MCQ Questions for Class 11 Engineering Maths Sets Quiz 1

If P is true and Q is false, then Bi-implication of p and q is

0%
True
0%
False

For any two sets

$A$ and

$B, A = B$ is equivalent to

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

Explanation

(A)

$A - B = B - A$

$A - A = A - A$ [

$\because A = B$ ]

(B)

$A\cup B$ =

$A \cap B$

$A\cup B$ =

$A\cap B$ [

$\because A = B$ ]

(C)

$A\cup C$ =

$B\cup C$

$A\cup C$ =

$A\cup C$ [

$\because A = B$ ]

$A\cap B$ =

$\phi$

$A\neq \phi$

$A\cap A=A$ [

$\because A = B$ ]

So (D) is not correct.

If A = {1, 2, 3, 4, 5, 6, 7, 8} and B = {1, 3, 5, 7}, then find

$A - B$ and

$A \cap B$

0%
{ 3 , 5 } and {2, 4, 6}
0%
{2, 4, 6} and (1, 5}
0%
{2, 4, 6, 7} and {1, 3, 5, 6}
0%
{2, 4, 6, 8} and {1, 3, 5, 7}

Explanation

$A = \{1, 2, 3, 4, 5, 6, 7, 8\}$

Very clearly

$A\cap B = \{1,3,5,7\}$

$A-B$ is nothing but

$A-A\cap B$

$A-B = \{2, 4, 6, 8\}$

Of course option D is the correct answer.

Suppose ${ A }_{ 1 },{ A }_{ 2 },...,{ A }_{ 30 }$ are thirty sets, each with five elements and ${ B }_{ 1 },{ B }_{ 2 },...,{ B }_{ 30 }$ are $n$ sets ecah with three elements. Let $\displaystyle \bigcup _{ i=1 }^{ 30 }{ { A }_{ i }= } \bigcup _{ j=1 }^{ n }{ { B }_{ j } } =S$

If each element of $S$ belongs to exactly ten of the ${ A }_{ i }'s$ and exactly none of the ${ B }_{ j }'s$ then $n=$

0%
45
0%
35
0%
40
0%
none of these

Explanation

Given

${ A }_{ i }$ 's are thirty sets with five elements each, so

$\displaystyle \sum _{ i=1 }^{ 30 }{ n\left( { A }_{ i } \right) } =5\times 30=150$ ...(1)

If there are

$m$ distinct elements in

$S$ and each element of

$S$ belongs to exactly

$10$ of the

${ A }_{ i }$ 's, we have

$\displaystyle \sum _{ i=1 }^{ 30 }{ n\left( { A }_{ i } \right) } =10m$ ...(2)

$\therefore$ from (1) and (2), we get

$10m=150\Rightarrow m=15$ ...(3)

Similarly

$\displaystyle \sum _{ j=1 }^{ 30 }{ n\left( { B }_{ j } \right) } =3n$ and

$\displaystyle \sum _{ j=1 }^{ 30 }{ n\left( { B }_{ j } \right) } =9m$

$\therefore 3n=9m \displaystyle \Rightarrow \frac { 9m }{ 3 } =3m=3\times 15=45$ (from (3))

Hence,

$n=45$

$\bigcup \phi =$

0%
$\phi$
0%
2
0%
A
0%
0

Explanation

We've,

$A\cup \phi$

$=A$ . [ Using direct formula (consistency property), since

$\phi \subset A$ ]

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

$A-B=$

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

Let $n$ be a fixed positive integer. Let a relation $R$ defined on $I$ (the set of all integers) as follows: $aRb$ iff $n/(a-b)$ , that is, iff $a-b$ is divisible by $n$ , then, the relation $R$ is

0%
Reflexive only
0%
Symmetric only
0%
Transitive only
0%
An equivalence relation

Explanation

$R$ is reflexive since for any integer

$a$ we have

$a-a=0$ and

$0$ is divisible by

$n$ .
Hence,

$aRa\quad \forall a\in I$ .

