Now we will go through some practical problems on sets related to everyday life.

**Application on Cardinality
of Sets:**

We
have learnt about the union, intersection, complement and difference of sets.

Now
we will go through some practical problems on sets related to everyday life.

**Results :**

If
*A* and *B* are two finite sets, then

(i) *n*(*A*âˆª*B*) =* n*(*A*)+*n*(*B*)â€“*
n*(*A*âˆ©*B*)

(ii) *n*(*A*â€“*B*)
=* n*(*A*) â€“* n*(*A*âˆ©*B*)

(iii)
*n*(*B*â€“*A*) =* n*(*B*)â€“* n*(*A*âˆ©*B*)

(iv)
*n*(*A*â€²) =* n*(U) â€“* n*(*A*)

**Note **

From the above results we may get,

*â€¢ n*(*A*âˆ©*B*) =* n*(*A*)+*n*(*B*)
â€“* n*(*A*âˆª*B*)

*â€¢ n*(U) =** ***n*(*A*)+*n*(*A*â€²)

*â€¢ *If *A* and *B* are disjoint sets then, *n*(*A*âˆª*B*) = *n*(*A*)+*n*(*B*).

**Example 1.27**

From the Venn diagram, verify that *n*(*A*âˆª*B*) =* n*(*A*)+*n*(*B*) â€“* n*(*A*âˆ©*B*)

** Solution **From the venn diagram,

*A
*= {5, 10, 15, 20}

*B
*= {10, 20, 30, 40, 50,}

Then

*A*âˆª*B *= {5, 10, 15, 20, 30, 40, 50}

*A*âˆ©*B
*= {10, 20}

*n*(*A*) = 4,* n*(*B*) =
5,* n*(*A*âˆª*B*) = 7,* n*(*A*âˆ©*B*) = 2

*n*(*A*âˆª*B*) = 7 â€¦â€¦..(1)

*n*(*A*)+*n*(*B*)â€“*n*(*A*âˆ©*B*)*
*= 4+5â€“2

=7 â€¦â€¦..(2)

From
(1) and (2), *n*(*A*âˆª*B*) = *n*(*A*)+*n*(*B*)â€“*n*(*A*âˆ©*B*).

**Example 1.28**

If *n*(*A*) = 36, *n*(*B*) = 10, *n*(*A*âˆª*B*)=40, and *n*(*A*â€²)=27 find *n*(U) and *n*(*A*âˆ©*B*).

*Solution**n*(*A*) = 36,* n*(*B*) =10,*
n*(*A*âˆª*B*)=40,* n*(*A*â€²)=27

(i) *n*(U)
=* n*(*A*)+*n*(*A*â€²) = 36+27 = 63

(ii) *n*(*A*âˆ©*B*) =* n*(*A*)+*n*(*B*)â€“*n*(*A*âˆª*B*) = 36+10-40 = 46-40 = 6

**Activity-4**

Fill in the blanks with appropriate cardinal numbers.

**Example 1.29**

Let *A*={*b*, *d*, *e*, *g*, *h*} and
*B* = {*a*, *e*, *c*, *h*}. Verify that *n*(*A***â€“***B*)
= *n*(*A*)**â€“***n*(*A*âˆ©*B*).

*Solution*

A=
{*b*, *d*, *e*, *g*, *h*}, *B* = {*a*, *e*,
*c*, *h*}

*A
***â€“*** B *= {*b*,* d*,* g*}

*n*(Aâ€“B) = 3 ... (1)

A
âˆ© B = {e, h}

n(A
âˆ© B) = 2 , n(A) = 5

*n*(*A*)* ***â€“*** n*(*A*âˆ©*B*) = 5-2

=
3
... (2)

Form
(1) and (2) we get *n*(*A***â€“***B*) = *n*(*A*)**â€“***n*(*A*âˆ©*B*).

**Example 1.30**

In a school, all students play either Hockey or Cricket or both.
300 play Hockey, 250 play Cricket and 110 play both games. Find

(i) the number of students who play only Hockey.

(ii) the number of students who play only Cricket.

(iii) the total number of students in the School.

*Solution:*

Let
*H* be the set of all students who play Hockey and *C* be the set of all
students who play Cricket.

