MCQ 11 Mark
If $A = (6, 7, 8, 9), B = (4, 6, 8, 10)$ and $C = \{x : x \in N : 2 < x ≤ 7\} ;$ find $: B − C$
- A
$\{4, 6\}$
- B
$\{4, 6, 8\}$
- C
$\{6, 8, 10\}$
- ✓
$\{8, 10\}$
AnswerCorrect option: D. $\{8, 10\}$
$C = \{3, 4, 5, 6, 7\}$
$B − C = \{4, 8, 10\}$
View full question & answer→MCQ 21 Mark
How many elements has $P(A),$ if $A = f ?$
View full question & answer→MCQ 31 Mark
In a class of $50$ students, $10$ did not opt for math, $15$ did not opt for science and $2$ did not opt for either. How many students of the class opted for both math and science.
AnswerTotal students $= 50$
Students who did not opt for math $= 10$
Students who did not opt for Science $= 15$
Students who did not opt for either maths or science $= 2$
Total of $40$ students in math and $13$ did not opt for science but did for math $= 40 - 13 = 27$
So, students of the class opted for both math and science is $27$
View full question & answer→MCQ 41 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.
View full question & answer→MCQ 51 Mark
The cardinality of the set $P(P(P(f)))$ is.
View full question & answer→MCQ 61 Mark
There are $19 $hockey players in a club. On a particular day $14$ were wearing the prescribed hockey shirts, while $11$ were wearing the prescribed hockey pants. None of then was without hockey pant or hockey shirt. How many of them were in complete hockey uniform?
AnswerWe can look at it in $2$ ways
First by set theory
$n(A ∩ B) = n(A) + n(B) − n(A ∪ B)$
$= 14 + 11 − 19$
$= 6$
Qualitatively, we know that $14$ people are wearing prescribed hockey shirts,which leaves us with $5$ players who must be wearing hockey pants.
So out of $11$ players who are wearing hockey pants,
$5$ are not wearing hockey shirts while the other $6$ are in complete uniform.
View full question & answer→MCQ 71 Mark
Choose the correct answers from the given four option: A survey shows that $63\%$ of the people watch a News Channel whereas $76\%$ watch another channel. If $x%$ of the people watch both channel, then
AnswerCorrect option: C. $39 \leq \text{x} \leq 63$
Let $p \%$ of the people watch a channel and $q \%$ of the people watch another channel
$\because \text{n}(\text{p}\cap\text{q})=\text{x}\% $ and $n(\text{p}\cup\text{q})\leq100$
So, $\text{n}(\text{p}\cap\text{q})\geq\text{n(p)}+\text{n(q)}-\text{n}(\text{p}\cap\text{q})$
$100\geq63+76-\text{x}$
$100\geq139-\text{x}$
$\Rightarrow\text{x}\geq139-100$
$\Rightarrow \text{x}\geq39$
Now $n(p) = 63$
$\therefore \text{n}(\text{p}\cap\text{q})\leq\text{n(p)}$
$\Rightarrow \text{x}\geq63$
So $39\leq\text{x}\geq63.$
Hence, the correcr option is $(c).$
View full question & answer→MCQ 81 Mark
Which of the following collections are sets?
AnswerCorrect option: A. The collection of all the days of a week
View full question & answer→MCQ 91 Mark
In a certain group of $36$ people, $18$ are wearing hats and $24$ are wearing sweaters. If six people are wearing neither a hat nor a sweater, then how many people are wearing both a hat and a sweater?
