MCQ 1511 Mark
If $A, B, C$ be three sets such that $\text{A } \cup \text{ B}=\text{A } \cup \text{ C}$ and $\text{A } \cap \text{ B}=\text{A } \cap \text{ C},$ then,
- ✓
$B = C$
- B
$A = C$
- C
$A = B = C$
- D
$A = B$
AnswerCorrect option: A. $B = C$
Given $A, B, C$ be three sets such that $\text{A } \cup \text{ B}=\text{A } \cup \text{ C}$ and $\text{A } \cap \text{ B}=\text{A } \cap \text{ C},$ then, $B = C$
View full question & answer→MCQ 1521 Mark
In any continuous class interval table $(a - b):$
- ✓
$A$ is included
- B
$B$ is included
- C
- D
AnswerCorrect option: A. $A$ is included
$A$ is included $b$ is included in the next interval.
View full question & answer→MCQ 1531 Mark
If $A − B = ∅,$ then relation between $A$ and $B$ is:
- A
$\text{A }\phi\text{ B}$
- B
$\text{B}\cup\text{A}$
- ✓
$\text{A}\cap\text{B}$
- D
$\text{A} = \text{B}$
AnswerCorrect option: C. $\text{A}\cap\text{B}$
If $A$ and $B$ are disjoint it would mean $A$ is a null set. Otherwise $A$ and $B$ must be equal to $A ∩ B$ at least
View full question & answer→MCQ 1541 Mark
Choose the correct answers from the given four option:
If $A = \{1, 3, 5, 7, 9, 11, 13, 15, 17\} B = \{2, 4, ....., 18\}$ and $N$ the set of natural numbers is the universal set, then $\text{A}' \cup (\text{A} \cup \text{B}) \cup \text{B}')$ is
AnswerGiven that:
$A = \{1, 3, 5, 7, 9, 11, 13, 15, 17\}$
$B = \{2, 4, ...., 18\}$
$U = N = \{1, 2, 3, 4, 5, .....\}$
$\text{A}'\cup(\text{A}\cup\text{B})\cap\text{B}'=\text{A}'\big[(\text{A}\cap\text{B}')\cup(\text{B}\cap\text{B}')\big]$
$=\text{A}'\cup(\text{A}\cap\text{B}')\cup\phi \ \big[\because \text{A}\cap\text{A}'=\phi\big]$
$=\text{A}'\cup(\text{A}\cap\text{B}')$
$=(\text{ A}'\cup\text{A})\cap(\text{A}'\cup\text{B}')$
$=\text{N}\cup(\text{A}'\cup\text{B}')\ \big[\because \text{A}'\cup\text{A}=\text{N}\big]$
$=\text{A}'\cup\text{B}'$
$=(\text{A}\cup\text{B}')=(\phi)'=\text{N} \ \big[\because \text{A}\cap\text{B}=\phi\big]$
Hence, the correct option is $(b).$
View full question & answer→MCQ 1551 Mark
If $A = \{1, 3, 5, B\}$ and $B = \{2, 4\},$ then:
Answer$(4\not\in\text{A) }(4\not\in\text{A})$
$\{4\}\not\subset\text{A}$
$\text{B}\not\subset\text{}A$
Thus, we can say that none of these options satisfy the given relation.
View full question & answer→MCQ 1561 Mark
If $A, B$ and $C$ are any three set, then $\ce{A ∪ (B ∩ C):}$
- A
$\ce{(A ∪ B) ∪ (A ∪ C)}$
- ✓
$\ce{(A ∪ B) ∩ (A ∪ C)}$
- C
$\ce{(A ∩ B) ∩ (A ∩ C)}$
- D
AnswerCorrect option: B. $\ce{(A ∪ B) ∩ (A ∪ C)}$
Using distributive law of sets Or it is the distributive law itself.
View full question & answer→MCQ 1571 Mark
The set of integers is closed with respect to which one of the following?
- A
- B
- ✓
By addition and multiplication
- D
AnswerCorrect option: C. By addition and multiplication
From group theory, integers are closed w.r.t. both addition & multiplication.
