Subject: Compulsory Maths

A set is a group of an object. It is a collection of things usually numbers. It is also defined as the " well-defined collection of objects". It is related with many objects or things. It is a collection of object which is different in nature from something else of a similar type. The objects in the set are called its elements. Objects can be anything; numbers, people, words etc. A set which does not contain any element is called an empty set.

**Examples of sets include:**

- Letters of English alphabet
- Natural numbers less than 10
- Five Places (Dharan, Biratnagar, Kathmandu, Butwal, Pokhara )

Usually, Capital letter is used to denote set and the small letter is used to denote an element of sets. For example: set A ={p, q, r, s, t}. The symbol '∈' refers set membership and symbol '∉' refers non-membership. For example:

In set: A = {a, b, c, d, e } and B= {f, g, h, i, j}, a ∈ A and f ∉ A, it means, that the element ′a′ belongs to set A but the element f does not belong to set A.

A set can be identified or membership of set can be point out in several ways. It is also the method to identify the sets and its elements.

**Two common ways among them are:**

**Listing or Tabulating Method**

In this method elements of a set are listed, separated by commas and enclosed in braces.

For example, V={a,e,i,o,u}**Description or rule Method**In this method, a set is specified by enclosing in braces, descriptive phrases or a rule.

For example, V = {letters of English alphabet} and N ={x:x is a natural number less than 10}, read as N is a set of all x such as x is a natural number less than 10.

The number of elements in a set is called its cardinal number. For example: if a = {a,b, c, d, e} then, the cardinal number of A is 5 and we write n(A) = 5.

A set having finite numbers of elements is called a finite set. For example: A = {1, 2, 3, 4} is a finite set. An infinite set is a set having infinite numbers of elements. For example: B = {1, 2, 3, 4............} is an infinite set.

Some of the set have only one element, the set having only one element is called Singleton set. For example:

C= {x : x is president of America} is a single set.

A set with no element is called the null set or empty set and is denoted by Φ (phi) or simply{ }. For example, the set of all prime numbers between 3 and 5 is a null set. The cardinal number of the null set is Zero.

The two sets, having, at least, one element common is said to be intersecting sets. For example; {3, 4, 5} and {5, 6, 7} are intersecting sets as number 5 is common to both sets.

The two sets, having no common elements are said to be disjoint sets. For example: {a ,b, c} and {1 ,2, 3, 4} are disjoint sets as they have no common elements.

The two sets having a same number of elements are said to be equivalent sets. For example: if A = {p, q, r} and B = {1, 2, 3} then A and B are equivalent sets as n(A) = n (B).

Two sets are said to be equal if both the sets have same elements.

For example: If C = {2, 3, 4} and D= {2, 3, 4} then C = D.

A set which lies between another set (element of any set lies in another set).

For example;

P ={ 1, 2, 3} and Q = {1, 2, 3, 4, 5}

We say that P is the subset of Q since every element of P is in Q. This is denoted by P ⊂ Q. In this case P ≠ Q. We say that P is a proper subset of Q.

Given: A = {1, 2, 3} and B = {2, 3, 4, 5, 6}, what is the relationship between these sets?

We can say A is not a subset of B since A ∉ B.

The statement "A is not a subset of B" is denoted by A ⊄ B.

It helps to show the relationship between these sets.

Given,

P = {1, 2, 3, 4, 5} and Q = {3, 1, 2, 4, 5}, what is the relationship between P and Q?

Recall that the order in which the elements appear in a set is not important. Looking at the elements of these sets, it is clear that:

P ⊂ Q

Q ⊂ P

∴ P = Q

Thus, for any set, if P ⊂ Q and Q ⊂ P, then P = Q. In this case, we say that P is improper subset of Q and we write P ⊆ Q.

**What are the differences between proper and improper sets?**

Proper subsets contain some elements but not the elements of the original set. An improper subset contains every element of the original cost.

A un.iversal set is a set which contains all the sets under consideration as its subsets.

A universal set is denoted by U. For example, If A = {6, 7, 8}, B = {1, 2, 3, 5, 7} and C = {4, 6, 8, 10} then the universal set of these sets may be taken as {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Note that the given sets, the choice of the universal set is not unique. For example, let A = {2, 4, 6, 7}, B = {3, 5, 7, 9} and C = {0, 4, 8, 12, 16} be the given sets. We can choose any of the following as a universal set:

{x : x ∈ W x ≤ 16}

or, {x : x ∈ W}

or, {x : x is an integer} etc.

