Mathematics Problem Solving Techniques for SETS
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### Problem solving Technique 1

Proving relationship between sets method I

using A-B = AcapB'

Replace - sign using above formula and than use Commutative , Associative distributive and De Morgan’s laws.

color{blue}(ul"Some Properties of the Operation of Union") :

(i) A ∪ B = B ∪ A (Commutative law)

(ii) ( A ∪ B ) ∪ C = A ∪ ( B ∪ C) (Associative law )

(iii)  A ∪ φ = A (Law of identity element, φ is the identity of ∪)

(iv) A ∪ A = A (Idempotent law)

(v) U ∪ A = U (Law of U)

color{green}(=>" Idempotent Law" ):

color{blue}(ul"De Morgan’s laws")

For any two finite sets A and B;

(i)\ \ (A U B)' = A' ∩ B' (which is a De Morgan's law of union).

(ii) \ \ (A ∩ B)' = A' U B' (which is a De Morgan's law of intersection).

Some more important Relationships

1. A-B = A∩B'

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 1774601556

For all sets A and B, A- (A cap B) = A - B.
NCERT Exemplar
Solution:

LHS = A- (A cap B)= A cap (A cap B)' quad [∵ A - B = A cap B']

 = A cap (A' cup B') quad [∵ (A cap B)'= A' cup B']

 = (A cap A') cup (A cap B') = phi cup (A cap B')

 = A cap B' quad [ ∵ phi cup A = A]

 = A - B = RHS
Q 1744601553

For all sets A and B, A cup (B - A) = A cup B.
NCERT Exemplar
Solution:

LHS = A cup (B - A)= A cup (B cap A') quad [∵ A - B = A cap B']

= (A cup B) cap (A cup A') = (A cup B) cap U quad [∵ A cup A' = U]

= A cup B = RHS quad [ ∵ A cap U = A]

### Problem Solving Technique 2

Proving relationship between sets method 2

eg. to Prove De Morgan’s laws.

(AcupB)' = A'cap B'

Let x in (Acup B)'

Prove x in A'cap B' => (AcupB)' subseteq A'capB' ...........................(1)

Now take y in A'cap B'

Prove Y in (Acup B)'

=> A'cap B' subseteq (A cup B)' ...............................................(2)

from (I) & (2) (A cupB)' = A' cap B'
Q 1764701655

If A, B and C be sets. Then, show that A cap (B cup C) = (A cap B) cup (A cap C).
NCERT Exemplar
Solution:

Let  x in A cap (B cup C)

 => x in A and x in (B cup C)

 => x in A and (x in B or x in C)

 => (x in A and x in B ) or (x in A and x in C )

 => x in A cap B or x in A cap C

 x in (A cap B) cup (A cap C)

 => A cap (B cup C) subset (A cap B) cup (A cap C) .........(1)

Again, let  y in (A cap B) cup (A cap C)

 => y in (A cap B) or y in (A cap C)

 => (y in A and y in B) or (y in A and y in C)

 => y in A and (y in B or y in C)

 => y in A and y in B cup C

 => y in A cap (B cup C)

 => (A cap B) cup (A cap C) subset A cap (B cup C) ............(2)

From Eqs.(1) and( 2),

 A cap (B cup C) = (A cap B) cup (A cap C)

### PST 4

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 2315767660

Out of 32 persons, 30 invest in National Savings
Certificates and 17 invest in shares. What is the
number of persons who invest in both?
NDA Paper 1 2010
(A)

13

(B)

15

(C)

17

(D)

19

Solution:

n(N cup S)= 32, n(N) = 30, n(S) = 17

We know that,

n(N cup S)= n(N) + n(S) - n(N cap S)

=> 32 = 30 + 17 - n (N cap S)

=> n(N cap S) = 47 - 32 = 15
Correct Answer is => (B) 15
Q 1762067835

In a survey of 25 students, it was found that 15 had taken
Mathematics, 12 had taken Physics and 11 had taken
Chemistry, 5 had taken Mathematics and Chemistry, 9 had
taken Mathematics and Physics, 4 had taken Physics and
Chemistry and 3 had taken all the three subjects.

The number of students who had taken only two subjects
is
NDA Paper 1 2014
(A)

7

(B)

8

(C)

9

(D)

10

Solution:

Given that,

Total number of surveyed students = 25

Number of students, who had taken all three subjects = 3

Number of students, who had taken Physics and Chemistry
 = 4

Number of students, who had taken Mathematics and

Physics = 9

Number of students who had taken Mathematics and

Chemistry = 5

Number of students, who had taken Chemistry = 11

Number of students, who had taken Physics = 12

Number of students, who had taken Mathematics = 15

The number of students who had taken only two

subjects = 6 + 2 + 1 = 9
Correct Answer is => (C) 9