MCQ Questions for Class 11 Engineering Maths Sets Quiz 13

Let R = {(1, 3), (4, 2), (2, 3), (3, 1)} be a relation on the set A = (1, 2, 3, 4). The relation R is

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Transitive
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Symmetric
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Reflexive
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None of these

$13$ Planar Surfaces, each of area

$1\ m^{2}$ are such that their total area is

$7\ m^{2}$ . If no three of more of these surfaces have any region in common; Then, the overlap of some Two of These surfaces has an Area.

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Not less than $\dfrac{7}{13} m^{2}$
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Not less than $\dfrac{2}{13} m^{2}$
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Not less than $\dfrac{1}{13} m^{2}$
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Not Less than $\dfrac{1}{17} m^{2}$

Explanation

Assume are of overlap of any

$2$ surfaces is less than

$\dfrac{1}{13}m^2$

Let us ignore the unit of area in our notation.

We know that,

$Area(A_{1}\cup A_{2}) = Area(A_{1}) + Area(A_{2}) - Area(A_{1}\cap A_{2})$

But,

$Area (A_{1}) = Area (A_{2}) = 1.$ (as given in question)
Also,

$Area(A_{1}\cap A_{2}) < \dfrac {1}{13}$ (as per our assumption)
Hence,

$Area(A_{1}\cup A_{2}) > 1 + 1 - \dfrac {1}{13}$

$Area(A_{1}\cup A_{2}) > 2 - \dfrac {1}{13}$

Similarly,

$Area (A_{1}\cup A_{2}\cup A_{3}) = Area (A_{1}) + Area(A_{2}) + Area(A_{3})$

$- Area(A_{1}\cap A_{2})- Area (A_{1}\cap A_{3}) - Area(A_{2} \cap A_{3})$

$+ Area (A_{1}\cap A_{2}\cap A_{3})$

But

$Area(A_{1}\cap A_{2}\cap A_{3}) = 0$ (As per the condition in Question)
Thus,

$Area(A_{1}\cup A_{2}\cup A_{3}) > 1 + 1 + 1 - \dfrac {1}{13} - \dfrac {1}{13} - \dfrac {1}{13} + 0$ .
i.e.

$Area(A_{1} \cup A_{2} \cup A_{3}) > 3 - \dfrac {3}{13}$ .

Similarly,

$Area(A_{1}\cup A_{2}\cup A_{3}\cup A_{4}) = Area(A_{1}) + Area (A_{2}) + Area(A_{3}) + Area(A_{4})$

$- Area(A_{1}\cap A_{2}) - Area(A_{1}\cap A_{3}) - Area (A_{1} \cap A_{4})$

$- Area(A_{2} \cap A_{3}) - Area(A_{2} \cap A_{4}) - Area (A_{3} \cap A_{4})$

$+ Area(A_{1}\cap A_{2}\cap A_{3})+ Area(A_{1}\cap A_{2}\cap A_{4})$

$+ Area (A_{2}\cap A_{3}\cap A_{4}) - Area (A_{1}\cap A_{2}\cap A_{3}\cap A_{4})$

But

$Area(A_{1}\cap A_{2}\cap A_{3}\cap A_{4}) = 0$

$Area (A_{1} \cap A_{2}\cap A_{3}) = 0$ , etc. (as per the condition in Question)
So,

$Area(A_{1}\cup A_{2}\cup A_{3}\cup A_{4}) > 1 + 1 + 1 + 1 - 6\left (\dfrac {1}{13}\right ) + 3(0) - 0$ .
i.e.

$Area(A_{1}\cup A_{2}\cup A_{3}\cup A_{4}) > 4 - \dfrac {6}{13}$

In general,

$Area(A_{1}\cup A_{2}\cup A_{3}\cup ...\cup A_{n}) > n - \dfrac {^{n}C_{2}}{13}$

Put

$n = 13$ ;

$Area(A_{1}\cup A_{2}\cup A_{3}\cup ....\cup A_{13}) > 13 - \dfrac {^{13}C_{2}}{13}$

That is

$Area(A_{1}\cup A_{2}\cup A_{3}\cup ....\cup A_{13}) > 7$

Hence, our assumption was wrong.

So, Area of overlap of any

$2$ surfaces is not less than

$\dfrac {1}{13}$ .