- The numbers of elements in a set are called its Cardinal number of Set.
- A set having a finite number of elements is called a Finite Set.
- A set having only one element is called Singleton Set.
- A set with no element is called the Null Set or Empty Set.
- The two sets having at least one element common is said to be Intersecting Set.
- The two sets having no element common are said to be Disjoint Set.
- The two sets having the same number of elements is said to be Equivalent Set.
- Two sets are said to be equal if they have same elements.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

If A = {2,4,6,8,10} and B = {4,8,10,12,14}, find A∪B.

Solution:

A∪B

= {2,4,6,8,10} ∪ {4,8,10,12,14}

= {2,4,6,8,10,12,14}

If A = {2,4,6,8,10} and B = {4,8,10,12,14}, find A∩B.

Solution:

A∩B

= {2,4,6,8,10} ∩ {4,8,10,12,14}

= {2,4,6,8,10}

If A = {2,4,6,8,10} and B = {4,8,10,12,14}, find A-B.

Solution:

A-B

= {x:x ∈ B and x ∉ B}

= {2,4,6,8,10} - {4,8,10,12,14}

= {2,6}

If A = {2,4,6,8,10} and B = {4,8,10,12,14}, find B-A

Solution:

B-A

= {x: x ∈ A and x ∉ A}

= {4,8,10,12,14} - {2,4,6,8,10}

= {12,14}

If A = {2,4,6,8,10} and B = {4,8,10,12,14}, find (A-B)∪(B-A).

Solution:

(A-B)∪(B-A)

= {2,6} ∪ {12,14}

= {2,6,12,14}

If n(A) = 30, n(B) = 40, n(A∩B) = 15 and n(\(\overline{A∪B}\)) = 10,find n(A∪B) and n(∪). Also,show these information in a venn diagram.

Soln:

Here,

n(A)=30, n(B)=40, n(A∩B)=15 and n(\(\overline{A∪B}\))=10

We have, n(A∪B) = n(A) + n(B)-n(A∩B) = 30+40-15 = 55

n(∪) = n(A∪B)+n(\(\overline{A∪B}\)) = 55+10 = 65

Then, n(A∩B)=15, write 15 in the portion common to both A and B.

Also, n(A-B) = n(A) - n(A∩B)

=30-15= 15, write 15 in the portion represented by horizontal lines.

Again, n(B-A)=n(B) - n(A∩B)

=40-15=25, write 25 in the portion represented by vertical lines.

In a class of 60 students,30 students take part in games,32 take part in dances and 7 take part in none of these two.

- How many take part in the both game?
- How many take part in games only?
- Show the above information in a Venn diagram.

Here, Let∪ be the universal set.

Let G and D denote the sets of students who take part in games and dances respectively.

Then, n(∪)=60, n(G)=30,n(D)=32 and n(\(\overline{G∪D}\))=7

Now, n(G∪D)=n(∪)- n(\(\overline{G∪D}\))=60-7=53

- we have n(\(\overline{G∪D}\))=n(G)+n(D)-n(G∪D)

=30+32-53

=62-53=9

- n(G-D)=n(G)-n(G∩D) =30-9=21

Hence, 21 students take part in the game only.

Out of 40 students, 14 are taking English composition and 29 are taking Science. If 5 student are in both classes, how many students are in neither classes? How many students are in neither classes? How many students are in either class? Represented the above information in a Venn diagram.

Solution:

Let U be the universal set.

Let E and S denote the sets of students taking English and Science respectively.

Then, n(U)=40, n(E)=14, n(S)=29 and n(E∩S)=5

We have, n(E∪S)=n(E)+n(S)-n(E∩S)

=14+29-5

=43-5

=38

Hence, there are 38 students in at least one of the classes.

Also,n\(\overline{E∪S}\)) =n(U) –n(E∪S)

40-38=2

Hence, 2 students are taking neither classes.

