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Union and intersection of sets - class-XI

Description: union and intersection of sets
Number of Questions: 48
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Tags: maths mathematics and statistics sets introduction to set sets and relations set language sets and functions sets, relations and functions set concepts sequences and sets
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Let $n$ be a fixed positive integer. Define a relation $R$ on $I$ (the set of all integers) as follows: a R b iff $n|(a-b)$ i.e., iff (a-b) is divisible by n. Show that $R$ is an equivalence relation on 1.

  1. $R$ is an equivalence relation on 1.

  2. $R$ is not an equivalence relation on 1.

  3. $R$ is a symjetric relation on 1.

  4. $R$ is an identity relation on 1.


Correct Option: A,C
Explanation:

If $A\subseteq B$ 
$\therefore A\cap B=A$ 
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, $aRb$. Then by definition of $R$, $a-b=nk$ where $k\in I$.
Hence $b-a=\left( -k \right) n$ where $-k\in I$ and so $bRa$.
Thus we 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 $b-a={ nk } _{ 2 }$
where ${ k } _{ 1 },{ k } _{ 2 }\in I$
It follow that $a-c=\left( a-b \right) +\left( b-c \right) ={ k } _{ 1 }n+{ k } _{ 2 }n=\left( { k } _{ 1 }+{ k } _{ 2 } \right) n$ 

If A = {5, 6, 7, 8} and B = {6, 8, 10} ; find : 
$A \cap B$
  1. ${5}$

  2. ${7,6}$

  3. ${6,8}$

  4. ${6,8,10}$


Correct Option: C
Explanation:

The only 2 common elements are 6 and 8.

A and B are two sets such that $A\displaystyle\cup B$ has $18$ elements If A has $8$ elements and B has $15$ elements then the number of elements in $A\displaystyle\cap  B$ will be: 

  1. $5$

  2. $8$

  3. $7$

  4. $4$


Correct Option: A
Explanation:

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

$n(A \cap B) = n(A)+n(B)-n(A \cup B)=8+15-18=5$

Let A = { even number} B = {prime numbers} Then A $\displaystyle\cap $ B equals: 

  1. {odd number}

  2. {composite number}

  3. {2}

  4. {whole numbers}


Correct Option: C
Explanation:

Given: A = ${2, 4, 6, ...}$
     B = ${2, 3, 5, ...}$
$\displaystyle \therefore $ 2 is the only even prime number $\displaystyle A\cap B=\left { 2 \right }$

Let $A = { x | x$ $\displaystyle \in $ $N$, $x$ is a multiple of 2$ }$
     $ B = { x | x$ $\displaystyle \in $ $N$, $x$ is a multiple of 5$}$
     $C = {x | x$ $\displaystyle \in $ $N$, $x$ is a multiple of 10$}$
The set $\displaystyle\left ( A\cap B \right )\cap C$ is equal to:

  1. $A$

  2. $\displaystyle A \cap C$

  3. $B$

  4. $C$


Correct Option: D
Explanation:
Given A = $\{ 2, 4, 6, 8, 10, 12, 14,...\}$  
B = $\{5, 10, 15, 20, 25,...\}$
C = $\{10, 20, 30, 40, ...\}$
$\displaystyle \Rightarrow $$\displaystyle A\cap B$ = $\{ 10, 20, 30, ...\}$ 
($\displaystyle A\cap B$) $\displaystyle \cap C=$ $\{10, 20, 30, ...\}$ = C

There are $19$ hockey players in a club. On a particular day $14$ were wearing the prescribed hockey shirts while $11$ were wearing the prescribed hockey paints. None of them was without a hockey pant or a hockey shirt. How many of them were in complete hockey uniform ?

  1. $8$

  2. $6$

  3. $9$

  4. $7$


Correct Option: B
Explanation:

Let $P$ and $S$ represents the sets of hockey player wearing the prescribed hockey pants and shirts respectively.
Then $n(P) = 11$, $n(S) = 14$, $n$$\displaystyle \left ( P\cup S \right )$ $= 19$
$\displaystyle \Rightarrow $ $\displaystyle \left ( P\cap S \right )$ $=$ no of people wearing both pantand shirt
$= n(P) + n(S) - n$$\displaystyle \left ( P\cup S \right )$
$= 11 + 14 - 19 = 6$

If $\displaystyle A\cap B=A$ and $\displaystyle B\cap C=B$ then $\displaystyle A\cap C$ is equal to :

