M.C.Q (1 Marks)

MCQ 11 Mark

If A and B are two sets such that $\text{n(A)}=70, \text{ n(B)}=60, \text{ n(A}\cup\text{B)}=110,$ then $\text{n(A}\cap\text{B)}$ is equal to:

A
240
B
50
C
40
✓
20.

Answer

Correct option: D.

20.

We have:
$\text{n(A}\cap\text{B) = n(A) + n(B)} - \text{n(A}\cup\text{B)}$
$=70+60-110$
$=20.$

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MCQ 21 Mark

Let $\text{A} = \{\text{x : x} \in \text{R}, \text{x > 4}\}$ and $\text{B}= \{\text{x}\in\text{R : x} < 5\}.$ Then, $\text{A}\cap\text{B}=$

A
(4, 5]
B
(4, 5)
✓
[4, 5)
D
[4, 5].

Answer

Correct option: C.

[4, 5)

$\text{A} = \{\text{x : x} \in \text{R}, \text{x > 4}\}$ and
$\text{B}= \{\text{x}\in\text{R : x} < 5\}$
$\text{A}\cap\text{B}=[4, 5).$

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MCQ 31 Mark

For any two sets A and B, $\text{A}\cap\text{(A}\cup\text{B)}=$

✓
A
B
B
C
$\phi$
D
None of these.

Answer

Correct option: A.

$\text{A}\cap\text{(A}\cup\text{B)}=\text{(A}\cap\text{A)}\cup\text{(A}\cap\text{B)}=\text{A}\cup\text{(A}\cap\text{B)}=\\\text{AA}\cap\text{(A}\cup\text{B)}=\text{(A}\cap\text{A)}\cup\text{(A}\cap\text{B)}=\text{A}\cup\text{(A}\cap\text{B)}=\text{A.}$

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MCQ 41 Mark

If A = {1, 3, 5, B} and B = {2, 4}, then:

A
$4\in\text{A}$
B
$\{4\}\subset\text{A}$
C
$\text{B}\subset\text{A}$
✓
None of these.

Answer

Correct option: D.

None of these.

$(4\not\in\text{A) }(4\not\in\text{A})$
$\{4\}\not\subset\text{A}$
$\text{B}\not\subset\text{}A$
Thus, we can say that none of these options satisfy the given relation.

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MCQ 51 Mark

Let A and B be two sets that $\text{n(A)} = 16, \text{ n(B)} = 14,\text{ n(A}\cup\text{B)}=25.$ Then, $\text{n(A}\cap\text{B)}$ is equal to:

A
30
B
50
✓
5
D
None of these.

Answer

Correct option: C.

We know:
$\text{n(A}\cup\text{B) = n(A) + n(B)} - \text{n(A}\cap\text{B)}$
Now,
$\text{n(A}\cap\text{B) = n(A) + n(B)} -\text{n(A}\cup\text{B)}$
$=16+14-25$
$=5.$

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MCQ 61 Mark

Which of the following statements is false:

A
$\text{A} - \text{B = A}\cap\text{B}'$
B
$\text{A} - \text{B = A} - \text{(A}\cap\text{B)}$
✓
$\text{A} - \text{B = A}-\text{B}'$
D
$\text{A} - \text{B = (A}\cup\text{B)}-\text{B.}$

Answer

Correct option: C.

$\text{A} - \text{B = A}-\text{B}'$

It includes all those elements of A which do not belongs to complement of B which is equal to $\text{A}\cap\text{B}$ but not equal to A - B.
$\therefore$ (c) ic false.

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MCQ 71 Mark

If $\text{A}\cap\text{B}=\text{B},$ then:

A
$\text{A}\subset\text{B}$
✓
$\text{B}\subset\text{A}$
C
$\text{A}=\phi$
D
$\text{B}=\phi.$

Answer

Correct option: B.

$\text{B}\subset\text{A}$

Only this case is possible.

