Solution:

Let `A = { –1, 0, 1 }`. The subset of A having no element is the empty
set `phi`. The subsets of A having one element are `{ –1 }, { 0 }, { 1 }`. The subsets of
A having two elements are `{–1, 0}, {–1, 1} ,{0, 1}.` The subset of A having three
elements of A is A itself. So, all the subsets of A are `phi, {–1}, {0}, {1}, {–1, 0}, {–1, 1},`
`{0, 1} and {–1, 0, 1}.`

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Solution:

Let `a ∈ A`. Then `a ∈ A ∪ B.` Since `A ∪ B = A ∩ B , a ∈ A ∩ B`. So `a ∈ B.`
Therefore, `A ⊂ B.` Similarly, if `b ∈ B`, then `b ∈ A ∪ B`. Since
`A ∪ B = A ∩ B, b ∈ A ∩ B. So, b ∈ A.` Therefore,` B ⊂ A`. Thus, `A = B`

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Solution:

Let X be the set of letters in “CATARACT”. Then
`X = { C, A, T, R }`
Let Y be the set of letters in “ TRACT”. Then
`Y = { T, R, A, C, T } = { T, R, A, C }`
Since every element in X is in Y and every element in Y is in X. It follows that X = Y.

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Solution:

Let `X ∈ P ( A ∩ B ).` Then `X ⊂ A ∩ B.` So, `X ⊂ A` and `X ⊂ B.` Therefore,
`X ∈ P ( A )` and `X ∈ P ( B )` which implies `X ∈ P ( A ) ∩ P ( B).` This gives `P ( A ∩ B )`
`⊂ P ( A ) ∩ P ( B ).` Let `Y ∈ P ( A ) ∩ P ( B ).` Then `Y ∈ P ( A) `and `Y ∈ P ( B ). `So,
`Y ⊂ A` and `Y ⊂ B. `Therefore, `Y ⊂ A ∩ B, `which implies `Y ∈ P ( A ∩ B ). `This gives
`P ( A ) ∩ P ( B ) ⊂ P ( A ∩ B)`
Hence `P ( A ∩ B ) = P ( A ) ∩ P ( B ).`

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Solution:

Let U be the set of consumers questioned, S be the set of consumers who liked the product A and T be the set of consumers who like the product B. Given that
`n ( U ) = 1000, n ( S ) = 720, n ( T ) = 450`
So `n ( S ∪ T ) = n ( S ) + n ( T ) – n ( S ∩ T )`
`= 720 + 450 – n (S ∩ T) = 1170 – n ( S ∩ T )`
Therefore, `n ( S ∪ T )` is maximum when `n ( S ∩ T )` is least. But `S ∪ T ⊂ U` implies
`n ( S ∪ T ) ≤ n ( U ) = 1000. `So, maximum values of `n ( S ∪ T ) is 1000.` Thus, the least
value of `n ( S ∩ T )` is 170. Hence, the least number of consumers who liked both products
is 170.

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Solution:

Let U be the set of car owners investigated, M be the set of persons who
owned car A and S be the set of persons who owned car B.
Given that `n ( U ) = 500, n (M ) = 400, n ( S ) = 200 and n ( S ∩ M ) = 50.`
Then `n ( S ∪ M ) = n ( S ) + n ( M ) – n ( S ∩ M ) = 200 + 400 – 50 = 550`
But `S ∪ M ⊂ U implies n ( S ∪ M ) ≤ n ( U ).`
This is a contradiction. So, the given data is incorrect.

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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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Solution:

We have `A = {2, 6}`,(since, only `x = 2`, and
`x = 6`, satisfy the equation `x^2- 8x + 12 = 0`)

`B= {2, 4, 6}, C = {2, 4, 6, 8 ........... }` and `D= {6}`

Every element of `A` is in `B` and `C`;

`:. A ⊂ B` and `A ⊂ C`

Again every element of `B` is in `C; :. B ⊂ C`

Also, every element of `D` is in `A, B` and `C`;

`:. D ⊂ A ,D ⊂ B` and `D ⊂ C`.

