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De morgan's law - class-XI

Description: de morgan's law
Number of Questions: 32
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State True or False:  $(A\cup B)'=A'\cap B'$

  1. True

  2. False


Correct Option: A
Explanation:

We have the De Morgan's theorem,

i) $(A\cup B)=A'\cap B'$
ii) $( A\cap B)=A' \cup B'$.
The given identity is the first one.
Hence the this statement is  true.

For any two sets A and  B $A-(B\cup C)=(A-B)\cap (A-C)$

  1. True

  2. False


Correct Option: A
Explanation:

As we know that 
$A-B=A\cap B^{C}$
For  the sets $A,B$ and $C$
$A-(B\cup C)=A\cap (B\cup C)^{C}=A\cap (B^{C}\cap C^{C})=(A\cap B^{C})\cap (A\cap C^{C})=(A-B)\cap (A-C)$

$AB=A$ and $BA=B$, then which of the following is not true?

  1. $A$ is idempotent matrix

  2. $B$ is idempotent matrix

  3. ${A}^{T}$ is idempotent matrix

  4. None of these


Correct Option: D
Explanation:
$AB=A\implies B=I;BA=B\implies A=I\implies A=B=I;A,B,A^{T}$ are idempotent.

For any two sets  A and B $A\cup B=A\cap B$ if A$=$B.

  1. True

  2. False


Correct Option: A
Explanation:

$A\cup B=B-A+A-B+A\cap B\\implies B-A=A-B=\phi \implies A=B$

If $A$ and $B$ are subsets of $U$ such that $n(U) = 700, n(A) = 200, n(B) = 300, n$$\displaystyle \left ( A\cap B \right )$ $= 100$, then find $n\displaystyle \left ( A'\cap B' \right )$

  1. $405$

  2. $305$

  3. $400$

  4. $300$


Correct Option: D
Explanation:

We know that $\displaystyle \left ( A\cup B \right )'=$ $\displaystyle A'\cap B'$ so we need to find $\displaystyle \left ( A\cup B \right ) $  first.
$n\displaystyle \left ( A\cup B \right ) =$ $n(A) + n(B) - n$$\displaystyle (A\cap B)$ $= 200 + 300 - 100 = 400$
$n\displaystyle \left ( A\cup B \right )' =$ $n(U) - n$$\displaystyle \left ( A\cup B \right )'=$$  700 - 400 = 300$
$\displaystyle \Rightarrow $ $n\displaystyle (A'\cap B')$$= 300$.

If A has 5 elements and B has 8 elements such that $\displaystyle A\subset B,$ then the number of elements in $\displaystyle A\cap  B,$ and $\displaystyle A\cup  B,$ are respectively :

  1. 8 , 5

  2. 3 , 3

  3. 5, 8

  4. 5, 13


Correct Option: C
Explanation:

Since $A\subset B$, $A\cap B$ will be all the elements of A. So, the number of elements will be 5.
Since $A\subset B$, $A\cup B$ will be all the elements of A & B. So, the number of elements will be 8.

While preparing the progress reports of the students, the class teacher found that $70$% of the students passed in Hindi, $80$% passed in English and only $65$% passed in both the subjects. Find out the percentage of students who failed in both the subjects

  1. $15$%

  2. $20$%

  3. $30$%

  4. $35$%


Correct Option: A
Explanation:

Students passed in Hindi $ = 70\% =n(H)$

Students passed in English $ = 80\% =n(E)$

Students passed in Both $ = 65\% =n(H\cap E)$

No. of students passed $=n(H \cup E)= n(H)+n(E)-n(H \cap E) = 70+80-65 =85\%$

No. of Students failed $= 100\% - n(H \cup E) = 100\% - 85\% = 15\%$

In a science talent examination, $50$% of the candidates fail in Mathematics and $50$% fail in Physics. If $20$% fail in both these subjects, then the percentage who pass in both Mathematics and Physics is

  1. $0$%

  2. $20$%

  3. $25$%

  4. $50$%


Correct Option: B
Explanation:

Fail: $n(M) =50\%$

$n(P) =50\%$

$n(M \cap P) =20\%$

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

$n(M\cup P) =50+50-20 = 80\%$

Pass:

n(Pass in both subjects) $=100\%$ -n(Fail in both the subjects)

$=100\% -80\% = 20\%$

$(A\cup B)^{'} = A^{'} \cap B^{'}$ is called ____________ law.

