$$\left \{b, c, f, h, i, j\right \}$$

$$\left \{b, c, d, h, i, j\right \}$$

$$\left \{b, c, d, g, i, j\right \}$$

$$\left \{b, c, f, g, i, j\right \}$$

MCQ Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 2

$S$ and

$T$ are two sets such that

$S$ has

$21$ elements,

$T$ has

$32$ elements and

$\displaystyle S\cap T$ has

$11$ elements, then
find the number of elements in

$\displaystyle S\cup T$ .

0%
$47$
0%
$42$
0%
$37$
0%
$52$

Explanation

Given

$n(S) = 21$ ,

$n(T) = 32,$

$n$ (

$\displaystyle S\cap T$ )

$= 11$
Now

$n(S) + n(T) = n($

$\displaystyle S\cap T$ )

$+ n($

$\displaystyle S\cup T$ )

$\displaystyle \Rightarrow$

$n($

$\displaystyle S\cup T$

$) = 21 + 32 - 11 = 42$ .

In a community of

$175$ persons,

$40$ read TOI,

$50$ read the Samachar Patrika and

$100$ do not read either. How many persons read both the papers?

0%
$16$
0%
$17$
0%
$15$
0%
$14$

Explanation

Number of people who read

$TOI$

$n(TOI)=40$

Number of people who read Samachar patrika, n(samachar patrika)

$=50$

Number of people who do not read newspaper

$=100$

Number of people who read newspaper

$(A\cup B)=$ Total number of people - people who don't read newspaper

$=175-100=75$

Number of persons who read both newspapers,

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

$n\left( A\cap B \right) =40+50-75$

$n\left( A\cap B \right) =90-75=15$

Number of persons who read both newspapers

$=15$

Let

$U=\{x: \in, W: 3<x< 12\}$ ,

$B=\{4,6,8,10\}$ .

$B'$

0%
$\{6,7,9,11,12\}$
0%
$\{5,7,9,11\}$
0%
$\{5,7,9,10,11,12 \}$
0%
None of the above

Explanation

Given universal set is

$\{x\in W ,3<x<12\}$

It can be written as a set

$\{4,5,6,7,8,9,10,11\}$

$B$ is given as

$B=\left\{4,6,8,10\right\}$

$\therefore B'=W-B=$ All the elements in the universal set but not in set

$B$

$=\{5,7,9,11\}$

Hence, option B is correct

M represents the children in a class who have no brothers and 8 represents the children who have no sisters.

$+$ denotes union,

$*$ denotes intersection, and

$(^\prime)$ denotes complement. The set of children who have no siblings is

0%
$S^\prime+M^\prime$
0%
$(S*M)^\prime$
0%
$(S+M)^\prime$
0%
$S*M$

Explanation

Solution

As per De Morgan's Law,

$\acute { \left( A\cup B \right) } =\acute { A } \cap \acute { B }$ .

Correct answer is C, that is Set of children having no children.

Which of the following has only one subset?

0%
$\{0,1\}$
0%
$\{1\}$
0%
$\{0\}$
0%
$\{\}$

$(A\cup B)'$

0%
$\left \{1,2,3,4,5,6\right \}$
0%
$\left \{7,8\right \}$
0%
$\left \{3,4,5\right \}$
0%
none

Explanation

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

$\therefore (A\cup B)^{1} = \cup -(A\cup B) = \left \{ 7,8 \right \}$

1125703_514102_ans_1a19f826c8ae4b5eb27f844222cdb576.jpg

$P\cap Q\cup R$

0%
$\left \{b, c, d, h, i, j\right \}$
0%
$\left \{b, c, d, g, i, j\right \}$
0%
$\left \{b, c, f, h, i, j\right \}$
0%
$\left \{b, c, f, g, i, j\right \}$

Explanation

$P = \left \{ a,b,c,e \right \}$

$Q = \left \{ b,c,d,f \right \}$

$\Rightarrow (P\cap Q) = \left \{ b,c \right \}$

$R = \left \{ c,f,h,i,j\right \}$

$\Rightarrow \boxed {(P\cap Q)\cup R = \left \{ b,f,h,i,j \right \}}$

1097481_514593_ans_5cc6368ad5da47e696571738b906b1fa.jpg

State whether following statement is true or false.