AnswerSince $6$ people are wearing neither hat nor sweater
$\ce{n(H ∪ S)} = 36 − 6 = 30$
By set theory
$\ce{n(H ∩ S) = n(H) + n(S) − n(H ∪ S)}$
$= 18 + 24 − 30$ $
$$= 12$
View full question & answer→MCQ 101 Mark
In an examination $80\%$ passed in English, $85\%$ in Maths, $75\%$ in both and $40$ students failed in both subjects. Then the number of students appeared are:
Answer$\text{n(E)}=80$
$\text{n(M)}=85$
$\text{n(E}\cap\text{M})=75$
$\text{n(E}\cup\text{M})=\text{n(E)}+\text{n(M)}-\text{n(E}\cap\text{m})$
$=80+85-75=90$
$\text{n(E}\cup\text{M})'=10$
Let $n$ be the total number of students appeared
$\frac{10}{100}\times\text{n}=40$
$\therefore\text{n}=400$
View full question & answer→MCQ 111 Mark
The equation $\text{x}+ \cos \text{x = a}$ has exactly one positive root. Complete set of values of $'a\ '$ is:
- A
$(0,1)$
- B
$(-\infty,1)$
- C
$(-1,1)$
- ✓
$(1,\infty)$
AnswerCorrect option: D. $(1,\infty)$
Let $\text{f(x) = x} + \cos \text{x a}$
$\Rightarrow \text{f}'(\text{x})=1-\sin\geq0\forall\text{x}\in\text{R}.$
Thus $f(x)$ is increasing in $(-\infty,\infty),$ as zero of $f'(x)$ don't for an interval. $f(0) = 1a$
For a positive root, $1 - a < 0$
$\Rightarrow a > 1$
View full question & answer→MCQ 121 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
$3, 7.$
AnswerCorrect option: C. $7, 4$
$\text{ATQ:}$
$ 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 $n = 4.$
View full question & answer→MCQ 131 Mark
Choose the correct answers from the given four option:
If sets $A$ and $B$ are defined as $\text{A}=\Big\{(\text{x},\text{y})|\text{y}=\frac{1}{\text{x}},0\neq\text{x}\in\text{R}\Big\}\ \text{B}=\{(\text{x},\text{y})|\text{y}=-\text{x},\text{x}\in\text{R}\},$ then
- A
$\text{A}\cap\text{B}=\text{A}$
- B
$\text{A}\cap\text{B}=\text{B}$
- ✓
$\text{A}\cap\text{B}=\phi$
- D
$\text{A}\cup\text{B}=\text{A}$
AnswerCorrect option: C. $\text{A}\cap\text{B}=\phi$
Given that: $\text{A}=\Big\{(\text{x},\text{y}|\text{y}=\frac{1}{\text{x}},0\neq\text{x}\in\text{R}\Big\},$
and $\text{B}=\big\{(\text{x},\text{y}|\text{y}=\text{x},\text{x}\in\text{R}\big\}$
It is very clear that $\text{y}=\frac{1}{\text{x}}$ and $y = -x$
$\because \frac{1}{\text{x}}\neq -\text{x}$
$\therefore \text{A}\cap\text{B}=\phi$
Hence, the correct option is $(c).$
View full question & answer→MCQ 141 Mark
If ${\text{A}\cup\text{ B}^1}$then $\text{n}(\text{A }\cup\text{ B})=?$
- ✓
$\text{n(A)} + \text{n(B)} - \text{n}(\text{A}\cap\text{B})$
- B
$\text{n(A)} - \text{n(B)} + \text{n}(\text{A}\cap\text{B})$
- C
$\text{n(A)} - \text{n(B)} - \text{n}(\text{A}\cap\text{B})$
- D
$\text{n(A)} + \text{n(B)} + \text{n}(\text{A}\cap\text{B})$
AnswerCorrect option: A. $\text{n(A)} + \text{n(B)} - \text{n}(\text{A}\cap\text{B})$
View full question & answer→MCQ 151 Mark
Choose the correct answers from the given four option:
In a class of $60$ students, $25$ students play cricket and $20$ students play tennis, and $10$ students play both the games. Then, the number of students who play neither is.
