MCQ Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 5

How many students in this group are not taking any of the three subjects?

0%
8
0%
9
0%
10
0%
11

Explanation

7 students are taking Physics only.

$7=n(P)-n(P\cap G)-n(P\cap E) +n(P\cap G \cap E)$

$\Rightarrow n(P)=7$

$+ n (P\cap G)+n(P\cap E) -n(P\cap G \cap E)$

$n(P)=7+24+30-16=45$

Similarly

$n(G)= 10+24+22-16=40$

$n(E)=5+22+30-16=41$

$n(P\cup G \cup E)= n(P) +n(G) +n(E)$

$-n(G\cap E)-n(P\cap G)-n(P\cap E) +n(P\cap G \cap E)$

$=45+40+41-22-24-30+16=66$

$75-66=9$ students are not taking any of the 3 subjects.

Let

$\displaystyle A= \left \{ 7,8,9,a,b,c \right \}$ and

$\displaystyle B= \left \{ 1,2,3,4 \right \}$ then number of universal relation from the set

$A$ to set

$B$ and set

$B$ to set

$A$ are

0%
equal in counting
0%
can not be equal in counting
0%
$24$
0%
$\displaystyle 2^{6}\times 2^{4}$

Explanation

Relation

$R$ is said to be universal relation from set

$A$ to set

$B$ or set

$B$ to set

$A$ if

$\displaystyle R= A\times B$

$\displaystyle \therefore$ Number of relations from

$A$ to

$B$ for

$\displaystyle R= A\times B$

$\displaystyle = n(A)\times n(B)= 6\times 4= 24$

Ans: C

If two intersecting circles with two points in common are drawn then how many common chords can be drawn? Draw the figure and write the answer

0%
Only one.
0%
Many.
0%
Three.
0%
None of the options.

Explanation

$We\quad know\quad that\quad the\quad common\quad chord\quad of\quad two\quad intersecting\quad circles\quad is\\ perpendicular\quad to\quad the\quad line\quad joining\quad the\quad centres\quad of\quad the\quad circles.\\ Here\quad are\quad two\quad intersecting\quad circles\quad with\quad centres\quad O\quad \& \quad O'.\\ The\quad common\quad chord\quad is\quad AB\quad which\quad is\quad perpendicular\quad to\quad OO'\quad at\quad P.\\ Now\quad if\quad the\quad circles\quad intersect\quad in\quad more\quad than\quad two\quad points\quad then\\ those\quad common\quad chords\quad will\quad have\quad different\quad incinations\quad with\quad the\quad \\ line\quad OO'\quad resulting\quad in\quad many\quad perpendiculars\quad on\quad OO'\quad with\quad \\ different\quad inclinations.\quad This\quad assumption\quad is\quad impossible.\\ So\quad AB\quad is\quad the\quad only\quad common\quad chord.\\ Ans-\quad Option\quad A.$

Which of the following sets is/are empty?

0%
$\displaystyle \left \{ x : x \in R ,x^{2} -4=0\right \}$
0%
$\displaystyle \left \{ x : x \in R ,x^{4} +4=0\right \}$
0%
$\displaystyle \left \{ x : x \in R ,x^{3} =1\right \}$
0%
$\displaystyle \left \{ x : x \in R ,x^{8}+x^{4}+1=0 \right \}$

Explanation

$x=2,-2$ is a solution

$x^{4}$ is positive always and 4 is also positive.The sum of two positive numbers is always positive.Hence x doesn't have a solution.

$x^{3}$ =1

$\Rightarrow x=1$ .

$x^{8}+x^{4}+1$ is always positive because the individual terms are positive.Hence x here doesn't have a solution.

The set of natural number is subset of set of real numbers.

