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 :
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 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
$[4, 5].$
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 31 Mark
For any two sets $A$ and $B, \text{A}\cap\text{(A}\cup\text{B)}=$
Answer$\text{A}\cap\text{(A}\cup\text{B)}=\text{(A}\cap\text{A)}\cup\text{(A}\cap\text{B)}=\text{A}\cup\text{(A}\cap\text{B)}=$
$\text{AA}\cap\text{(A}\cup\text{B)}=\text{(A}\cap\text{A)}\cup\text{(A}\cap\text{B)}=\text{A}\cup\text{(A}\cap\text{B)}=\text{A.}$
View full question & answer→MCQ 41 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 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 :
AnswerWe know:
$\text{n(A}\cup\text{B) = n(A) + n(B)} - \text{n(A}\cap\text{B)}$
Now,
$\text{n(A}\cap\text{B) = n(A) + n(B)} -\text{n(A}\cup\text{B)}$
$=16+14-25$
$=5.$
View full question & answer→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
$\text{A} - \text{B = (A}\cup\text{B)}-\text{B.}$
AnswerCorrect 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.
View full question & answer→MCQ 71 Mark
If $\text{A}\cap\text{B}=\text{B},$ then :
AnswerCorrect option: B. $\text{B}\subset\text{A}$
Only this case is possible.
View full question & answer→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
AnswerCorrect option: C. $\text{A}\cap\text{B}'$
$A - B $ belongs to those elements of $A$ that do not belong to $B$.
$\therefore\text{A} - \text{B = A}\cap\text{B}'.$
View full question & answer→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
$7, 7.$
AnswerCorrect 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 $(b).$
View full question & answer→MCQ 101 Mark
For two sets $\text{A}\cap\text{B = A}$ iff:
AnswerCorrect 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 $\text{B}\subseteq\text{A}.$
View full question & answer→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
$\text{A}\cap\text{B}.$
AnswerCorrect 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 $(c)$.
View full question & answer→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
$\text{(A}\cup\text{B)}\cap\text{(A}\cap\text{B)}.$
AnswerCorrect 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$]$
$=[\text{(A}\cup\text{B)}\cup\text{(U)}]\cap[\text{(U)}\cap\text{(B}'\cup\text{A}')]$
$[\text{A}\cup\text{A'= U = B}'\cup\text{B}]$
$=[\text{A}\cup\text{B}]\cap[\text{B}'\cup\text{A}']$
$=[\text{A}\cup\text{B}]\cap[\text{B}'\cup\text{A}']$ $\begin{bmatrix}\text{(A}\cup\text{B)}\cap\text{(U)}=\text{(A}\cup\text{B)}\\\text{ and (U)}\cap\text{(B}'\cup\text{A)}'=\text{(B}'\cup\text{A}')]\end{bmatrix}$
$=[\text{A}\cup\text{B}]\cap[\text{(A}\cap\text{B)}']$
$[\text{(A}\cap\text{B)}'=\text{B}'\cup\text{A}']$
$=[\text{A}\cup\text{B}]\cap[\text{(A}\cup\text{B)}-\text{(A}\cap\text{B)}]$
$=[\text{(A}\cup\text{B)}-\text{(A}\cap\text{B)}].$
View full question & answer→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
$\{\text{(A}\cup\text{B)}-\text{A\}}\cup\{\text{(A}\cup\text{B)} - \text{B}\}.$
AnswerCorrect option: B. $\text{(A} - \text{B)}\cup\text{(B}- \text{A)}$
The symmetric difference of $A$ and $B$ is given by $:-$
$\text{(A} - \text{B)}\cup\text{(B}- \text{A)}.$
View full question & answer→MCQ 141 Mark
In set $-$ builder method the null set is represented by :
AnswerCorrect option: C. $\{\text{x : x} \not=\text{x}\}$
$\{\text{x : x}\not=\text{x}\}.$
View full question & answer→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
$\text{F}_2\cup\text{F}_3$
- ✓
$\text{F}_2\cup\text{F}_3\cup\text{F}_4\cup\text{F}_1.$
AnswerCorrect 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 $(d).$
View full question & answer→MCQ 161 Mark
The symmetric difference of $A = \{1, 2, 3\}$ and $B = \{3, 4, 5\}$ is :
AnswerHere,
$\text{A} = \{1, 2, 3\}$ 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:
$\text{(A} - \text{B)}= \{1, 2\}$
$\text{(B} - \text{A)}=\{4, 5\}$
$\text{(A}-\text{B)}\cup\text{(B}-\text{A)}=\{1, 2, 4, 5\}.$
View full question & answer→MCQ 171 Mark
If $A = \{x : x \ \text{is a multiple of} \ 3\}$ and $B = \{x : x \text{ is a multiple of} \ 5\},$ 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
$\overline{\text{A}\cap{\text{B}}}.$
AnswerCorrect option: B. $\text{A}\cap\overline{\text{B}}$
$A = \{x : x\ \text{ is a multiple of} \ 3\}$
$A = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, .....$
$B = \{x : x\ \text{ is a multiple of}\ 5\}$
$B = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, ......$
Now, we have:
$A - B = 3, 6, 9, 12, 18, 21, 24, 27, 30, 33,36, 39, 42, ....$
$=\text{A}\cap\overline{\text{B}}.$
View full question & answer→MCQ 181 Mark
If $A = \{1, 2, 3, 4, 5\},$ then the number of proper subsets of $A$ is :
AnswerThe 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 $A = 25 - 1 = 31.$
View full question & answer→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 :
Answersolve for None :
$80 \ +$ None $= 100$
None $= 20$.
