Sets and Relations theory

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1. If every element of a set P is also an element of set Q then set P is a subset of Q then the set Q is known as

(a)        Set of P
(b)        Sub set of P
(c)        Super set of P
(d)        Any of these.

Every subset is a subset of a super set
Hence Option (c) is the correct answer

2. The empty set is also known as

(a)        Zero set
(b)        Null set
(c)        Void set
(d)        (b) or (c).

Empty set is also known as null set or void set.
Hence Option (d) is the correct answer

3. Inter section of two set A & B is known as $\displaystyle \begin{array}{*{20}{l}} {(a)A\subset B} \\ {(b)A\cap B} \\ {(c)A\cup B} \\ {(d)A\not\subset B} \end{array}$

Intersection of two set A & B is known as ( $\displaystyle {A\cap B}$)
Hence Option (b) is the correct answer

4. If n(A) = n(B), for two finite set A & B is known

(a)        Singleton set
(b)        Equal sets
(c)        Equivalent sets
(d)        None of these.

Two finite sets are called equivalent if they have same number of element. i.e. n (A) = n (B)
Hence Option (c) is the correct answer

5. The objects of a set are called its

(a)        Universal set
(b)        Sub set
(c)        Unit
(d)        Elements of set.

Objects of a set are called its element
Hence Option (d) is the correct answer

6. The methods of describing a set

(a)        Roster from
(b)        Set Builder from
(c)        Either (a) or (b)
(d)        None of the above

There are two method of describing a set
(1) Roster from
(2) Set builder from
Hence Option (c) is the correct answer

7. Universal set is denoted by _____________ .

(a) { }
(b) $\displaystyle \phi$
(c) U
(d) None of these

Universal set is denoted by $\displaystyle U$
Hence Option (c) is the correct answer

8. _____________is also known as Tabular from.

(a)        Rule method
(b)        Roster form
(c)        Set-Builder form
(d)        Any of the above

Another name for Roster from is Tabular form. In such form, the elements are being separated by commas and are enclosed within braces { }.
Hence Option (b) is the correct answer

9.    ____________ is also known as Rule Method.

(a)        Rule method
(b)        Roster from
(c)        Set-Builder from
(d)        None.

Another name for Set Builder form is Rule Method. It is a statement or an expression to represent all the elements of a set.
Hence Option (c) is the correct answer

10. Under  _______________ , we just make a list of all elements of the set and put them under { }

(a)        Roster method
(b)        Set-Builder form
(c)        Both (a) & (b)
(d)        None of the above.

Under Roster method we make a list of elements separated by comma and we put it under { }
Hence Option (a) is the correct answer

11. $\displaystyle \mathbf{If}\text{ }\mathbf{A}\text{ }\subset \text{ }\mathbf{B},\text{ }\mathbf{B}\text{ }\subset \text{ }\mathbf{A}\text{ }\mathbf{then}\text{ }\mathbf{two}\text{ }\mathbf{sets}\text{ }\mathbf{A}\text{ }\And \text{ }\mathbf{B}\text{ }\mathbf{are}~$                      .

(a)        Proper sub-set
(b)        Equal set
(c)        Equivalent sets
(d)        Power set.

( $\displaystyle If\text{ }A\subset B\text{ }and\text{ }B\subset A\text{ }then\text{ }two\text{ }setsA\text{ }\And \text{ }B\text{ }are\text{ }equal\text{ }set$)
Hence Option (b) is the correct answer

12. Set A and set B be two sets, the  ______________of A and B is $\displaystyle \mathbf{A}\text{ }\cap \text{ }\mathbf{B}~~$.

(a)        Intersection
(b)        Equivalent
(c)        Union
(d)        Void

Intersection of set A & set B = ( $\displaystyle A\cap B$)
Hence Option (a) is the correct answer

13. Set A = {1, 2, 3} &  set  B = {2, 1, 3}.

(a)        A & B are Equal
(b)        A & B is disjoint
(c)        A and B is NOT Equal
(d)        A is proper sub set of B

Set A and set B are equal set because these sets have same elements.
Hence Option (a) is the correct answer

14. If A = {2, 5, 7} and B = {3, 5, 7, 8}, the value of A $\displaystyle \cup$ B is

(a)        {5, 7, 8}
(b)        {3, 8}
(c)        {2, 3, 4, 5, 6, 7, 8}
(d)        {0}

Set A = {2, 5, 7} B = {3, 5, 7, 8}
A $\displaystyle \cup$ B (A union B)
= {2, 3, 5, 7, 8}
Hence Option (c) is the correct answer

15. If A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}, the value of A – {B $\displaystyle \cup$ C } is

(a)        {1, 2, 3}
(b)        {2, 3, 4, 5}
(c)        {1}
(d)        {0}.