State true or false:

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

$x-5<0\\ \Rightarrow \quad x\quad <\quad 5\\ Since\quad x\quad is\quad a\quad natural\quad no\\ x\quad =\quad \{ 1,2,3,4\} \\$

$\displaystyle \left \{ \left ( 1, 2 \right ) \right \}, and B=\left \{ 1, 3 \right \}, then\left ( A\times B \right )\cup \left ( B\times A \right )$

$\displaystyle =\left \{ \left ( 1, 3 \right ), \left ( 2, 3 \right ), \left ( 3, 1 \right ), \left ( 3, 2 \right ), \left ( 1, 1 \right ), \left ( 1, 2 \right ), \left ( 2, 1 \right ),\right \}$

0%
True
0%
False

Explanation

$A=\left\{ 1,2 \right\} \\ B=\left\{ 1,3 \right\} \\ A\times B=\left\{ \left( 1,1 \right) ,\left( 1,3 \right) ,\left( 2,1 \right) ,\left( 2,3 \right) \right\} \\ B\times A=\left\{ \left( 1,1 \right) ,\left( 1,2 \right) ,\left( 3,1 \right) ,\left( 3,2 \right) \right\} \\ \therefore \left( A\times B \right) \cup \left( B\times A \right) =\left\{ \left( 1,1 \right) ,\left( 1,3 \right) ,\left( 1,2 \right) ,\left( 2,1 \right) ,\left( 2,3 \right) ,\left( 3,1 \right) \left( 3,2 \right) \right\}$

True or false :

$\displaystyle A\times \left ( B\cap C \right )=\left ( A\times B \right )\cap \left ( A\times C \right )$

0%
True
0%
False

Explanation

$A\times (B\cap C)=(A\times B)\cap (A\times C)$

Let

$(a, b)$ been arbitrary elements of

$A\times (B\cap C)$ , then

$(a, b)\in A\times (B\cap C)$

$\Rightarrow a\in A$ and

$b\in (B\cap C)$

$\Rightarrow a\in A$ and

$(b\in B$ and

$b\in C)$

$\Rightarrow (a\in A$ and

$b\in B)$ and

$(a\in A$ and

$b\in C)$

$\Rightarrow (a, b)\in A\times B$ and

$(a, b)\in A\times C$

$\Rightarrow (a, b)\in (A\times B)\cap (A\times C)$

$\therefore A\times (B\cap C)\subset(A\times B)\cap (A\times C)$ .....

$(1)$

Now, let

$(x, y)$ bean arbitrary elements of

$(A\times B)\cap (A\times C)$ , then

$(x, y)\in (A\times B)\cap (A\times C)$

$\Rightarrow (x, y)\in (A\times B)$ and

$(x, y)\in (A\times C)$

$\Rightarrow (x\in A$ and

$y\in B)$ and

$(x\in A$ and

$y\in C)$

$\Rightarrow x\in A$ and

$(y\in B$ and

$y\in C)$

$\Rightarrow x\in A$ and

$y\in(B\cap C)$

$\Rightarrow (x, y)\in A\times (B\cap C)$

$\therefore (A\times B)\cap (A\times C)\subset A\times (B\cap C)$ .......

$(2)$

Hence from equation

$(1)$ and

$(2)$

we get

$A\times (B\cap C)=(A\times B)\cap (A\times C)$

so this equation is true.

1123081_183268_ans_1fd838363bba4979bd27f99082a272f6.jpg

Given the universal set

$B = \{-7,-3,-1,0,5,6,8,9\}$ , find :

$B = \{x : -4 < x < 6\}$

0%
$\{ -7, 0, 5,6\}$
0%
$\{5,6,8,9\}$
0%
$\{-3, -1, 0, 5\}$
0%
$\{0,5\}$

$A=\left\{ x:x\neq x \right\}$ represents-

0%
$\left\{ 0 \right\}$
0%
$\left\{ \right\}$
0%
$\left\{ 1 \right\}$
0%
$\left\{ x \right\}$

Explanation

It is fundamental concept that a void set is defined as,

$A =\{x : x\neq x\}=\{\}$

Which of the following statements is false?