View full question & answer→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 :
AnswerCorrect 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,
$\text{(A}\cap\text{A}\cap\text{B)}^\text{c}=\text{(A}\cap\text{B)}^\text{c}.$
View full question & answer→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, $\text{n(A}'\cap\text{B}')=$
Answer$\text{n(A}'\cap\text{B}')=\text{n(A}\cup\text{B}')$
$=\text{n(U)}-\text{n(A}\cup\text{B})$
$=700 - 200 + 300 - 100 = 300.$
View full question & answer→MCQ 221 Mark
For any set $A, (A')'$ is equal to :
AnswerThe complement of the complement of a set is the set itself.
View full question & answer→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:
AnswerIt 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 $(c).$
View full question & answer→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?
AnswerLet $M, P$ 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 $60$.
View full question & answer→MCQ 251 Mark
The number of subsets of a set containing $n$ elements is$:$
- A
$n$
- B
$2^n - 1$
- C
$n^2$
- ✓
$2^n.$
AnswerCorrect option: D. $2^n.$
The total number of subsets of a finite set consisting of $n$ elements is $2^n$.
View full question & answer→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
$\text{A}\cup\text{(B} - \text{C)}=\text{(A}\cup\text{B)}\cap\text{(A}\cup\text{C}')$
- ✓
$\text{A}\cup\text{(B} - \text{C)}=\text{(A}\cup\text{B)}-\text{(A}\cup\text{C}).$
AnswerCorrect option: D. $\text{A}\cup\text{(B} - \text{C)}=\text{(A}\cup\text{B)}-\text{(A}\cup\text{C}).$
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 $\text{(x}\in {B}$ and ${x}\not\in\text{C)}$
$\Rightarrow\text{x}\in {A}$ and ${x}\in\text{B}$ and
$\Rightarrow\text{X}\in {A}$ and ${x}\not\in\text{C}$
$\Rightarrow\text{x(A}\cap\text{B)}$ 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, $\text{A}\cup\text{(B} - \text{C)}=\text{(A}\cup\text{B)}-\text{(A}\cup\text{C}).$
View full question & answer→MCQ 271 Mark
If $A$ and $B$ are two disjoint sets, then $\text{n(A}\cup\text{B)}$ is equal to :
AnswerCorrect option: A. $\text{n(A) + n(B)}$
Two sets are disjoint if they do not have a common element in them,
i.e., $\text{A}\cap\text{B}=\phi.$
$\therefore\text{n(A}\cup\text{B) = n(A) + n(B)}.$
View full question & answer→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
$70\%.$
AnswerCorrect option: C. $60\%$
Suppose $C$ and $B$ represents the population travels by car and bus respectively.
$\text{n(C}\cup\text{B) = n(C) + n(B)} -\text{n(B}\cap\text{C)}$
$=0.20+0.50-0.10$
$=0.6$ or $60\%.$
View full question & answer→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
$7, 6$
- ✓
$6, 4$
- C
$7, 4$
- D
$3, 7.$
AnswerCorrect option: B. $6, 4$
$2^m - 1 = 48 + 2^n - 1$
$\Rightarrow 2^m - 2^n = 48$
$\Rightarrow 2^m - 2^n = 2^6 - 2^4$
By comparing we get:
$m = 6$ and $n = 4$.
View full question & answer→MCQ 301 Mark
The set $\text{(A}\cup\text{B}')'\cup\text{B}\cap\text{C}$ is equal to :
AnswerCorrect 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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