Finally, we represented the above information in a Venn-diagram

If n(U) = 40, n(A) = 25, n(B) = 10 and n(A\(\cap\)B) = 5 find:

i. n(A\(\cup\)B)

ii. n(\(\overline{A}\))

iii. n_{o}(A)

iv. n_{o}(B)

Solution:

n(U) = 40, n(A) = 25, n(B) = 10 and n(A∩B) = 5

n(A∪B) = n(A) + n(B) - n(A∩B)= 25 + 10 - 5 = 30

n(\(\overline{A}\)) = n(U) – n(A) = 40 - 25 = 15

n_{o}(A) = n(A) - n(A∩B) = 25 - 5 = 20

n_{o}(B) = n(B) - n(A∩B) = 10 - 5 = 5

Let U ={1,2,.....,8,9}, A={1,2}, B={1,2,3,4,5}. Verify the following:

\(\overline{A∪B}\) =\(\overline{A∩B}\)

\(\overline{A∩B}\) =\(\overline{A∪B}\)

Solution:

∪={1,2,.....,8,9}, A={1,2} and B{1,2,3,4,5}

\(\overline{A}\)=U-A ={3,4,5,6,7,8,9}

\(\overline{B}\)=U-B={6,7,8,9}

A∪B={1,2,3,4,5}

\(\overline{A∪B}\)={6,7,8,9}

\(\overline{A}\)∩\(\overline{B}\)={6,7,8,9}

Hence,\(\overline{A \cup B}\)=\(\overline{A}\)∩\(\overline{B}\)

A∩B={1,2}

\(\overline{A∩B}\)={3,4,5,6,7,8,9}

\(\overline{A}\)∪\(\overline{B}\)={3,4,5,6,7,8,9}

Hence,\(\overline{A∩B}\)=\(\overline{A}\)∪\(\overline{B}\)

There are 20 people in a room. Of these,15 are holding newspaper and 8 are wearing glasses.Everyone wears glasses or holds newspapers. How many people are wearing glasses and holding a newspaper?

Soln:, Let n(N)= and (G) be the sets of people who are holding the newspaper and wearing glasses.

Here, n(u)=20 ,n(n)=15, n(G)=8 n(N∪G)=20, n(N∩G)=?

We know,

n(N∩G) =n_{o}(n)+n_{o}(G)-n(FUG)

=15+8-20

= 15+8-20

=23-20

= 3.Ans.

In a class there are 30 students, 21 students like Maths,16 students like English, 6 students don't like like Maths or English.How many students like both Maths and English?

Soln: Let n(M) and n(E) be the number of students who like maths and English respectively.

n(U)=30, n(M)=21 , n(E)=16 n(\(\overline{M∪E}\))=6

We know, n(M∪E)=n(U)-n(\(\overline{M∪E}\))

= 30-6

=24

Now, n(M∪E)=n(M)+n(E)-n(M∩E)

=21+16-24

= 37-24

=13 Ans.

Determine whether the following sets is null set or not.

- X = {x: x
^{2 }= 9, 2x = 4} - Z = {x: x + 8 = 8}

Solution:

- There is no number which satisfies both x
^{2}= 9 and 2x = 4. Hence, x = Φ. - The number zero satisfies x + 8 = 8. Hence, Z = {0} and Z ≠ Φ.

Let A = {x :3x = 6}. Does A = 2?

Solution:

A = {x : 3x = 6} = {2}

A is the set which consists of the single element 2 i.e A = {2}. The number 2 belongs to A, it does not equal to A. There is a basic difference between an element p and the singleton set {p}.

In a Class, there are 8 students who play football and hockey, 7 students who don't play football or hockey and13 students play hockey, 19 students play football. How many students are there in class?

Solution:

Given information,

n (F) = 19

n (H) = 13

n (\(\overline{FUH}\)) = 7

Now,

n (FUH) =n(F) + n(H) -n(F∩H)

= 19 + 13 - 8

= 32 - 8

= 24

n (U) = n (FUH) + n(\(\overline{FUH}\))

= 24 + 7

= 31

Let, U = { 1,2,3........9}, A={2,3,4,5} and B ={4,5,6,7}, Find AUB, A∩B, A-B, B-A,\(\overline{A}\) and \(\overline{B}\).

Solution:

Given information,

U = {1,2,3,4,5,6,7,8,9}

A = {2,3,4,5}

B = {4,5,6,7}

Now,

AUB ={2,3,4,5} U {4,5,6,7}

= {2,3,4,5,6,7}

A∩B = {2,3,4,5}∩ {4,5,6,7}

= {4,5}

A - B = {2,3,4,5} - {4,5,6,7}

= {2,3}

B-A = {4,5,6,7} -{2,3,4,5}

={6,7}

\(\overline{A}\) = U - B

= {1,2,3,4,5,6,7,8,9} - {2,3,4,5}

= {1,6,7,8,9}

\(\overline{B}\) = {1,2,3,4,5,6,7,8,9} -{4,5,6,7}

= {1,2,3,8,9}

If P={factors of 12} and Q= {factors of 8} Find.

