MCQ Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 8

$A/B=\left\{ a,b \right\} ,B\setminus A=\left\{ c,d \right\}$ and

$A\cap B=\left\{ e,f \right\}$ then the set

$B$ is equal to

0%
$\left\{ a,b,c,d \right\}$
0%
$\left\{ e,f,c,d \right\}$
0%
$\left\{ a,b,e,f \right\}$
0%
$\left\{ c,d,a,e \right\}$
0%
$\left\{ d,e,a,b \right\}$

Explanation

Given,

$A/B=\left\{ a,b \right\} ,B\setminus A=\left\{ c,d \right\}$

and

$A\cap B=\left\{ e,f \right\}$

$\therefore A-B=\left\{ a,b \right\} ...(i)\quad$

$B-A=\left\{ c,d \right\} ...(ii)$

$A\cap B=\left\{ e,f \right\} ...(iii)$
From Eqs. (i),(ii),(iii) we get

$B=\left\{ c,d,e,f \right\}$

If $a.N = \left\{ ax\thinspace :\thinspace x\in N \right\}$ then $3N\cap 7N=$

0%
$21\ N$
0%
$10\ N$
0%
$4\ N$
0%
$none$

Explanation

Clearly the set

$aN$ will include all multiples of

$a$ .

Hence

$3N$ and

$7N$ contain multiples of 3 and 7 respectively.

Their intersection must have the multiples of their L.C.M,i.e,21

Hence

$3N\cap 7N=21N$

State whether the following statement is true or false.
Whole numbers

$\subseteq$ natural numbers

0%
True
0%
False

Explanation

We have to state whether the statement "

$whole \:numbers \subseteq natural \:numbers$ " is true or false.

We know that, whole numbers start with

$0$ . That is

$0,1,2,3,...$ and natural numbers start with

$1$ . That is

$1,2,3,...$

Thus

$0 \notin natural \: numbers$ .

$\Rightarrow whole \:numbers \nsubseteq natural \:numbers$ .

Hence the given statement is false.

Let

$A$ and

$B$ be two events such that

$P\left( \overline { A\cup B } \right) =\dfrac { 1 }{ 6 } ,$

$P\left( A\cap B \right) =\dfrac { 1 }{ 4 }$ and

$P\left( \overline { A } \right) =\dfrac { 1 }{ 4 }$ , where,

$\overline { A }$ stands for complement of event

$A$ . Then, event

$A$ and

$B$ are

0%
Mutually exclusive and independent
0%
Independent but not equally likely
0%
Equally likely but not independent
0%
Equally likely and mutually exclusive

Explanation

Given that,

$P\left( \overline { A\cup B } \right) =\dfrac { 1 }{ 6 } ,$

$P\left( A\cap B \right) =\dfrac { 1 }{ 4 } ,$

$P\left( \overline { A } \right) =\dfrac { 1 }{ 4 }$

$P\left( \overline { A\cup B } \right) =\dfrac { 1 }{ 6 }$

$\Rightarrow 1-P\left( A\cup B \right) =\dfrac { 1 }{ 6 }$

$\Rightarrow \left[ 1-P\left( A \right) \right] -P\left( B \right) +P\left( A\cap B \right) =\dfrac { 1 }{ 6 }$

$\Rightarrow P\left( \overline { A } \right) -P\left( B \right) +P\left( A\cap B \right) =\dfrac { 1 }{ 6 }$

$\Rightarrow \dfrac { 1 }{ 4 } -P\left( B \right) +\dfrac { 1 }{ 4 } =\dfrac { 1 }{ 6 }$

$\Rightarrow P\left( B \right) =\dfrac { 2 }{ 4 } -\dfrac { 1 }{ 6 }$

$\Rightarrow P\left( B \right) =\dfrac { 1 }{ 3 }$
and

$P\left( A \right) =1-P\left( \overline { A } \right) =1-\dfrac { 1 }{ 4 } =\dfrac { 3 }{ 4 }$

Since,

$P\left( A\cap B \right) =P\left( A \right) \cdot P\left( B \right)$ , so the events

$A$ and

$B$ are independent events but not equal likely.

