MCQ Questions for Class 11 Engineering Maths Sets Quiz 7

The sets

$\displaystyle S_{x}$ are defined to be

$(x, x + 1, x + 2, x + 3, x + 4)$ where

$x=1, 2, 3,.....80$ . How many of these sets contain

$6$ or its multiple?

0%
$65$
0%
$66$
0%
$59$
0%
$60$

Explanation

There are

$14$ multiples of

$6$ till

$84$ .

Since

$5$ consecutive no.s are chosen only one set in

$6$ consecutive sets will not have a multiple of

$6$ . So till

$78$ sets there are

$78-\dfrac{78}6=78-13=65$ sets containing 6 or multiples of 6.

$S_{79}$ does not contain any multiple of 6

Hence

$S_{80}$ must contain a multiple of 6.

Answer

$=66$ sets

$A=\{\dots,-6,-4,-2,0,2,4,6,\dots\}$ , then

0%
$10\notin A$
0%
$-10\notin A$
0%
$5\in A$
0%
$50\in A$

$\displaystyle Q=\left((x|x=\frac{1}{y}\:\ \text{wher} \ e\:y\in N\right)$ , then

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

Explanation

When

$y=1$ ,

$\displaystyle x=\frac{1}{1}=1$ .

If A and B are disjoint then

$\displaystyle \left ( A\cap B \right ){}'=$ _______

0%
$A$
0%
$B$
0%
$\displaystyle \phi$
0%
$\displaystyle \mu$

Explanation

When A and B are disjoint sets, they do not have any overlapping region as shown in the figure.

So,

$(A \cap B) = \phi$

And

${(A \cap B)}^{`} = \mu - \phi = \mu$ will be the complete region of the universal set as there is no overlapping region of sets A and B

For any two sets

$A$ and

$B$ ,

$A-\left( A-B \right)$ equals

0%
$B$
0%
$A-B$
0%
$A\cap B$
0%
${ A }^{ C }\cap { B }^{ C }$

Explanation

Now,

$A-\left( A-B \right) =A-\left( A\cap { B }^{ C } \right)$

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

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

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

$=A\cap B$

Three sets

$A, B, C$ are such that

$\displaystyle A=B\cap C$ and

$\displaystyle B=C\cap A$ , then

0%
$\displaystyle A\subset B$
0%
$\displaystyle A\supset B$
0%
$\displaystyle A\equiv B$
0%
$\displaystyle A\subset { B }^{ \prime }$

Explanation

Since,

$\displaystyle A=B\cap C$ and

$\displaystyle B=C\cap A$ , then

$\displaystyle A\equiv B$

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

P, Q and R are three sets and

$\xi = P\cup Q\cup R$ . Given that

$n(\xi) = 60, n (P\cap Q) = 5, n(Q\cap R) = 10, n(P) = 20$ and

$n(Q) = 23$ , find

$n(P\cup R)$

0%
$37$
0%
$38$
0%
$45$
0%
$52$

Explanation

$n(U) =60 (Universal \; Set)$

$n(P \cap Q ) = 5$

$n(Q \cap R ) = 10$

$n(P ) = 20$

$n(Q)=23$

$n(P \cup Q ) = n(P) + n(Q)- n(P \cap Q ) = 20+23-5 = 38$

$n(R)=n(U)-n(P \cup Q ) +n(Q\cap R)= 60-38+10 = 32$

$n(P \cup R ) = n(P) + n(R)- n(P \cap R ) = 20+32-0 =52$

If n is a member of both set A

$=\left\{\displaystyle\frac{4}{7}, 1, \frac{5}{2}, 4, \frac{1}{2}, 7\right\}$ and set B

$=\left\{\displaystyle\frac{4}{7}, \frac{7}{4}, 4, 7\right\}$ , which of the following must be true?
I. n is an integer.
II.

$4n$ is an integer.
III.

$n=4$

0%
None
0%
II only
0%
I and II only
0%
I and III only
0%
I, II, and III

Explanation

None of the criteria satisfy all the elements in the intersection.

$\dfrac47$ is not an integer nor is

$\dfrac {4\times 4}{7}$ an integer.

Of course 4 is not the only element in the intersection. Hence 3 is untrue.

$A - B =$ _____

0%
$A \cup (A\cap B)$
0%
$B - A$
0%
$A - (A\cap B)$
0%
$A'\cap B$

Explanation

A-B means everything in A except for anything in $A \cap B$

Pick an element x. Let $x\in(A-B)$

$\therefore x \in A \; but \; x \notin B$

$\therefore x\in A , x\in B'$

$x\in (A \cap B') = A- (A \cap B)$

or, $x\in (A- B)$

Hence, $A-B = (A \cap B') = A- (A \cap B)$

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

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

$A$ and

$B$ are any two sets, then state the following statement is true/false

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

0%
True
0%
False

Explanation

$L.H.S=A-(A-B)$

$=A\cap (A-B)'$ [ Set difference ]

$=A\cap (A\cap B')'$ [ Set difference ]

$=A\cap(A\cup(B')')$ [ De-Morgan's Law ]

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

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

$=\phi \cup (A\cap B)$ [ Since, A set and its complement are disjoint ]

$=A\cap B$

$=R.H.S$

$\therefore$ Given statement is true.

In a class,

$20$ opted for Physics,

$17$ for Maths,

$5$ for both and

$10$ for other subjects. The class contains how many students?

0%
$35$
0%
$42$
0%
$52$
0%
$60$

Explanation

$n(P) = 20$

$n(M) = 17$

$n(M \cap P) = 5$

$n(other \; subjects)=10$

$n(M\cup P) =n(M)+n(P)-n(M\cap P)$

$n(M\cup P) =17+20-5 = 32$

Total students $= n(P\cup M)+n (other \; subjects )$

$32+10=42$

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

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,

$\{5\}\subset A$

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

$X=\left \{ a,\left \{ b,c \right \},d \right \}$ , 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.

The number of subsets of the set

$\left \{ 10,11,12 \right \}$ is

0%
$3$
0%
$8$
0%
$6$
0%
$7$

Explanation

No. of subsets

$= 2^3=8$ .

We have 2 choices with each of the elements: either put them in a subset or to not put them in a subset. Hence there are 8 subsets.

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

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

$A= \{ p, q, r, s \}$ ,

$B = \{ r, s, t, u \}$ , then

$A /B$ is:

0%
{ p, q }
0%
{ r, s }
0%
{ t, u }
0%
{p, q, t, u }

Explanation

The given sets are

$A=\left\{ p,q,r,s \right\}$ and

$B=\left\{ r,s,t,u \right\}$

$A/setminus B$ have those elements that belongs to set

$A$ but does not belongs to set

$B$ .

Let us find

$A/setminus B$ as follows:

$Asetminus B=\left\{ p,q,r,s \right\} /setminus \left\{ r,s,t,u \right\} =\left\{ p,q \right\}$

Hence,

$A/setminus B=\left\{ p,q \right\}$

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

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

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

$n(A\cap B)=140$

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

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.

CBSE Questions for Class 11 Engineering Maths Sets Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Sets Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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