$R$ is symmetric, let

$aRb$ .

Then by definition of

$R$ ,

$a-b=nk$ where

$k\in I$ .

Hence

$b-a=(-k)n$ where

$-k\in I$ and so

$bRa$ .

Thus we have shown that

$aRb\Rightarrow bRa$ .

$R$ is transitive, let

$aRb$ and

$bRc$ .

Then by definition of

$R$ , we have

$a-b={ k }_{ 1 }n$ and

$b-c={ k }_{ 2 }n$ , where

${ k }_{ 1 },{ k }_{ 2 }\in I$ .

It then follows that

$a-c=\left( a-b \right) +\left( b-c \right) ={ k }_{ 1 }n+{ k }_{ 2 }n=\left( { k }_{ 1 }+{ k }_{ 2 } \right) n$

where

${ k }_{ 1 }+{ k }_{ 2 }\in I$

Let

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

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

$A \triangle\ B$ is equal to

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

Explanation

We have,

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

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

We know that,

$A\Delta B =(B-A)\cup(A-B)$

Therefore,

$A\Delta B =\{5, 6\} \cup \{1\}$

$A\Delta B =\{1, 5, 6\}$

Hence, this is the answer.

If X

$=$ (multiples of 2), Y

$=$ (multiples of 5), Z

$=$ (multiples of 10), then

$X \cap ( Y \cap Z )$ is equal to

0%
multiples of 10
0%
multiples of 5
0%
multiples of 2
0%
multiples of 7

Explanation

$We\quad can\quad write\quad the\quad given\quad sets\quad as\quad \\ X=\quad \left\{ 2,4,6,8,10,-----18,20,----28,30,---- \right\} \\ Y=\quad \left\{ 5,10,15,20,----30----40---- \right\} \\ Z=\quad \left\{ 10,20,30,40,----- \right\} \\ \therefore \quad Y\cap Z=\left\{ 10,20,30,---- \right\} \\ \therefore \quad X\cap \left( Y\cap Z \right) =\left\{ 10,20,30,----- \right\} =Z\\ \therefore \quad X\cap \left( Y\cap Z \right) =Z\quad \quad (Ans)$

In a locality two-thirds of the people have cable TV one-fifth have Dish TV and one-tenth have both What is the fraction of people having either cable TV or Dish TV?

0%
$\displaystyle \frac{2}{3}$
0%
$\dfrac13$
0%
$\dfrac45$
0%
$\dfrac35$

Explanation

Fraction of people who watch cable only

$\displaystyle =\left ( \frac{2}{3}-\frac{1}{10} \right )=\frac{17}{30}$
Fraction of people who watch Dish TV only

$\displaystyle =\left ( \frac{1}{5}-\frac{1}{10} \right )=\frac{1}{10}$

$\displaystyle\therefore$ Fraction of people who watch either cable of Dish TV

$\displaystyle =\frac{17}{30}+\frac{1}{10}=\frac{20}{30}=\frac{2}{3}$

Let

$P =$ Set of all integral multiples of

$3$ ;

$Q =$ Set of integral multiples of

$4$ ;

$R =$ Set of all integral multiples of

$6$ . Consider the following relations :

$1$

$\displaystyle P\cup Q=R$

$2.$

$\displaystyle P\subset R$

$3.$

$\displaystyle R\subset \left ( P\cup Q \right )$
Which of the relations given above is/are correct ?

0%
only $1$
0%
only $2$
0%
only $3$
0%
both $2$ and $3$

Explanation

Given that P =

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

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

$\{6, 12, 18,...\}$
Considering the choices given :
1. P

$\displaystyle \cup$ Q =

$\{3, 4, 6, 8, 9, 12, 16, 18,...\}$

$\displaystyle \neq$ R
2. All the element of P are not in R so P

$\displaystyle \nsubseteq$ R
3. P

$\displaystyle \cup$ Q =

$\{3, 4, 6, 8, 9, 12, 15, 16, 18, ...\}$

$\displaystyle \Rightarrow$ All the element of R are in P

$\displaystyle \cup$ Q

$\displaystyle \Rightarrow$ R

$\displaystyle \subset$ (P

$\displaystyle \cup$ Q)

State true or false.