Let

$X$ and

$Y$ be two non-empty sets such that

$X\cap A=Y\cap A=\phi$ and

$X\cup A=Y\cup A$ for some non-empty set

$A$ . Then which of the following is true?

0%
$X$ is a proper subset of $Y$
0%
$Y$ is a proper subset of $X$
0%
$X=Y$
0%
$X$ and $Y$ are disjoint sets
0%
${ X }/{ A }=\phi$

Explanation

We have,

$X\cup A=Y\cup A$

$\Rightarrow X\cap \left( X\cup A \right) =X\cap \left( Y\cup A \right)$

$\Rightarrow X=\left( X\cap Y \right) \cup \left( X\cap A \right)$

$\left[ \because X\cap \left( X\cup A \right) =X \right]$

$\Rightarrow X=\left( X\cap Y \right) \cup \phi$

$\left[\because X\cap A=\phi \right]$

$\Rightarrow X=X\cap Y$ .....(i)
Again,

$X\cup A=Y\cup A$

$\Rightarrow Y\cap \left( X\cup A \right) =Y\cap \left( Y\cup A \right)$

$\Rightarrow \left( Y\cap X \right) \cup \left( Y\cap A \right) =Y$

$\Rightarrow \left( Y\cap X \right) \cup \phi =Y$

$\Rightarrow Y\cap X=Y$

$\Rightarrow X\cap Y=Y$ ....(ii)
From equations (i) and (ii), we get

$X=Y$

$X = \left \{1, 2, 3, ..., 10\right \}$ and

$A = \left \{1, 2, 3, 4, 5\right \}$ . Then, the number of subsets

$B$ of

$X$ such that

$A - B = \left \{4\right \}$ is

0%
$2^{5}$
0%
$2^{4}$
0%
$2^{5} - 1$
0%
$1$
0%
$2^{4} - 1$

Explanation

Given,

$X = \left \{1, 2, 3, ...., 10\right \}$
and

$A = \left \{1, 2, 3, 4, 5\right \}$
Now,

$A - B = \left \{4\right \}$

$\Rightarrow B = A - \left \{4\right \} = \left \{1, 2, 3, 4, 5\right \} - \left \{4\right \}$

$= \left \{1, 2, 3, 5\right \}$

$\therefore X - \left \{4\right \} - B$

$= \left \{1, 2, 3, ..., 10\right \} - \left \{4\right \} - \left \{1, 2, 3, 5\right \}$

$= \left \{6, 7, 8, 9, 10\right \}$
Therefore,number of subsets

$B$ of

$X$ i.e.,

$\left \{6, 7, 8, 9, 10\right \}$

$= 2^{5}$ .

If the sets

$A$ and

$B$ are as follows:

$A=\left\{ 1,2,3,4 \right\} ,B=\left\{ 3,4,5,6 \right\}$ , then

0%
$A-B=\left\{ 1,2 \right\}$
0%
$B-A=\left\{ 5,6 \right\}$
0%
$\left[ \left( A-B \right) -\left( B-A \right) \right] \cap A=\left\{ 1,2 \right\}$
0%
$\left[ \left( A-B \right) -\left( B-A \right) \right] \cup A=\left\{ 3,4 \right\}$

Explanation

Given,

$A=\left\{ 1,2,3,4 \right\} ,B=\left\{ 3,4,5,6 \right\}$

$\therefore A-B=\left\{ 1,2 \right\} ;\quad B-A=\left\{ 5,6 \right\} \quad$

$\left[ (A-B)-(B-A) \right] \cap A=\left[ \left\{ 1,2 \right\} -\left\{ 5,6 \right\} \right] \cap \left\{ 12,3,4, \right\} =\left\{ 1,2 \right\}$

$n(A)=10,n(B)=6$ and

$(C)=5$ for three disjoint sets

$A,B,C$ then

$n(A\cup B\cup C)$ equals

0%
$21$
0%
$11$
0%
$1$
0%
$9$

Explanation

Since,

$A,B,C$ are disjoint sets

$\therefore n(A\cup B\cup C)=n(A)+n(B)+n(C)$

$=10+6+5=21$

The total number of subsets of {1, 2, 6, 7} are?

0%
$16$
0%
$8$
0%
$64$
0%
$32$

Explanation

We have to find the total number of subsets of

$\{1,2,6,7\}$ .