AnswerTotal number of students $= 60$
Number of students who play cricket $= 25$
Number of students who play tennis $= 20$
Number of students who play cricket and tennis both $= 10$
$\Rightarrow \text{n}(\text{C}\cap\text{T})=10$
$\therefore \text{n}(\text{C}\cap\text{T}) =\text{n(C)}+\text{n(T)}-\text{n}(\text{C}\cap\text{T})$
$=25+20-10=45-10=35$
$\therefore \text{n}(\text{C}'\cap\text{T}') = \text{n(U)}-\text{n}(\text{C}\cap\text{T})$
$=60-35=25$
Hence, the correct option is $(b).$
View full question & answer→MCQ 161 Mark
A market research group conducted a survey of $1000$ consumers and reported that $720$ consumers like product $A$ and $420$ consumers like product $B.$Then, the least number of consumers that must have liked both the products is:
AnswerTotal consumers $= 1000$
Like product $A = n(A) = 720$
Like product $B = n(B) 420$
$n (A ∩ B) ($Both the products$) = n(A) + n(B) − n(A ∪ B)$
$= 720 + 420 − 1000$
$= 140$
View full question & answer→MCQ 171 Mark
If $A = \{2, 4, 6, 8, 10\}, B = \{1, 3, 5, 7, 9\},$ then $A - B =............$
- A
$\{\}$
- ✓
$\{2, 4, 6, 8, 10\}$
- C
$\{1, 3, 5, 7, 9\}$
- D
AnswerCorrect option: B. $\{2, 4, 6, 8, 10\}$
$A = \{2, 4, 6, 8, 10\}$
$B = \{1, 3, 5, 7, 9\}$
$A - B =\{2, 4, 6, 8, 10\} - \{1, 3, 5, 7, 9\}$
$= \{2, 4, 6, 8, 10\}$
View full question & answer→MCQ 181 Mark
IF $R = \{(2, 1), (4, 3), (4, 5)\},$ then range of the function is?
- A
Range $R = \{2, 4\}$
- ✓
Range $R = \{1, 3, 5\}$
- C
Range $R = \{2, 3, 4, 5\}$
- D
Range $R= \{1, 1, 4, 5\}$
AnswerCorrect option: B. Range $R = \{1, 3, 5\}$
Given $R = \{(2, 1), (4, 3), (4, 5)\}$
then Range $(R) = \{1, 3, 5\}$
View full question & answer→MCQ 191 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 201 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 211 Mark
For any three sets A, B and C:
- ✓
$\text{A}\cap\text{(B} -\text{C)}=\text{(A}\cap\text{B)} - \text{(A}\cap\text{C)}$
- B
$\text{A}\cap\text{(B} -\text{C)}=\text{(A}\cap\text{B)}- \text{C}$
- C
$\text{A}\cup\text{(B} - \text{C)}=\text{(A}\cup\text{B)}\cap\text{(A}\cup\text{C}')$
- D
$\text{A}\cup\text{(B} - \text{C)}=\text{(A}\cup\text{B)}-\text{(A}\cup\text{C}).$
AnswerCorrect option: A. $\text{A}\cap\text{(B} -\text{C)}=\text{(A}\cap\text{B)} - \text{(A}\cap\text{C)}$
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 $\text{(x}\in\text{B and x}\not\in\text{C)}$
$\Rightarrow\text{x}\in\text{A and x}\in\text{B}$ and $\Rightarrow\text{X}\in\text{A and x}\not\in\text{C}$
$\Rightarrow\text{x(A}\cap\text{B)}$ 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, $\text{A}\cap\text{(B} - \text{C)}=\text{(A}\cap\text{B)} - \text{(A}\cap\text{C)}.$
View full question & answer→MCQ 221 Mark
Out of $800$ boys in a school $224$ played cricket, $240$ played hockey and $236$ played basketball. Of the total $64$ played both basketball and hockey, $80$ played cricket and basketball and $40$ played cricket and hockey, $24$ players all the three games. The number of boys who did not play any game is:
AnswerNo. of players who played at least one game is:
By set theory
$\ce{n(C ∪ H ∪ B) = n(C) + n(H) + n(B) − n(B ∩ H) − n(C ∩ B) − n(C ∩ H) + n(C ∩ H ∩ B)}$
$= 224 + 240 + 236 − 64 − 80 − 40 + 24$
$= 540$
View full question & answer→MCQ 231 Mark
The number of subsets of a set containing $n$ elements is:
- A
$n$
- B
$2^n - 1$
- C
$n^2$
- ✓
$2^n$
AnswerThe total number of subsets of a finite set consisting of $n$ elements is $2^n$
View full question & answer→MCQ 241 Mark
In a class of $80$ children, $35\%$ children can play only cricket, $45\%$ children can play only table$-$tennis and the remaining children can play both the games. In all, how many children can play cricket?
