MCQ

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

7, 4.

Solution:
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.

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

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

Solution:
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 $A_1, A_2, \ldots, A_{30}$ are thirty sets each having 5 elements and $B_1, B_2, \ldots, B_n$ 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 $A_i^{\prime s}$ and exactly 9 of the $B_j{ }^{\text {se }}$, then $n$ is equal to:

A
15
B
3
✓
45
D
35

Answer

Correct option: C.

Solution:
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
$\mathrm{n}^2$
✓
$2^n$.

Answer

Correct option: D.

$2^n$.

$2^n$.

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

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MCQ 261 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 271 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 281 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

6, 4.

Solution:
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 291 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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