0%
$2$ $\in$ {first five counting numbers.}
0%
$f$ ${\notin }$ {consonants}
0%
Rose ${\notin }$ {the set of all fruits.}
0%
Asia ${\in }$ {Continents}

Explanation

$\textbf{Step-1: Observing the Option 1}$

$\text{2 belongs to the sets of first five counting numbers}$

$\text{so option (A) is True.}$

$\textbf{Step-2: Observing other option}$

$\text{f belongs to the set of consonants so option B is false}$

$\text{Rose belongs to the set of all fruits so option C is True}$

$\text{Asia belongs to the sets of continents so option D is also True}$

$\textbf{Hence, the false statement is option B}$

Let

$A = \{2,3,5,7,8,11\}$ then which among the following is true?

0%
${7\notin A}$
0%
$\{2,3\} \not\subset \ of \ A$
0%
$2 \in A$
0%
None of the above

Explanation

$2\in A$ means 2 is element of A which is true.

Which of the following statements is true

0%
$3\subseteq \left\{ 1,3,5 \right\}$
0%
$3\in \left\{ 1,3,5 \right\}$
0%
$\left\{ 3 \right\} \in \left\{ 1,3,5 \right\}$
0%
$\left\{ 3,5 \right\} \in \left\{ 1,3,5 \right\}$

Explanation

Element/ number three is denoted by

$3$ and not

$\{3\}$ .

$\{3\}$ denotes a subset containing element 3. Hence

$\{3\}\subseteq \{1,3,5\}$
And

$3\in \{1,3,5\}$

Let

$P = \{ x | x$ is a multiple of

$3$ and less than

$100$ ,

$x$

$\displaystyle \in$

$N \}$

$Q = \{ x | x$ is a multiple of

$10$ and less than

$100$ ,

$x$

$\displaystyle \in$

$N\}$

0%
$\displaystyle Q\subset P$
0%
$\displaystyle P\cup Q=$ $\{ x | x$ is multiple of 30 $;$ $\displaystyle x \in N$ $\}$
0%
$\displaystyle P\cap Q=\phi$
0%
$\displaystyle P\cap Q=$ $\{ x | x$ is a multiple of 30 $;$ $\displaystyle x\in N$ $\}$

Explanation

P =

$\{ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 ..., 60, ...,90, ....,99\}$
Q =

$\{10, 20, 30, 40, 50, 60, 70, 80, 90 \}$
Then looking at the multiple choices we get

$P \displaystyle \cap Q$ =

$\{30, 60, 90\}$

$\displaystyle \Rightarrow$ P

$\displaystyle \cap$ Q = {x

$\displaystyle \mid$ x is a multiple of 30 x

$\displaystyle \epsilon$ N}

The set of real numbers r satisfying

$\displaystyle \frac{3 r^2- 8r+5}{4r^2-3r+7}>0$ is

0%
the set of all real numbers
0%
the set of all positive real numbers
0%
the set of all real numbers strictly between 1 and 5/3
0%
the set of all real numbers which are either less than 1 or greater than 5/3

Explanation

$\displaystyle \frac{3.r^2 - 8.r + 5 }{4. r^2 - 3.r + 7} > 0$

$\Rightarrow 3r^2 -3r -5r + 5 > 0$ and

$4r^2 - 3r + 7 > 0$ as D < 0

$\Rightarrow 3r(r-1) -5(r -1) > 0$

$\Rightarrow (3r - 5) (r - 1) > 0$

$(3r - 5) (r - 1) > 0$

$r \varepsilon (- \infty, 1) \cup (5/3, \infty)$

Find the set of all solutions of the equation

$2^{\left | y \right |}-\left | 2^{y-1}-1 \right |=2^{y-1}+1$ , the solution includes

0%
$y =-1$
0%
$y>1$
0%
$y=1$
0%
$y<1$

Explanation

Here,

$2^{\left | y \right |}-\left | 2^{y-1}-1 \right |=2^{y-1}+1$
We know to define modulus, we have three cases as
Case I: y< 0

$\Rightarrow$

$2^{-y}+\left ( 2^{y-1}-1 \right )=2^{y-1}+1$

$\Rightarrow$

$2^{-y}=2^{1}$ (as when y< 0 |y|=-y and

$\left | 2^{y-1}-1 \right |=-\left ( 2^{y-1}-1 \right )$ )
Hence, y=-1, which is true when y< 0 (i)
Case II:

$0\leq y< 1$

$\Rightarrow$

$2^{y}+\left ( 2^{y-1}-1 \right )=2^{y-1}+1$

$\Rightarrow$

$2^{y}=2$ (as when

$0\leq y< 1$ |y|=-y and

$\left | 2^{y-1}-1 \right |=-\left ( 2^{y-1}-1 \right )$ )

$\Rightarrow$

$y=1$ , which shows no solution as,

$0\leq y< 1$ (ii)
Case III:

$y\geq 1$

$\Rightarrow$

$2^{y}-\left ( 2^{y-1}-1 \right )=2^{y-1}+1$

$\Rightarrow$

$2^{y}=2^{y-1}+2^{y-1}$

$\Rightarrow$

$2^{y}=2.2^{y-1}$ (as when

$y\geq 0$ |y|=y and

$\left | 2^{y-1}-1 \right |=\left ( 2^{y-1}-1 \right )$ )

$\Rightarrow$

$2^{y}=2^{y}$ , which is an identity therefor, it is true

$\forall y\geq 1$ (iii)
Hence, from Eqs. (i), (ii), (iii) the solution of set is

$\left \{ y: y\geq 1\cup y=-1 \right \}$ .

If X and Y are two sets then

$\displaystyle X\cap (Y\cup X)'$ equals:

0%
$X$
0%
$Y$
0%
$\displaystyle \phi$
0%
$\{ 0 \}$

Explanation

We have

$\displaystyle X\cap \left ( Y\cup X \right )'=X\cap \left ( Y'\cap X' \right )$
=

$\displaystyle X\cap Y'\cap X'=X\cap X'\cap Y'$
=

$\displaystyle \phi \cap Y'= \phi$

$Q=\{ x:x=\cfrac { 1 }{ y },$ where

$y\in N \}$ , then

0%
$0\in Q$
0%
$1\in Q$
0%
$2\in Q$
0%
$\cfrac{2}{3} \in Q$

Explanation

Since

$\cfrac{1}{y}\ne 0$ ,

$\cfrac{1}{y}\ne 2$ ,

$\cfrac{1}{y}\ne \cfrac{-2}{3}$ [

$\because y\in N$ ]

$\therefore \cfrac{1}{y}$ can be

$1$

Ans: B

reflexive, symmetric and transitive.

0%
$\displaystyle R_{3}= \left \{ \left ( 1,1 \right ), \left ( 2,2 \right ), \left ( 3,3 \right ), \left ( 4,4 \right ), \left ( 1,2 \right ), \left ( 2,1 \right ) \right \}$
0%
$\displaystyle R_{3}= \left \{ \left ( 1,1 \right ), \left ( 2,2 \right ), \left ( 3,3 \right ), \left ( 4,4 \right ), \left ( 1,2 \right ), \left ( 2,1 \right ),\left ( 1,3 \right ),\left ( 3,1 \right ),\left ( 4,1 \right ),\left ( 1,4 \right ) \right \}$
0%
$\displaystyle R_{3}= \left \{ \left ( 1,1 \right ), \left ( 2,2 \right ), \left ( 3,3 \right ), \left ( 4,4 \right ) \right \}$
0%
none of these

Explanation

We define

$\displaystyle R_{3}$ as follows:

$\displaystyle R_{3}= \left \{ \left ( 1,1 \right ), \left ( 2,2 \right ), \left ( 3,3 \right ), \left ( 4,4 \right ), \left ( 1,2 \right ), \left ( 2,1 \right ) \right \}.$

Then evidently

$\displaystyle R_{3}$ is reflexive, symmetric and transitive, that is,

$\displaystyle R_{3}$ is an equivalence relation on A.