We know that, for a set containing

$n$ elements, the total number of subsets is

$2^n$ .

Consider

$\{1,2,6,7\}$ , wich has

$4$ elements.

$\therefore$ here

$n=4$

Hence total number of subsets is

$2^4=16$ .

Thus the total number of subsets of

$\{1,2,6,7\}$ is

$16$ .

If U = {1, 3, 5, 7, 9, 11, 13}, then which of the following are subsets of U.
B = {2, 4}
A = {0}
C = {1, 9, 5, 13}
D = {5, 11, 1}
E = {13, 7, 9, 11, 5, 3, 1}
F = {2, 3, 4, 5}

0%
$A$ and $B$
0%
$C$ , $D$ and $E$
0%
$A$ and $D$
0%
only $A$

Explanation

Given

$U=\{1,3,5,7,9,11,13\}, A=\{0\}, B=\{2,4\}, C=\{1,9,5,13\}, \\ D=\{5,11,1\}, E=\{13,7,9,11,5,3,1\}, F=\{2,3,4,5\}$

Now we have to find the subsets of

$U$

Consider

$A=\{0\}$
Since

$0\notin U, A\nsubseteq U$

Consider

$B=\{2,4\}$
Since

$2,4\notin U, B\nsubseteq U$

Consider

$C=\{1,9,5,13\}$
Since

$1,9,5,13\in U, C\subset U$

Consider

$D=\{5,11,1\}$
Since

$5,11,1\in U, D\subset U$

Consider

$E=\{13,7,9,11,5,3,1\}$
Since

$13,7,9,11,5,3,1\in U, E\subseteq U$

Consider

$F=\{2,3,4,5\}$
Since

$2,3,4\notin U, F\nsubseteq U$

Therefore

$C,D,E$ are subsets of

$U$ .

Let

$S = \left \{(a, b): a, b\epsilon Z, 0\leq a, b\leq 18\right \}$ . The number of elements

$(x, y)$ in

$S$ such that

$3x + 4y + 5$ is divided by

$19$ is

0%
$38$
0%
$19$
0%
$18$
0%
$1$

Explanation

Maximum value of

$3x+4y+5$ will occur at

$(18,18)$

$\Rightarrow$ Maximum value

$=3\times18+4\times18+5=131$

And, the minimum value will occur at

$0,0$

$\Rightarrow$ Minimum value

$=3\times0+4\times0+5=5$

Now, the multiples of

$19$ that lie between minimum and maximum value of

$3x+4y+5$ are

$19,38,57,76,95,114$

$\mathbf{Case 1-}$ When

$3x+4y+5=19$

By trial and error the solutions for this case is-

$(2,2)$

$\Rightarrow$ The number of elements

$(x,y)$ in

$S$ for this case is

$1$

$\mathbf{Case 2-}$ When

$3x+4y+5=38$

By trial and error the solutions for this case are-

$(11,0);(7,3);(3,6)$

$\Rightarrow$ The number of elements

$(x,y)$ in

$S$ for this case are

$3$

$\mathbf{Case 3-}$ When

$3x+4y+5=57$

By trial and error the solutions for this case are-

$(16,1);(12,4);(8,7);(4,10);(0,13)$

$\Rightarrow$ The number of elements

$(x,y)$ in

$S$ for this case are

$5$

$\mathbf{Case 4-}$ When

$3x+4y+5=76$

By trial and error the solutions for this case are-

$(17,5);(13,8);(9,11);(5,14);(1,17)$

$\Rightarrow$ The number of elements

$(x,y)$ in

$S$ for this case are

$5$

$\mathbf{Case 5-}$ When

$3x+4y+5=95$

By trial and error the solutions for this case are-

$(18,9);(14,12);(10,15);(6,18)$

$\Rightarrow$ The number of elements

$(x,y)$ in

$S$ for this case are

$4$

$\mathbf{Case 6-}$ When

$3x+4y+5=114$

By trial and error the solutions for this case are-

$(15,16)$

$\Rightarrow$ The number of elements

$(x,y)$ in

$S$ for this case are

$1$

Therefore, the total number of elements

$(x,y)$ in

$S$ which are divisible by 19

$=1+3+5+5+4+1=19$

Hence the correct answer is option

$B$

State whether true or false.
Quadrilateral

$\subseteq$ polygon

0%
True
0%
False

Explanation

We have to state whether the statement "

$Quadrilateral \subseteq polygon$ " is true or false.