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MCQ 81 Mark

Let A and B be two sets in the same universal set. Then, A - B =

A
$\text{A}\cap\text{B}$
B
$\text{A}'\cap\text{B}$
✓
$\text{A}\cap\text{B}'$
D
None of these.

Answer

Correct option: C.

$\text{A}\cap\text{B}'$

A - B belongs to those elements of A that do not belong to B.
$\therefore\text{A} - \text{B = A}\cap\text{B}'.$

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MCQ 91 Mark

Two finite sets have $m$ and $n$ elements. The number of subsets of the first set is $112$ more than that of the second. The values of $m$ and $n$ are respectively:

A
$4, 7$
✓
$7, 4$
C
$4, 4$
D
$7, 7.$

Answer

Correct option: B.

$7, 4$

We know that if a set $X$ contains $k$ elements, then the number of subsets of $X$ are $2^k.$
It is given that the number of subsets of a set containing $m$ elements is $112$ more than the number of subsets of set containing $n$ elements.
$\therefore 2^\text{m}-2^\text{n}=112$
$\Rightarrow2^\text{n}(2^\text{m - n}-1)=2\times2\times2\times2\times7$
$\Rightarrow2^\text{n}(2^{\text{m}-\text{n}}-1)=2^4(2^3-1)$
$\Rightarrow\text{n}=4$ and $\text{m}-\text{n}=3$
$\therefore\text{ m}-4=3$
$\Rightarrow\text{m}=7$
Thus, the values of $m$ and $n$ are $7$ and $4$, respectively.
Hence, the correct answer is option $(b).$

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MCQ 101 Mark

For two sets $\text{A}\cap\text{B = A}$ iff:

✓
$\text{B}\subseteq\text{A}$
B
$\text{A}\subseteq\text{B}$
C
$\text{A}\not=\text{B}$
D
$\text{A}=\text{B}.$

Answer

Correct option: A.

$\text{B}\subseteq\text{A}$

The union of two sets is a set of all those elements that belong to A or to B or to both A and B.
If $\text{A}\cup\text{B = A},$ then $\text{B}\subseteq\text{A}.$

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MCQ 111 Mark

For any two sets A and B, $\text{A}\cap\text{(A}\cup\text{B)}'$ is equal to:

A
$\text{A}$
B
$\text{B}$
✓
$\phi$
D
$\text{A}\cap\text{B}.$

Answer

Correct option: C.

$\phi$

$\text{A}\cap\text{(A}\cup\text{B)}'$
$=\text{A}\cap\text{(A}'\cup\text{B}')$ (De Morgen Law)
$=\text{(A}\cap\text{A}')\cap\text{B}'$
$=\phi\cap\text{B}'$
$=\phi$
Hence, the correct answer is option (c).

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MCQ 121 Mark

For any two sets A and $\text{B, A - B}\cup\text{B}=\text{A}=$

A
$\text{(A - B)}\cup\text{A}$
B
$\text{(B - A)}\cup\text{B}$
✓
$\text{(A}\cup\text{B)}-\text{(A}\cap\text{B)}$
D
$\text{(A}\cup\text{B)}\cap\text{(A}\cap\text{B)}.$

Answer

Correct option: C.

