MCQ Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 7

$A = \{ a, b, p, d\} B = \{ p, d, e\} C = \{p, e, f, g\}$ then find

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

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

In a class of 250 students, 175 take mathematics and 142 take science. How many take both mathematics

and science? (All take math and/or science.)

0%
67
0%
75
0%
33
0%
184
0%
cannot be determined from information given

If X is a finite set. Let

$P(X)$ denote the set of all subsets of X and let

$n(X)$ denote the number of elements in X. If for two finite subsets

$A, B, n(P(A)) = n(P(B)) + 15$ then

$n(B) =$ ____,

$n(A) =$ _____

0%
$n(A) = 4, n(B) = 0$
0%
$n(A) = 5, n(B) = 0$
0%
$n(A) = 4, n(B) = 2$
0%
$n(A) = 6, n(B) = 0$

Explanation

If X is a finite set.

Let,

Number of element in a subset

$A$ be

$n(A)$ =

$m$ and

Number of elements in subset

$B$ be

$n(B)$ =

$n$

Then total number of subsets of finite set contaning say

$n$ elements is

$2^n$ .

Therefore,

$n[P(A)]=2^m$ ,

$n[P(B)]=2^n$

Substituting in equations , we get:-

$2^m=2^n+15$

$2^m-2^n=15$

$m=4,n=0$

Option

$\textbf A$ is correct

If A and B are two sets, where A has more elements than B. Calculate the least possible value of n(A) + n(B), where n(A) is the number of elements in A and n(B) is the number of elements in B.

0%
$n(A)$
0%
$n(B)$
0%
$n(A+B)$
0%
None of these

Solve the following inequalities:

$\displaystyle\frac{1}{2-|x|}\geq 1$ .

0%
$x\epsilon (-8, -1]\cup [1,2)$
0%
$x\epsilon (-2, -1]\cup [1,2)$
0%
$x\epsilon (-2, -1]\cup [1,6)$
0%
$x\epsilon (-2, 0]\cup [1,2)$

Let

$S = \left \{(a, b, c)\epsilon N\times N\times N : a + b + c =a \leq b\leq c\right \}$ and

$T = \left \{a, b, c)\epsilon N\times N\times N : a, b, c,\ are\ in\ A.P.\right \}$ , where

$N$ is the set of all natural numbers. Then the number of elements in the set

$S\cap T$ is

0%
$6$
0%
$7$
0%
$13$
0%
$14$

Explanation

$a + b + c = 21$ and

$b = \dfrac {a + c}{2}$

$\Rightarrow a + c = 14$ and

$b = 7$
So, a can take values from

$1$ to

$6$ , when

$c$ ranges from

$13$ to

$8$ , or

$a = b = c = 7$
So,

$7$ triplets

$A=\left \{ 5,\left \{ 5,6 \right \},7 \right \}$ , which of the following is correct?

0%
$\left \{ 5,6 \right \}\in A$
0%
$\left \{ 5 \right \}\in A$
0%
$\left \{ 7 \right \}\in A$
0%
$\left \{ 6 \right \}\in A$

Explanation

$A =\{5,\{5,6\},7\}$ then the elements of

$A$ are

$5, \{5,6\},7$

Hence

$5\in A, \{5,6\}\in A$ and

$7\in A$

Note:

$5\ne\{5\}$

5 is an element of

$A$

i.e,

$5\in A$

where as

$\{5\}$ is a subset of

$A$

i.e,

$\{5\}\subset A$

$X=\left \{ a,\left \{ b,c \right \},d \right \}$ , which of the following is a subset of

$X$ ?

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

Explanation

Given

$X = \{a,\{b,c\},d\}$ . Hence the elements of

$X$ are

$a, \{b,c\}$ and

$d$ .

Subset of

$X$ are formed using elements of

$X$ .

Considering

$\{a,b\}, a \in X$ but

$b \notin X$ . Hence

$\{a,b\}$ is not a subset of

$X$ .

Considering

$\{b,c\}, b \notin X$ and

$c \notin X$ . Hence

$\{b,c\}$ is not a subset of

$X$ .

Considering

$\{c,d\}, c \notin X$ and

$d \in X$ . Hence

$\{c,d\}$ is not a subset of

$X$ .

Considering

$\{a,d\}, a \in X$ and

$d \in X$ . Hence

$\{a,d\}$ is a subset of

$X$ .

