# Practical problems on Union and Intersection of two sets( Set theory part 8 NCERT)

0
1026

Practical problems on Union and Intersection of two sets

In this topic, we will go through some practical problems related to our daily life. Let A and B be finite sets. If A ∩ B = φ, then
(i) n ( A ∪ B ) = n ( A ) + n ( B ) … (1)
The elements in A ∪ B are either in A or in B but not in both as A ∩ B = φ. So, (1) follows immediately.
In general, if A and B are finite sets, then
(ii) n ( A ∪ B ) = n ( A ) + n ( B ) – n ( A ∩ B ) … (2)
Note that the sets A – B, A ∩ B and B – A are disjoint and their union is A ∪ B (Fig 1.11). Therefore
n ( A ∪ B) = n ( A – B) + n ( A ∩ B ) + n ( B – A )
= n ( A – B) + n ( A ∩ B ) + n ( B – A ) + n ( A ∩ B ) – n ( A ∩ B)
= n ( A ) + n ( B ) – n ( A ∩ B), which verifies (2)
(iii) 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 ) … (3)
In fact, we have n ( A ∪ B ∪ C ) = n (A) + n ( B ∪ C ) – n [ A ∩ ( B ∪ C ) ] [ by (2) ]
= n (A) + n ( B ) + n ( C ) – n ( B ∩ C ) – n [ A ∩ ( B ∪ C ) ] [ by (2) ]
Since A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ), we get
n [ A ∩ ( B ∪ C ) ] = n ( A ∩ B ) + n ( A ∩ C ) – n [ ( A ∩ B ) ∩ (A ∩ C)]
= n ( A ∩ B ) + n ( A ∩ C ) – n (A ∩ B ∩ C)
Therefore
n ( A ∪ B ∪ C ) = n (A) + n ( B ) + n ( C ) – n ( A ∩ B ) – n ( B ∩ C)
– n ( A ∩ C ) + n ( A ∩ B ∩ C )
This proves (3).

Example 23 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
Alternatively, suppose n ( X ∩ Y ) = k, then
n ( X – Y ) = 28 – k , n ( Y – X ) = 32 – k (by Venn diagram in Fig 1.12 )
This gives 50 = n ( X ∪ Y ) = n (X – Y) + n (X ∩ Y) + n ( Y – X)
= ( 28 – k ) + k + (32 – k )
Hence k = 10.

Example 24 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.

Example 25 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.

Example 26 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.

Example 27 There are 200 individuals with a skin disorder, 120 had been exposed to the chemical C1, 50 to chemical C2, and 30 to both the chemicals C1 and C2. Find the number of individuals exposed to
(i) Chemical C1 but not chemical C2       (ii) Chemical C2 but not chemical C1
(iii) Chemical C1 or chemical C2

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 C1 and B denote the set of individuals exposed to the chemical C2
Here n (U) = 200, n ( A ) = 120, n ( B ) = 50 and n ( A ∩ B ) = 30 (i) From the Venn diagram given in Fig 1.13, 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 C1 but not to chemical C2 is 90.
(ii) From the Fig 1.13, 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 C2 and not to chemical C1 is 20.
(iii) The number of individuals exposed either to chemical C1 or to chemical C2, i.e.,
n ( A ∪ B ) = n ( A ) + n ( B ) – n ( A ∩ B )
= 120 + 50 – 30 = 140.