We know that, "A polygon is any 2-dimensional shape formed with any number of straight lines. "

Also we know that "A quadrilateral is a closed figure with four straight lines. "

Hence

$Quadrilateral \subseteq polygon$

Therefore the given statement is true.

$a, b, c, d$ are four distinct numbers chosen from the set

$\left \{1, 2, 3, ..., 9\right \}$ , then the minimum value of

$\dfrac {a}{b} + \dfrac {c}{d}$ is

0%
$\dfrac {3}{8}$
0%
$\dfrac {1}{3}$
0%
$\dfrac {13}{36}$
0%
$\dfrac {25}{72}$

Explanation

To minimize the value of

$\dfrac{a}{b}+\dfrac{c}{d}$ ,

$b$ and

$d$ must highest numbers chosen from the set and

$a$ and

$c$ must be the lowest numbers of given set.

Therefore,

$b=9,\, d=8,\, a=2$ and

$c=1$ .

Thus, the minimum value for

$\dfrac{a}{b}+\dfrac{c}{d}$

$=\dfrac{2}{9}+\dfrac{1}{8}=\dfrac{25}{72}$

Let

$A,B$ and

$C$ be three events such that

$P(A)=0.3,P(B)=0.4,P(C)=0.8,P(A\cup B)=0.08,\quad P(A\cap C)=0.28,P(A\cap B\cap C)=0.09$ . If

$P(A\cup B\cup C)\ge 0.75$ , then

$P(B\cap C)$ satisfies

0%
$P(B\cap C)\le 0.23$
0%
$P(B\cap C)\le 0.48\quad$
0%
$0.23\le P(B\cap C)\le 0.48$
0%
$0.23\le P(B\cap C)\ge 0.48$

Explanation

Since

$P(A\cup B\cup C)\ge 0.75\quad$

$\quad \therefore 0.75\le P(A\cup B\cup C)\le 1$

$\Rightarrow 0.75\le P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)\le 1\quad \quad$

$\Rightarrow 0.75\le 1.23-P(B\cap C)\le 1\Rightarrow -0.48\le -P(B\cap C)\le -0.23$

$\Rightarrow 0.23\le P(B\cap C)\le 0.48\quad$

A set of

$n$ numbers has the sum

$s$ . Each number of the set is increased by

$20$ , then multiplied by

$5$ , and then decreased by

$20$ . The sum of the numbers in the new set thus obtained is:

0%
$s+20n$
0%
$5s+80n$
0%
$s$
0%
$5s$

Explanation

$s={ a }_{ 1 }+{ a }_{ 2 }+....+{ a }_{ n }$

$s'=5({ a }_{ 1 }+20)-20+5({ a }_{ 2 }+20)-20+.....=5{ a }_{ 1 }+80+5{ a }_{ 2 }+80+....+5{ a }_{ n }+80$

$=5\left( { a }_{ 1 }+{ a }_{ 2 }+....+{ a }_{ n } \right) +80n=5s+80n$
or

$\sum _{ i=1 }^{ n }{ { a }_{ i } } =s,\sum _{ i=1 }^{ n }{ { [5(a }_{ i }+20)-20] } =5\sum _{ i=1 }^{ n }{ { a }_{ i } } +\sum _{ i=1 }^{ n }{ 80 } =5s+80n\quad \quad$