A = { 1, 2, 3, 4} B = {2, 4, 6, 8} C = {3, 4, 5, 6}
B $\displaystyle \cup$ C = {2, 3, 4, 5, 6, 8}
A – {B $\displaystyle \cup$ C } = {1, 2, 3, 4} – {2, 3, 4, 5, 6, 8} = {1}
Hence Option (c) is the correct answer

16. Which of the following is an empty set ?

(a) {0}
(b) { $\displaystyle \phi$}
(c) 0
(d) $\displaystyle \phi$ .

{0} is not an empty set because 0 is an element.
{ $\displaystyle \phi$} is not an empty set because $\displaystyle \phi$ is an element. $\displaystyle \phi$ $\displaystyle \to$ is an empty set because it has no element
Hence Option (d) is the correct answer

17. The subsets of {0} are

(a) 1
(b) {0}
(c) $\displaystyle \phi$
(d) $\displaystyle \phi$ and {0} . $\displaystyle \phi \text{ and }\left\{ 0 \right\}\text{ }are\text{ }subset\text{ }of\text{ }every\text{ }set$
Hence Option (d) is the correct answer

18. If A = {1, 2, 3} and B = {2, 4}, the value of A $\displaystyle \cap ~~$ B is

(a) {1, 2, 4}
(b) {2}
(c) {1, 2}
(d) $\displaystyle \phi$ $\displaystyle \begin{array}{*{20}{l}} {A=\left\{ {1,2,3} \right\}\text{ }and\text{ }B=\left\{ {2,4} \right\},} \\ {\therefore A\cap B=\left\{ 2 \right\}~\left( {only\text{ }common\text{ }elements.} \right)} \end{array}$
Hence Option (b) is the correct answer

19. $\displaystyle \mathbf{A}\text{ }=\phi ~~~~$, which one of the following statements is correct ? $\displaystyle \begin{array}{*{20}{l}} {\left( a \right)} & {\phi \in A} \\ {\left( b \right)} & {A\in \phi } \\ {\left( c \right)} & {A\not\subset \phi } \\ {\left( d \right)} & {\phi \subset A} \end{array}$

If A = $\displaystyle \phi$ then $\displaystyle \phi$ $\displaystyle \subset$ A . Because f is a subset of every set.
Hence Option (d) is the correct answer

20. If A ={1, 2, 3}, B = {3, 4} and C = {4, 5, 6} then $\displaystyle \mathbf{A}\cup (\mathbf{B}\cap \mathbf{C})$ is equal to

(a)        {1, 2, 3,4}
(b)        {3}
(c)        {1, 2, 3, 4, 5, 6}
(d)        {1, 2} . $\displaystyle \begin{array}{*{20}{l}} {A=\left\{ {1,\text{ }2,\text{ }3} \right\},\text{ }B\text{ }=\text{ }\left\{ {3,\text{ }4} \right\},~~~~~~C\text{ }=\text{ }\left\{ {4,\text{ }5,\text{ }6} \right\}} \\ {(B\cap C)\text{ }\left\{ 4 \right\}.\text{ }\left[ {\text{ }only\text{ }common\text{ }elements\text{ }of\text{ }set\text{ }B\text{ }and\text{ }set\text{ }C} \right]} \\ {A\cup (B\cap C)=\left\{ {1,\text{ }2,\text{ }3,\text{ }4} \right\}} \end{array}$
Hence Option (a) is the correct answer

21. Of the following sets, the one having exactly eight subsets is

(a)        {1, 2}
(b)        {0, 1, 2}
(c)        {8}
(d)        {1, 2, 3, 4, 5, 6, 7, 8} .

