Mathematics Miscellaneous Examples and Miscellaneous Exercise

### Miscellaneous Examples

Q 3087578487

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 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}.
Q 3027678581

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
Q 3037578482

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.
Q 3077678586

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. 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 ).
Q 3057778684

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.
Q 3007778688

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 ).
This is a contradiction. So, the given data is incorrect.
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 ?

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.

### Miscellaneous Exercise

Q 2625856761

Decide, among the following sets which sets
are subsets of one and another

A= { x : x in R and x satisfy x^2 -8x+ 12 = 0},

B = {2,4,6},C= {2, 4,6, 8, ......... },D=, {6}
Class 11 Exercise 1.mis Q.No. 1
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.
Q 2655056864

Let A, B, and C be the sets such that A ∪ B

=A ∪ C and A ∩ B =A ∩ C . Show that B = C

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
Q 2665056865

Show that the following four conditions are equivalent:

(i) A ⊂ B

(ii) A -B= phi

(iii) A ∪ B = B

(iv) A ∩ B =A

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.
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
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
Q 2675467366

Using properties of sets, show that

(i) A ∪ (A ∩ B) =A .

(ii) A ∩ (A ∪ B) =A
Class 11 Exercise 1.mis Q.No. 9
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 ]
Q 2655567464

Show that A ∩ B = A ∩ C need not imply
B=C.
Class 11 Exercise 1.mis Q.No. 10
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
Q 2615567469

Let A and B be sets, If A ∩ X= B ∩ X = phi and

A ∪ X= B ∪ X for some set X, show that
A=B.
Class 11 Exercise 1.mis Q.No. 11
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
Q 2675667566

Find sets A, B, and C such that A ∩ B, B ∩ C
and A ∩ C arc not-empty sets and A ∩ B ∩ C = phi
Class 11 Exercise 1.mis Q.No. 12
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
Q 2615667569

In a survey of 600 students in a school, 150 students were found to be taking tea and 225
taking coffee, 100 were taking both tea and coffee. Find bow many students were taking
neither tea nor coffee?
Class 11 Exercise 1.mis Q.No. 13
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
Q 2615767660

In a group of students, 100 students know Hindi, 50 know English and 25 know both.
Each of the students knows either Hindi or English. How many students are there in the
group'?
Class 11 Exercise 1.mis Q.No. 14
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
Q 2635767662

In a survey of 60 people, it was found that 25 people read newspaper H, 26 read Newspaper
T, 26 read newspaper I, 9 read both H and I, 11 react both H and T, 8 read both T and I, 3 read
all three newspapers. Find :

(i) the number of people who read at leastone of the newspapers.
(ii) the number of people who read exactly one newspaper.
Class 11 Exercise 1.mis Q.No. 15
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
Q 2645767663

In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked
product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people
liked products B and C and 8 liked all the three products. Find how many liked product C only.
Class 11 Exercise 1.mis Q.No. 16
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