Which one of the following is a finite set?

0%
$\left \{ x:x\in Z,x< 5 \right \}$
0%
$\left \{ x:x\in W,x\geq 5 \right \}$
0%
$\left \{ x:x\in N,x> 10 \right \}$
0%
{ $x:x$ is an even prime number }

Explanation

(A)

Let

$A=\{x:x\in Z, x<5\}$

$A$ can be represented in roster form as

$A=\{........,-3,-2,..,3,4\}$ which is an infinite set

(B)

Let

$B=\{x:x\in W, x\ge 5\}$

$B$ can be represented in roster form as

$B=\{5,6,7,......\}$ which is an infinite set

(C)

Let

$C=\{x:x\in N, x>10\}$

$C$ can be represented in roster form as

$C=\{11,12,13,14,........\}$ which is an infinite set

(D)

Let

$D=\{x:x \text{ is an even prime number}\}$

The only prime number which is even is

$2$

So,

$D$ can be represented in roster form as

$D=\{2\}$ which is a finite set

Hence, the set in option D is a finite set.

Which one of the following is correct?

0%
$\left \{ x:x^{2}=-1,x\in Z \right \}=\phi$
0%
$\phi=0$
0%
$\phi=\left \{ 0 \right \}$
0%
$\phi=\left \{ \phi \right \}$

Explanation

$x^2=-1$ is satisfied by

$i$ and

$-i$ none of which are integers. Hence required set is null.

$n(A) = 20, n(B) = 30$ and

$n(A \cup B)= 40$ , then

$n(A \cap B)$ is equal to:

0%
$50$
0%
$10$
0%
$40$
0%
$70$

Explanation

We know that for any two finite sets

$A$ and

$B$ ,

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

Here, it is given that

$n(A)=20,n(B)=30$ and

$n(A\cup B)=40$ , therefore,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)\\ \Rightarrow 40=20+30-n(A\cap B)\\ \Rightarrow 40=50-n(A\cap B)\\ \Rightarrow n(A\cap B)=50-40\\ \Rightarrow n(A\cap B)=10$

Hence,

$n(A\cap B)=10$

For any three sets , A B and C,

$B\setminus (A \cup C)$ is:

0%
$(A\setminus B)\cap (A\setminus C)$
0%
$(B\setminus A)\cap (B \setminus C)$
0%
$(B \setminus A)\cap (A\setminus C)$
0%
$(A\setminus B)\cap (B \setminus C)$

Explanation

In the problem statement we are taking union of

$A$ and

$C$ and then taking its difference from

$B$

This means

$B$ \

$(A\cup C)$ will contain elements that are in

$B$ but not in both

$B$

$\textbf{and}$

$C$ .

$\therefore B$ \

$(A\cup C)$ is equivalent to taking difference of

$A,B$ and

$A,C$ and then taking there intersection i.e.

$(B$ \

$A)\cap (B$ \

$C)$

Hence option (

$b$ ) is correct.

Let U = {

$x \in N : 1 \le x \le 10$ } be the universal set,

$N$ being the set of natural numbers. If

$A = \{1, 2, 3, 4\}$ and

$B = \{2, 3, 6, 10\}$ then what is the complement of

$(A - B)$ ?

0%
$\{6, 10\}$
0%
$\{1, 4\}$
0%
$\{2, 3, 5, 6, 7, 8, 9, 10\}$
0%
$\{5, 6, 7, 8, 9, 10\}$

Explanation

Now

$U=\{ 1,2,3,4,5,6,7,8,9,10\}$

and

$(A-B)=\{ 1,4\}$

So the compliment of

$(A-B)$ considering

$U$ as the universal set is

$\{ 2,3,5,6,7,8,9,10\}$

Out of 500 first year students, 260 passed in the first semester and 21 0 passed in the second semester. If 170 did not pass in either semester, how many passed in both semesters ?

0%
$30$
0%
$40$
0%
$70$
0%
$140$

Explanation

Let A be the set of students who passed first semester so

$n(A)=260$

and B be the set of students who passed second semester so

$n(B)=210$ .

Now

$170$ did not passed any semester

So,

$(500-170=330)$ students passed atleast one of the semesters

$\therefore n(A\cup B)=330$

Now

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

$330=260+210-n(A\cap B)$

$n(A\cap B)=140$

Which one of the following is not true?