The number of subsets (called Power Set) of a set of n elements contains 2n-1 subsets. Here {0, 1, 2} has 3 elements. It contains 23-1 = 8 subsets : $\displaystyle {\left\{ 0 \right\},\text{ }\left\{ 1 \right\},\text{ }\left\{ 2 \right\},\text{ }\left\{ {0,\text{ }1} \right\},\text{ }\left\{ {0,\text{ }2} \right\}\text{ }\left\{ {1,\text{ }2} \right\}\text{ }\left\{ {0,\text{ }1,\text{ }2} \right\},\phi }$
Hence Option (b) is the correct answer

22. If A = {1, 2, 3, 4} and B = {5, 6, 7}, then $\displaystyle \mathbf{A}\cap \mathbf{B}$ is $\displaystyle \begin{array}{l}(a)\left\{ {1,2} \right\}\\(b)\left\{ {6,7} \right\}\\(c)\left\{ 3 \right\}\\(d)\phi \end{array}$

A = {1, 2, 3, 4} B = {5, 6, 7} $\displaystyle A\cap B\text{ }=\phi$ (as there is no common elements.)
Hence Option (d) is the correct answer

23. The number of subsets of the set {a , b, c} is

(a)        12
(b)        6
(c)        8
(d)        9.

The number of subsets (called Power Set) of a set of n elements contains 2n-1 subsets.. Here {0, 1, 2} contains 23-1 = 8 subsets $\displaystyle Subsets\text{ }of\text{ }set\text{ }\left\{ {a,\text{ }b,\text{ }c} \right\}\text{ }are\text{ }\left\{ a \right\},\text{ }\left\{ b \right\},\text{ }\left\{ c \right\},\text{ }\left\{ {a,\text{ }c} \right\},\text{ }\left\{ {a,\text{ }b} \right\},\text{ }\left\{ {a,\text{ }b,\text{ }c} \right\},\text{ }\left\{ {b,\text{ }c} \right\},\text{ }\phi$
Hence Option (c) is the correct answer

24. {1, 2, 3} $\displaystyle \cup$ {2, 3, 4} is equal to

(a) {1, 2}
(b) $\displaystyle \phi$
(c) {1, 2}
(d) {1, 2, 3, 4} .

= {1, 2, 3} $\displaystyle \cup$ {2, 3, 4}
= {1, 2, 3, 4}
Hence Option (d) is the correct answer

25. If A = {x : x2 3x + 2 = 0} and B = {x : x2 + 4x 12 = 0}, then A B is equal to $\displaystyle \begin{array}{l}(a)\left\{ 3 \right\}\\(b)\left\{ {1,6} \right\}\\(c)\left\{ 1 \right\}\\(d)\phi \end{array}$

Eq A : x2 – 3x + 2 = 0                    Eq B :  x2 + 4x – 12 = 0

Or, x2 – x – 2x + 2 = 0                   Or, x2 + 6x – 2x – 12 = 0

Or, x (x – 1) – 2 (x – 1) = 0            Or, x (x + 6) – 2 (x + 6) = 0

Or, (x – 1) (x – 2) = 0                    Or, (x – 2) (x + 6) = 0 $\displaystyle \begin{array}{*{20}{l}} {\therefore x=1\text{ }Or,\text{ }2} & {\therefore x\text{ }=\text{ }2\text{ }Or,-6} \\ {\therefore A=\left\{ {1,2} \right\}~B=\{-6,2\}} & ~ \\ {\therefore A-B\text{ }=\text{ }\left\{ 1 \right\}~\left[ {items\text{ }of\text{ }set\text{ }A\text{ }not\text{ }available\text{ }in\text{ }set\text{ }B} \right]} & {} \end{array}$

Hence Option (c) is the correct answer

26. Write down the power set of the set {0}

(a) $\displaystyle \phi$
(b) {0}
(c) { $\displaystyle \phi$}
(d) { $\displaystyle \phi$, {0} }. $\displaystyle Power\text{ }set\text{ }is\text{ }the\text{ }set\text{ }of\text{ }all\text{ }subsets.\text{ }\phi \text{ }is\text{ }subset\text{ }of\text{ }each\text{ }set.\text{ }So,\text{ }Power\text{ }set\text{ }of\text{ }the\text{ }set\text{ }\left\{ 0 \right\}\text{ }is\text{ }\{\phi ,\text{ }\left\{ 0 \right\}\text{ }\}.~$