Then
*n*(*H*) = 300, *n*(*C*) = 250 and *n*(*H* âˆ©
*C*) = 110

Using Venn
diagram,

From
the Venn diagram,

(i)
The number of students who play only Hockey = 190

(ii)
The number of students who play only Cricket = 140

(iii)
The total number of students in the school = 190+110+140 =440

**Aliter**

(i)
The number of students who play only Hockey

*n*(*H*â€“*C *) =* n*(*H*)
â€“* n*(*H*âˆ©*C*)

=300 â€“110 = 190

(ii)
The number of students who play only Cricket

*n*(*C*â€“*H *) =* n*(*C*)
â€“* n*(*H*âˆ©*C*)

= 250 â€“ 110 = 140

(iii)
The total number of students in the school

*n*(*H*U*C*) =* n*(*H*)
+* n*(*C*) â€“* n*(*H*âˆ©*C*)

=
300+250 â€“ 110 = 440

**Example 1.31**

In a party of 60 people, 35 had Vanilla ice cream, 30 had Chocolate
ice cream. All the people had at least one ice cream. Then how many of them had,

(i) both Vanilla and Chocolate ice cream.

(ii) only Vanilla ice cream.

(iii) only Chocolate ice cream.

*Solution :*

Let *V*
be the set of people who had Vanilla ice cream and *C* be the set of people
who had Chocolate ice cream.

Then *n*(*V*)
= 35, *n*(*C*) = 30, *n*(*V*âˆª*C*) = 60,

Let
*x* be the number of people who had both ice creams.

From
the Venn diagram

35
â€“ *x* + *x* +30 â€“ *x* = 60

65
â€“ *x* = 60

*x*= 5

Hence
5 people had both ice creams.

(i)
Number of people who had only Vanilla ice cream = 35 â€“ *x*

=35â€“5=30

(ii)
Number of people who had only Chocolate ice cream = 30 â€“ *x*

=30â€“5=25

We
have learnt to solve problems involving two sets using the formula *n *(*A
*âˆª* B *)* *=*
n*(*A*)* *+*
n*(*B *)* *âˆ’*n*(*A *âˆ©*
B*).* *Suppose we have three sets,
we can apply this formula* *to get a similar formula for three sets.

For any three finite sets *A*, *B* and *C*

*n *(*A *âˆª* B *âˆª*C*)* *=* n *(*A*)* *+* n *(*B*)* *+* n *(*C*)* *âˆ’*n *(*A *âˆ©* B *)* *âˆ’* n *(*B *âˆ©*C*)* *âˆ’* n *(*A *âˆ©*C *)* *+* n *(*A *âˆ©* B *âˆ©*C*)

**Note**

Let us consider the following results which will be useful in solving
problems using Venn diagram. Let three sets *A*, *B* and *C* represent
the students. From the Venn diagram,

Number of students in only set *A* = *a*, only set *B* = *b*, only set *C* = *c* .

**â€¢ **Total number of students in only one set** **=** **(*a***
**+*b*** **+*c*)** **

**â€¢ **Total number of students in only two sets** **= (*x* + *y* + *z*)

**â€¢ **Number of students exactly in three sets** **=** ***r*

**â€¢ **Total number of students in atleast two sets (two or more sets) =
*x* + *y* +*z* + *r*

**â€¢ **Total number of students in 3 sets = (*a* + *b* + *c* + *x* + *y* + *z* + *r*)

**Example 1.32**

In
a college, 240 students play cricket, 180 students play football, 164 students play
hockey, 42 play both cricket and football, 38 play both football and hockey, 40
play both cricket and hockey and 16 play all the three games. If each student participate
in atleast one game, then find (i) the number of students in the college (ii) the
number of students who play only one game.

*Solution*

Let
*C*, *F* and *H* represent sets of students who play Cricket, Football
and Hockey respectively.

Then
, *n* (*C* ) = 240, *n* (*F*) =
180, *n* (*H* ) = 164, *n* (*C* âˆ©
*F*) = 42,

*n
*(*F *âˆ©*
H *)* *=*
*38,* n *(*C *âˆ©*
H*)* *=*
*40,* n *(*C *âˆ©*
F *âˆ©*
H*)* *=*
*16.

Let
us represent the given data in a Venn diagram.

(i)
The number of students in the college

=174+26+116+22+102+24+16
= 480

(ii)
The number of students who play only one game

=
174+116+102 = 392

**Example 1.33**

In
a residential area with 600 families 3/5 owned scooter, 1/3 owned car, 1/4 owned
bicycle, 120 families owned scooter and car, 86 owned car and bicylce while 90 families
owned scooter and bicylce. If 2/15 of families owned all the three types of vehicles,
then find (i) the number of families owned atleast two types of vehicle. (ii) the
number of families owned no vehicle.