  1. $B$

  2. $C$

  3. $\displaystyle B\cup C$

  4. $A$


Correct Option: D
Explanation:

Given:-

$A\cap B=A$ and $B\cap C$
So,$A$ is subset of $B$.
B is a subset of C.Since $B\cap C =B$
$A$ is a subset of $B$ and $B$ is subset of $C$.
So, $A$ and $B$ is subset of $C$.
So, $A\cap C=A$

In a group of $500$ people $200$ can speck Hindi alone while only $125$ can speck English alone The number of people can speck both Hindi and English is

  1. $175$

  2. $325$

  3. $300$

  4. $375$


Correct Option: A
Explanation:

Total no. of people$=500$

People who speak Hindi only$=200$
People who speak English only$=125$
$\therefore$ The number of people who can speak both Hindi and English$=500-(200+125)$
$\Rightarrow 500-325=175$

Given $A={a,b,c,d,e,f,g,h}$ and $B={a,e,i,o,u}$ then $A\cap B$ is equal to

  1. ${a,e}$

  2. ${f,g}$

  3. ${g,h}$

  4. ${i,u}$


Correct Option: A
Explanation:
Given, $A=\{a,b,c,d,e,f,g,h\}$ and $B=\{a,e,i,o,u\}$

$ A$ intersection $B $, which means a new set can be constructed by determining which members are common among the two sets.

So as per the question:-
$A\cap B=\left\{ a,e \right\}$

If X=(multiples of $2$ ), Y = ( multiples of $5$) , Z= (multiples of $10$), then $ \displaystyle X \cap(Y\cap Z)    $ is equal to 

  1. Multiples of $10$

  2. Multiples of $5$

  3. Multiples of $2$

  4. Multiples of $7$


Correct Option: A
Explanation:
$x= (multiples\ of\ 2\ is\ 2, 4, 6, 8, 10.............)$
$y=(multiples\ of\ 5\ is\ 5, 10,15,20,25.........)$
$z=(multiples\ of\ 10\ is\ 10,20,30,40...........)$
Then, $ X\cap(Y\cap Z)$

Apply the value
$=(2, 4, 6, 8, 10,.......)\cap[ (5, 10, 15, 20, 25,........)\cap (10, 20, 30, 40,.......)]$
$=(2, 4, 6, 8, 10,.......)\cap (10, 20, 30,........)$
$=(10, 20, 30,........)$

Hence, this is multiple of $10$.


Hence, this is the answer.

If $A={x:x^2-3x+2=0}$ and $B={x:x^2+4x-5=0}$ then the value of $A-B$ is

  1. ${1,2}$

  2. ${2}$

  3. ${1}$

  4. ${5,2}$


Correct Option: B
Explanation:

$A={x:x^2-3x+2=0}\implies A={1,2}$
$B={x:x^2+4x-5=0}\implies A={1,-5}$
$\therefore A-B={2}$

There are $19$ hockey players in a club. On a particular day $14$ were wearing the prescribed hockey shirts, while $11$ were wearing the prescribed hockey pants. None of them was without hockey pant or hockey shirt. How many of them were in complete hockey uniform?

  1. $8$

  2. $6$

  3. $9$

  4. $7$


Correct Option: B
Explanation:

$n(S\cup P) = 19$

$n(S)=14 $
$ n(P)= 11 $
$n(S\cup P)= n(S)+n(P)-n(S\cap P)$
$19=14+11-n(S\cap P)$
$n(S\cap P)= 25-19 = 6$

Out of $450$ students in a school, $193$ students read Science Today, $200$ students read Junior Statesman, while $80$ students read neither. How many students read both the magazines?

  1. $137$

  2. $80$

  3. $57$

  4. $23$


Correct Option: D
Explanation:

$n(U)=450$

n(Students read Science Today) $=193 =n(S) $

n(Students read Junior Statesman) $=200=n(J) $

n(students read neither) $=80$

$n(S \cup J) $= n(U)-n(students read neither) $= 450-80=370$

Also, $n(S \cup J) = n(S) + n(J)-n(S\cap J) $

$370 = 193+200-n(S\cap J) $

$n(S\cap J)=393-370 =23 $

In a community of $175$ persons, $40$ read the Times, $50$ reads the Samachar and $100$ do not read any. How many persons read both the papers?

