, S =
the set of people who speaks Spanish
Given:
n(F) = 60, n(S) = 30, n(F∩S) = 20
n(F∪S) = ?
As we know that n(S∪F)=n(S)+n(F)–n(S∩F)
n(F∪S) = 30 + 60 – 20
n(F∪S) = 70
Therefore, there are 70
people who can speak at least one of the two languages.
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