`star` Important Formula on operation of Sets

`star` Formula of union and Intersection

`star` Formula of union and Intersection

1. `A-B = A∩B'`

Proof of this property is shown in Figure with the help of venn diagram

Similarly

2. `A-(B∩C) = (A-B) ∪ (A-C)`

3. `A-(B∪C) = (A-B) ∩ (A-C)`

4.`A∩(B-C) = (A∩B) - (A∩C)`

Proof of this property is shown in Figure with the help of venn diagram

Similarly

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 2655267164

Show that for any sets `A` and `B`

`A = (A ∩ B) ∪ (A - B)`

and `A ∪ (B-A)=(A ∪ B)`

Class 11 Exercise 1.mis Q.No. 8

`A = (A ∩ B) ∪ (A - B)`

and `A ∪ (B-A)=(A ∪ B)`

Class 11 Exercise 1.mis Q.No. 8

`(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`

- 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)]`

` 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 3017378280

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 ?

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`

Q 3047478383

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 ?

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.

Q 3067478385

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 ?

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.

Q 3077478386

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.

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.

Q 3007478388

There are 200 individuals with a skin disorder, 120 had been exposed to the chemical `C_1,` 50 to chemical `C_2`, and 30 to both the chemicals `C_1` and `C_2`. Find the number of individuals exposed to

(i) Chemical `C_1` but not chemical `C_2` (ii) Chemical `C_2` but not chemical `C_1`

(iii) Chemical `C_1` or chemical `C_2`

(i) Chemical `C_1` but not chemical `C_2` (ii) Chemical `C_2` but not chemical `C_1`

(iii) Chemical `C_1` or chemical `C_2`

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.`

Q 3057878784

A college warded 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 ?

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.