Consider that

$n(S)$ represented the number of elements in set S. If

$n(A\cup B\cup C)=40, n(A\cap B'\cap C')=5, n(B\cap A'\cap C')=10, n(C\cap B' \cap A')=6$ then number of element which belongs to at least two of the set is

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Less than $19$
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More than $21$ but less than $40$
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$19$
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$20$

Explanation

Given,

$n(A\cup B\cup C)=40,n(A'\cap B'\cap C')=5$

$n(B\cap A'\cap C')=10,n(C\cap B'\cap A')=6$

Then number of element Which belongs to at least two of the sets is-

$\le n(A\cup B\cup C)- n(A'\cap B'\cap C')-n(B\cap A'\cap C')-n(C\cap B'\cap A')$

$\le 40-5-10-6$

$\le 19$

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

$n(A\cup B)$ is equal to

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373
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165
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370
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none of these

Explanation

$n(A)=115$

$n(B)=326$

$n(A-B)=47$

$\therefore$

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

$=115-47=68$

$\therefore$

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

$=115+326-68$

$=326+47$

$=373$

Option A.

For 3 sets A,B,C if A

$\subset B,B\subset C$ then

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A $\cup B\subset C$
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C $\subset A\cup B$
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A-B=C
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None of these

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

$\underset{i = 1}{\overset{30}{\cup}} A_i = \underset{j = 1}{\overset{n}{\cup}} B_j = S$ and each element of S belongs to exactly 10 of the

$A_i's$ and exactly 9 of the

$B_j'S$ . then n is equal to

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15
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3
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45
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35

Explanation

If elements are not repeated then the number of elements in

$A_1\cup A_2\cup A_3\cup A_{30}$ is

$30\times 5$

But each elements is used 10 times

so,

$S=\cfrac{30\times 5}{10}=15$

If the elements in

$B_1,B_2,.....,B_n$ are not repeated.

then total number of element in

$B_1\cup B_2\cup B_3.....\cup B_n$ is 3n

but each elements is repeated 9 times

so,

$S=\cfrac{3n}{9}=15=\cfrac{3n}{9}$

$n=45$

Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set

$A \times B$ , each having at least three elements is............

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$275$
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$510$
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$219$
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$256$

Explanation

Set A has

$4$ elements

Set B has

$2$ elements

Number of elements in set

$\left( A\times B\right)=4\times 2=8$

Total number of subsets of

$\left( A\times B\right) =2^8=256$

Number of subsets having

$0$ elements

$=^{ 8 }{ { C }_{ 0 } }=1$

Number of subsets having

$1$ elements each

$=^{ 8 }{ { C }_{ 1 } }=8$

Number of subsets having

$2$ elements each

$=^{ 8 }{ { C }_{ 2 } }=\dfrac{8!}{2!6!}=28$

Number of subsets having atleast

$3$ elements

$=256-1-8-28$

$=256-37$

$=219$

Hence, the answer is

$219.$

The value of

$\left( {A \cup B \cup C} \right) \cap \left( {A \cap {B^c} \cap {C^c}} \right) \cap {C^c}$ is

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$B \cap {C^c}$
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${B^c} \cap {C^c}$
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$B \cap {C}$
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$A \cap {B^c} \cap {C^c}$

If A =[a,b],B={c,d} and C={d,e}, then { (a,c),(a,d),(a,e),(b,c),(b,d),(b,e) is equal to

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

The number of subsets

$R\,of\,P\, = \left\{ {1,2,3,....8} \right\}$ which satisfies the property 'There exist integers

$a < \,b < \,c$ with

$a\, \in \,R,b\, \notin \,R,c\, \in \,R''\,$ is

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215
0%
219
0%
222
0%
223

Given

$n(U) =20, n(A) =12, n(B) =9, n(A \cap B) =4$ , where U is the universal set, A and B are subset of U, then

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

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$17$
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$9$
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$11$
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$3$

if S is a set of p(x) is polynomial of degree <2 such that p(0)=, P(1)=1, p(x)>0

$\forall \quad x\quad \varepsilon$ (0, 1) then

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S=0
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$S=ax+91-1){ x }^{ 2 }\forall a\varepsilon (0,\infty )$
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$S=ax+(1-a){ x }^{ 2 }\forall a\varepsilon R$
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$S=ax+(1-a){ x }^{ 2 }\forall a\varepsilon R(0,\quad 2)$

Let

$A,B,C$ finite sets. Suppose then

$n(A)=10, n(B)=15, n(C)=20, n(A\cap B)=8$ and

$n(B\cap C)=9$ . Then the possible value of

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

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$26$
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$27$
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$28$
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Any of the three values $26, 27, 28$ is possible