0%
$A\setminus B=A\cap B'$
0%
$A\setminus B=A\cap B$
0%
$A\setminus B=(A\cup B)\cap B'$
0%
$A\setminus B=(A\cup B)\setminus B$

Explanation

$A$ \

$B$ contains elements that are in set

$A$ and not in set

$B$

$A\cap B$ contains elements that are both in

$A$ and

$B$

Therefore option (

$b$ ) contains the incorrect statement.

If A and B are finite sets and

$A \subset B$ , then

0%
$n(A \cap B) = \phi$
0%
$n(A \cup B) = n(B)$
0%
$n(A \cap B) = n (B)$
0%
$n(A \cup B) = n (A)$

Explanation

A and B are finite set and

$A\subset B$

$A\subset B$ means A is a subset of B
If A is subset of B, then all elements of A are present in B

$\therefore n\left( A\bigcup { B } \right) =n\left( B \right)$

The set

$(A\cup B\cup C) \cap (A\cap B'\cap C')\cap C'$ is equal to?

0%
$B\cap C'$
0%
$A\cap C$
0%
$B' \cap C'$
0%
None of these

Explanation

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

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

$= [(A\cap A')\cup (B\cup C)]\cap C'$

$= (\phi \cup B\cup C)\cap C' = (B\cup C)\cap C'$

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

$= (B\cap C') \cup \phi = B\cap C'$ .

Let X be a set of

$5$ elements. The number d of ordered pairs (A, B) of subsets of X such that

$A\neq \Phi, B\neq \Phi, A\cap B=\Phi$ satisfies.

0%
$50\leq d \leq 100$
0%
$101\leq d \leq 150$
0%
$151 \leq d \leq 200$
0%
$201 \leq d$

Explanation

Total no. of ordered pairs is

${ =C }_{ 2 }^{ 5 }\times 2!+{ C }_{ 3 }^{ 5 }\left( \frac { 3! }{ 1!\times 2! } \times 2! \right) +{ C }_{ 4 }^{ 5 }\left( \frac { 4! }{ 1!\times 3! } \times 2!+\frac { 4! }{ 2!\times 2! } \times \frac { 2! }{ 2! } \right) +{ C }_{ 5 }^{ 5 }\left( \frac { 5! }{ 1!\times 4! } \times 2!+\frac { 5! }{ 2!\times 3! } \times 2! \right) \\ =10(2)+10(6)+5(8+6)(10+20)\\ =180\\$

So option

$C$ is correct

$A = \left \{(x, y) : x^{2} + y^{2} \leq 1; x, y \epsilon R\right \}$ and

$B = \left \{(x, y) : x^{2} + y^{2} \geq 4; x, y \epsilon R\right \}$ , then

0%
$A - B = \phi$
0%
$B - A = \phi$
0%
$A\cap B \neq \phi$
0%
$A\cap B = \phi$

Explanation

$A$ is the set of all points on the inner circle

$x^{2} + y^{2} = 1. B$ is the set of all points on the outer circle

$x^{2} + y^{2} = 4$ .

$\therefore A - B = A, B - A = B, A\cap B = \phi$ .

Consider the following:

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

$A\cap \left( B\cup C \right) =\left( A\cup B \right) \cap \left( A\cup C \right)$
Which of the above is/are correct?

0%
1 only
0%
2 only
0%
Both 1 and 2
0%
Neither 1 nor 2

Explanation

Figure

$1$ represents

$A\cup (B\cap C)$ and figure

$2$ represents

$(A\cap B)\cup (A\cap C)$ . We ca see both of them are not equal.

Figure

$3$ represents

$A\cap (B\cup C)$ and figure

$4$ represents

$(A\cup B)\cap (A\cup C)$ . We can see both of them are not equal.

So, both the equations are incorrect.

What is the percentage of persons who read only two papers ?