Solution:

Given information,

a) PUQ

Solution:

PUQ = {1,2,3,4,6,12} U {1,2,4,8}

= {1,2,3,4,6,8,12}

b) P∩Q

Solution:

P∩Q ={1,2,3,4,6,12}∩ {1,2,4,8}

= {1,2,4}

c) Q-P

Solution:

Q-P = {1,2,4,8} - {1,2,3,4,6,12}

= {8}

d) P-Q

Solution:

P-Q = {1,2,3,4,6,12} - {1,2,4,8}

={3,6,12}

Given that A={0,1,2,3,4,5}, B={0,2,4,6,8} & C={0,3,6,9}.Show that

AU(BUC) = (AUB)UC

Solution:

Given information,

L.H.S = AU(BUC)

= {0,1,2,3,4,5} U[ {0,2,4,6,8} U {0,3,6,9} ]

= {0,1,2,3,4,5} U {0,2,3,4,6,8,9}

= {0,1,2,3,4,5,6,7,8,9}

R.H.S = (AUB)UC

=[{0,1,2,3,4,5} U {0,2,4,6,8} U {0,3,6,9}]

= {0,1,2,3,4,5,6,8} U {0,3,6,9}

= {0,1,2,3,4,5,6,8,9}

∴ L.H.S =R.H.S

Proved

If A = {0, 1, 2, 3, 4, 5}, B = {0, 2, 4, 6, 8} and C= {0, 3, 6, 9}. Prove, AU(B∩C) = (AUB)∩(AUC).

Solution:

L.H.S = AU(B∩C)

= {0,1,2,3,4,5} U [{0,2,4,6,8}∩ {0,3,6,9}]

= {0,1,2,3,4,5} U {0,6}

= {0,1,2,3,4,5,6}

R.H.S = (AUB) ∩ (AUC)

= [{0,1,2,3,4,5} U {0,2,4,6,8}] ∩ [{0,1,2,3,4,5} U {0,3,6,9}]

= {0,1,2,3,4,5,6,8} ∩ {0,1,2,3,4,5,6,9}

= {0,1,2,3,4,5,6}

∴ L.H.S =R.H.S Proved

If A = {a,b,c,d,e} and B= {c,d,e,f,g}. Find A - B.

Solution:

A - B

= {a,b,c,d,e} - {c,d,e,f,g}

= {a,b}

From the given diagram, Show the relationship between A and B.

a) n(A)

b) n(B)

c) n(A∩B)

d) n(AUB)

e) n_{0}(A)

f) n_{0}(B)

g) n(\(\overline{A}\))

i) n(\(\overline{B}\))

Solution:

Given information,

Here,

a)

A = {a,b,c,d,e,f,g}

n(A) =7

b)

B = {b, g, h, i, j, l, m, n, o}

n(B) =9

c)

A?B ={b,g}

n(A?B) =2

d)

AUB = {a,b, c, d, e, f, g, ,h i, j, l, m, n, o}

n(AUB) = 14

e)

n_{0}(A) = n(A) -n(A?B) = 7 - 2 =5

f)

n_{0}(B) =n(B) - n(A?B) = 9 -2 =7.

g)

n_{0}(\(\overline{A}\)) = n(U) - n(A) = 15 - 7 = 8

h)

n_{0}(\(\overline{B}\)) = n(U) - n(B0 = 15 - 9 = 6

If A = {2, 4, 6, 8, 10} and B = {4, 8, 10,12,14}. Find

(A-B) U (B-A).

Solution:

Given information,

A ={2, 4, 6, 8, 10}, B = {4, 8, 10, 12, 14}

A-B = {2,6}

B-A={12,14}

Now, (A-B) U (B-A) = {2,6} U {12,14}

= {2,6,12,14}

If, U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}, A={0, 1, 2, 3, 4, 5}, B={4, 5, 6, 7, 8} and C={2, 3, 4, 8, 9}. Find

a) A∪B∪C

b) \(\overline{A∪B∪C}\)

Solution:

Given information,

a) A∪B∪C = {0, 1, 2, 3, 4, 5} ∪ {4, 5, 6, 7, 8} ∪ {2, 3, 4, 8, 9}

= {0, 1, 2, 3, 4, 5, 6, 7, 8} ∪ {2, 3, 4, 8, 9}

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

b) ( \(\overline {A∪B∪C}\)) = U - A∪B∪C

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14} - {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

= {10, 11, 12, 13, 14}

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