Given

$\mu = \left \{x / x \text {is an integer}, -4\leq x \leq 6\right \}$ ,

$P' = \left \{-4, -3, 0, 1, 2, 4, 6\right \}$

$Q' = \left \{-3, -2, 1, 3, 6\right \}$

The elements of the set

$P$ are

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

0%
True
0%
False

Explanation

$\cup = \left \{ -4,-3,-2,-1,0,1,2,3,4,5,6 \right \}$

$P^{1} = \left \{ -4,-3,0,1,2,4,6 \right \}$

$Q^{1} = \left \{ -3,-2,1,3,6 \right \}$

$\boxed {(P) = \cup -P^{1} = \left \{ 2,-1,+3,+5 \right \}}$

1103187_514567_ans_cd9397bfa5524439b89c1157b1d69bd5.jpg

$A, B$ and

$C$ are any three set, then

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

0%
$(A \cup B) \cup (A\cup C)$
0%
$(A \cup B) \cap (A\cup C)$
0%
$(A \cap B) \cap (A\cap C)$
0%
none

The set

$\displaystyle A=\left\{ x:x\in { x }^{ 2 }=16\quad and\quad 2x=6 \right\}$ equals:

0%
Null set
0%
Singleton set
0%
Infinite set
0%
Not a Well Defined Collection

Explanation

Since,

$\displaystyle { x }^{ 2 }=16\Rightarrow x=\pm 4$
and

$\displaystyle 2x=6\Rightarrow x=3$
Hence, no value of

$x$ is satisfied.

$\displaystyle \therefore A=\phi$

$(P\cap Q)\cup (Q\cap R)$

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

Explanation

$P = \left \{ a,b,c,e \right \};Q = \left \{ b_{n}c_{n}d_{n}f \right \}$

$\Rightarrow (P\cap Q) = \left \{ b,c \right \}$

$R = \left \{ c,f,n,i,j \right \} \Rightarrow (Q\cap R) = \left \{ c,f \right \}$

$\boxed { (P\cap Q)\cup (Q\cap R) = \left \{ b,c,f \right \}}$

1097522_514596_ans_815ad612f7f4421092a77691f51b7cbc.jpg

Let

$A$ and

$B$ be two sets such that

$A\cap B=\phi$ . Find the value of

$(A\cup B')=$

0%
$A'$
0%
$B$
0%
$\phi$
0%
none of these

Explanation

Given,

$A\cap B=\phi$ .

Now,

$(A\cup B')$

$=(A'\cap B)'$ [ Using De Morgan's law]

$=B'$ . [ As

$A'\cap B=B-(A\cap B)=B$ since

$A\cap B=\phi$ ]

In a school with an envolment of

$950$ students, each student must join either the lions club or the country club or both. Given that

$646$ students are members of the lions club and

$532$ are members of the country club, calculate the number of students who are members of both clubs

0%
$228$
0%
$230$
0%
$232$
0%
$234$

Explanation

Total number of students

$=950$

Number of students who are members of lion club

$=646$

Number of students who are members of country club

$=532$

Number of students who are members of both club

$=(646+532)-950$

$=1178-950$

$=228$

$\therefore$ There are

$228$ students who joined both clubs.

A survey was carried out to find out the types of shampoo that a group of

$150$ women have tried. It was found that

$84$ women have used brand A shampoo,

$93$ have used brand B, and

$69$ have used brand C of these women,

$45$ have tried brands A and B,

$25$ have tried brands A and C and

$40$ have tried brand B and C. Determine the number of women who have tried (a) all three brands, (b) only brand A

0%
(a) $14$ (b) $28$
0%
(a) $15$ (b) $28$
0%
(a) $14$ (b) $29$
0%
(a) $16$ (b) $29$

Explanation

$n(A\cup B\cup C)=150$

$n(A)=84$

$n(B)=93$

$n(C)=69$

$n(A\cap C)=25$

$n(A\cap B)=45$

$n(B\cap C)=40$

$(a)$ The number of women who have tried all three brands

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

$=150-84-93-69+45+25+40$

$=14$

$(b)$ The number of women who have tried only brand

$A$

The number of women who have tried only brand

$A$

$=n(A)-\left[n(A\cap B)+ n(A\cap C)\right]+n(A\cap B\cap C)$

$=84-[45+25]+14$

$=84-70+14$

$=28$

$A = \left \{1, 2, 3, 4\right \}$ , what is the number of subsets of A with at least three elements?