  1. Associative

  2. Commutative

  3. De Morgan's

  4. Distribute


Correct Option: C
Explanation:

$(A\cup B)' = A'\cap B'$ is called De Morgan's law.

In a survey, it was fond that $65$% of the people watched news on TV, $40$% read in newspaper, $25$% read newspaper and watched TV. What percentage of people neither watched TV nor read newspaper?

  1. $0$%

  2. $5$%

  3. $10$%

  4. $20$%


Correct Option: D
Explanation:

n(Watch new on TV) $=65\%=n(T)$

n(Reads newspaper) $=40\%=n(N)$

n(Watch news on TV & read newspapers ) $=25\%=n(T\cap N)$

$n(N \cup T)= n(T)+n(N)-n(N\cap T)$

$n(N \cup T)= 65+40-25 = 80\%$

$n(N\cap T)^{c}=100-80=20\%$

Comment true or false  on the following statements
$ A\cap \left( B-C \right) =\left( A\cap B \right) -\left( A\cap C \right)$

  1. True

  2. False


Correct Option: A
Explanation:

Let $x\in A\cap (B - C)$

$\Rightarrow x\in A$ and $x\in (B-C)$
$\Rightarrow x\in A$ and $(x\in B \text{ and } x \notin C)$
$\Rightarrow( x\in A \text{ and } x\in B)$ and $(x\in A \text{ and } x \notin C)$
$\Rightarrow (A\cap B) - (A\cap C)$
Hence true.

A survey on a sample of $25$ new cars being sold at a local auto dealer was conducted to see which of the three popular options - air-conditioning, radio and  power windows - were already installed.
The survey found:
$15$ had air-conditioning
$2$ had air-conditioning and power windows but no radios.
$12$ had power windows
$6$ had air-conditioning and radio but no power windows.
$11$ had radio.
$4$ had radio and power windows.
$3$ had all three options.
What is the number of cars that had none of the options?

  1. $4$

  2. $3$

  3. $1$

  4. $2$


Correct Option: D
Explanation:

$n(A\cup R\cup P)= n(A) +n(R) +n(P)-n(A\cap R)-n(R\cap P) -n(P\cap A) + n(A\cap R \cap P)$


$2= n(A\cap P)-n(A\cap P\cap R)$

$\Rightarrow n(A\cap P) = 2+3=5$

$6= n(A\cap R)-n(A\cap P\cap R)$

$\Rightarrow n(A\cap R) = 6+3=9$

$n(A\cup R\cup P)= n(A) +n(R) +n(P)-n(A\cap R)-n(R\cap P) -n(P\cap A) + n(A\cap R \cap P)$

$=15+11+12-9-5-4+3$

So answer $= 25-23 = 2$

With usual notations $n\left( A\cup B\cup C \right) =20,n\left( A\cap B\cap C\prime  \right) =2,n\left( B\cap C\cap A\prime  \right) =n\left( A\cap C\cap B\prime  \right) =4\quad$

$ and\quad n\left( A\cap B\cap C \right) =1$, then the number of elements belonging to exactly one of the sets is

  1. $9$

  2. $13$

  3. $14$

  4. $16$


Correct Option: A

Let $n(u)=700,n(A)=200,n(B)=300$
$n\left( A\cap B \right) =100,n\left( A^{\prime} \cap B^{\prime}  \right) =$

  1. $400$

  2. $600$

  3. $300$

  4. $None$


Correct Option: C
Explanation:

Ans. $(c). n(A\cap B)=n(A\cup B)$
$=n(u)-n(A\cup B)$
$=n(u)-\left{ n\left( A \right) +n\left( B \right) -n\left( A\cap B \right)  \right}$
$=700-\left{ 200+300-100 \right} = 300$

The value of $(A\cup B\cup C)\cap {(A\cap {B}^{c}\cap {C}^{c})}^{c}\cap {C}^{c}$

  1. $B\cap {C}^{c}$

  2. ${B}^{c}\cap {C}^{c}$

  3. $B\cap C$

  4. $A\cap B\cap C$


Correct Option: A

Given that the universal set,$ \xi =$ {x : 1 < x < 12 and x is an integer} and the sets P = {x : x is a prime number}, Q = {x : x is a multiple of 4} and R = {2, 3, 8, 9} the elements of the set $(Q \cup R)' \cap P$ are:

  1. {2, 3}

  2. {2, 3, 5}

  3. {5, 7, 11}

  4. {1, 5, 7, 11}


Correct Option: C
Explanation:

$Q={4,8};$     $P={2,3,5,7,11}$;     


$Q\cup R ={2.3.4.8.9}$

$(Q\cup R)' = {5,6,7,10,11}$

$(Q\cup R)'\cap P = {5,7,11}$

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

  1. $A$

  2. $A'$

  3. $B$

  4. $B'$


Correct Option: B
Explanation:

$(A'-B)\cup (B-A)$
$=A'\cap B'\cup B\cap A'$      $[\because A-B=A\cup B']$
$=A'\cap B'\cup A'\cap B$      $[\because A\cup B=B\cup A]$
$=A'\cap (B'\cup B)$
$=A'\cap U$
$=A'$
answer is B

if $X' = Y$ then $\displaystyle \left (X \cap Y  \right )'$ is equal to 

  1. $\displaystyle \phi $

  2. $X$

  3. $U$

  4. $Y$


Correct Option: C
Explanation:

Given 

$X' = Y$
$X$ $\cap$ $Y = X$ $\cap$  $X'$
$(X$ $\cup$ $X')' $ $= U $(Universal set)
Complement  of null set is Universal set $U$.

If among natural numbers $A={5,6,7}$ and $B={8,9,10}$ , then 

  1. $A \cap B =$ null

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

  3. $A$ $\cap$ $B = {2,3,4}$

  4. None of these


Correct Option: A,B
Explanation:

Since, no elements of A and B are the same.
Therefore, the Intersection is empty.
i.e. $A \cap B = $ null
We know, $U = $  { $ N $ }
$A = $  { $ 5,6,7 $ }
$B = $  { $ 8,9,10 $ } 
($A \cup B$) $=$ {$5,6,7$) $ \cup $ {$8,9,10$}
               $=$ {$5,6,7,8,9,10$}
Therefore,  ($A \cup B$)'$= $ All natural numbers except $ {5,6,7,8,9,10} \dots (i)$
Now, $A'= $ All Natural numbers except ${5,6,7}$
and $B'=$ All Natural numbers except ${8,9,10}$
Therefore, $A' \cap B' = $ All Natural numbers except ${5,6,7,8,9,10}\dots (ii)$
By comparing $(i) $ and $(ii)$ we get,
$(A \cup B) = A' \cap B'$

Given $A={x\in N :x<6} ,B={3,6,9}$ and $C={x \in N: 2x-5\le 8}$

  1. $A \cup $ (B $ \cap $C)=(A $ \cap  $B) $\cap $(A $ \cap$ C)

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

  3. A $\cup$B=null set

  4. None of the above


Correct Option: B
Explanation:

$A = { x \in N : x < 6 }$

$\therefore A = { 1,2,3,4,5 }$
$B = { 3,6,9 }$
$C = { x \in N : 2x - 5 \leq 8 }$
$\Rightarrow C = { 1,2,3,4,5,6 }$
Option A. $A \cup (B \cap C) = A \cup {3, 6 } = { 1,2,3,4,5,6 }$
RHS $= (A \cap B) \cap ( A \cup C) = { 3 } \cap {1,2,3,4,5,6} = {3}$
LHS $\neq$ RHS

Option B. $(A \cup B)' = A' \cap B'$ is always true by De Morgan's Law  . 
$A\cup B={1,2,3,4,5,6,9}$
$(A\cup B)'={7,8,10,11,...}$
$A'={x\in N:x\geq 7}$ and $B'={1,2,4,5,7,8,10,11,12...}$
$A'\cap B'={7,8,10,11,12...}$
$\therefore (A\cup B)'=A'\cap B'$

Option C. $A \cap B = { 3 } \neq$ null set
Hence, only B is correct.