View full question & answer→MCQ 1581 Mark
For two sets $A$ and $\ce{B, A ∩ (A ∪ B) =}$
Answer$\ce{(A ∩ A) ∪ (A ∩ B) = A ∪ (A ∩ B) = A}$
$\therefore\text{A}\cap\text{B}\subset\text{A}$
View full question & answer→MCQ 1591 Mark
For any two sets $A$ and $B, \text{A}\cap\text{(A}\cup\text{B)}=$
Answer$\text{A}\cap\text{(A}\cup\text{B)}=\text{(A}\cap\text{A)}\cup\text{(A}\cap\text{B)}=\text{A}\cup\text{(A}\cap\text{B)}=\text{A}$
View full question & answer→MCQ 1601 Mark
Let $S$ be a non$-$empty subset of $R.$ Consider the following statement:
$p :$ There is a rational number $x$ such that $x > 0$.
Which of the following statements is the negation of the statement $P?$
- A
There is no rational number $x\in S$ such that $x \leq 0$
- ✓
Every rational number $x \in S$ satisfies $x \leq 0$
- C
$x \in S$ and $x \leq 0 = x$ is not rational
- D
There is a rational number $x \in S$ such that $x \leq 0$
AnswerCorrect option: B. Every rational number $x \in S$ satisfies $x \leq 0$
$P :$ there is a rational number $x \in S$ such that $x > 0$
$\sim P:$ Every rational number $x \in S$ satisfies $x \leq 0$
View full question & answer→MCQ 1611 Mark
Let $A = \{a, b\}, B = \{a, b, c\}.$ What is $\text{A }\cup\text{ B }?$
- A
$\{a, b\}$
- B
$\{a, c\}$
- ✓
$\{a, b, c\}$
- D
$\{b, c\}$
AnswerCorrect option: C. $\{a, b, c\}$
View full question & answer→MCQ 1621 Mark
Set $A$ has $3$ elements and set $B$ has $6$ elements. What can be the minimum number of elements in $\ce{A ∪ B}\ ?$
Answer$\ce{A ∪ B}$ must contain all the elements of the bigger set.
View full question & answer→MCQ 1631 Mark
If $A = \{a, b, c\},B = \{c, d, e\}, C\{a, d, f\},$ then $\ce{A \times (B ∪ C)}$ is:
- A
$\{(a, d),(a, e),(a, c)\}$
- B
$\{(a, d),(b, d),(c, d)\}$
- C
$\{(d, a),(d, b),(d, c)\}$
- ✓
Answer$A \times (B ∪ C) = \{a, b, c\} \times \{a, c, d, e, f\}.$
The above set will consist of $15$ ordered pairs and not $3.$
View full question & answer→MCQ 1641 Mark
Out of $450$ students in a school, $193$ students read Science Today, $200$ students read Junior Statesman, while $80$ students read neither. How many students read both the magazines?
AnswerSince $80$ do not read any
$\ce{n(S ∪ J)} = 450 − 80 = 370..........(S =$ Science Today; $J =$ Junior Statesman$)$
By set theory
$\ce{n(J ∩ S) = n(J) + n(S) − n(J ∪ S)}$
$= 200 + 193 − 370$
$= 23$
View full question & answer→MCQ 1651 Mark
The members of the set $S = \{x | x$ is the square of an integer and $x < 100\}$ is.
- A
$\{0, 2, 4, 5, 9, 58, 49, 56, 99, 12\}$
- ✓
$\{0, 1, 4, 9, 16, 25, 36, 49, 64, 81\}$
- C
$\{1, 4, 9, 16, 25, 36, 64, 81, 85, 99\}$
- D
$\{0, 1, 4, 9, 16, 25, 36, 49, 64, 121\}$
AnswerCorrect option: B. $\{0, 1, 4, 9, 16, 25, 36, 49, 64, 81\}$
The set $S$ consists of the square of an integer less than $100$
So, $S = \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81\}$
View full question & answer→MCQ 1661 Mark
$(A’)’ = ?$
- A
$\cup-\text{A}$
- B
$\text{A}'$
- C
$\cup$
- ✓
$\text{A}$
AnswerCorrect option: D. $\text{A}$
$(A’)’ = A$
View full question & answer→MCQ 1671 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
$[4, 5].$
AnswerCorrect option: C. $[4, 5)$
$\text{A} = \{\text{x : x} \in \text{R}, \text{x > 4}\}$ and
$\text{B}= \{\text{x}\in\text{R : x} < 5\}$
$\text{A}\cap\text{B}=[4, 5).$
View full question & answer→MCQ 1681 Mark
Given the universal set $B = \{−7, −3, −1, 0, 5, 6, 8, 9\},$ find: $B = \{x: − 4 < x < 6\}$
- A
$\{−7, 0, 5, 6\}$
- B
$\{5, 6, 8, 9\}$
- ✓
$\{−3, −1, 0, 5\}$
- D
$\{0, 5\}$
AnswerCorrect option: C. $\{−3, −1, 0, 5\}$
The only $4$ no.s that lie in the given range are $-3, 0. -1$ and $5.$
View full question & answer→MCQ 1691 Mark
If $A = (6, 7, 8, 9), B = (4, 6, 8, 10)$ and $C = \{x : x \in N : 2 < x ≤ 7\}$ ; find : $B − B$
- ✓
$ϕ$
- B
$\{0\}$
- C
$\{6,7\}$
- D
$\{4\}$
AnswerGiven: $A = (6, 7, 8, 9), B = (4, 6, 8, 10)$ and $C = \{x : x \in N : 2 < x ≤ 7\}$
$B − B$ will always be a null set it will contain elements of $B$ which are not in $B$
i.e. no elements.