** Solution** Let

Given,
n( U) = 600* *

*n*(S) = 3/5 Ã— 600 = 360

*n* (C ) = 1/3 Ã— 600 = 200*
*

*n *(B) = 1/4 Ã— 600 = 150

n
(S âˆ© C âˆ© B) = 2/15 Ã— 600 = 80

From
Venn diagram,

(i)
The number of families owned atleast two types of vehicles = 40+6+10+80 = 136

(ii)
The number of families owned no vehicle

= 600 â€“ (owned atleast one vehicle)

= 600âˆ’(230+40+74 +6+54 +10+80)

= 600 âˆ’ 494 = 106

**Example 1.34**

In a group
of 100 students, 85 students speak Tamil, 40 students speak English, 20 students
speak French, 32 speak Tamil and English, 13 speak English and French and 10 speak
Tamil and French. If each student knows atleast any one of these languages, then
find the number of students who speak all these three languages.

** Solution** Let

Given, *n *(*A *âˆª*
B *âˆª*C*)* *= 100,* n *(*A*)* *=*
*85,* n*(B)* *=*
*40,* n*(C)* *=*
*20,

*n *(Aâˆ©* B*)* *=*
*32,* n *(Bâˆ©*
C *)* *=*
*13,* n *(Aâˆ©*
C *)* *=*
*10* *.

We
know that,

*n
*(*A *âˆª* B *âˆª*C*)* *=* n *(*A*)* *+*
n *(*B*)* *+*
n *(C)* *âˆ’*
n *(*A *âˆ©*
B*)* *âˆ’*
n *(*B *âˆ©*C
*)* *â€“ *n
*(*A *âˆ©*C*)* *+*
n *(*A *âˆ©*
B *âˆ©*C*)

100=
85 +
40 +
20 âˆ’
32 âˆ’13
âˆ’
10 +
*n* (*A* âˆ© *B* âˆ©*C*)

Then,
*n* (*A* âˆ© *B* âˆ©*C*) = 100 âˆ’ 90 = 10

Therefore,
10 students speak all the three languages.

**Example 1.35**

A
survey was conducted among 200 magazine subscribers of three different magazines
*A*, *B* and *C*. It was found that 75 members do not subscribe magazine
*A*, 100 members do not subscribe magazine *B*, 50 members do not subscribe
magazine C and 125 subscribe atleast two of the three magazines. Find

(i)
Number of members who subscribe exactly two magazines.

(ii)
Number of members who subscribe only one magazine.

*Solution*

Total
number of subscribers = 200

From
the Venn diagram,

Number
of members who subscribe only one magazine = *a* +*b* +*c*

Number
of members who subscribe exactly two magazines = *x* +
*y* + *z* and 125 members subscribe atleast two magazines.

That
is, *x* + *y* + *z* +
*r* = 125 ... (1)

Now,
*n* (*A* âˆª
*B* âˆª*C*) = 200 , *n*(*A*) = 125, *n*(*B*)
= 100, *n*(*C*) = 150, *n*(*A* âˆ©
*B*) = *x* + *r*

*n*(*B *âˆ©*
C*)* *=* y *+* r, n*(*A
*âˆ©*
C*)* *=* z *+* r,* *n*(*A *âˆ©* B *âˆ©*
C*)* *=* r*

We
know that,

*n*(*A *âˆª* B *âˆª* C) = n*(*A*)+* n*(*B*)+* n*(*C*)â€“*
n*(*A *âˆ©* B*)* *â€“* n*(*B *âˆ©* C*)â€“* n*(*A *âˆ©*
C*)+* n*(*A *âˆ©*
B *âˆ©*
C*)

200
= 125+100+150â€“*x*â€“*r*â€“*y*â€“*r*â€“*z*â€“*r*+*r*

=
375â€“(*x*+*y*+*z*+*r*)â€“*r*

=
375â€“125â€“*r *[x +* y *+ z + r =
125]

200
= 250â€“*r* â‡’ *r* = 50

From (1)
*x* + *y* + *z* +
50 = 125

We get, *x *+* y *+*
z *= 75

Therefore,
number of members who subscribe exactly two magazines = 75.

From
Venn diagram,

(*a*
+
*b* + *c* ) + (*x* +
*y* + *z* + *r*) = 200 ...
(2)

substitute
(1) in (2),

*a *+*
b *+*
c *+*
*125* *= 200

*a *+*
b *+*
c *= 75

Therefore,
number of members who subscribe only one magazine = 75.

Tags : Set Language | Maths , 9th Maths : UNIT 1 : Set Language

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9th Maths : UNIT 1 : Set Language : Application on Cardinality of Sets | Set Language | Maths

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