MCQ Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 12

State true or false for each of the following. Correct the wrong statement

0%
True
0%
False

Explanation

Given

$n (φ) = 1$
The statement given here is false
Correct statement:

$n (φ) = 0$

$n(P)= n(M)$ , then

$P \rightarrow M$

0%
True
0%
False

Explanation

$n(P)= n(M)$ means the number of elements of set P = Number of elements of set M.

$\therefore$ P and M are equivalent.
Hence, the given statement is true.

$\cup$ N = {1, 2, 3, 4, 5, 6} and M = {1, 2, 4} then which of the following represent set N?

0%
{1, 2, 3}
0%
{3, 4, 5, 6}
0%
{2, 5, 6}
0%
{4, 5, 6}

Explanation

We have, M

$\cup$ N = {1, 2, 3, 4, 5, 6} and M = {1, 2, 4}
Since, {1, 2, 3}

$\cup$ {1, 2, 4} = {1, 2, 3, 4}

$\neq$ M

$\cup$ N = {1, 2, 3, 4, 5, 6};
{3, 4, 5, 6}

$\cup$ {1, 2, 4} = {1, 2, 3, 4, 5, 6} = M

$\cup$ N = {1, 2, 3, 4, 5, 6};
{2, 5, 6}

$\cup$ {1, 2, 4} = {1, 2, 4, 5, 6}

$\neq$ M

$\cup$ N = {1, 2, 3, 4, 5, 6}; and
{4, 5, 6}

$\cup$ { 1, 2, 4} = {1, 2, 4, 5, 6}

$\neq$ M

$\cup$ N = {1, 2, 3, 4, 5, 6}

$P \ \subseteq \ M$ , then Which of the following set represent

$P \ \cap \ (P \ \cup \ M)$ ?

0%
P
0%
M
0%
P $\cup$ M
0%
P' $\cap$ M

Find the correct option for the given question.
Which of the following collections is a set?

0%
Colours of the rainbow
0%
Tall trees in the school campus
0%
Rich people in the village
0%
Easy examples in the book

Examine whether the following statements are true or false:

$\left\{a,e \right\} \subset \left\{x : x \ is\ a\ vowel\ in\ the\ English\ alphabet \right\}$

0%
True
0%
False

In a town of

$10,000$ families it was found that

$40\%$ families buy newspaper

$A$ ,

$20\%$ families buy newspaper

$B$ and

$10\%$ families buy newspaper

$C$ .

$5\%$ families buy

$A$ and

$B$ ,

$3\%$ buy

$B$ and

$C$ and

$4\%$ buy

$A$ and

$C$ . If

$2\%$ families buy all the three newspaper, find the number of families which buy (i)

$A$ only (ii)

$B$ only (iii) none of

$A, B$ and

$C$

0%
(i) $3000$ (ii) $1800$ (iii) $4600$
0%
(i) $3300$ (ii) $1400$ (iii) $4000$
0%
(i) $3500$ (ii) $1600$ (iii) $3800$
0%
none

Examine whether the following statements are true or false:

$\left\{x : x\ is\ an\ even\ natural\ number\ less\ than\ 6 \right\} \subset \left\{x : x is\ a\ natural\ number\ which\ divides\ 36 \right\}$

0%
True
0%
False

Examine whether the following statements are true or false:

$\left\{1,2,3 \right\} \subset \left\{1,3,5 \right\}$

0%
True
0%
False

Examine whether the following statements are true or false:

$\left\{a,b \right\} \not \subset \left\{b,c,a \right\}$

0%
True
0%
False

In each of the following, determine whether the statements is true or false if it is true prove it if it false given an example.
If

$A \subset B$ and

$B \subset C$ , then

$A \subset C$

0%
True
0%
False

Examine the following statements:
{x : x is an even natural number less then 6}

$\subset$ { x : x is natural number which divide 36 }

0%
True
0%
False

Explanation

{x : x is an even natural number less then 6}

$\subset$ { x : x is natural number which divide 36 }

$\left \{ 2, 4\right \} \subset \left \{ 1, 2, 3, 4 , 6, 9, 12, 18, 36\right \}$
Hence, the statement is true