$\text{(A}\cup\text{B)}-\text{(A}\cap\text{B)}$

$\text{(A}-\text{B)}\cup\text{(B}-\text{A)}=\text{(A}\cap\text{B}')\cup\text{(B}\cap\text{A}')$
$=[\text{A}\cup\text{(B}\cup\text{A}')]\cap[\text{B}'\cup\text{(B}\cap\text{A}')]$ [Using distribution law]
$=[\text{(A}\cup\text{B})\cap\text{(A}\cup\text{A}')]\cap[\text{(B}'\cup\text{B})\cap\text{(B}'\cup\text{A}')]$ [Using distribution law]
$=[\text{(A}\cup\text{B)}\cup\text{(U)}]\cap[\text{(U)}\cap\text{(B}'\cup\text{A}')]$ $[\text{A}\cup\text{A'= U = B}'\cup\text{B}]$
$=[\text{A}\cup\text{B}]\cap[\text{B}'\cup\text{A}']$ $\begin{bmatrix}\text{(A}\cup\text{B)}\cap\text{(U)}=\text{(A}\cup\text{B)}\\\text{ and (U)}\cap\text{(B}'\cup\text{A)}'=\text{(B}'\cup\text{A}')]\end{bmatrix}$
$=[\text{A}\cup\text{B}]\cap[\text{(A}\cap\text{B)}']$ $[\text{(A}\cap\text{B)}'=\text{B}'\cup\text{A}']$
$=[\text{A}\cup\text{B}]\cap[\text{(A}\cup\text{B)}-\text{(A}\cap\text{B)}]$
$=[\text{(A}\cup\text{B)}-\text{(A}\cap\text{B)}].$

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MCQ 131 Mark

The symmetric difference of A and B is not equal to:

A
$\text{(A} - \text{B)}\cap\text{(B} -\text{A)}$
✓
$\text{(A} - \text{B)}\cup\text{(B}- \text{A)}$
C
$\text{(A}\cup\text{B)}-\text{(B}\cap\text{A)}$
D
$\{\text{(A}\cup\text{B)}-\text{A\}}\cup\{\text{(A}\cup\text{B)} - \text{B}\}.$

Answer

Correct option: B.

$\text{(A} - \text{B)}\cup\text{(B}- \text{A)}$

The symmetric difference of A and B is given by:-
$\text{(A} - \text{B)}\cup\text{(B}- \text{A)}.$

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MCQ 141 Mark

In set-builder method the null set is represented by:

A
$\{\}$
B
$\phi$
✓
$\{\text{x : x} \not=\text{x}\}$
D
$\{\text{x : x} =\text{x}\}.$

Answer

Correct option: C.

$\{\text{x : x} \not=\text{x}\}$

$\{\text{x : x}\not=\text{x}\}.$

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MCQ 151 Mark

Let $F_1$ be the set of all parallelograms, $F_2$ the set of all rectangles, $F_3$ the set of all rhombuses, $F_4$ the set of all squares and $F_5$ the set of trapeziums in a plane. Then $F_1$ may be equal to:

A
$\text{F}_2\cap\text{F}_3$
B
$\text{F}_3\cap\text{F}_4$
C
$\text{F}_2\cup\text{F}_3$
✓
$\text{F}_2\cup\text{F}_3\cup\text{F}_4\cup\text{F}_1.$

Answer

Correct option: D.

$\text{F}_2\cup\text{F}_3\cup\text{F}_4\cup\text{F}_1.$

We know that every rectangle, rhombus and square in a plane is a parallelogram but every trapezium is not a parallelogram.
So, $F_1$ is either of $F_1$ or $F_2$ or $F_3$ or $F_4.$
$\therefore\text{F}_1=\text{F}_1\cup\text{F}_2\cup\text{F}_3\cup\text{F}_4$
Hence, the correct answer is option $(d).$

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MCQ 161 Mark

The symmetric difference of A = {1, 2, 3} and B = {3, 4, 5} is:

A
{1, 2}
✓
{1, 2, 4, 5}
C
{4, 3}
D
{2, 5, 1, 4, 3}.

Answer

Correct option: B.