AnswerClearly $35\%$ children can play cricket.
Also $20\%$ can play both.
So $55\%$ children can play cricket
Total no. of kids $= 0.55 \times 80 = 44$
View full question & answer→MCQ 251 Mark
Choose the correct answers from the given four option:
If $X$ and $Y$ are two sets and $X′$ denotes the complement of $X$, then $\text{X}\cap(\text{X}\cup\text{Y})'$ is equal to.
- A
$\text{X}.$
- B
$\text{Y}.$
- ✓
$\phi.$
- D
$\text{X}\cap\text{Y}.$
AnswerCorrect option: C. $\phi.$
Let $\text{x}\in\text{X}\cap(\text{X}\cup\text{Y})'$
$\Rightarrow \text{x}\in\text{X}\cap(\text{X}'\cup\text{Y})'$
$\Rightarrow \text{x}\in(\text{X}\cap\text{X})\cap(\text{X}\cap\text{Y}')$
$\Rightarrow \text{x}\in\phi\cap(\text{x}\cap\text{Y}')\ \big[\because \text{A}\cap\text{A}'=\phi\big]$
$\Rightarrow \text{x}\in\phi$
Hence, the correct option is $(c).$
View full question & answer→MCQ 261 Mark
The set of all those elements of $A$ and $B$ which are common to both is called:
AnswerThe set of all those elements of $A$ and $B$ which are common to both is called $A$ intersection $B = A ∩ B.$
View full question & answer→MCQ 271 Mark
Which of the following properties are associative law?
- A
$\text{A } \cup \text{ B}=\text{B } \cup \text{ A}$
- B
$\text{A } \cup \text{ C}=\text{C } \cup \text{ A}$
- C
$\text{A } \cup \text{ D}=\text{D } \cup \text{ A}$
- ✓
$(\text{A } \cup \text{ B})\cup\text{C}=\text{A} \cup (\text{B }\cup\text{ C})$
AnswerCorrect option: D. $(\text{A } \cup \text{ B})\cup\text{C}=\text{A} \cup (\text{B }\cup\text{ C})$
View full question & answer→MCQ 281 Mark
Which one is different from the others?
MCQ 291 Mark
Consider the following equations:
$1.\ \text{A - B}=\text{A}-(\text{A }\cap \text{ B})$
$2.\ \text{A}=({\text{A }\cap \text{ B}})\cup(\text{A }-\text{ B})$
$3.\ \text{A}-(\text{B }\cup\text{ C})=(\text{A - B})\cup(\text{A - C})$
Which of these is/are correct?