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Solution:

We have, `A ∪ B = A ∪ C`

`=> ( A ∪ B ) ∩ C=(A ∪ C) ∩ C`

`=> (A ∩ C) ∪ ( B ∩ C) =C`

`[ :. ( A ∪ C ) ∩ C= C ]`

`=> ( A ∩B ) ∪ ( B ∩ C)=C` ... ( i )

`[ :. A ∩ C =A ∩ B ]`

Again `A ∪ B =A ∪ C`

`=> ( A ∪ B ) ∩ B=( A ∪ C) ∩ B`

`=> B = ( A ∩ B ) ∪ (C ∩ B)`

`[ :. ( A ∪ B) ∩ B = B]`

`( A ∩ B) ∪ (B ∩ C)= B` ... (ii)

`[ :. C ∩ B '= B ∩ C]`

From (i) and (ii) we get `B = C`

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Solution:

(i) `⇔` (ii); `A ⊂ B ⇔` All elements of `A` are in

`B ⇔ A-B= phi`

(ii) ` ⇔ ` (iii ); `A-B=phi ⇔ `AII elements of

`A` are in `B ⇔ A ∪ B = B`

(iii) ` ⇔` (iv); `A ∪ B = B ⇔ `All elements of
`A` are in `B`


` ⇔ `All the elements of `A` are common in

`A` and `B ⇔ A ∩ B = A`

Thus, all the four given conditions are
equivalent.

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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:

(i) `A ∪ (A ∩ B)=(A ∪ A) ∩ (A ∪ B)`


[By distributive law]

`= A ∩ (A ∪ B) ` ` [ :. A ∪ A= A]`

`= A ` `[ :. A ⊂ A ∪ B ]`


(ii) `A ∩ (A ∪ B) = (A ∩ A) ∪ ( A ∩ B)`

`= A ∪ (A ∩ B )`

`= A ` `[ :. A ∩ B ⊂ A ]`

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Solution:

Let `A= { 1, 2}, B = { 1, 3}` and `C = { 1, 4}`

Now, `A ∩ B = { 1,2 } ∩ { 1,3}={ 1} `;

`A ∩ C = { 1,2} ∩ {1,4} ={ 1 }`

`:. A ∩ B =A ∩ C => B ≠ C`

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Solution:

We have `A ∪ X= B ∪ X` for some set `X`

` = > A ∩ ( A ∪ X ) = A ∩ ( B ∪ X)`

` = > A = (A ∩ B) ∪ (A ∩ X)`

`[ :. A ∩ ( A ∪ X )= A ]`

` = >A = ( A∩ B) ∪ phi`

`[ :. A ∩ X = phi` (given)]

` => A= A ∩ B = > A ⊂ B` ... (i)

Again, ` A ∪ X= B ∪ X`

`=> B ∩ ( A ∪ X)= B ∩ ( B ∪ X)`

`[:. B ∩ (B ∪ X)= B]`

`=> ( B ∩ A) ∪ ( B ∩ X) = B`

`[ :. B ∩ X = phi` (given)]

`=> ( B ∩ A) ∪ phi = B => B ∩ A= B`

`=> A ∩ B = B => B ⊂ A` ... (ii)

From (i) and (ii) we get, `A = B`

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Solution:

Let `A= { 1,2}, 8= { 2, 3 }` and `C={ 1 ,3}`,


`A ∩ B = {2}, B ∩ C = {3}` and `A ∩ C ={ 1 }`

i.e., `A ∩ B, B ∩ C` and `A ∩ C` are non-empty
sets.


`:. (A ∩ B) ∩ C = {2} ∩ { 1, 3} = phi`

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Solution:

We have, `n(T) = 150, n(C) = 225`

and `n(T ∩ C) = 100`

We know that

`n(T ∪ C)= n(T) + n( C)-n(T ∩ C)`

`= 150+225-100 =275`

Total no. of students `= 600`

No. of students who neither take tea nor coffee

`= 600- n ( T ∪ C)=600-275 =325`

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Solution:

We have `n(H) = I 00, n(E) =50` and

`n (H ∩ E)= 25`

We know that

`n( H ∪ E) = n( H)+ n( E)- n( H ∩ E)`

`= 100+50-25= 125`

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Solution:

(i) No. of people who read at least one newspaper


`= 8 +8+10+6+3+5+12= 52`


(ii) No. of people who read exactly one newspaper

`= 52-(8 + 3 +6 + 5)=52-22 = 30`

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Solution:

`n ( A ∩ B) = 14 , n( A ∩ C) = 12` ,

`n( B ∩ C) = 14` and `n( A ∩ B ∩ C) = 8`

`n( C text (only ) ) = 29 - 4- 8- 6 = 29- 18 = 11`

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