If $A, B$ be any two sets, then $(A\cup B)'$ is equal to

  1. $A'\cup B'$

  2. $A'\cap B'$

  3. $A\cap B$

  4. $A\cup B$


Correct Option: B
Explanation:

It is De Morgan's law that $(A\cup B)' = A'\cap B'$

If $U = {3, 4, 5, 6, 7, 8, 9}, X = {3, 4}, Y = {5, 6}$ and $Z = {7, 8, 9}$, then $\displaystyle Y'\cap \left ( X\cap Z \right )'$ is equal to 

  1. $\displaystyle X\cup Y$

  2. $\displaystyle Y\cup Z$

  3. $\displaystyle X'\cap Y'$

  4. $X\cup Z$


Correct Option: D
Explanation:

$ { Y ' } = U - Y = { 3, 4, 7, 8, 9} $

$ X\cap Z = {0} $ as there is no element common in sets X and Z

$\Rightarrow (X\cap Z) ' = U - (X\cap Z) =  {3, 4, 5, 6, 7, 8, 9} $

Now, $Y' \cap (X\cap Z)'  = { 3, 4, 7, 8, 9} \cap {3, 4, 5, 6, 7, 8, 9} ={ 3, 4, 7, 8, 9} $

This is equal to $ X \cup Z = {3,4,7,8,9} $

If $U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$, $A = {0, 3, 4, 7}$ ,$B = {1, 2, 8, 9}$
then $(A U B)'$ is

  1. ${2, 5}$

  2. ${5, 6}$

  3. ${8, 9}$

  4. ${6, 7}$


Correct Option: B
Explanation:

Given, $U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$, $A = {0, 3, 4, 7}$ ,$B = {1, 2, 8, 9}$


$A^c={1,2,5,6,8,9}, B^c= {0,3,4,5,6,7} $

By De Morgan's law

$(A\cup B)^c = A^c\cap B^c$

$=\{1,2,5,6,8,9\}\cap \{0,3,4,5,6,7\}$

$=\{5,6\}$

Out of 800 boys in a school 224 played cricket, 240 played hockey and 236 played basketball. Of the total 64 played both basketball and hockey, 80 played cricket and basketball and 40 played cricket and hockey, 24 players all the three games. The number of boys who did not play any game is

  1. $128$

  2. $216$

  3. $240$

  4. $260$


Correct Option: D
Explanation:

No. of players who played at least one game is:


By set theory

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

$=224+240+236-64-80-40+24=540$

Hence $260$ players do not play any game.

Find the De Morgan's law of intersection.

  1. $(A\cap B)^{'} = A \cup B^{'}$

  2. $(A\cap B)^{'} = A^{'} \cup B^{'}$

  3. $(A\cup B)^{'} = A^{'} \cup B^{'}$

  4. $(A\cap B)^{'} = A^{'} \cap B^{'}$


Correct Option: B
Explanation:

De Morgan's law of intersection is $(A\cap B )' = A' \cup B' $

Find the De Morgan's law of union.

  1. $(A\cap B)^{'} = A^{'} \cap B^{'}$

  2. $(A\cup B)^{'} = A^{'} \cap B^{'}$

  3. $(A\cup B)^{'} = A^{'} \cup B^{'}$

  4. $(A\cup B)^{'} = A\cap B^{'}$


Correct Option: B
Explanation:

De Morgan's law of union is 

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

In order to draw a graph of $f(x) = ax^{2} + bx + c$, a table of values was constructed. These values of the function for a set of equally spaced increasing values of $x$ were $3844, 4096, 4227, 4356, 4489, 4624$, and $4761$. The one which is incorrect is

  1. $4096$

  2. $4356$

  3. $4489$

  4. $4761$

  5. None of these


Correct Option: E
Explanation:

We are told that the values of $f(x)$ listed correspond to
7$f(x), f(x + h), f(x + 2h), ...., f(x + 7h)$.
Observe that the difference between successive values is given by
$f(x + h) - f(x) = a(x + h)^{2} + b(x + h) + c - (ax^{2} + bx + c)$
$= 2ahx + ah^{2} + bh$.
Since the difference is a linear function of $x$, it must change by the same amount whenever $x$ is increased by $h$. But the successive differences of the listed values
$3844\ 3969\ 4096\ 4227\ 4356\ 4489\ 4624\ 4761$
are $125\ 127\ 131\ 129\ 133\ 135\ 137$
so that, if only one value is incorrect; $4227$ is the value.

Let the universal set, $\xi$ = {$x : 1 \leq  x \leq  15$ and x is an  integer} set H = {x : x is a multiple of 3} and set K = {x : x is an even number}. Find $n(H' \cap K)$.

  1. $2$

  2. $5$

  3. $7$

  4. $13$


Correct Option: B
Explanation:

H' includes all numbers from 1 to 15 that are not multiples of 3.


$H'\cap K ={2,4,8,10,14}$

$n(H'\cap K) =5$

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