$A= \{x:x$ is a multiple of

$2\}, \,\,B= \{x:x$ is a multiple of

$5\}$ and

$C = \{x:x$ is a multiple of 10

$\}$ , then

$A\cap (B\cap C)$ is equal to

0%
$A$
0%
$B$
0%
$C$
0%
$\{x:x$ is a multiple of $100\}$

Explanation

$A= \{x:x$ is a multiple of

$2\}=\{2,4,6,8,10,12,14,....\}$

$B= \{x:x$ is a multiple of

$5\}=\{5,10,15,20,25,....\}$ and

$C = \{x:x$ is a multiple of

$10\}=\{10,20,30,40,....\}$

Here,

$C \subset A$ and

$C \subset B$

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

$= A \cap C = C$

$X=\left\{ { 4 }^{ n }-3n-1;n\in R \right\}$ and

$Y=\left\{ 9\left( n-1 \right) ;n\in N \right\}$ , then

$X\cap Y=$

0%
$X$
0%
$Y$
0%
$\phi$
0%
$\left\{ 0 \right\}$

Explanation

Let us restrict the domain of

$X$ to natural numbers,

$\Longrightarrow { 4 }^{ n }-3n-1\\ \Longrightarrow { \left( 1+3 \right) }^{ n }-3n-1\\ \Longrightarrow 1+3n+{ n }_{ { C }_{ 2 } }{ 3 }^{ 2 }+...+{ 3 }^{ n }-3n-1\\ \Longrightarrow 9k$

Where

$k$ is some natural number,

So,the elements of

$X$ are multiples of

$9$ ,

The elements of the set

$Y$ are also multiples of

$9$ ,

But if we extend the domain of

$X$ to the set of real numbers there would be other elements which are not present in

$Y$ as the elements of

$Y$ are only natural numbers where as the elements of

$X$ can also be rational and irrational,

So the elements of

$Y$ are contained in

$X$ for all natural numbers,

$\therefore X\cap Y=Y$ .

In the equation

${ \left( x-m \right) }^{ 2 }-{ \left( x-n \right) }^{ 2 }={ \left( m-n \right) }^{ 2 }$ ,

$m$ is a fixed positive number, and

$n$ is a fixed negative number. The set of values

$x$ satisfying the equation is:

0%
$x\ge 0$
0%
$x\le n$
0%
$x=0$
0%
the set of all real numbers
0%
none of these

The number of binary operations on the set

$\{1, 2, 3\}$ is _________.

0%
$3^9$
0%
$9^3$
0%
$27$
0%
$3!$

Explanation

Let us denote this set by

$S$ , then

$|S| = 3$ .

A binary relation defined on the elements of

$S$ maps all elements in

$S\times S$ to elements in

$S$ by definition.

In this case any binary relation will thus have

$3^2 =9$ inputs each of which is an ordered pair of elements from

$S$ and only

$3$ number of possible outputs.

If all possible binary operations are considered then it is possible to assign any of the

$3$ outputs to any of the

$9$ inputs. So the number of all binary operations would exactly be

$3^9$ .

So option A is the correct answer.

Choose the correct answers from the alternatives given.
Directions for questions

$71$ to

$75$ : Study the pie chart carefully to answer the questions that follow.
Percentage distribution of students in different courses.
What is the value of half of the difference between the number of students in MBA and MBBS?
The Total number of student =

$6500$

0%
$800$
0%
$1600$
0%
$1300$
0%
$650$

The relation

$S=\{(3, 3), (4, 4)\}$ on the set

$A=\{3, 4, 5\}$ is __________.

0%
Not reflexive but symmetric and transitive
0%
Reflexive only
0%
Symmetric only
0%
An equivalence relation

Explanation

given, relation

$S=\{(3,3),(4,4)\}$ on the set

$A=\{3,4,5\}$

It is not reflexive because if every element of

$A$ is not related to itself (i.e.,)

$(5,5)$ is absent.

$A$ is symmetric because if for all

$a$ and

$b$ in

$A$ that

$a$ is related to

$b$ if and only if

$b$ is related to

$a$ .

$A$ is transitive because if

$a$ is related to

$b$ and

$b$ is related to

$c$ then

$a$ is also related to

$c$ .

Hence, not reflexive but symmetric and transitive.

$M = \left\{ {x:x \geqslant 7\,\,{\text{and}}\,x \in N} \right\}$ for universal set of natural numbers, then

$M'$ is

0%
$\left\{ {1,2,3,4,5} \right\}$
0%
$\left\{ {1,2,3,4,5,6,7} \right\}$
0%
$\left\{ {1,2,3,4,5,6} \right\}$
0%
$\left\{ {0,1,2,3,4,5,6} \right\}$

Explanation

Given

$M =\{x:x\ge 7 , x\in N\}$ .