Hence Option (d) is the correct answer

27. If U = { 2, 3, 4, 5, 6, 7, 8, 9,10, 11}, A = {2, 4, 7}, B = {3, 5, 7, 9, 11} and C = {7, 8, 9, 10, 11}, compute : $\displaystyle (\mathbf{A}\cap \mathbf{U})\cap (\mathbf{B}\cup \mathbf{C})$

(a) {7}
(b) $\displaystyle \phi$
(c) {2}
(d) {10}

U = { 2, 3, 4, 5, 6, 7, 8, 9,10, 11},
A = {2, 4, 7},
B = {3, 5, 7, 9, 11}
C = {7, 8, 9, 10, 11}, $\displaystyle \begin{array}{*{20}{l}} {A\cap U=\left\{ {2,\text{ }4,\text{ }7} \right\},~~~~~~~~~~~~~~~~~~\left( {common\text{ }in\text{ }both\text{ }U\text{ }\And \text{ }A} \right)} \\ {B\cup C=\left\{ {3,\text{ }5,\text{ }7,\text{ }8,\text{ }9,\text{ }11} \right\}} \\ {(A\cap U)\cap (B\cup C)\text{ }=\text{ }\left\{ 7 \right\}} \end{array}$

Hence Option (a) is the correct answer

28. If U = n {a, b, c, d, e, f}, A = {a, b, c}, find (U $\displaystyle \cup$ A)

(a) U
(b) A
(c) $\displaystyle \phi$
(d) {a, b, c} $\displaystyle \begin{array}{*{20}{l}} {U=\left\{ {a,\text{ }b,\text{ }c,\text{ }d,\text{ }e,\text{ }f} \right\},} \\ {A=\left\{ {a,\text{ }b,\text{ }c} \right\}} \\ {(U\cup A)=\left\{ {a,\text{ }b,\text{ }c,\text{ }d,\text{ }e,\text{ }f} \right\}=U} \end{array}$

Hence Option (a) is the correct answer

29. If U = {a, b, c, d, e, f}, A = {a, b, c}, B = {c, d, e, f} and C = {c, d, e}, find $\displaystyle (\mathbf{A}\cup \mathbf{B})\cup \mathbf{C}~$

(a)        {a, b, c}
(b)        B
(c)        {c, d, e}
(d)        U .

A = {a, b, c}
B = {c, d, e, f}
C = {c, d, e} $\displaystyle \begin{array}{*{20}{l}} {(A\cup B)=\left\{ {a,\text{ }b,\text{ }c,\text{ }d,\text{ }e,\text{ }f} \right\}} \\ {(A\cup B)\cup C=\left\{ {a,\text{ }b,\text{ }c,\text{ }d,\text{ }e,\text{ }f} \right\}=U} \end{array}$

Hence Option (d) is the correct answer

30. If U = {a, b, c, d, e, f}, A = {a, b, c}, B = {c, d, e, f}, C = {c, d, e}, find $\displaystyle (\mathbf{A}\cap \mathbf{B})\cup (\mathbf{A}\cap \mathbf{C})$

(a)        {c}
(b)        {a}
(c)        {f}
(d)        {e}

A = {a, b, c}
B = {c, d, e, f}
C = {c, d, e} $\displaystyle \begin{array}{*{20}{l}} {A\cap B\text{ }=\text{ }\left\{ c \right\}} \\ {A\cap c\text{ }=\text{ }\left\{ c \right\}} \\ {(A\cup B)\cup (A\cup C)\text{ }=\text{ }\left\{ c \right\}} \end{array}$

Hence Option (a) is the correct answer

31. If S and T are two sets such that S has 21 elements, T has 32 elements, and S $\displaystyle \cap$ T has 11 elements, how many elements does S $\displaystyle \cup$ T have ?

(a)        48
(b)        26
(c)        42
(d)        36

S has 21 elements
T has 32 elements $\displaystyle S\text{ }\cap \text{ }T\text{ }has\text{ }11\text{ }elements$

So, there are 11 common elements in S & T
So, uncommon elements in S = 21 – 11 = 10
Un common elements in T = 32 – 11 = 21 $\displaystyle S\cup T=11+10+21=42$

We may also calculate it as 21 (S) + 32 (T) – 11 common elements = 53-11=42
Hence Option (c) is the correct answer

32. In a group of 1,000 people, there are 750 people who can speak Hindi and 400 who can speak English. How many can speak Hindi only ?