X and Y are two sets and

$f:X\rightarrow Y$ . If

$f(c)=\left\{ y;c\subset X,y\subset Y \right\}$ and

${ f }^{ 1 }(d)=\left\{ x;d\subset Y,x\subset X \right\}$ , then the true statement is

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$f({ f }^{ 1 }(b))=b$
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$f({ f }^{ 1 }(a))=a$
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$f({ f }^{ 1 }(b))=b,b\subset y$
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$f({ f }^{ 1 }(a))=a,a\subset x$

A = {n/n is a digits in the number 33591} and

$B=\left\{ n/n\in N,n<10 \right\} ,$ then

$B-A =$

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$\left\{ 2,4,6,8 \right\}$
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$\left\{ 7,2,4,8,6 \right\}$
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$\left\{ 1,3,5,7 \right\}$
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$\left\{ \left( 1,2 \right) ,\left( 1,3 \right) ,\left( 2,3 \right) \right\}$

Let A, B, C be three seta such that A

$\cup$ B

$\cup$ C =

$U$ , where

$U$ is the universal set then ,
[(A B)

$\cup$ (B C)

$\cup$ (C A)] is equal to

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

Let

$n\left( U \right) =700,n\left( A \right) =200,n\left( B \right) =300$ and

$n\left( A\cap B \right) =100,$ then

$n\left( { A }^{ c }\cap { B }^{ c } \right) =$

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400
0%
600
0%
300
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200

Which set is most like the given set?
(8, 18, 37)

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(4, 9, 20)
0%
(16, 33, 67)
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(13, 30, 67)
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(5, 12, 25)

$A = \{ 4,5,8,12 \} , B = \{ 1,4,6,9 \} \text { and } C = \{ 1,2,3,4 \}$ then

$A - ( C - B ) =$

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$A$
0%
$B$
0%
$C$
0%
$\phi$

Tell whether set A is a subset of set B.

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set A : whole numbers less than 8
set B : whole numbers less than 10
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set A : prime numbers
set B : odd numbers
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set A : numbers divisible by 6
set B : numbers divisible by 3
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set A : set of letters in the word 'FLAT'
set B : set of letters in the word 'PLATE'

The solution set of

$x^{2}+5x+6=0$ is ........

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{2,3}
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{-2,-3}
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{2,-3}
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{-2,3}

$P = \{ x:x < 3,x \in N\}$ and

$Q = \{ x:x \le 2,x \in W\}$ , where W is the set of whole numbers then the set of whole numbers then the set

$(P \cup Q) \times (P \cap Q)$ is

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${(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)}$
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$(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)$
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$(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)$
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$(1,1),(1,2),(2,1),(2,2)$

$\text{A} = \{ 1,2,3,4,5\} \,,\text{B} = \{ 2,3,5,6\}$ and

$\text{C} = \{ 4,5,6,7\}$
Then, which of the following is true?

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$A \cup \left( {B \cap C} \right) = \left( {A \cup B} \right) \cap (A \cup C)$
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${A \cup B} =\{1,2,3,4,5\}$
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${A \cup B} =\{1,2,3,4,5,6\}$
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Both $\text{A}$ and $\text{C}$

$n\left( {P \cap Q} \right) = 23,\,\,n\left( {P \cup Q} \right) = 57,$ and

$n\left( {Q - P} \right) = 26,$ then

$n\left( {P - Q} \right) = \_\_\_\_\_\_\_\_.$

0%
24
0%
54
0%
8
0%
14

if P and Q are two sets having 5 elements in com- mon, then how many elenents do P

$\times$ Q and Q

$\times$ P have in common?

0%
5
0%
10
0%
25
0%
20

$If\,\,A = \left\{ {x:{x^2} = 1} \right\}\,\,\,and\,B = \left\{ {x:{x^4} = 1} \right\},\,then\,A\Delta B\,\,\,is\,\,equal\,to\,$

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$\left\{ {i, - i} \right\}\,$
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$\,\left\{ { - 1,1} \right\}\,$
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$\left\{ { - 1,1,i - i} \right\}$
0%
$\,\left\{ {1,i} \right\}$

Let

$A$ and

$B$ be two sets. The

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

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

Examine whether the following statements are true or false:

$\left\{a\right\}\subset \left\{\{a\},\, b \right\}$

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True
0%
False

CBSE Questions for Class 11 Engineering Maths Sets Quiz 13 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Sets Quiz 13 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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