0%
$19\%$
0%
$31\%$
0%
$44\%$
0%
None of the above

Explanation

Let total numbers of people be 100, So for newspapers

$n(A)=42, n(B)=51, n(C)=68, n(A\cap B)=30$

$n(B\cap C)=28$

$n(A\cap C)=36$

Also given,

$n(\overline { A\cup B\cup C) } =8$

$n(A\cup B\cup C)=100-n(\overline { A\cup B\cup C) } =100-8=92$

Also,

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

$92=42+51+68-30-28-36+n(A\cap B\cap C)$

$n(A\cap B\cap C)=25$

Now, for people who read only two papers

$=n(A\cap B)+n(B\cap C)+n(A\cap C)-3n(A\cap B\cap C)$

$=30+28+36-75$

$=19$

Thus,

$19\%$ people read only two newspapers.

Hence, A is correct.

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.

0%
$140$
0%
$180$
0%
$210$
0%
$190$

Explanation

Total consumers

$=1000$
Like product

$A=n(A)=720$
Like product

$B=n(B)420$

$n(A\cap { B })$ (Both the products)

$=n(A)+n(B)-n(A\cup { B })$

$=720+420-1000$

$=140$

$A={1 , 11 , 21 , 31 ,....... , 541 , 551}$ . B is a subset of A such that

$x+y\neq552$ , for any

$x , y \epsilon B.$ The maximum number of elements in B is

0%
$26$
0%
$30$
0%
$29$
0%
$28$

In a group of

$50$ people, two tests were conducted, one for diabetes and one for blood pressure.

$30$ people were diagnosed with diabetes and

$40$ people were diagnosed with high blood pressure. What is the minimum number of people who were having diabetes and high blood pressure?

0%
$0$
0%
$10$
0%
$20$
0%
$30$

Explanation

Let

$A$ be the set containing people diagnosed with diabetes and

$B$ be the set containing people diagnosed with Blood pressure.

We know set formula

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

$n(A\cap B)=x$ are people having both

$n(A\cup B)$ are total people

$n(A)$ is people having diabetes and

$n(B)$ is having blood pressure.

$\Rightarrow 50+x=40+30\\ x=20$

Thus, there should be minimum of 20 people having both.

$A : \left \{x : x \text { is a multiple of } 3\right \}$ and

$B = \left \{x : x \text { is a multiple of } 4\right \}$ and

$C = \left \{x : x \text { is a multiple of } 12\right \}$ , then which one of the following is a null set?

0%
$(A / B) \cup C$
0%
$(A / B) / C$
0%
$(A \cap B)\cap C$
0%
$(A \cap B)/ C$

Explanation

$A : \left \{x : x \text { is a multiple of } 3\right \}$

$A=\{ 3,6,9,12,15,18,21,....\}$

$B = \left \{x : x \text { is a multiple of } 4\right \}$

$B=\{ 4,8,12,16,20,24,......\}$

$C = \left \{x : x \text { is a multiple of } 12\right \}$

$C=\{ 12,24,36,48,60,......\}$

Then,

$A\cap B=\{ 12,24,36,48,60,.....\}$

$\Longrightarrow (A\cap B)/C$ is a null set.

In a city, three daily newspapers A, B, C are published,

$42\%$ read A;

$51\%$ read B;

$68\%$ read C;

$30\%$ read A and B;

$28\%$ read B and C;

$36\%$ read A and C;

$8\%$ do not read any of the three newspapers.

What is the percentage of persons who read only one paper ?

0%
$38\%$
0%
$48\%$
0%
$51\%$
0%
None of the above.

Explanation

Let total numbers of people be 100, So for newspapers

$n(A)=42, n(B)=51, n(C)=68, n(A\cap B)=30$

$n(B\cap C)=28$

$n(A\cap C)=36$

Also given,

$n(\overline { A\cup B\cup C) } =8$

$n(A\cup B\cup C)=100-n(\overline { A\cup B\cup C) } =100-8=92$

Also,

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

$92=42+51+68-30-28-36+n(A\cap B\cap C)$

$n(A\cap B\cap C)=25$

Now,

People who read only one paper

$=n(A)+n(B)+n(C)-2n(A\cap B)-2n(B\cap C)-2n(A\cap C)+3n(A\cap B\cap C)$

$=42+51+68-60-56-72+75$

$=48$

Hence, B is correct.