  1. $10$

  2. $15$

  3. $20$

  4. $25$


Correct Option: B
Explanation:

$n(U)=175$

n(read Times) $=40 =n(T) $

n(read the samachar) $=50 = n(S) $

n(do not read any) $= 100$

$n(T\cup S)=  n(U)-$ n(do not read any) $= 175-100 =75$

$\therefore n(T \cup S)= n(T)+ n(S)- n(T\cap S) $

$n(T\cap S) =90-75 =15$

In a group of $15, 7$ have studied, German, $8$ have studied French, and $3$ have not studied either. How many of these have studied both German and French?

  1. $0$

  2. $3$

  3. $4$

  4. $5$


Correct Option: B
Explanation:

$n(U)=15$

$n(German) =7 =n(G) $

$n(French) =8 =n(F) $

n(students who have studied neither) $=3$

$n(G \cup F) = n(U)-$ n(students who studied neither) $= 15-3=12 $

$n(G \cup F) = n(G) + n(F)-n(G\cap F) $

$12 = 7+8-n(G\cap F) $

$n(G\cap F)=15-12=3 $

In a class consisting of $100$ students, $20$ know English and $20$ do not know Hindi and $10$ know neither English nor Hindi. The number of students knowing both Hindi and English is

  1. $5$

  2. $10$

  3. $15$

  4. $20$


Correct Option: B
Explanation:

$n(U)=100$

$n(English) =20 =n(E) $

$n(Hindi) =100- 20= 80 =n(H) $

n(students who have studied neither Hindi nor English) $=10$

$n(E \cup H) = n(U)-$ n(students who studied neither) $= 100-10 =90 $

$n(E \cup H) = n(E) + n(H)-n(E\cap H) $

$90 = 20+80-n(E\cap H) $

$n(E\cap H)= 10$

If $A = \left {1, 2, 3, 4, 5, 6, 7, 8\right }$ and $B \left {1, 3, 5, 7\right }$, then find $A - B$ and $A \cap B$

  1. $\left {3, 5\right }$ and $\left {2, 4, 6\right }$

  2. $\left {2, 4, 6\right }$ and $\left {1, 5\right }$

  3. $\left {2, 4, 6, 8\right }$ and $\left {1, 3, 5, 7\right }$

  4. $\left {1, 3, 5, 8\right }$


Correct Option: C
Explanation:

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

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

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

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

In a certain group of $36$ people, $18$ are wearing hats and $24$ are wearing sweaters. If six people are wearing neither a hat nor a sweater, then how many people are wearing both a hat and a sweater?

  1. $30$

  2. $22$

  3. $12$

  4. $8$


Correct Option: C
Explanation:

$n(U)=36$

$n(Hats) =18 =n(H) $

$n(Sweaters) =24 =n(S) $

n(Wearing neither hat nor Sweater) =6

$n(S \cup H) = n(U)-$ n(Wearing neither hat nor sweater) $= 36-6 = 30 $

$n(S \cup H) = n(S) + n(H)-n(S\cap H) $

$30 = 24+18-n(S\cap H) $

$n(S\cap H)=42-30 = 12 $

In a class of $80$ children, $35$% children can play only cricket, $45$% children can play only table-tennis and the remaining children can play both the games. In all, how many children can play cricket?

  1. $55$

  2. $44$

  3. $36$

  4. $28$


Correct Option: B
Explanation:

n(Children who can play only cricket) $ = 35\% = 80\times \dfrac{35}{100} = 28$

n(Children who can play only table tennis) $ = 45\% = 80\times \dfrac{45}{100} = 36$

$n(C \cap T) = 80-28-36 = 16$

n(Children can play cricket) $= 28+16 =44$

If x belongs to set of integers, A is the solution set of $2(x-1)< 3x-1$ and B is the solution set of $4x-3\leq 8+x$, find A$\cap$B.

  1. $\left{0, 1, 2\right}$

  2. $\left{1, 2, 3\right}$

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

  4. $\left{0, 2, 4\right}$


Correct Option: C
Explanation:
$2(x-1)< 3x-1$
$\Rightarrow 2x-2< 3x-1$
$\Rightarrow 2x-3x< -1+2$
$\Rightarrow -x< 1$
$\Rightarrow x> -1$

$4x-3\leq 8+x$
$4x-3-x+3\leq 8+x-x+3$
$\Rightarrow 4x-x\leq 8+3$
$\Rightarrow 3x\leq 11$
$x\leq \dfrac{11}{3}\Rightarrow x\le 3.67$
Then $-1< x\leq 3.67$
Then, $A\cap B$ is $ \{0,1,2,3 \}$