- A
$1$ and $3$
- B
$2$ only
- C
$2$ and $3$
- ✓
$1$ and $2$
AnswerCorrect option: D. $1$ and $2$
View full question & answer→MCQ 301 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 311 Mark
The solution set of $3x − 4 < 8$ over the set of non$-$negative square numbers is:
- A
$\{1, 2, 3\}$
- B
$\{1,4\}$
- ✓
$\{1\}$
- D
$\{16\}$
AnswerCorrect option: C. $\{1\}$
$3x − 4 < 8$
$3x < 12$
$x < 4$
Hence set of non$-$negative square numbers belonging to the above set is $\{1\}.$
View full question & answer→MCQ 321 Mark
Let $n(A) = 28,n(A ∩ B) =8, n(A∪B) =52,$ then $n(A ∩ B′):$
AnswerGiven $n(A) = 28, n(A ∩ B) = 8.$
We have $A ∩ B′ = A - A ∩ B.$
This give $n(A ∩ B′) = n(A) -n(A ∩ B)$
or, $n(A ∩ B′) = 28 - 8 = 20.$
View full question & answer→MCQ 331 Mark
Let $A$ and $B$ be two sets such that $n(A) = 16, n(B) = 12,$ and $n(A ∩ B) = 8.$Then $n(A∪B)$ equals:
Answer$n(A ∪ B) = n(A) + n(B) − n(A ∩ B)$
$= 16 + 12 − 8$
$= 20$
View full question & answer→MCQ 341 Mark
If $A$ and $B$ are finite sets, then which one of the following is the correct equation?
- A
$\ce{n (A - B) = n (A) - n (B)}$
- B
$\ce{n (A - B) = n (B - A)}$
- C
$\ce{n (A - B) = n (A) - n (\text{A }\cup \text{ B})}$
- ✓
$\ce{n (A - B) = n (B) - n (A\cap B)}$
AnswerCorrect option: D. $\ce{n (A - B) = n (B) - n (A\cap B)}$
View full question & answer→MCQ 351 Mark
Choose the correct answers from the given four option:
In a town of $840$ persons, $450$ persons read Hindi, $300$ read English and $200$ read both. Then the number of persons who read neither is,
AnswerLet $H$ be the set of persons who read Hindi and $E$ be the ser of persons who read English.
Then, $\ce{n(U) = 840, n(H) = 450, n(E) 300, n(H \cap E)}=200$
Number of persons who read neither $=\text{n}(\text{H}'\cap\text{F}')$
$=\text{n}(\text{H}\cup\text{E})'=\text{n(U)}-\text{n}(\text{H}\cup\text{E})$
$=840-\big[\text{n(H)}+\text{n(E)}-\text{n}(\text{H}\cap\text{E})\big]$
$=840-(450+300-200)$
$=290$
View full question & answer→MCQ 361 Mark
If $A$ and $B$ are two sets such that $\ce{n(A) = 17, n(B) = 23, n(A ∪ B)} = 38,$ find $\ce{n(A ∩ B):}$
AnswerWe know,
$\ce{n(A ∩ B) = n(A) + n(B) - n(A ∪ B)}$
$\ce{n(A ∩ B)} = 17 + 23 - 38$
$= 2$
View full question & answer→MCQ 371 Mark
If $A = \{1, 2, 3, 4\},$ what is the number of subsets of $A$ with at least three elements?
AnswerA subset containing $3$ elements $= \{1, 2, 3\};\{1, 3, 4\};\{1, 2, 4\}$ and ${2, 3, 4}$
A subset containing $4$ elements $=\{1, 2, 3, 4\}$
$\therefore$ there are five subsets containing at least $3$ elements.
View full question & answer→MCQ 381 Mark
Choose the correct answers from the given four option:
The set $(\text{A} \cap \text{B}')' \cup (\text{B} \cap \text{C})$ is equal to.