(a)        600
(b)        700
(c)        850
(d)        900

No. of people can speak in Hindi = 750
No. of people can speak in English = 400
No. of people can speak in both Hindi & English
= 750 + 400 – 1000 = 1150 – 1000 = 150
No. of people can speak in only Hindi = 750 – 150 = 600

Hence Option (a) is the correct answer

33. In a class of 50 students, 35 opted for Mathematics and 37 opted for Biology. How many have opted for both Mathematics and Biology ? How many have opted for only Mathematics ?

(Assume that each student has to opt for at least one of the subjects).

(a)        20
(b)        21
(c)        13
(d)        18

No. of students opted for Mathematics = 35
No. of students opted for Biology = 37
No. of students opted for Mathematics & Biology
= 35 + 37 – 50 = 72 – 50 = 22
No. of students opted only for Mathematic (35 – 22) = 13

Hence Option (c) is the correct answer

34. In a town with a population of 5,000, 3,200 people are egg-eaters, 2,500 meat-eaters and 1,500 eat both egg and meat. How many are pure vegetarians ?

(a)        400
(b)        800
(c)        700
(d)        750

Both egg-eater & meat-eater = 1500
Only egg-eater = 3200 – 1500 = 1700
Only meat-eater = 2500 – 1500 = 1000
vegetarians = 5000 – (1700 + 1500 + 100) = 5000 – 4200 = 800

Hence Option (b) is the correct answer

35. If S = {x : x2 + 1 = 0, x real}, then S is $\displaystyle \begin{array}{*{20}{l}} {\left( a \right)} & 0 \\ {\left( b \right)} & {\left\{ 0 \right\}} \\ {\left( c \right)} & {\{\phi \}} \\ {\left( d \right)} & \phi \end{array}$ $\displaystyle \begin{array}{*{20}{l}} {~{{x}^{2}}+1=0~~~~~~~~{{x}^{2}}=-1~(x\text{ }is\text{ }not\text{ }real\text{ }number,\text{ }imaginary\text{ }number} \\ {As\text{ }per\text{ }problem,\text{ }x\text{ }is\text{ }real.\text{ }There\text{ }is\text{ }no\text{ }such\text{ }real\text{ }number\text{ }(\surd -1\text{ }is\text{ }imaginary\text{ }number)} \\ {\therefore s=\phi } \end{array}$

Hence Option (d) is the correct answer

36. Which of the following is correct statement ? $\displaystyle \begin{array}{*{20}{l}} {\left( a \right)} & {\phi =\text{ }\left\{ \text{ } \right\}} \\ {\left( b \right)} & {\phi =\text{ }\left\{ 0 \right\}} \\ {\left( c \right)} & {\phi =\text{ }\{\phi \}} \\ {\left( d \right)} & {All\text{ }of\text{ }the\text{ }above} \end{array}$ $\displaystyle \phi \text{ }is\text{ }indicated\text{ }by\text{ }\left\{ \text{ } \right\}$

Hence Option (a) is the correct answer

37. If A = {1, 2, 3}, B = {2, 3, 4}, then A $\displaystyle \cap$ B is

(a)        {2, 3}
(b)        {2}
(c)        {3,2}
(d)        Both a & c $\displaystyle \begin{array}{*{20}{l}} {A\text{ }=\text{ }\left\{ {1,\text{ }2,\text{ }3} \right\},} \\ {B\text{ }=\text{ }\left\{ {2,\text{ }3,\text{ }4} \right\},} \\ {A\cap B\text{ }=\text{ }\left\{ {2,\text{ }3} \right\}~or\text{ }\left\{ {3,2} \right\}\text{ }\left( {common\text{ }elements\text{ }of\text{ }both\text{ }sets} \right)} \end{array}$

Hence Option (d) is the correct answer

38. Let A = {0, 1, 3, 4}, B = {5, 6, 1, 3, 9} and C = {1, 2, 3, 4, 9, 13}, then $\displaystyle (\mathbf{A}\cap \mathbf{B})\cup \mathbf{C}$ is