If A = {1, 2, 3}, B = {3, 4, 5}, C = {4, 6}, then A x

$( B \cup C)$ =

0%
{(1, 3) (1, 4) (1, 5) (1, 6) (2, 3) (2, 4) (2, 5) (2, 6) (3,3) (3, 4) (3,5) (3, 6)}
0%
A x $(B \cap C)$
0%
B x $(A \cap C)$
0%
all the above

Which of the following statements is true (if N, W and I are sets of Natural, Whole and Integer numbers respectively ?

0%
$N\, \subset \, W\, \subset \, I$
0%
$I\, \subset \, N\, \subset \, W$
0%
$W\, \subset \, N\, \subset \, I$
0%
$I\, \subset \, W\, \subset \, N$

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

$\displaystyle \sum$ max (A), where the sum is taken over all 28 elements subsets A to S.

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

There are 6 boxes numbered 1, 2, ...Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is:

0%
5
0%
21
0%
33
0%
60

The set

$\{x/| x L|< K\}$ is the same for all

$K > 0$ and for all L, as

0%
$\{x/0 < x < L + K\}$
0%
$\{x/L K < x < L + K\}$
0%
$\{x/|L K| < x <|L + K|\}$
0%
$\{x/|L x| > K\}$
0%
$\{x/ K < x < L\}$

$S$ represents the set of all real numbers

$x$ such that

$1\le x \le 3$ and

$T$ represents the set of all real numbers

$x$ such that

$2 \le x \le 5$ , the set represented by

$S \cap T$ is

0%
$2 \le x \le 3$
0%
$1 \le x \le 5$
0%
$x \le 5$
0%
$x\ge 5$
0%
none of these

Explanation

$S\cap T$ (common part of S and T) =

$2\leq x\leq 3$

1122899_536658_ans_bee7ddda151c4383a7f05502ca0bfa30.png

Suman is given an aptitude test containing 80 problems, each carrying I mark to be tackled in 60 minutes. The problems are of 2 types; the easy ones and the difficult ones. Suman can solve the easy problems in half a minute each and the difficult ones in 2 minutes each. (The two type of problems alternate in the test). Before solving a problem, Suman must spend one-fourth of a minute for reading it. What is the maximum score that Suman can get if he solves all the problems that he attempts?

0%
30
0%
40
0%
53
0%
60

Explanation

$Assuming\quad that\quad all\quad the\quad problems\quad suman\quad solves\quad are\quad right,\\ marks\quad earned\quad per\quad minute\quad solving\quad easy\quad problem=\dfrac { 1 }{ 3/4 } =\dfrac { 4 }{ 3 } \\ marks\quad earned\quad per\quad minute\quad solving\quad difficult\quad problem=\frac { 1 }{ 9/4 } =\frac { 4 }{ 9 } \\ since\quad we\quad have\quad to\quad maximize\quad the\quad marks,\quad we\quad will\quad solve\quad all\quad the\quad easy\quad problems\quad first.\\ marks\quad earned\quad by\quad solving\quad easy\quad problems=40,\\ time\quad spent=\frac { 3 }{ 4 } *40=30\quad minutes\\ time\quad left\quad for\quad difficult\quad problems=30\quad minutes,\\ problems\quad solved=\frac { 30 }{ 9/4 } =13,\\ marks\quad earned=13,\\ total\quad marks=40+13=53$

Union set is defined as

0%
a collection of sets is the set of all elements in the collection
0%
It is one of the fundamental operations through which sets can be combined and related to each other.
0%
Set whole each element is an element of all the present set
0%
None of these