0%
$3$
0%
$4$
0%
$5$
0%
$10$

Explanation

A subset containing

$3$ elements

$= \left \{1, 2, 3\right \}; \left \{1, 3, 4\right \}; \left \{1, 2, 4\right \}$ and

$\left \{2, 3, 4\right \}$
A subset containing

$4$ elements

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

$\therefore$ there are five subsets containing at least

$3$ elements.

For any two sets A and B,

$A = B$ is equivalent to

0%
$A - B = B - A$
0%
$A\cup B = A\cap B$
0%
$A\cup C = B\cup C$ and $A\cap C = B\cap C$ for any set C
0%
$A\cap B = \oslash$

Explanation

For the sets A and B , where A=B

(a) $A-B = B-A$ . If the two sets are same i.e. having same elements then subtracting A from B or vice versa, we get the same results.

(b) $A \cup B = A \cap B$ . Since $A=B,$ $A \cup B = A \; or \; B$ and $A \cap B = A$ or $B$

(c) $A\cup C = B \cup C$ for any set C. If $A= B,$ and we are adding any other set C, it will be same as for the other set.

$A\cap C = B \cap C$ for any set C. If A= B, Both the intersection will be the same for any set C.

(d) $A\cap B\ne \phi$

Hence, D is not equivalent.

The set of integers is closed with respect to which one of the following?

0%
Addition only
0%
Multiplication only
0%
By addition and multiplication
0%
Division

In the Venn diagram, the numbers represent the number of elements in the subsets. Given that

$\xi = F\cup G\cup H$ and

$n(\xi) = 42$ , find

$n(G'\cup H)$

0%
$18$
0%
$28$
0%
$30$
0%
$38$

Explanation

$n(\xi) = 42$

$n(G) = 12, n(F) = 17, n(H) = 8$

$n(F \cap G) = 5, n(G \cap H) = 0, n(F \cap H) = 0$

$n(G’)= 42-7-5 =30$

$n(H)= 8, n(G’ \cup H) = n(G')+n(H)-n(G'\cap H)=30+8-8 = 30$

If A and B be two sets containing

$4$ and

$8$ elements respectively, what can be the maximum number of elements in

$A\cup B$ ? Find also, the minimum number of elements in

$(A\cup B)$ ?

0%
Maximum number of elements $= 12$
Minimum number of elements $= 8$
0%
Maximum number of elements $= 14$
Minimum number of elements $= 8$
0%
Maximum number of elements $= 12$
Minimum number of elements $= 9$
0%
Maximum number of elements $= 14$
Minimum number of elements $= 7$

Explanation

${\textbf{Step -1: Find maximum number of elements.}}$

${\text{Given,}}$

$n(A) =4$

$n(B) =8$

${\text{Minimum number of elements in}}$

$A\cup B=8.$

${\text{A is subset of B.}}$

${\text{Maximum number of elements in}}$

$A\cup B=4+8=12.$

${\textbf{[All element taken in both set A and B and elements in both sets are different.]}}$

${\textbf{Hence , maximum number of elements in}}$

$\mathbf{(A \cup B) =12.}$

How many of these students are taking physics?

0%
$n(P) = 45$
0%
$n(P) = 40$
0%
$n(P) = 50$
0%
$n(P) = 55$

Explanation

$n(P\cap G\cap E)=16$

$n(P\cap G)=24$

$n(P\cap E)=30$

$n(G\cap E)=22$

Only

$7$ students are taking Physics.

$\therefore$

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

$\Rightarrow$

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

$\Rightarrow$

$7=n(P)-54+16$

$\Rightarrow$

$n(P)=7+54-16$

$\therefore$

$n(P)=45$

Which one of the following is incorrect?