{1, 2, 4, 5}

Here,
$\text{A} = \{1, 2, 3\}$ and
$\text{B} = \{3, 4, 5\}$
The symmetric difference of A and B is given by:-
$\text{(A} - \text{B)}\cup\text{(B} -\text{A)}$
Now, are have:
$\text{(A} - \text{B)}= \{1, 2\}$
$\text{(B} - \text{A)}=\{4, 5\}$
$\text{(A}-\text{B)}\cup\text{(B}-\text{A)}=\{1, 2, 4, 5\}.$

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MCQ 171 Mark

If A = {x : x is a multiple of 3} and B = {x : x is a multiple of 5}, then A - B is:

A
$\text{A}\cap\text{B}$
✓
$\text{A}\cap\overline{\text{B}}$
C
$\overline{\text{A}}\cap\overline{\text{B}}$
D
$\overline{\text{A}\cap{\text{B}}}.$

Answer

Correct option: B.

$\text{A}\cap\overline{\text{B}}$

A = {x : x is a multiple of 3}
A = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, .....
B = {x : x is a multiple of 5}
B = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, ......
Now, we have:
A - B = 3, 6, 9, 12, 18, 21, 24, 27, 30, 33,36, 39, 42, ....
$=\text{A}\cap\overline{\text{B}}.$

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MCQ 181 Mark

If A = {1, 2, 3, 4, 5}, then the number of proper subsets of A is:

A
120
B
30
✓
31
D
32.

Answer

Correct option: C.

The number of proper subsets of any set is given by the formula 2n - 1, where n is the number of elements in the set.
Here,
n = 5
$\therefore$ Number of proper subsets of A = 25 - 1 = 31.

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MCQ 191 Mark

An investigator interviewed 100 students to determine the performance of three drinks: milk, coffee and tea. The investigator reported that 10 students take all three drinks milk, coffee and tea; 20 students take milk and coffee; 25 students take milk and tea; 12 students take milk only; 5 students take coffee only and 8 students take tea only. Then the number of students who did not take any of three drinks is:

A
10
✓
20
C
25
D
30.

Answer

Correct option: B.

solve for None:
80 + None = 100
None = 20.

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MCQ 201 Mark

If A and B are two given sets, then $\text{A}\cap\text{(A}\cap\text{B})^\text{c}$ is equal to:

A
$\text{A}$
B
$\text{B}$
C
$\phi$
✓
$\text{A}\cap\text{B}^\text{c}.$

Answer

Correct option: D.

$\text{A}\cap\text{B}^\text{c}.$

A and B are two sets.
$\text{A}\cap\text{B}$ is the common region in both the sets.
$\text{A}\cap\text{B}^\text{c}$ is all the region in the universal set except $\text{A}\cap\text{B}.$
Now,
$\text{(A}\cap\text{A}\cap\text{B)}^\text{c}=\text{(A}\cap\text{B)}^\text{c}.$

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MCQ 211 Mark

Let U be the universal set containing 700 elements. If A, B are subsets of U such that $\text{n(A)}=200,\text{ n(B)}=300$ and $\text{n(A}\cap\text{B)}=100.$ Then, $\text{n(A}'\cap\text{B}')=$

A
400
B
600
✓
300
D
None of these.

Answer

Correct option: C.

300

$\text{n(A}'\cap\text{B}')=\text{n(A}\cup\text{B}')$
$=\text{n(U)}-\text{n(A}\cup\text{B})$
$=700 - 200 + 300 - 100 = 300.$

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MCQ 221 Mark

For any set A, (A')' is equal to:

A
A'
✓
A
C
$\phi$
D
None of these.

Answer

Correct option: B.

The complement of the complement of a set is the set itself.

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MCQ 231 Mark

Suppose $A_1, A_2, ..., A_{30}$ are thirty sets each having $5$ elements and $B_1, B_2, ..., B_n$ are $n$ sets each with $3$ elements. Let $\bigcup\limits^{30}_\text{i = 1}\text{A}_\text{i}=\bigcup\limits^{\text{n}}_\text{j = 1}\text{B}_\text{j}=\text{S}$ and each element of $S$ belong to exactly $10$ of the ${A_i}^{'s}$ and exactly $9$ of the ${B_j}^{'s},$ then $n$ is equal to:

A
$15$
B
$3$
✓
$45$
D
$35.$

Answer

Correct option: C.

