Mathematics Important Formula on operation of Sets and Formula of union and Intersection
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### Topics Covered

star Important Formula on operation of Sets
star Formula of union and Intersection

### Important Formula on operation of Sets

1. A-B = A∩B'

Proof of this property is shown in Figure with the help of venn diagram

Similarly

2. A-(B∩C) = (A-B) ∪ (A-C)
3. A-(B∪C) = (A-B) ∩ (A-C)
4.A∩(B-C) = (A∩B) - (A∩C)
Q 2655267164

Show that for any sets A and B

A = (A ∩ B) ∪ (A - B)

and A ∪ (B-A)=(A ∪ B)
Class 11 Exercise 1.mis Q.No. 8
Solution:

(A ∩ B) ∪ (A- B)= (A ∩ B) ∪ (A ∩ B')

[ :. A-B= A ∩ B' ]

 = A ∩ (B ∪ B' )

[by distributive law)

= A ∩ X

[ X is a universal set]

=A

A ∪ (B- A)= A ∪ (B ∩ A')

[ :. B-A=B ∩ A']

=-(A ∪ B) ∩ (A ∪ A') [by distributive law]

= (A ∪ B ) ∩ X

[:. X= A ∪ A' is universal set ]

= A ∪ B

### Formula of union and Intersection :

- If A and B are finite sets, then
 n ( A ∪ B ) = n ( A ) + n ( B ) – n ( A ∩ B )

- If A ∩ B = φ, then

 n ( A ∪ B ) = n ( A ) + n ( B )

 n ( A ∪ B) = n ( A – B) + n ( A ∩ B ) + n ( B – A )

- If A, B and C are finite sets, then

n ( A ∪ B ∪ C ) = n ( A ) + n ( B ) + n ( C ) – n ( A ∩ B ) – n ( B ∩ C) – n ( A ∩ C ) + n ( A ∩ B ∩ C )

n [ A ∩ ( B ∪ C ) ] = n ( A ∩ B ) + n ( A ∩ C ) – n [ ( A ∩ B ) ∩ (A ∩ C)]
Q 3017378280

If X and Y are two sets such that X ∪ Y has 50 elements, X has 28 elements and Y has 32 elements, how many elements does X ∩ Y have ?

Solution:

Given that
n ( X ∪ Y ) = 50, n ( X ) = 28, n ( Y ) = 32, n (X ∩ Y) = ?
By using the formula

n ( X ∪ Y ) = n ( X ) + n ( Y ) – n ( X ∩ Y ),
we find that
n ( X ∩ Y ) = n ( X ) + n ( Y ) – n ( X ∪ Y )
= 28 + 32 – 50 = 10
Q 3047478383

In a school there are 20 teachers who teach mathematics or physics. Of these, 12 teach mathematics and 4 teach both physics and mathematics. How many teach physics ?

Solution:

Let M denote the set of teachers who teach mathematics and P denote the set of teachers who teach physics. In the statement of the problem, the word ‘or’ gives us a clue of union and the word ‘and’ gives us a clue of intersection. We, therefore, have

n ( M ∪ P ) = 20 , n ( M ) = 12 and n ( M ∩ P ) = 4

We wish to determine n ( P ).
Using the result
n ( M ∪ P ) = n ( M ) + n ( P ) – n ( M ∩ P ),
we obtain
20 = 12 + n ( P ) – 4
Thus n ( P ) = 12
Hence 12 teachers teach physics.
Q 3067478385

In a class of 35 students, 24 like to play cricket and 16 like to play football. Also, each student likes to play at least one of the two games. How many students like to play both cricket and football ?

Solution:

Let X be the set of students who like to play cricket and Y be the set of
students who like to play football. Then X ∪ Y is the set of students who like to play
at least one game, and X ∩ Y is the set of students who like to play both games.
Given n ( X) = 24, n ( Y ) = 16, n ( X ∪ Y ) = 35, n (X ∩ Y) = ?
Using the formula n ( X ∪ Y ) = n ( X ) + n ( Y ) – n ( X ∩ Y ), we get
35 = 24 + 16 – n (X ∩ Y)

Thus, n (X ∩ Y) = 5
i.e., 5 students like to play both games.
Q 3077478386

In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and 75 were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice nor orange juice.

Solution:

Let U denote the set of surveyed students and A denote the set of students
taking apple juice and B denote the set of students taking orange juice. Then
n (U) = 400, n (A) = 100, n (B) = 150 and n (A ∩ B) = 75.
Now n (A′ ∩ B′) = n (A ∪ B )′
= n (U) – n (A ∪ B)
= n (U) – n (A) – n (B) + n (A ∩ B)
= 400 – 100 – 150 + 75 = 225
Hence 225 students were taking neither apple juice nor orange juice.
Q 3007478388

There are 200 individuals with a skin disorder, 120 had been exposed to the chemical C_1, 50 to chemical C_2, and 30 to both the chemicals C_1 and C_2. Find the number of individuals exposed to

(i) Chemical C_1 but not chemical C_2 (ii) Chemical C_2 but not chemical C_1
(iii) Chemical C_1 or chemical C_2

Solution:

Let U denote the universal set consisting of individuals suffering from the
skin disorder, A denote the set of individuals exposed to the chemical C_1 and B denote
the set of individuals exposed to the chemical C_2.
Here n ( U) = 200, n ( A ) = 120, n ( B ) = 50 and n ( A ∩ B ) = 30
(i) From the Venn diagram given in Fig we have
A = ( A – B ) ∪ ( A ∩ B ).
n (A) = n( A – B ) + n( A ∩ B ) (Since A – B) and A ∩ B are disjoint.)
or n ( A – B ) = n ( A ) – n ( A ∩ B ) = 120 –30 = 90
Hence, the number of individuals exposed to
chemical C_1 but not to chemical C_2 is 90.
(ii) From the Fig we have
B = ( B – A) ∪ ( A ∩ B).
and so, n (B) = n (B – A) + n ( A ∩ B)
(Since B – A and A ∩B are disjoint.)
or n ( B – A ) = n ( B ) – n ( A ∩ B )
= 50 – 30 = 20

Thus, the number of individuals exposed to chemical C_2 and not to chemical C_1 is 20.
(iii) The number of individuals exposed either to chemical C_1 or to chemical C_2, i.e.,
n ( A ∪ B ) = n ( A ) + n ( B ) – n ( A ∩ B )
= 120 + 50 – 30 = 140.
Q 3057878784

A college warded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to a total of 58 men and only three men got medals in all the three sports, how many received medals in exactly two of the three sports ?

Solution:

Let F, B and C denote the set of men who
respectively.
Then n ( F ) = 38, n ( B ) = 15, n ( C ) = 20
n (F ∪ B ∪ C ) = 58 and n (F ∩ B ∩ C ) = 3
Therefore, n (F ∪ B ∪ C ) = n ( F ) + n ( B ) + n ( C ) – n (F ∩ B ) – n (F ∩ C ) – n (B ∩ C ) + n ( F ∩ B ∩ C ),
gives n ( F ∩ B ) + n ( F ∩ C ) + n ( B ∩ C ) = 18
Consider the Venn diagram as given in Fig
Here, a denotes the number of men who got medals in football and basketball only, b
denotes the number of men who got medals in football and cricket only, c denotes the
number of men who got medals in basket ball and cricket only and d denotes the
number of men who got medal in all the three. Thus, d = n ( F ∩ B ∩ C ) = 3 and a +
d + b + d + c + d = 18
Therefore a + b + c = 9,
which is the number of people who got medals in exactly two of the three sports.