Given universal set=

$\displaystyle =\left \{ -6,-5\frac{3}{4}, -\sqrt{4}, -\frac{3}{5}, -\frac{3}{8}, 0, \frac{4}{5}, 1, 1\frac{2}{3}, \sqrt{8}, 3.01, \pi , 8.47 \right \}$

From the given set, find set of non-negative integers is

$\displaystyle \left \{0,1 \right \}$ .

0%
True
0%
False

Explanation

An integer is a number that has no fractional part, and no digits after the decimal point. An integer can be positive, negative or zero.

$0$ and

$1$ are the only non-negative integers, from the given set.
set of non-negative integers is

$\{0,1\}$ .

The given statement is true.

set of irrational numbers is

$\displaystyle \left \{ \sqrt{8}, \pi \right \}$

0%
True
0%
False

Explanation

Rational numbers are those numbers which can be expressed in the form $\frac {p}{q}$ , where p and q are integers and $q \neq 0$

From the given set of numbers, $-6, -5\frac{3}{4}, -\frac{3}{5}, -\frac{3}{8}, 0, \frac {4}{5}, 1, 1, \frac{2}{3}, 3.01, 8.47$ are clearly rational numbers are they can be written in $\frac {p}{q}$ form.
Now, $- \sqrt {4} = -2$ which is also a rational number.
And, $\sqrt {8} = 2\sqrt {2}$ is not a rational number.
Also, $\pi$ is not a rational number.
Hence, the irrational numbers in the given set are { $\sqrt {8}, \pi$ }

set of rational numbers is

$\displaystyle \left \{ -6,-5\frac{3}{4}, -\sqrt{4}, -\frac{3}{5}, -\frac{3}{8}, 0, \frac{4}{5}, 1, 1\frac{2}{3}, 3.01, 8.47 \right \}$

0%
True
0%
False

Explanation

Rational numbers are those numbers which can be expressed in the form

$\frac {p}{q}$ , where p and q are integers and

$q \neq 0$
From the given set of numbers,

$-6, -5\frac{3}{4}, -\frac{3}{5}, -\frac{3}{8}, 0, \frac {4}{5}, 1, 1, \frac{2}{3}, 3.01, 8.47$ are clearly rational numbers are they can be written in

$\frac {p}{q}$ form.

Now,

$- \sqrt {4} = -2$ which is also a rational number.
And,

$\sqrt {8} = 2\sqrt {2}$ is not a rational number.
Also,

$\pi$ is not a rational number.
Hence, the rational numbers in the given set are {

$-6, -5\frac{3}{4}, - \sqrt {4}, -\frac{3}{5}, -\frac{3}{8}, 0, \frac {4}{5}, 1, 1, \frac{2}{3}, 3.01, 8.47$ }

State true or false.

Given universal set=

$\displaystyle =\left \{ -6,-5\frac{3}{4}, -\sqrt{4}, -\frac{3}{5}, -\frac{3}{8}, 0, \frac{4}{5}, 1, 1\frac{2}{3}, \sqrt{8}, 3.01, \pi , 8.47 \right \}$

From the given set, the set of integers is

$\displaystyle \left \{ -6,-\sqrt{4}, 0,1 \right \}$ .

0%
True
0%
False

Explanation

An integer is a number that has no fractional part, and no digits after the decimal point. An integer can be positive, negative or zero.

All integers from given set are

$-6,-\sqrt{4}(-2),0,1$
set of integers

$=\displaystyle \left \{ -6,-\sqrt{4}, 0,1 \right \}$

The given statement is true.