(a)        C
(b)        {0, 1, 2, 3, 4, 9, 13}
(c)        A
(d)        B

A = {0, 1, 3, 4}
B = {5, 6, 1, 3, 9}
C = {1, 2, 3, 4, 9, 13} $\displaystyle \begin{array}{*{20}{l}} {A\cap B\text{ }=\text{ }\left\{ 3 \right\}} \\ {A\cap B\cup C\text{ }=~\left\{ {1,\text{ }2,\text{ }3,\text{ }4,\text{ }9,\text{ }13} \right\}\text{ }=\text{ }\left\{ C \right\}} \end{array}$

Hence Option (a) is the correct answer

39. Let A = {1, 2, 3,4,5,6,7,8,9}, B = {2, 4, 6, 7, 8} and C = {3, 4, 5, 8, 9, 10}, then (A B)

(a)        {1, 3, 4, 5, 8, 9, 10}
(b)        {1, 2, 3, 4, 5,  9}
(c)        {6, 7, 8}

(d)        {1, 3, 4}

A = {1, 2, 3, 4,5,6,7,8,9},
B = {2, 4, 6, 7, 8}
C = {3, 4, 5, 8, 9, 10},

A – B = {1, 3, 5, 9} $\displaystyle (A-B)\cup C\text{ }=\text{ }\left\{ {1,\text{ }3,\text{ }4,\text{ }5,\text{ }8,\text{ }9,\text{ }10} \right\}$

Hence Option (a) is the correct answer

40. Which one of the following is a null set ?

(a) {x : x + 1 = 1} $\displaystyle {\begin{array}{*{20}{l}} {\left( b \right)} & {{\phi }} \end{array}}$
(c) {x : x > 1 and x < 1} $\displaystyle {\left( d \right)\text{ }{x\text{ }:\text{ }x\ge 1\text{ }and\text{ }x\le 1}}$

Hence Option (c) is the correct answer

41. In a group of 75 persons, 20 take tea but not coffee and 30 take tea. How many take coffee but not tea ?

(a)        45
(b)        32
(c)        50
(d)        60

Only tea = 20
both coffee & tea = 30 – 20 = 10
Only coffee = 75 – (20 + 10) = 45

Hence Option (a) is the correct answer

42. Set the positive integers, less than 3 is

(a)        {-1}
(b)        {0, 1}
(c)        {1, 2, 3}
(d)        {1, 2}

Positive integers less 3 are 1, 2.

Hence Option (d) is the correct answer

43. A = {2, 3, 5, 7}, B = {4, 6, 8, 10} then A $\displaystyle \cap$ B can be written as $\displaystyle \begin{array}{*{20}{l}} {\left( a \right)} & {\left\{ \text{ } \right\}} \\ {\left( b \right)} & {\{\phi \}} \\ {\left( c \right)} & {A\text{ }\cup \text{ }B} \\ {\left( d \right)} & {\left\{ 0 \right\}} \end{array}$ $\displaystyle \begin{array}{*{20}{l}} {A\text{ = }\left\{ {2,\text{ }3,\text{ }5,\text{ }7} \right\}} & {} \\ {B=\left\{ {4,\text{ }6,\text{ }8,\text{ }10} \right\}} & {(Because\text{ }there\text{ }is\text{ }no\text{ }common\text{ }Element.} \\ {A\cap B=\left\{ \text{ } \right\}} & {So,\text{ }A\cap B\text{ }is\text{ }a\text{ }null\text{ }set.)} \\ {\{\phi \}\text{ }is\text{ }not\text{ }a\text{ }null\text{ }set,\text{ }it\text{ }is\text{ }a\text{ }singleton\text{ }set} & {} \end{array}$

Hence Option (a) is the correct answer

44. By Roster method to express integers greater than 5 and less than 8. $\displaystyle \begin{array}{*{20}{l}} {\left( a \right)} & {\left\{ {5,\text{ }6,\text{ }7} \right\}} \\ {\left( b \right)} & {\left\{ {5,\text{ }6} \right\}} \\ {\left( c \right)} & \phi \\ {\left( d \right)} & {\left\{ {6,\text{ }7} \right\}} \end{array}$

Integers greater than 5 and less than or equal to 8 are 6, 7,

Hence Option (d) is the correct answer

45. Write the set containing all days of the week beginning with S.

(a)        {Sunday, Monday}
(b)        {Saturday, Sunday}
(c)        {Friday, Saturday}
(d)        none