$45$

It is given that each set $\text{A}_\text{j}(1\leq\text{i}\leq30)$ contains $5$ elements and $\bigcup\limits^{30}_\text{i = 1}\text{A}_\text{i}=\text{S}.$
$\therefore\text{n(S)}=30\times5=150$
But, it is given that each element of $S$ belong to exactly $10$ of the ${A_i}^{'s}.$
$\therefore$ Number of distinct elements in $\text{S}=\frac{150}{10}=15......(1)$
It is also given that each set $\text{B}_\text{j}(1\leq\text{j}\leq\text{n})$ contains $3$ elements and $\bigcup\limits^{\text{n}}_\text{j = 1}\text{B}_\text{j}=\text{S}.$
$\therefore\text{ n(S)}=\text{n}\times3=\text{3n}$
Also, each element of $S$ belong to eactly $9$ of ${B_j}^{'s}.$
$\therefore$ Number of distinct elements in $\text{S}=\frac{\text{3n}}{9}......(2)$
From $(1)$ and $(2)$, we have
$\frac{\text{3n}}{9}=15$
$\Rightarrow\text{n} = 45.$
Hence, the correct answer is option $(c).$

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MCQ 241 Mark

In a class of 175 students the following data shows the number of students opting one or more subjects. Mathematics 100; Physics 70; Chemistry 40; Mathematics and Physics 30; Mathematics and Chemistry 28; Physics and Chemistry 23; Mathematics, Physics and Chemistry 18. How many students have offered Mathematics alone?

A
35
B
48
✓
60
D
22.

Answer

Correct option: C.

Let M, P and C denote the sets of students who have opted for mathematics, physics, and chemistry, respectively.
Here,
$\text{n(M)}= 100, \text{ n( P)} = 70, \text{ n(C)} = 40$
Now,
$\text{n(M}\cap\text{P)}=30,\text{n(M}\cap\text{C)}=28,\\\text{n(P}\cap\text{C)}=23,\text{n(M}\cap\text{P}\cap\text{C)}=18$
Number of students who opted for only mathematics:
$\text{n(M}\cap\text{P}'\cap\text{C)}'=\{\text{M}\cap\text{(P}\cap\text{C})'\}$
$=\text{n(M)}-\text{n}\{\text{M}\cap\text{(P}\cap\text{C})\}$
$=\text{n(M)}-\text{n}\{\text{(M}\cap\text{P)}\cup\text{(M}\cap\text{C})\}$
$=\text{n(M)}-\{\text{n(M}\cap\text{P)}+\text{n(M}\cap\text{C})-\text{n(M}\cap\text{P}\cap\text{C}\}$
$=100-(30+28-18)$
$=60$
$\therefore$ the number of students who opted for mathematics alone is 60.

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MCQ 251 Mark

The number of subsets of a set containing $n$ elements is:

A
$n$
B
$2^n - 1$
C
$n^2$
✓
$2^n.$

Answer

Correct option: D.

$2^n.$

The total number of subsets of a finite set consisting of $n$ elements is $2^n.$

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MCQ 261 Mark

For any three sets $A, B$ and $C:$

A
$\text{A}\cap\text{(B} -\text{C)}=\text{(A}\cap\text{B)} - \text{(A}\cap\text{C)}$
B
$\text{A}\cap\text{(B} -\text{C)}=\text{(A}\cap\text{B)}- \text{C}$
C
$\text{A}\cup\text{(B} - \text{C)}=\text{(A}\cup\text{B)}\cap\text{(A}\cup\text{C}')$
✓
All of above

Answer

Correct option: D.