$\displaystyle \left ( 1,2 \right ) \epsilon R_{3}, \left ( 2,1 \right ) \epsilon R_{3}\Rightarrow \left ( 1,1 \right ) \epsilon R_{3}$

Given

$\displaystyle \xi$ = {x : x is a natural number}
A = {x : x is an even number x

$\displaystyle \in$ N}
B = {x : x is an odd number, x

$\displaystyle \in$ N}
Then

$\displaystyle (B\cap A)-(x-A)=....$

0%
$\displaystyle \phi$
0%
$A$
0%
$B$
0%
$A-B$

Explanation

$A = \{2, 4, 6, 8, ........\}$

$B = \{1, 3, 5, 7, .......\}$

$\displaystyle B\cap A=\left \{ 2,4,6,8,..... \right \}\cap \left \{ 1,3,5,7,.... \right \}=\phi$

$\displaystyle \xi -A=\left \{ 1,2,3,4,5,6,.... \right \}-\left \{ 2,4,6,8,.... \right \}=\left \{ 1,3,5,7,.... \right \}$

$\displaystyle \therefore B\cap A-(\xi -A)=\phi -\left \{ 1,3,5,7,.... \right \}=\phi$

Which of the following statements is true ?

0%
$\displaystyle 3\quad \subseteq \quad \left\{ 1,3,5 \right\}$
0%
$\displaystyle 3\quad \in \quad \left\{ 1,3,5 \right\}$
0%
$\displaystyle \{ 3\} \quad \in \quad \left\{ 1,3,5 \right\}$
0%
$\displaystyle \{ 3,5\} \quad \in \quad \left\{ 1,3,5 \right\}$

Explanation

$\displaystyle 3\quad \subseteq \quad \left\{ 1,3,5 \right\}$

An element cannot be a subset of another set. Hence,false.

$\displaystyle 3\quad \in \quad \left\{ 1,3,5 \right\}$

$3$ lies in the set. Hence, true.

$\displaystyle \{ 3\} \quad \in \quad \left\{ 1,3,5 \right\}$

$\{ 3\}$ doesnot lie in the set. Hence, false.

$\displaystyle \{ 3,5\} \quad \in \quad \left\{ 1,3,5 \right\}$

$\{ 3,5\}$ doesnot lie in the set. Hence, false.

Hence, option B.

$\displaystyle A=\left\{ x:x\neq x \right\}$ represents:

0%
${0}$
0%
$\phi$
0%
${1}$
0%
${x}$

Explanation

$x=x$ for all

$x$
Hence, there is no element in A.
A is an empty set .

$A=\{ \}$

If n (A) = 120, N(B) = 250 and n (A - B) = 52, then find

$\displaystyle n(A\cup B)$

0%
$302$
0%
$250$
0%
$368$
0%
None of the above

Explanation

$\displaystyle n(A-B)=n(A)-n(A\cap B)$

$\displaystyle \Rightarrow 52=120-n(A\cap B)$

$\displaystyle \Rightarrow n(A\cap B)=120-52=68$
Now

$\displaystyle n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$\displaystyle =120+250-68$

$=302$

The solution set of

$x+2<9$ over a set of positive even integers is

0%
$\displaystyle \left \{ 8,10,12,... \right \}$
0%
$\displaystyle \left \{ 2,4,6 \right \}$
0%
$\displaystyle \left \{ 1,2,3,4,5,6 \right \}$
0%
$\displaystyle \left \{ 2,4,6,8 \right \}$

Explanation

$x+2<9$

$x<7$
Since

$x$ is even positive integer then

$x\in [2,7)$ .

Hence the set of positive even integers which fall in the above set is

$\{2,4,6\}$ .

The solution set of

$3 x - 4 < 8$ over the set of non-negative square numbers is

0%
$\displaystyle \left \{ 1,2,3 \right \}$
0%
$\displaystyle \left \{ 1,4 \right \}$
0%
$\displaystyle \left \{ 1 \right \}$
0%
$\displaystyle \left \{ 16 \right \}$

Explanation

$3x-4<8$

$3x<12$

$x<4$
Hence set of non-negative square numbers belonging to the above set is

$\{1\}$ .

Let

$P$ and

$Q$ be two sets then what is

$\displaystyle (P\cap Q')\cup (P\cup Q)'$ equal to ?