If $A = {1, 3, 5, 7, 8, 6}$, $B = {2, 4, 6, 8, 9}$ .Find $A\cap B$

  1. ${8}$

  2. ${9}$

  3. ${10}$

  4. None of these


Correct Option: A

If $A = \left {2, 3, 4, 8, 10\right }, B = \left (3, 4, 5, 10, 12\right }, C = \left {4, 5, 6, 12, 14\right }$, then $(A\cap B)\cup (A\cap C)$ is equal to

  1. $\left {3, 4, 10\right }$

  2. $\left {2, 9, 10\right }$

  3. $\left {4, 5, 6\right }$

  4. $\left {3, 5, 14\right }$


Correct Option: A
Explanation:

$ A={2,3,4,8,10} , B={3,4,5,10,12} , C={4,5,6,12,14}$


To find : $(A\cap B)$ $\cup  (A\cap C)$

$A\cap B$ = elements common in sets $A$ and $B$. 
$A\cap C$ = elements common in sets $A$ and $C$ 

$A\cap C = \{4\}$ 
$(A\cap B)\cup (A\cap C)=$elements in $(A\cap B)$ and $(A\cap C)$
$(A\cap B) \cup  (A\cap C) = \{3,4,10\} \cup \{4\} $
$(A\cap B) \cup  (A\cap C)={3,4,10}$

If U = {1, 2, 3, 4, 5, 6}; A = {3, 5}; B = {2, 3, 4} C = {4, 5}, find then A $\cap$ (B $\cup$ C).

  1. {1, 2, 3, 4, 5, 6}

  2. {3, 5}

  3. {2, 3, 4}

  4. {4, 5, 6}


Correct Option: B
Explanation:
Given, 
$U=\{1, 2, 3, 4, 5, 6\}$
$A=\{3, 5\}$
$B=\{2, 3, 4\}$
$C=\{4, 5\}$

$B\cup C=\{2, 3, 4, 5\}$

$A\cap (B\cup C)=\{3, 5\}$.

Which is the simplified representation of 
$\left( {{A^/} \cap \,{B^/} \cap \,C} \right) \cup \left( {B\, \cap \,C} \right) \cup \left( {A \cap \,C} \right)$  

where A,B,C are subsets of X

  1. A

  2. B

  3. C

  4. $X \cap \,\left( {A \cup B \cup C} \right)$


Correct Option: D

If $aN=\left{ ax:x\epsilon N \right}$, then the set $3N\cap 7N$ is

  1. $21\ N$

  2. $10\ N$

  3. $4\ N$

  4. None of these


Correct Option: A

If $n(U)= 60, n(A)= 35, n(B)= 24$ and $n(A \cup B)' = 10$ ,then $n(A \cap B)$ is

  1. $9$

  2. $8$

  3. $6$

  4. $7$


Correct Option: A
Explanation:
we have formula,

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

$n(A\cup B)=60-10=50$

Now,

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

$50=35+24-n(A\cap B)$

$\therefore n(A\cap B)=9$

A is a set containing $n$ elements. $A$ subset $P$ of $A$ is chosen. the set $A$ is reconstructed by replacing the elements of $P.A$ subset $Q$ of $A$ is again chosen. the number of ways of choosing $P$ and $Q$ so that $P \cap Q$

  1. $9. ^{n}C _{2}$

  2. $3^{n}- ^{n}C _{2}$

  3. $^{n}C _{2}.3^{n-2}$

  4. $4^{n}-3^{n}$


Correct Option: A

Let $A=\left{ a,b,c,d \right} ,B=\left{ b,c,d,e \right}$. Then $n\left[ \left( A\times B \right) \cap \left( B\times A \right)  \right]$ is equal to 

  1. $3$

  2. $6$

  3. $9$

  4. $none$


Correct Option: A

The set of all points where the function $f(x)=||x|$ is twice differentiable is

  1. $(-\infty, \infty)$

  2. $(-\infty, 0)\cup (0, \infty)$

  3. $(0, \infty)$

  4. $[0, \infty)$


Correct Option: A

Let $S=\left{ \left( x,y \right) :\dfrac { y\left( 3x-1 \right)  }{ x\left( 3x-2 \right)  } <0 \right}$ and $S'=\left{ \left( x,y \right) \in A\times B;\ -1\le A\le 1,-1\le B\le 1 \right} $ There area of $S\cap S'$ is

  1. $1$

  2. $3$

  3. $2$

  4. $4$


Correct Option: A

Let A={1, 2, 3, 4), B={2, 3, 4, 5}, then $n{ (A\times B)\cap (B\times A)} =$?