0%
Every subset of a finite set is finite
0%
$P=\left \{ x:x-8=-8 \right \}$ is a singleton set
0%
Every set has a proper set
0%
Every non-empty set has at least two subsets, $\phi$ and the set itself

The number of elements of the set

$\left \{ x:x\in Z,x^{2}=1 \right \}$ is :

0%
$3$
0%
$2$
0%
$1$
0%
$0$

Explanation

$x^2=1\Rightarrow x=1,-1$

Since both solutions are integers the set has

$2$ elements

State whether the following statement is True or False
If

$U=\left\{1,2,3,4,5,6,7\right\}$ and

$A=\left\{5,6,7\right\}$ , then

$U$ is the subset of

$A$ .

0%
True
0%
False

Explanation

Given,

$U=\left \{1,2,3,4,5,6,7\right\}$ and

$A=\left\{5,6,7\right\}$

We can say

$A$ is the subset of

$U$ .

Since, every element of

$A$ is also the element of

$U$ .

Let

$n(A)=28$ ,

$n(A\cap B)=8$ ,

$n(A\cup B)=52$ , then

$n(A\cap B')=$ .

0%
$30$
0%
$32$
0%
$20$
0%
none of these

Explanation

Given

$n(A)=28$ ,

$n(A\cap B)=8$ .

We have

$A\cap B'=A-A\cap B$ .

This give

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

or,

$n(A\cap B')=28-8=20$ .

From among the given alternatives select the one in which the set of numbers is most like the set of numbers given in the question.
Given set

$:$

$(7, 15, 31)$

0%
$7, 13, 28$
0%
$5, 13, 28$
0%
$9, 13, 26$
0%
$5, 13, 29$

Explanation

Let us find the Relation between the numbers of the set

$(7 , 15 , 31)$ .

$(7)\times 2 +1 = (15)$

$(15)\times 2 +1 = (31)$

Options

$A,B,C$ are not of similar type of above set

This similar type of relation is shown by option D

$(7 , 15 , 31)$ .

$(5)\times 2 +3 = (13)$

$(13)\times 2 +3 = (29)$

Hence, option D is correct.

$A \subset B$ , then

$A \cap B$ is

0%
$B$
0%
$A\setminus B$
0%
$A$
0%
$B \setminus A$

Explanation

We are given that

$A$ is the subset of

$B$

$\Rightarrow$ Every element of

$A$ is an element of

$B$ .

Therefore, the intersection elements of sets

$A$ and

$B$ are

$A\cap B=A$ .

Upper limit of class

$'41 - 50'$ is __________.

0%
$41$
0%
$50$
0%
$45$
0%
$91$

Explanation

Upper Limit of class

$'41-50'=50$

Part (B) is correct answer.

Let

$A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ . Then the number of subsets of

$A$ containing exactly two elements is

0%
$20$
0%
$40$
0%
$45$
0%
$90$

Explanation

Number 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

${}^{ 10 }{ C }_{2}=\dfrac { 10\times 9 }{ 2 } =45$

$\therefore$ Number of subsets of

$A$ containing exactly two elements

$=45$

State the following statement is True or False.
If we have two sets,

$A$ and

$B$ and every element of the set

$A$ is also the element of the set

$B$ . then we can say

$A$ is subset of the set

$B$ .

0%
True
0%
False

Explanation

If every element of set

$A$ is also the element of set

$B$ , then

$A$ is the subset of

$B$ .

Thus given statement is true.

In a class of

$60$ students,

$45$ students like music,

$50$ students like dancing,

$5$ students like neither. Then the number of students in the class who like both music and dancing is

0%
$35$
0%
$40$
0%
$50$
0%
$55$

Explanation

Total number of students in class

$=60$

Students who like music

$=X+Y=45$

Students who like dancing

$=Y+Z=50$

Students who like nothing

$=5$

Students who like both music and dancing

$=Y$

We know that

$X+Y+Z+5=60$

$\Rightarrow 45+Z+5=60$

$\Rightarrow Z=10$

$\Rightarrow Y=40$

Hence,

$40$ students like both music and dancing.

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 2 - 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 2 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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