The dual of

$-p\wedge (q\vee \sim r)$ is

0%
$p\vee (\sim q \wedge r)$
0%
$\sim p\vee (q\wedge r)$
0%
$p\wedge (q\wedge r)$
0%
$\sim q\vee(q\wedge \sim r)$

$20$ % of three subsets (i.e., subsets containing exactly three elements) of the set

$A = \left \{a_{1}, a_{2}, ...., a_{n}\right \}$ contain

$a_{2}$ , then the value of

$n$ is

0%
$15$
0%
$16$
0%
$17$
0%
$18$

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

$\displaystyle \bigcup _{ i=1 }^{ 30 }{ { A }_{ i } } =\bigcup _{ j=1 }^{ n }{ { B }_{ j } =S}$ 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

0%
$15$
0%
$3$
0%
$45$
0%
$None\ of\ these$

Explanation

$(c)$ .
Since each

$A_{ i }$ has

$5$ elements, we have

$\overset { 30 }{ \underset { i=1 }{ \Sigma } } \ n\left( { A }_{ i } \right) =5\times 30=150$ . . .

$(1)$
Let

$S$ consist of

$m$ distinct elements. Since each elements of

$S$ belongs to exactly

$12$ of the

$A_{ i }$

$s$ we also have

$\overset { 30 }{ \underset { i=1 }{ \Sigma } } \quad n\left( { A }_{ i } \right) =10m$ . . .

$(2)$
Hence from

$(1)$ and

$(2)$ ,

$10m=150$ or

$m=15$ . Again since each

$B_{ i }$ has

$3$ elements and each element of

$S$ belongs to exactly

$9$ of the

$B_{ j }$

$s$ we have

$\overset { 30 }{ \underset { j=1 }{ \Sigma } } \quad n\left( { B }_{ j } \right) =3n$ and

$\overset { 30 }{ \underset { j=1 }{ \Sigma } } \quad n\left( { B }_{ j } \right) =9m$
It follows that

$3n=9m=9\times 15$
. . .

$[\therefore m=15]$
This gives

$n=45$ .

All the permissible values of

$b$ , if

$a=0$ and

${S}_{2}$ is a subset of

$\left( 0,\pi \right)$

0%
$b\in \left( -n\pi ,2n\pi \right) ;\in Z$
0%
$b\in \left( -n\pi ,2\pi -n\pi \right) ;\in Z$
0%
$b\in \left( -n\pi ,n\pi \right) ;\in Z$
0%
none of these

$S_k=\begin{bmatrix} 1 & k \\ 0 & 1\end{bmatrix}$ , k

$\in N^+$ , where N is the set of all natural numbers, then

$(S_2)^n(S_k)^{-1}$ for n

$\in$ N is?

0%
$S_{2n-k}$
0%
$S_{2n+k-1}$
0%
$S_{2^n+k-1}$
0%
$S_{2^n+k+1}$

$13$ 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.

0%
Not less than $\dfrac{7}{13} m^{2}$
0%
Not less than $\dfrac{2}{13} m^{2}$
0%
Not less than $\dfrac{1}{13} m^{2}$
0%
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

$\dfrac {1}{13}$ .

State which of the following is total number of reflexive relations form set

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

$B = \left \{d, e\right \}$ is

0%
$2^{6}$
0%
$2^{8}$
0%
$2^{4}$
0%
$2^{15}$

Explanation

Number of reflexive relation from

$A=[a,b,c]$ and

$B=[d,e]$ is

$=2^{n^2-n}=2^{3^2-3}=2^{9-3}=2^6$

Sets

$A$ and

$B$ have

$5$ and

$6$ elements respectively and

$\left( A\triangle B \right) =C$ then the number of elements in set

$\left( A-\left( B\triangle C \right) \right)$ is

0%
$5$
0%
$6$ $
0%
$0$
0%
$4$

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

0%
15
0%
3
0%
45
0%
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............

0%
$275$
0%
$510$
0%
$219$
0%
$256$

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

$219.$

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 12 - 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 12 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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