In an examination

$70\%$ students passed both in Mathematics and Physics

$85\%$ passed in Mathematics and

$80\%$ passed in Physics If

$30$ students have failed in both the subjects then the total number of students who appeared in the examination is equal to :

0%
$900$
0%
$600$
0%
$150$
0%
$100$

Explanation

Student passed in atleast one subject

$= n$

$\displaystyle \left ( P\cup M \right )= n\left ( P \right )+n\left ( M \right )-n\left ( P\cup M \right )$

$= 80 + 85 - 70 = 95$

$\displaystyle \therefore$

$5\%$ student failed in both the subjects

$\displaystyle \Rightarrow$

$5\%$ of total students

$= 30$

$\displaystyle \Rightarrow$ Total students =

$\displaystyle \frac{30\times 100}{5}= 600$

In a class of

$50$ students

$35$ opted for Mathematics and

$37$ opted for Biology How may have opted for only Mathematics? ( Assume that each student has to opt for at least one of the subjects)

0%
$15$
0%
$17$
0%
$13$
0%
$19$

Explanation

Here

$n$

$\displaystyle \left ( M\cup B \right )$

$= 50$ ,

$n(M) = 35$ ,

$n(B) = 37$

$\displaystyle \therefore n\left ( M\cap B \right )=n\left ( M \right )+n(B)-n\left ( M\cup B \right )$

$= 35 + 37 - 50 = 22$

$\displaystyle \Rightarrow$

$22$ student have opted for both Mathematics and Biology.

Now the number of students who have opted for Mathematics only

$= n(M) - n$

$\displaystyle \left ( M\cap B \right )$

$= 35 - 22 = 13$

$n(A) = 65, n(B) = 32$ and

$\displaystyle n\left ( A\cap B \right )=14$ , then

$\displaystyle n\left ( A\Delta B \right )$ equals

0%
$65$
0%
$47$
0%
$97$
0%
$69$

Explanation

$\displaystyle n(A\Delta B)=n (A-B)+n(B-A)$

$\displaystyle \therefore A\Delta B=(A-B)\cup (B-A)$

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

$\Rightarrow A\Delta B =n(A)+n(B)-2n(A\cap B)=65+32-2\times 14=69$

Hence,

$A\Delta B=69$

$n(A) = 115$ ,

$n(B) = 326$ ,

$n(A - B) = 47$ then

$\displaystyle n\left ( A\cup B \right )$ is equal to

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

Explanation

$n(A)=115,n(B)=326$

$n(A-B)=47$

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

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

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

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

$\Rightarrow n(A\cup B)=115+326-68$

$\Rightarrow n(A\cup B)=373$

$X$ and

$Y$ are any two non empty sets then what is

$\displaystyle \left ( X-Y \right )'$ equal to?

0%
$X' - Y'$
0%
$\displaystyle {X}'\cap Y$
0%
$\displaystyle {X}'\cup Y$
0%
$X - Y'$

Explanation

$X - Y$

$\displaystyle =\left \{ x :x \in X,x\notin Y \right \}$

$\displaystyle =\left \{ x :x \in X,x\in Y'\right \}$

$\displaystyle \Rightarrow \left \{ x : x\in X\cap Y' \right \}$

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

$\displaystyle X'\cup (Y')=X'\cup Y$

$A$ and

$B$ are non empty sets and A' and B' represents their compliments respectively then

0%
$A - B = A' - B'$
0%
$A - A' = B - B'$
0%
$A - B = B' - A'$
0%
$A - B' = A' - B$

Explanation

Let

$U \to$ Universal set

$X\to U - (A+B)$

$B'=X+A$

$A'=X+B$

$B'-A'= X+A-(X+B)$

$=X+A-X-B$

$B'-A'=A-B$

$\displaystyle \xi =\left \{ 2,3,4,5,6,7,8,9,10,11 \right \}$

$\displaystyle A =\left \{ 3,5,7,9,11 \right \}$

$\displaystyle B =\left \{ 7,8,9,10,11 \right \}$ , then find

$(A - B)'$

0%
$(A - B)' = \{2,4,6,7,8,9,10,11\}$
0%
$(A - B)' = \{2,4,6,7,8,9,11\}$
0%
$(A - B)' = \{2,4,6,7,9,10,11\}$
0%
$(A - B)' = \{2,4,6,8,9,10,11\}$