$x$ is a natural number such that

$x\ge 7$ . Hence

$x= \{7,8,9,10,11,12,.......\infty\}$

$\therefore M' = N- M$

$=\{1,2,3,4,5,6\}$

$X$ and

$Y$ are two sets, then

$X\cap \left( X\cup Y \right)$ equals

0%
$X$
0%
$Y$
0%
$\phi$
0%
$None\ of\ these$

Explanation

Ans.

$(a).$
If

$X\subset Y$ , then

$X\cup Y=Y$ and

$X\cap Y=X$
If

$Y\subset X$ , then

$X\cup Y=X$ and

$X\cap X=X$

Let

$A=\left\{ \left( x,y \right) :y={ e }^{ x },x\in R \right\}$

$B=\left\{ \left( x,y \right) :y={ e }^{ -x },x\in R \right\}$ . Then

0%
$A\cap B=\phi$
0%
$A\cap B\neq \phi$
0%
$A\cup B={ R }^{ 2 }$
0%
$none\ of\ these$

Explanation

Ans.

$(b)$ .

$y={e}^{x},\ y={e}^{-x},$ meet where

${ e }^{ x }=\ { e }^{ -x }$ or

$e^{2x}=1$

$\therefore x=0$ and hence

$y=1$ . These curve meet at

$(0,1)$ so that

$A\cap B=(0,1)$ .
In other words

$A\cap B\neq \phi$ .

If a relation from a finite set A having m elements to a finite set B having n elements, then the number of relations from A to B is

0%
$2^{mn}$
0%
$2^{mn}-1$
0%
2 mn
0%
$m^n$

$X$ and

$Y$ are two sets, then

$X\cap \left( Y\cup X \right)'$ equals

0%
$X$
0%
$Y$
0%
$\phi$
0%
$None\ of\ these$

Explanation

$X\cap \left( X' \cap Y' \right) =\left( X\cap X' \right) \cap Y' =\phi \cap Y' =\phi$

If X = {

$4^n \,-\,3n\,-\,1\,:\,\epsilon N$ } and
Y = {

$9(n\,-\,1) \,:\,\epsilon N$ }
then

$X\subset Y$

0%
True
0%
False

Explanation

Given

$X=\left\{ { 4 }^{ n }-3n-1;n\in N \right\}$

For

$n=0$ ,

$X={ 4 }^{ 0 }-0-1$

$\therefore X=0$

For

$n=1$ ,

$X={ 4 }^{ 1 }-3\left( 1 \right) -1$

$\therefore X=0$

For

$n=2$ ,

$X={ 4 }^{ 2 }-3\left( 2 \right) -1$

$\therefore X=16-6-1=9$

For

$n=3$ ,

$X={ 4 }^{ 3 }-3\left( 3 \right) -1$

$\therefore X=64-9-1=54$

For

$n=4$ ,

$X={ 4 }^{ 4 }-3\left( 4 \right) -1$

$\therefore X=256-12-1=243$

$\therefore X=\left\{ 0,0,9,54,243....... \right\}$ (1)

Consider second set

$Y=\left\{ 9\left( n-1 \right) ;n\in N \right\}$

For

$n=0$ ,

$Y=9\left( 0-1 \right)$

$\therefore Y=-9$

For

$n=1$ ,

$Y=9\left( 1-1 \right)$

$\therefore Y=0$

For

$n=2$ ,

$Y=9\left( 2-1 \right)$

$\therefore Y=9$

For

$n=3$ ,

$Y=9\left( 3-1 \right)$

$\therefore Y=18$

For

$n=4$ ,

$Y=9\left( 4-1 \right)$

$\therefore Y=27$

$\therefore Y=\left\{ -9,0,9,18,27,36,45,54,63,72......... \right\}$

$\therefore Y=\left\{ y:y\quad is\quad multiple\quad of\quad 9\quad and\quad y\ge -9 \right\}$ (2)

From (1) and (2),

$X\subset Y$

Given : A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}
Find :