Days $\displaystyle \to$  Monday , Wednesday, Tuesday, Thursday, Friday, Saturday, Sunday
So, all days of the week beginning with S are Saturday, Sunday.
Required set = {Saturday, Sunday}

Hence Option (b) is the correct answer

46.  A $\displaystyle \cap$ A is equal to

(a) A
(b) U
(c) $\displaystyle \phi$
(d) A’ $\displaystyle A\cap A\text{ }=\text{ }A~~\left( {Always} \right)$

Hence Option (a) is the correct answer

47.  A $\displaystyle \cap$ A’ is equal to

(a) U
(b) $\displaystyle \phi$
(c) A
(d) A $\displaystyle \cup$ A’ $\displaystyle \begin{array}{*{20}{l}} {Set\text{ }A'\text{ }contain\text{ }elements\text{ }not\text{ }available\text{ }in\text{ }set\text{ }A.} \\ {So,\text{ }there\text{ }is\text{ }no\text{ }common\text{ }element\text{ }between\text{ }set\text{ }A\text{ }and\text{ }set\text{ }A'} \\ {\therefore A\cap A'=\phi } \end{array}$

Hence Option (b) is the correct answer

48.  A = {1, 2, 3, 4, 5}, B = {4, 6, 7, 5}, C = {2, 3, 8, 9}, then $\displaystyle \mathbf{A}\cap (\mathbf{B}\cup \mathbf{C})$ is

(a)        Set A
(b)        Set B
(c)        Set C
(d)        {2, 3, 4, 5} $\displaystyle \begin{array}{*{20}{l}} {A\text{ }=\text{ }\left\{ {1,\text{ }2,\text{ }3,\text{ }4,\text{ }5} \right\}} \\ {B\text{ }=\text{ }\left\{ {4,\text{ }6,\text{ }7,\text{ }5} \right\},\text{ }C\text{ }=\text{ }\left\{ {2,\text{ }3,\text{ }8,\text{ }9} \right\},} \\ {B\cup C\text{ }=\text{ }\left\{ {2,\text{ }3,\text{ }4,\text{ }5,\text{ }6,\text{ }7,\text{ }8,\text{ }9} \right\}} \\ {A\cap (B\cup C)\text{ }=\text{ }\left\{ {2,\text{ }3,\text{ }4,\text{ }5} \right\}} \end{array}$

Hence Option (d) is the correct answer

49.  V = {x : x is vowel}, Y = {x : x is ‘a’ or any letter before ‘e’ in the alphabet}, then V $\displaystyle \cap$ Y is

(a) {a, o, u}
(b) $\displaystyle \phi$
(c) {a, e, i, o, u}
(d) {a} $\displaystyle \begin{array}{*{20}{l}} {V\text{ }=\text{ }\left\{ {a,\text{ }e,\text{ }i,\text{ }o,\text{ }u} \right\}} \\ {Y\text{ }=\text{ }\left\{ {a,\text{ }b,\text{ }c,\text{ }d,} \right\}} \\ {V\cap Y\text{ }=\text{ }\left\{ a \right\}} \end{array}$

Hence Option (d) is the correct answer

50. If A has 32 elements, B has 42 elements and A $\displaystyle \cup$ B has 62 elements. The number of elements in  A $\displaystyle \cap$ B is

(a)        12
(b)        70
(c)        20
(d)        50 $\displaystyle \begin{array}{*{20}{l}} {A\text{ }=\text{ }32\text{ }elements} \\ {B\text{ }=\text{ }42\text{ }elements} \\ {A\cup B\text{ }=\text{ }62\text{ }elements} \\ {A\cup B\text{ }=\text{ }A\text{ }+\text{ }B-A\cap B} \\ {62\text{ }=\text{ }32\text{ }+\text{ }42-A\cap B} \\ {Or,\text{ }A\cap B\text{ }=\text{ }74-62\text{ }=\text{ }12} \end{array}$

Hence Option (a) is the correct answer

51. In a group of 20 children, 8 drink tea but not coffee and 13 like tea. The number of children drinking coffee but not tea is

(a)        5
(b)        7
(c)        3
(d)        4.