All of above

Let $x$ be any arbitrary element of $\text{A}\cap\text{B}-\text{C.}$
Thus, we have,
$\text{x}\in\text{A}\cap\text{(B - C)}\Rightarrow\text{x}\in\text{A}$ and $\text{x}\in\text{B}-\text{C}$
$\Rightarrow\text{x}\in\text{A}$ and $\text{(x}\in\text{B}$ and $\text{x}\not\in\text{C)}$
$\Rightarrow\text{x}\in\text{A}$ and $\text{x}\in\text{B}$ and $\Rightarrow\text{X}\in\text{A}$ and $\text{x}\not\in\text{C}$
$\Rightarrow\text{x(A}\cap\text{B)}$ and $\text{x}\not\in\text{(A}\cap\text{C)}$
$\Rightarrow\text{x}\in[\text{(A}\cap\text{B)}-\text{(A}\cap\text{C)}]$
$\Rightarrow\text{A}\cap\text{(B}-\text{C)}\subseteq\text{(A}\cap\text{B)} - \text{(A}\cap\text{C)}$
Similarly, $\text{(A}\cap\text{B)}-\text{(A} - \text{C)}\subseteq\text{(A}\cap\text{(B}-\text{C)}$
Hence, $\text{A}\cap\text{(B} - \text{C)}=\text{(A}\cap\text{B)} - \text{(A}\cap\text{C)}.$

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MCQ 271 Mark

If A and B are two disjoint sets, then $\text{n(A}\cup\text{B)}$ is equal to:

✓
$\text{n(A) + n(B)}$
B
$\text{n(A) + n(B)} - \text{n(A}\cap\text{B)}$
C
$\text{n(A) + n(B) + n(A}\cap\text{B)}$
D
$\text{n(A) n(B)}.$

Answer

Correct option: A.

$\text{n(A) + n(B)}$

Two sets are disjoint if they do not have a common element in them, i.e., $\text{A}\cap\text{B}=\phi.$
$\therefore\text{n(A}\cup\text{B) = n(A) + n(B)}.$

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MCQ 281 Mark

In a city 20% of the population travels by car 50% travels by bus and 10% travels by both car and bus. Then, persons travelling by car or bus is:

A
80%
B
40%
✓
60%
D
70%.

Answer

Correct option: C.

60%

Suppose C and B represents the population travels by car and bus respectively.
$\text{n(C}\cup\text{B) = n(C) + n(B)} -\text{n(B}\cap\text{C)}$
$=0.20+0.50-0.10$
$=0.6\text{ or }60\%.$

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MCQ 291 Mark

Two finite sets have $m$ and $n$ elements. The number of elements in the power set of first set is $48$ more than the total number of elements in power set of the second set. Then, the values of $m$ and $n$ are:

A
$7, 6$
B
$6, 3$
✓
$7, 4$
D
$3, 7.$

Answer

Correct option: C.

$7, 4$

$\ce{ATQ}:$
$2^m - 1 = 48 + 2^n - 1$
$\Rightarrow 2^m - 2^n = 48$
$\Rightarrow 2^m - 2^n = 2^6 - 2^4$
By comparing we get:
$m = 6$ and $n = 4.$

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MCQ 301 Mark

The set $\text{(A}\cup\text{B}')'\cup\text{B}\cap\text{C}$ is equal to:

A
$\text{A}'\cup\text{B}\cup\text{C}$
✓
$\text{A}'\cup\text{B}$
C
$\text{A}'\cup\text{C}'$
D
$\text{A}'\cap\text{B}.$

Answer

Correct option: B.

$\text{A}'\cup\text{B}$

$\text{(A}\cup\text{B}')'\cup\text{(B}\cap\text{C})$
$=[\text{A}\cap\text{(B}')']\cup\text{(B}\cap\text{C})$ (De Morgen law)
$=\text{(A}'\cap\text{B})\cup\text{(B}\cap\text{C})$
$=\text{(A}'\cup\text{C})\cup\text{B}$ (Distributive law)
Disclimer: The question seems to be incorrect or there is some printing mistake in the question. The options given in the question does not match with the answer.

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Maths M.C.Q (1 Marks)

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