MCQ Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 6

Suppose

$\displaystyle A_{1},A_{2}.....A_{30}$ are thirty sets having 5 elements and

$\displaystyle B_{1},B_{2}....B_{n}$ are n sets each with 3 elements. Let

$\displaystyle \bigcup_{i=1}^{30}Ai=\bigcup_{i=1}^{n}Bj=S$ and each elements of S belongs to exactly 10 of the Ai's and exactly 9 of the Bj's. Then n is equal to

0%
35
0%
45
0%
55
0%
65

Explanation

Since each element of S belongs to

$10$

$A_{i}s$ , there are only

$15$ elements in S

$\left[\dfrac{30\times 5}{10}\right]$

Since each of the

$15$ elements belongs to 9

$B_j s$ there are 135 elements in the second union.

No. of sets

$=\dfrac{135}{3}=45$

S = {1, 2, 3, 5, 8, 13, 21, 34}. Find

$\displaystyle \sum max\left ( A \right )$ where the sum is taken over all 28 two elements subsets A to S

0%
844
0%
480
0%
484
0%
488

Given n(A) = 11, n(B) = 13, n(C) = 16,

$\displaystyle n\left ( A\cap B \right )=3,n\left ( B\cap C \right )=6,n\left ( A\cap C \right )=5\: \: and\: \: n\left ( A\cap B\cap C \right )=2$ then the value of

$\displaystyle n [ A\cup B \cup C ]=$

0%
$24$
0%
$27$
0%
$25$
0%
$28$

Explanation

We know,

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

$=11+13+16-3-6-5+2=28$

In a group, if 60% of people drink tea and 70% drink coffee. What is the maximum possible percentage of people drinking either tea or coffee but not both?

0%
$100\%$
0%
$70\%$
0%
$30\%$
0%
$10\%$

Explanation

To find maximum possible percentage of people drinking either coffee or tea, we can assume everyone drinks at least either of the options.

Hence

$a+b=100$

$a+2b=60+70=130$

$b=30$

hence

$a=70$ which is the required area of the venn diagram.

If A and B are two disjoint sets and N is universal set, then

$\displaystyle A^{\circ}\cup \left [ \left ( A\cup B \right )\cap B^{\circ} \right ]$ is

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

Explanation

In the Venn Diagram, we can see that

$(A \cup B) \cap B'$ covers the pink coloured region inside

$N$ except

$B$ .
And

$A'$ is the region which is shown using striking lines , that is the region outside

$A$
And if we perform

$A' \cup (A \cup B) \cap B'$ , it will cover the pink region as well s the striken region, which is nothing but the complete set

$N$ .

If out of 150 students who read at least one newspaper The Times of India, The Hindustan Times and The Hindu. There are 65 who read The Times of India, 41 who read The Hindu and 50 who read The Hindustan Times. What is the maximum possible number of students who read all the three newspaper?

0%
$7$
0%
$42$
0%
$3$
0%
Cannot be determined

Explanation

From venn diagram

$a+b+c=150$

$a+2b+3c=156$

Hence

$b+2c=6$

To maximise c we take minimum value of b that is 0.

Hence

$c=3$

Find the number of students that had taken only mathematics

0%
2
0%
3
0%
4
0%
5

Explanation

Let M=Set of students who took Mathematics

C=Set of students who took Chemistry

P=Set of student who took Physics

$\therefore n(M)=15,n(P)=12,n(C)=11,n(M\cap C)=5,n(M\cap P)=9,n(P\cap C)=4,n(M\cap P \cap C)=3$

No.of students that had taken only mathematics

$\Rightarrow n(M-(P\cup C))=n(M)-n(M \cap(P\cup C))$

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

$= n(M)-[n(M\cap P)+n(M\cap C)-n((M\cap P)\cap(M\cap C))]$

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

$= 15-9-5+3=4$

Sets A and B have 3 and 6 elements respectively. What can be the minimum number of elements in

$A\cup B$ ?

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

Explanation

Let

$A$ be the left circle and

$B$ be the right circle.There are

$3$ elements in

$A$ and

$6$ elements in

$B$ .

The union of

$A$ and

$B$ contains elements that are in either circle.Thus,the union of

$A$ and

$B$ will be all of the elements in

$A$ along with all of the element

$B$ .However,you have to subtract the elements that are in the overlapping area because you are counting twice.

$A$ and

$B$ don't overlap at all,then the union will ontain

$9$ elements.If

$A$ is completely inside

$B$ then the union will only contain

$6$ elements,which is the minimum no. of elements in the union of

$A$ and

$B$ .

let

$A=1,2, B=2,3$

$\therefore A\cup B=1,2,3$ which is

$3$ elements.

$\therefore A$ has

$2$ elements ,

$B$ has

$2$ elements,and there is

$1$ element overlapping.

$\therefore 2+2-1=3!=3\times 2\times 1=6$

1032996_325611_ans_d8adf9fc368f46ff86ec6d8234101010.png

How many subsets can be formed from the set

$\left \{p, q, r\right \}$ is?

0%
6
0%
7
0%
8
0%
9

Explanation

$\left\{p,q,r \right\}$ It has 3 elemente

$\therefore 2^n=2^3=8 Subset$

$A=\left \{1, 2, 3, .....9\right \}$ and

$B=\left \{2, 3, 4, 5, 7, 8\right \}$ , then A-B is given by

0%
$\left \{1, 6, 7, 8\right \}$
0%
$\left \{1, 6, 9\right \}$
0%
$\left \{1, 9\right \}$
0%
$\left \{6, 9\right \}$

Explanation

$A=\left\{1,2,3,4,5,6,7,8,9 \right\}$

$B=\left\{2,3,4,5,7,8 \right\}$

A-B is a set of all those elements of A which are not in B.