So, $B − B = ϕ$
View full question & answer→MCQ 1701 Mark
Which set is the subset of the set containing all the whole numbers?
- A
$\{1, 2, 3, 4, ....\}$
- B
$\{1\}$
- C
$\{0\}$
- ✓
AnswerNull set is the subset of all given sets as it can lie in all sets.
View full question & answer→MCQ 1711 Mark
Which of the following is set?
- ✓
The collection of months having names starting with J.
- B
The collection of smart boys in your class.
- C
The collection of most talented persons.
- D
The collection of sand grains in a Earth.
AnswerCorrect option: A. The collection of months having names starting with J.
As the collection of months having names starting with $J$ is well defined.
So, it's a set. Rest are not well defined , hence are not set.
View full question & answer→MCQ 1721 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:
AnswerWe have:
$\text{n(A}\cap\text{B) = n(A) + n(B)} - \text{n(A}\cup\text{B)}$
$=70+60-110$
$=20.$
View full question & answer→MCQ 1731 Mark
Let $n$ be a fixed positive integer. Let a relation $R$ defined on $I ($the set of all integers$)$ as follows: $aRb$ iff $ \frac{\text{n}}{(\text{a}-\text{b}})$, that is, iff $a − b$ is divisible by $n$, then, the relation $R$ is:
Answer$R$ is reflexive since for any integer a we have $a - a = 0$ and $0$ is divisible by $n.$
Hence, $\text{aRa} \forall a \in I.$
$R$ is symmetric, let $\text{aRb.}$
Then by definition of $R, a - b = nk$ where $k \in I.$
Hence $b - a = (-k)n$ where $-k \in I$ and so $\text{bRa.}$
Thus we have shown that $\text{aRb} \Rightarrow \text{bRa}.$
$R$ is transitive, let $\text{aRb}$ and $\text{bRc.}$
Then by definition of $R$, we have $a - b = k_1n$ and $b - c = k_2n,$ where $k_1, k_2 \in I.$
It then follows that
$a - c = (a - b) + (b - c) = k_1n + k_2n = (k_1 + k_2)n$
where $k_1 + k_2 \in I$
View full question & answer→MCQ 1741 Mark
If $n(A) = 10, n(B) = 6$ and $(C) = 5$ for three disjoint sets $\text{A, B, C}$ then $\text{n(A ∪ B ∪ C)}$ equals:
AnswerSince, $\text{A, B, C}$ are disjoint sets
$\therefore \text{n(A ∪ B ∪ C) = n(A) + n(B) + n(C)}$
$= 10 + 6 + 5$
$= 21$
View full question & answer→MCQ 1751 Mark
The smallest set $A$ such that $A ∪ \{1, 2\} = \{1, 2, 3, 5, 9\}$ is:
- A
$\{2, 3, 5\}$
- ✓
$\{3, 5, 9\}$
- C
$\{1, 2, 5, 9\}$
- D
$\{1, 2\}$
AnswerCorrect option: B. $\{3, 5, 9\}$
$A ∪ \{1, 2\} = \{1, 2, 3, 5, 9\}$
Thus,
$A = \{1, 2, 3, 5, 9\} − \{1, 2\}$
View full question & answer→