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`

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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`

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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.

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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.

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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.

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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.`

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Solution:

Let F, B and C denote the set of men who
received medals in football, basketball and cricket,
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.

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