$a\mathbb{N}=(an:n\in \mathbb{N})$ and

$b\mathbb{N}\cap c\mathbb{N}=d\mathbb{N}$ , where

$a,b,c\in \mathbb{N}$ and

$b,c$ are coprime, then

0%
$b=cd$
0%
$c=bd$
0%
$d=bc$
0%
None of these

Explanation

Given :

$a\mathbb{N}=(an:n\in \mathbb{N})$ and

$b\mathbb{N}\cap c\mathbb{N}=d\mathbb{N}$ , where

$a,b,c\in \mathbb{N}$ and

$b,c$ are coprime

We have,

$b\mathbb{N}=(bn:n\in \mathbb{N})$ ,

$c\mathbb{N}=(cn:n\in \mathbb{N})$

and

$d\mathbb{N}=(dn:n\in \mathbb{N})$

Also,

$bc\in b\mathbb{N} \cap c\mathbb{N}=d\mathbb{N}$

$\therefore$

$bc=d$ .....

$[\because b,c$ are coprime

$]$ .

A college awarded

$38$ medals in football,

$15$ in basketball and

$20$ in cricket. If these medals went to a total of

$58$ men and only three men got medals in all the three sports. Then the number of students who received medals in exactly two of the three sports, is

0%
$18$
0%
$15$
0%
$9$
0%
$6$

Explanation

$\textbf{Step – 1: Formulating the information to use Inclusion-Exclusion Rule.}$

$\text{Let number of men who have got medals in football, basketball, and cricket be represented as}$

$n(F),n(B)$ and

$n(C)$ .

$\text{Therefore the number of men who won in all the three sports will be}$ $n(F \cap B \cap C)$

$\text{and the number of people who got medals in two of the three sports is}$

$n(F \cap B) + n(B \cap C) + n(C \cap F)$

$\text{and the total number of men is represented as}$ $n(F \cup B \cup C)$ .

$\text{From the given data we have:}$

$n(F) = 38$

$n(B) = 15$

$n(C) = 20$

$n(F \cup B \cup C) = 58$

$n(F \cap B \cap C) = 3$

$\textbf{Step – 2: Apply the Inclusion-Exclusion Rule}$ .

$\text{Using the Inclusion-Exclusion Rule we get:}$

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

$\Rightarrow 58 = 38 + 15 + 20 - n(F \cap B) - n(B \cap C) - n(C \cap F) + 3$

$\Rightarrow n(F \cap B) + n(B \cap C) + n(C \cap F) = 38 + 15 + 20 + 3 - 58$

$\Rightarrow n(F \cap B) + n(B \cap C) + n(C \cap F) = 18$

$\therefore\text{ the number of students who have received medals in exactly two of the three sports is}$

${= n(F \cap B) + n(B \cap C) + n(C \cap F) - 3 \times n(F \cap B \cap C) }$

${= 18 - 3 \times 3 }$

${= 18 - 9}$

${= 9}$

$\textbf{Hence, the number of students who have received medals in exactly two of the three sports is 9}$

$S$ is a set with

$10$ elements and

$A = \left \{(x, y) : x, y\epsilon S, x\neq y\right \}$ , then number of elements in

$A$ is

0%
$100$
0%
$90$
0%
$50$
0%
$45$

Explanation

Given number of elements in

$S = 10$
and

$A = \left \{(x, y) : x, y\epsilon S, x\neq y\right \}$

$\therefore$ Number of elements in

$A = 10\times 9 = 90$ .

$A=\left\{ a,b,c \right\} ,$

$B=\left\{ b,c,d \right\}$ and

$C=\left\{ a,d,c \right\}$ , then

$\left( A-B \right) \times \left( B\cap C \right)$ is equal to

0%
$\left\{ \left( a,c \right) ,\left( a,d \right) \right\}$
0%
$\left\{ \left( a,b \right) ,\left( c,d \right) \right\}$
0%
$\left\{ \left( c,a \right) ,\left( d,a \right) \right\}$
0%
$\left\{ \left( a,c \right) ,\left( a,d \right) ,\left( b,d \right) \right\}$

Explanation

Given that;

$A=\left\{ a,b,c \right\} ,$

$B=\left\{ b,c,d \right\}$ and

$C=\left\{ a,d,c \right\}$

Now,

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

$B\cap C=\left\{ b,c,d \right\} \cap \left\{ a,d,c \right\}$

$=\left\{ c,d \right\}$

Therefore,

$\left( A-B \right) \times \left( B\cap C \right) =\left\{ a \right\} \times \left\{ c,d \right\}$

$=\left\{ \left( a,c \right) ,\left( a,d \right) \right\}$

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 7 - 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 7 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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