$\therefore A-B=\left\{1,6,9 \right\}$

Let

$S=\left\{1,2,3,.....40\right\}$ and let

$A$ be a subset of

$S$ such that no two elements in

$A$ have their sum divisible by

$5$ What is the maximum number of elements possible in

$A$ ?

0%
$10$
0%
$13$
0%
$17$
0%
$20$

Explanation

There are

$20$ maximum number of elements possible in

$A$

$A=(1,2,5,6,7,11,12,15,16,17,21,22,25,26,27,31,32,35,36,37)$

Look at the set of numbers below.
Set :

$\left \{6, 12, 30, 48\right\}$
Which statement about all the numbers in this set is NOT true?

0%
They are all multiples of $3$
0%
They are all even numbers
0%
They are all factors of $48$
0%
They are all divisible by $2$

Let

$A$ and

$B$ be two sets such that

$n(A)=16$ ,

$n(B)=12$ , and

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

$n(A\cup B)$ equals

0%
$28$
0%
$20$
0%
$36$
0%
$12$

Explanation

$n(A\cup B)=n(A)+n(B)-n(A\cap B)=16+12-8=20$

If A={a,b,c,d,e}, B={a,c,e,g} and C={b,d,e,g} then which of the following is true?

0%
$\displaystyle C\subset \left ( A\cup B \right )$
0%
$\displaystyle C\subset \left ( A\cap B \right )$
0%
$\displaystyle A\cup B=A\cup C$
0%
Both(1) and (3)

Explanation

For the given sets,

$(A \cup B ) =$ {

$a,b,c,d,e,g$ } (Combination of all elements of both sets)

Clearly, elements of C are a part of

$A \cup B$ as well,
So,

$C\subset (A\cup B)$

And,

$(A \cap B ) =$ {

$a,c,e$ } (Common elements of both sets )
As the elements of C are not completely a part of

$A \cap B$ , Option B is not True.

Also,

$(A \cup C ) =$ {

$a,b,c,d,e,g$ }

Clearly,

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

Hence, both first option and third options are True.

Let

$A=\{1,2,3,4,5,6\}$ . How many subsets of

$A$ can be formed with just two elements, one even and one odd?

0%
$6$
0%
$8$
0%
$9$
0%
$10$

Explanation

$\{1,2\}$ ,

$\{1,4\}$ ,

$\{1,6\}$ ,

$\{2,3\}$ ,

$\{2,5\}$ ,

$\{3,4\}$ ,

$\{3,6\}$ ,

$\{4,5\}$ ,

$\{5,6\}$ ,

In a class 60% of the students were boys and 30% of them had I class. If 50%of the students in the class had I class, find the fraction of the girls in the class who did not have a I class.

0%
$1/5$
0%
$4/5$
0%
$1/4$
0%
$1/3$

Explanation

$Let\quad the\quad number\quad of\quad students\quad be\quad x,\\ boys\quad with\quad first\quad class=0.3*0.6x=0.18x,\\ number\quad of\quad girls=0.4x,\\ students\quad with\quad first\quad class=0.5x,\\ girls\quad with\quad first\quad class=0.5x-0.18x=0.32x,\\ fraction\quad of\quad girls\quad without\quad first\quad class=\dfrac { 0.4-0.32x }{ 0.4x } =\dfrac { 1 }{ 5 }$

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 the universal set is U =

$\displaystyle \left \{ 1^{2},2^{2},3^{2},4^{2},5^{2},6^{2} \right \}$ What is the complement of the intersection of set A =

$\displaystyle \left \{ 2^{2},4^{2},6^{2} \right \}$ and set B=

$\displaystyle \left \{ 2^{2},3^{2},4^{2} \right \}$ ?

0%
$\displaystyle \left \{ 2^{2},4^{2} \right \}$
0%
$\displaystyle \left \{ 1^{2},5^{2} \right \}$
0%
$\displaystyle \left \{ 1^{2},5^{2},6 ^{2} \right \}$
0%
$\displaystyle \left \{ 1^{2},3^{2},5^{2},6^{2} \right \}$
0%
Answer required

Explanation

$A\cap B=\{2^2,4^2\}$

$\bar{A\cap B}=U-(A\cap B)=\{1^2,3^2,5^2,6^2\}$
Option D is correct.

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

$\displaystyle A\equiv B$

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.

If A is a finite set, let

$P(A)$ denote the set all subsets of A and

$n(A)$ denote the number of elements in A. If for two finite sets X and Y,

$n[P(X)] = n[P(Y)] + 15$ then find

$n(X)$ and

$n(Y)$

0%
$n(X) = 4; n(Y) = 0$
0%
$n(X) = 5; n(Y) = 0$
0%
$n(X) = 6; n(Y) = 0$
0%
$n(X) = 7; n(Y) = 0$

Explanation

$X$ is a finite set.

Let,

Number of elements in subset

$A$ be

$n(A)=m$

Number of elements in subset

$B$ be

$n(B)=n$

Then total number of subsets of finitr set containing 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 the universal set

$\{x\in W ,3<x≤12\} ,A=\{5,7,9\}$ , then

$A'=$

0%
$\{3,6,8,10,11,12\}$
0%
$\{4,6,8,10,11,12\}$
0%
$\{6,8,10,11,12\}$
0%
None of the above

Explanation

Given universal set is

$\{x\in W ,3<x≤12\}$

It can be written as a set

$\{4,5,6,7,8,9,10,11,12\}$

$A$ is given as

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

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

$A$

$=\{4,6,8,10,11,12\}$

Hence, option B is correct

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

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

The number of subsets of the set

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

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

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.

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,

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

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

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

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

$=A\cap B$

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

$32+10=42$

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 6 - 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 6 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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