AnswerCorrect option: B. $\text{A}'\cup\text{B}$
We konw that: $(\text{A}\cap\text{B})'=\text{A}'\cup\text{B}' [$De Morgan's law$]$
$\therefore (\text{A}\cap\text{B}')'\cup(\text{B}\cap\text{C})=\big[\text{A}'\cup(\text{B}')\big]\cup(\text{B}\cap\text{C})$
$=(\text{A}'\cap\text{B})\cup(\text{B}\cap\text{C})\big[\because (\text{B}')'=\text{B}\big]$
$=\text{A}'\cup\text{B}$
Hence, the correct optiom is $(b).$
View full question & answer→MCQ 391 Mark
Choose the correct answers from the given four option: Two finite sets have $m$ and $n$ elements. The number of subsets of the first set is $112$ more than that of the second set. The values of $m$ and $n$ are, respectively,
- A
$4, 7$
- ✓
$7, 4$
- C
$4, 4$
- D
$7, 7$
AnswerCorrect option: B. $7, 4$
According to the question,
$ \Rightarrow 2^m-2^n=12 $
$ \Rightarrow 2^n\left(2^{m-n}-1\right) 2^4 \cdot 7 $
$ \Rightarrow 2 n=2^4$ and $2^{m-n}-1=7 $
$ \Rightarrow n=2$ and $2^{m-n}=8 $
$ \Rightarrow 2^{m-n}=2^3$
$\Rightarrow m-n=3$
$\Rightarrow m-4=3$
$\Rightarrow m=7 $
View full question & answer→MCQ 401 Mark
In a class $60\%$ of the students were boys and $30\%$ of them had $I$ class. If $50\%$ of the students in the class had $I$ class, find the fraction of the girls in the class who did not have a $I$ class:
- ✓
$\frac{1}{5}$
- B
$\frac{4}{5}$
- C
$\frac{1}{4}$
- D
$\frac{1}{3}$
AnswerCorrect option: A. $\frac{1}{5}$
View full question & answer→MCQ 411 Mark
The range of the function $f(x) = 3x - 2‚$ is.
- ✓
$(-\infty,\infty)$
- B
$\text{R}-(3)$
- C
$(-\infty,0)$
- D
$(0,- \infty)$
AnswerCorrect option: A. $(-\infty,\infty)$
Let the given function is
$y = 3x - 2$
$\Rightarrow y + 2 = 3x$
$\Rightarrow \text{x} =\frac{(\text{y}+2)}{3}$
Now $x$ is satisfied by all values.
So, Range ${f(x)} = R =(-\infty,\infty)$
View full question & answer→MCQ 421 Mark
$\{ (A, B) : A^2 + B^2 = 1\}$ on the sets has the following relation.
AnswerGiven ${(a, b) : a^2 + b^2 = 1}$ on the set $S.$
Now $a^2 + b^2 = b^2 + a^2 = 1$
So, the given relation is symmetric.
View full question & answer→MCQ 431 Mark
While preparing the progress reports of the students, the class teacher found that $70\%$ of the students passed in Hindi, $80\%$ passed in English and only $65\%$ passed in both the subjects. Find out the percentage of students who failed in both the subjects.
- ✓
$15\%$
- B
$20\%$
- C
$30\%$
- D
$35\%$
AnswerCorrect option: A. $15\%$
The sets $E$ and $H$ represent the students failing in the respective subjects.
$\ce{n(H ∪ E)} = 1− 0.65 = 0.35$
By set theory
$\ce{n(H ∩ E) = n(H) + n(E) − n(N ∪ E)}$
$= 0.3 + 0.2 − 0.35 = 0.15$
Hence $15\%$ of students failed in both subjects.
View full question & answer→MCQ 441 Mark
Let $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}.$ Then the number of subsets of $A$ containing exactly two elements is:
AnswerNumber of elements in $A = 10$
Number of subsets of $A$ containing exactly two elements
$=$ Number of ways we can select $2$ elements from $10$ elements
$10\text{c}_2=\frac{10\times9}{2}=45$
$\therefore$ Number of subsets of $A$ containing exactly two elements $= 45$
View full question & answer→MCQ 451 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 $(c).$
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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 471 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$\text{n(A}'\cap\text{B}')=\text{n(A}\cup\text{B}')$
$=\text{n(U)}-\text{n(A}\cup\text{B})$
$=700 - 200 + 300 - 100$
$= 300.$
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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 491 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 501 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 $A = 25 - 1 = 31.$
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