Total number of children = 20
Drink only tea = 8
Like tea          = 13
Drink both tea & coffee (13 – 8) = 5
Drink only coffee = (20 – 8 – 5) = 7

Hence Option (b) is the correct answer

52. The number of subsets of the {6, 8, 11} is

(a)        7
(b)        5
(c)        8
(d)        3

Number of subsets = 2n-1, where n is the number of elements in the given set.  The number of elements in the set is 3. So, the number of subsets =  23 -1 =8, having elements as : $\displaystyle \left\{ 6 \right\},\text{ }\left\{ 8 \right\},\text{ }\left\{ {11} \right\},\text{ }\left\{ {6,\text{ }8} \right\},\text{ }\left\{ {6,\text{ }11} \right\},\text{ }\left\{ {11,\text{ }8} \right\},\text{ }\left\{ {6,\text{ }8,\text{ }11} \right\},\text{ }\phi$

Hence Option (c) is the correct answer

53. If A = {a, b, c, d, e, f} B= {a, e, i, o, u} and C = {m, n, o, p, q, r, s, t, u} then A $\displaystyle \cup$ B is $\displaystyle \begin{array}{*{20}{l}} {\left( a \right)} & {\left\{ {a,\text{ }b,\text{ }c,\text{ }d,\text{ }e,\text{ }f,\text{ }i,\text{ }o,\text{ }u} \right\}} \\ {\left( b \right)} & {\left\{ {a,\text{ }b,\text{ }c} \right\}} \\ {\left( c \right)} & {\left\{ {d,\text{ }e,\text{ }f} \right\}} \\ {\left( d \right)} & \phi \end{array}$ $\displaystyle \begin{array}{*{20}{l}} {A\text{ }=\left\{ {a,\text{ }b,\text{ }c,\text{ }d,\text{ }e,\text{ }f} \right\}} \\ {B\text{ }=\text{ }\left\{ {a,\text{ }e,\text{ }i,\text{ }o,\text{ }u,} \right\}} \\ {C\text{ }=\text{ }\left\{ {m,\text{ }n,\text{ }o,\text{ }p,\text{ }q,\text{ }r,\text{ }s,\text{ }t,\text{ }u} \right\}} \\ {A\cup B\text{ }=\text{ }\left\{ {a,\text{ }b,\text{ }c,\text{ }d,\text{ }e,\text{ }f,\text{ }i,\text{ }o,\text{ }u} \right\}} \end{array}$

Hence Option (a) is the correct answer

54. On a survey of 100 boys it was found that 50 used white shirt, 40 used red and 30 used blue. 20 were habituated in using both white and red shirts, while 15 used both red and blue shirts and 10 used  blue and white shirts. Find the number of boys using all the colours.

(a)        22
(b)        25
(c)        45
(d)        40. $\displaystyle \begin{array}{*{20}{l}} {~W=\text{ }50~~~~~~~~~~~~~~~~~~R\text{ }=\text{ }40~~~~~~~~~~~~~~~~~~~~~B\text{ }=\text{ }30} \\ {W\cap R\text{ }=\text{ }20~~~~~~~~~R\cap B\text{ }=\text{ }15~~~~~~~~~~~W\cap B\text{ }=\text{ }10} \\ {W\cup R\cup B\text{ }=\text{ }100} \\ {W\cap R\cap B\text{ }=\text{ }?} \end{array}$ $\displaystyle \begin{array}{*{20}{l}} {We\text{ }know} \\ {W\cup R\cup B\text{ }=\text{ }W\text{ }+\text{ }R\text{ }+\text{ }B-W\cap R-R\cap B~-W\cap B} \\ {+\text{ }W\cap R\cap B} \end{array}$ $\displaystyle \begin{array}{*{20}{l}} {Or,\text{ }100\text{ }=\text{ }50\text{ }+\text{ }40\text{ }+\text{ }30-20-15-10\text{ }+\text{ }W\cap R\cap B} \\ {Or,\text{ }100\text{ }=\text{ }120-45\text{ }+\text{ }W\cap R\cap B} \\ {Or,\text{ }100\text{ }=\text{ }75\text{ }+\text{ }W\cap R\cap B} \\ {Or,\text{ }W\cap R\cap B\text{ }=\text{ }100-75\text{ }=\text{ }25} \end{array}$

Hence Option (b) is the correct answer