Explanation

$A - B = \{3, 5\}$

$(A - B)' = \{2,4,6,7,8,9,10,11\}$

$\displaystyle Q=\left\{ x:x=\frac { 1 }{ y } ,where\ \ y\ \in \ N \right\}$ , then find the correct one.

0%
$\displaystyle 0\quad \in \quad Q$
0%
$\displaystyle 1\quad \in \quad Q$
0%
$\displaystyle 2\quad \in \quad Q$
0%
$\displaystyle \frac { 2 }{ 3 } \quad \in \quad Q$

Explanation

(A) Since

$\displaystyle y\neq\frac{1}{0}$ then

$0\notin Q$

(B) Since

$\displaystyle y=\frac{1}{2}$ as

$1\in N$ then

$1\in Q$

$\displaystyle y\neq\frac{1}{2}$ as

$\displaystyle\frac{1}{2}\notin N$ then

$2\notin Q$

(D) Since

$\displaystyle y\neq\frac{3}{2}$ as

$\displaystyle\frac{3}{2}\notin N$ then

$\displaystyle\frac{2}{3}\notin Q$

Which set is the subset of the set containing all the whole numbers?

0%
$\{1,2,3,4,.......\}$
0%
$\{1\}$
0%
$\{0\}$
0%
All of the above

The set of all those elements of A and B which are common to both is called

0%
union of two sets
0%
intersection of two sets
0%
disjoint sets
0%
none of these

Explanation

The set of all those elements of A and B which are common to both is called A intersection B=

$A\cap B$

In Question some relationship have been expressed through symbols as defined below : (N-83)
+ - x

$\displaystyle\div$ = > <
V

$\displaystyle\Lambda$ ( ) U

$\displaystyle\cap$ 0
In question only one of the five relationship is correct Find the correct one and encircle its serial number on the space provided against the question ______________.

0%
24 $\displaystyle\cap$ 3 ( 4 V 2 ) 8
0%
24 U 3 V 4 $\displaystyle\Lambda$ 2 ) 8
0%
24 ( 3 0 4 ) 2 V 8
0%
24 0 3 ( 4 $\displaystyle\Lambda$ 2 V 8
0%
24 $\displaystyle\Lambda$ 3 V 4 U 2 ( 8

Set

$A$ has

$3$ elements and set

$B$ has

$6$ elements. What can be the minimum number of elements in

$A\cup B$ ?

0%
$6$
0%
$3$
0%
$9$
0%
$18$

Explanation

$A\cup B$ must contain all the elements of the bigger set B.

Given

$A=\{a,b,c,d,e,f,g,h\}$ and

$B=\{a,e,i,o,u\}$

then

$B-A$ is equal to

0%
$\{i,o,u\}$
0%
$\{a,b,c\}$
0%
$\{c,d,e\}$
0%
$\{a,,i,z\}$

Explanation

The sets

$A=\{a,b,c,d,e,f,g,h\}$ and

$B=\{a,e,i,o,u\},$ in order to find the difference between the two sets as

$B-A,$ we begin by writing all the elements of

$B$ and then take away every element of

$A$ which is also the element of

$B$ . Since

$B$ share the elements

$a,e$ with

$A$ , so

$B-A=\{i,o,u\}.$

Let

$A$

$=$ set of all cuboids and B

$=$ set of all cubes. Which of the following is true?

0%
$A\subset B$
0%
$B\subset A$
0%
$B=A$
0%
$B\in A$

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.

CBSE Questions for Class 11 Engineering Maths Sets Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Sets Quiz 1 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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