0%
$\displaystyle (P\cap Q')\cup (P\cup Q)'=\xi \cap Q'=\xi \cap Q'=\xi$
0%
$\displaystyle (P\cup Q')\cup (P\cup Q)'=\xi \cap Q'=\xi \cap Q'=\xi$
0%
$\displaystyle (P\cap Q')\cup (P\cap Q)'=\xi \cap Q'=\xi \cap Q'=\xi$
0%
none of the above

Explanation

$\displaystyle (P\cap Q')\cup (P\cup Q)'=(P\cap Q')\cup (P'\cap Q')$

$=$

$\displaystyle (P\cup P')\cap (P\cup Q')\cap (Q'\cup P')\cap (Q'\cup Q')$

$=$

$\displaystyle \xi \cap \left \{ Q'\cup (P\cap P') \right \}\cap Q'$

$=\displaystyle \xi \cap \{Q'\cup \xi )\cap Q'$

$=$

$\displaystyle \xi \cap Q'\cap Q'=\xi \cap Q'=\xi$

$A$ and

$B$ are finite sets which of the following is the correct statement?

0%
$n(A - B) = n(A) - n(B)$
0%
$n(A - B) = n(B - A)$
0%
$n(A - B) = n(A) -$ $\displaystyle n\left ( A\cap B \right )$
0%
$n(A - B) = n(B) -$ $\displaystyle n\left ( A\cap B \right )$

U is a universal set and n(U) =A, B and C are subset of U. If n(A) = 50, n(B) = 70,

$\displaystyle n\left ( B\cup C \right )=\Phi$ ,

$\displaystyle n\left ( B\cap C \right )=15$ and

$\displaystyle A\cup B\cup C=U$ . then n(C) equals

0%
40
0%
50
0%
55
0%
60

$n (A) = 115, n(B) = 326, n(A - B) = 47$ , then

$\displaystyle n(A+B)$ is equal to

0%
373
0%
165
0%
370
0%
None

Explanation

$n(A-B)=47\Rightarrow n(A)-n(A\cap B) = 47\Rightarrow n(A\cap B)=115-47=68$

Hence

$n(A+B)=115+326-68=373$

If A and B are two disjoint sets and N is the universal set then

$\displaystyle A^{c}\cup \left [ \left ( A\cup B \right )\cap B^{c} \right ]$ is

0%
$\displaystyle \phi$
0%
A
0%
B
0%
N

Explanation

Since

$A$ and

$B$ are disjoint sets

$B^c\cap(A\cup B)$ will be only

$A$ as there is no intersection between

$A$ and

$B$ .

Of course

$A^c\cup A=N$

Suppose

$\displaystyle A_{1},A_{2},....,A_{30}$ are thirty sets each having 5 elements and

$\displaystyle B_{1},B_{2},....,B_{n}$ are n sets each with 3 elements. Let

$\displaystyle \bigcup_{i=1}^{30}A_{i} = \bigcup_{j=1}^{n}B_{j}=S$ and each elements of S belongs to exactly 10 of the

$\displaystyle A_{i}$ and exactly 9 of the

$\displaystyle B_{j}$ . Then n is equal to-

0%
35
0%
45
0%
55
0%
65

Explanation

$A_0,A_1,.............,A_{30}\implies$ each of 5 elements

$B_1,B_2,B_3...........n\implies$ each of 3 elements

The number of elements in the union of the A sets is

$5(30)-r$ where 'r' is the number repeats likewise the number of elements in the B sets

$3n-rB$

Each element in the union (in5) is repeated 10 times in A which means if x was the real number of elements in A (not counting repeats) then q out of those 10 should be thrown away or 9x .likewise on the B side 8x of those elements should be thrown away So,

$\implies 150-9x=3n-8x$

$n=50-3x$

Now in figure out what x is we need to use the fact that the union of a group of sets contains every member of each sets . If every element in 'S' is repeated 10 times that means every element in the union of the n's is repeated 10 times .

This means that

$10/10\implies 15$ is the number of in the A's without repeats counted (same for the B's aswell ) So now

$\cfrac{50-15}{3}=n$

$n=45$

Subset:- A proper subset is nothing but it contain atleast one more element of main set .

Ex:

$\{3,4,5\}$ is a set then the possible subsets are

$\{3\},\{4\},\{5\},\{1,5\},\{3,4\}$

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 5 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 5 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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