﻿ Class 11 NCERT Math Solution
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TOPICS
Miscellaneous

Example-1 :-  Show that the set of letters needed to spell “ CATARACT ” and the set of letters needed to spell “ TRACT” are equal.

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.
```

Example-2 :-  List all the subsets of the set { –1, 0, 1 }.

Solution :-
```  Let  A = { –1, 0, 1 }. The subset of  A having no element is the empty set φ.
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 φ, {–1}, {0}, {1}, {–1, 0}, {–1, 1}, {0, 1} and {–1, 0, 1}.
```

Example-3 :-  Show that A ∪ B = A ∩ B implies A = B .

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
```

Example-4 :-  For any sets A and B, show that P ( A ∩ B ) = P ( A ) ∩ P ( B ).

Solution :-
```  Let X ∈ P ( A ∩ B ).
Then X ⊂  A ∩ B.
So, X ⊂  A and X ⊂ 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 ).
```

Example-5 :- A market research group conducted a survey of 1000 consumers and reported that 720 consumers like product A and 450 consumers like product B, what is the least number that must have liked both products?

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
```

Example-6 :-  Out of 500 car owners investigated, 400 owned car A and 200 owned car B, 50 owned both A and B cars. Is this data correct?

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 ).
So, the given data is incorrect.
```

Example-7 :-  A college awarded 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 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

By Venn diagram, 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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