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

0%
{4,4}
0%
{3,4}
0%
{3,4}, {3,3}
0%
{3,3}

Explanation

Given

$A=\left\{ 1,2,3 \right\}$ ,

$B=\left\{ 3,4 \right\}$ and

$C=\left\{ 4,5,6 \right\}$

$A\times B=\left\{ 1,2,3 \right\}\times \left\{ 3,4 \right\}$

$\therefore A\times B=\left\{ \left( 1,3 \right) ,\left( 1,4 \right) ,\left( 2,3 \right) ,\left( 2,4 \right) ,\left( 3,3 \right) ,\left( 3,4 \right) \right\}$

$B\times C=\left\{ 3,4 \right\}\times \left\{ 4,5,6 \right\}$

$\therefore B\times C=\left\{ \left( 3,4 \right) ,\left( 3,5 \right) ,\left( 3,6 \right) ,\left( 4,4 \right) ,\left( 4,5 \right) ,\left( 4,6 \right) \right\}$

$\therefore \left( A\times B \right) \cap \left( B\times C \right) =\left\{ \left( 1,3 \right) ,\left( 1,4 \right) ,\left( 2,3 \right) ,\left( 2,4 \right) ,\left( 3,3 \right) ,\left( 3,4 \right) \right\} \cap \left\{ \left( 3,4 \right) ,\left( 3,5 \right) ,\left( 3,6 \right) ,\left( 4,4 \right) ,\left( 4,5 \right) ,\left( 4,6 \right) \right\}$

$\therefore \left( A\times B \right) \cap \left( B\times C \right) =\left\{ \left( 3,4 \right) \right\}$

Let

$A=\left\{ x:x\ \in\ R,\left| x \right| <1 \right\}$

$B=\left\{ x:x\ \in\ R,\left| x-1 \right| \ge 1 \right\}$
and

$A\cup B=R-D$ , then set

$D$ is

0%
$\left\{ x:1 < x \le 2 \right\}$
0%
$\left\{ x:1\le x<2 \right\}$
0%
$\left\{ x:1\le x\le 2 \right\}$
0%
None of these

Explanation

Given

$A=\left\{ x:x\in R\left| x \right| <1 \right\}$

$x$ is positive,

$\left| x \right| =x$

$\therefore x<1$

$x$ is negative,

$\left| x \right| =-x$

$\therefore -x<1$

$\therefore x>-1$

$\therefore A=\left\{ -1<x<1 \right\}$ (1)

Given

$B=\left\{ x:x\in R,\left| x-1 \right| \ge 1 \right\}$

$\left| x-1 \right|$ is positive,

$\left| x-1 \right| =x-1$

$\therefore x-1\ge 1$

$\therefore x\ge 2$

$\left| x-1 \right|$ is negative,

$\left| x-1 \right| =-\left( x-1 \right)$

$\therefore -\left( x-1 \right) \ge 1$

$\therefore \left( x-1 \right) \le -1$

$\therefore x\le 0$

$\therefore B=\left\{ x\le 0\quad and\quad x\ge 2 \right\}$ (2)

Thus,

$A\cup B=\left\{ -1<x<1 \right\} \cup \left\{ x\le 0\quad and\quad x\ge 2 \right\}$

The region which is not included in this domain has to be subtracted from remaining domain of real numbers.

$\therefore A\cup B=R-D$

Thus, set D is set of excluded region which is the region between 1 and 2 including 1.

$\therefore D:-1\le x<2$

If X and Y are two sets such that n(X)=17, n(Y)=23 and n(X

$\cup$ Y)=38, find n(X

$\cap$ Y).

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

$\textbf{Step 1: Label the given information and use relevant formula}$

${\text{It is given that n(X) = 17,n(Y) = 23,n(X}} \cup {\text{Y) = 38}}$

${\text{As we know that, n(X}} \cup {\text{Y) = n(X) + n(Y) - n(X}} \cap {\text{Y)}}$

${\text{Using the formula we get}}$

${\text{n(X) + n(Y) - n(X}} \cap {\text{Y) = (X}} \cup {\text{Y)}}$

$\therefore {\text{17 + 23 - n(X}} \cap {\text{Y) = 38}}$

$\Rightarrow {\text{n(X}} \cap {\text{Y) = 40 - 38 = 2}}$

${\textbf{Hence, the value is 2.}}$

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 8 - 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 8 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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