  1. 13

  2. 16

  3. 9

  4. 10


Correct Option: A

Let $P={ \theta :sin\theta -cos\theta =\sqrt { 2 } cos\theta } $ and $Q={ sin\theta + cos\theta =\sqrt { 2 } sin\theta } $ be two sets. Then:

  1. $P\subset Q\quad and\quad Q-P\neq \emptyset $

  2. $Q\subset P$

  3. $P\subset Q$

  4. $P=Q$


Correct Option: D
Explanation:

$P = \left{ {\theta :\sin \theta  - \cos \theta  = \sqrt 2 \cos \theta } \right}$

$Q = \left{ {\theta :\sin \theta  + \cos \theta  = \sqrt 2 \sin \theta } \right}$
From $P$
$\sin \theta  - \cos \theta  = \sqrt 2 \cos \theta $
$\sin \theta  = \left( {\sqrt 2  + 1} \right)\cos \theta $
$\frac{{\sin \theta }}{{\cos \theta }} = \left( {\sqrt 2  + 1} \right)$
$\tan \theta  = \left( {\sqrt 2  + 1} \right)$
from $Q$
$\sin \theta  + \cos \theta  = \sqrt 2 \sin \theta $
$\sin \theta \left( {\sqrt 2  - 1} \right) = \cos \theta $
$\frac{{\sin \theta }}{{\cos \theta }} = \left( {\sqrt 2  - 1} \right)$
$\tan \theta  = \left( {\sqrt 2  - 1} \right)$
$\tan \theta  = \frac{1}{{\sqrt 2  - 1}} \times \frac{{\sqrt 2  + 1}}{{\sqrt 2  + 1}} = \sqrt 2  + 1$
$\therefore P = Q$
Hence,
option $(D)$ is correct answer.

If $A = {1, 2, 3, 4, 5}, B = {2, 4, 6, 8}$ and C= ${3,4,5,6}$, 

then verify : $A - (B \cup C) = (A - B) \cap (A - C)$.

  1. True

  2. False


Correct Option: A
Explanation:
Given, $A = \{1, 2, 3, 4, 5\}, B = \{2, 4, 6, 8\}$ and $C=\{3,4,5,6\}$

For the LHS:

Union of two sets will have the elements of both sets.

So, $ B \cup C = \{2,3,4,5,6,8 \}$ 

$ A - (B \cup C) $ will have elements of $A$ which are not in $ (B \cup C) $

So, $ A - (B \cup C) = \{ 1 \}$ ..... $(1)$

For the RHS:

$ A - B $ will have elements of $A$ which are not in $B$.

So, $ A - B = \{ 1,3,5 \}$  

$ A - C $ will have elements of $A$ which are not in $C$.

So, $ A - C = \{ 1,2 \}$  

Intersection of two sets has the common elements of both the sets. 

$\Rightarrow (A - B) \cap (A - C) = \{1\}$ ..... $(2)$

From $(1)$ and $(2),$ we have

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

Hence, the given expression is true.

Let $A = {$ multiples of $3$ less than $20 }$
      $B = {$ multiples of $5$ less than $20}$
Then  $A$ $\displaystyle\cap$ $B$ is

  1. ${3, 5}$

  2. ${5, 9}$

  3. ${15}$

  4. $\displaystyle\phi $


Correct Option: C
Explanation:

$A = {$ multiples of $ 3 $ less than $20}$

    $= {3,6,9,12,15,18}$
$B={ $ multiples of $5$ less than $20}$
    $= {5,10,15}$

$A \cap B = 15$

Let $P _1$ be the set of all prime numbers, i.e., $P _1=\left {2, 3, 5, 7, 11, ....\right }$, Let $Pn=\left {np|p\epsilon P _1|\right }$, i.e., the set of all prime multiples of n. Then which of the following sets is non empty?

  1. $P _1\cap P _{23}$

  2. $P _7\cap P _{21}$

  3. $P _{12}\cap P _{20}$

  4. $P _{20}\cap P _{24}$


Correct Option: C
Explanation:

Check by option
$P _{12}=\left {24, 36, 60, 84, ....\right }$
$P _{20}=\left {40, 60, 100, .....\right }$
$P _{12}\cap P _{20}$ has common element.

If $A={a,b,c,d,e,f}$ and $B={c,e,f,g,h}$, then the number of elements of $(A-B)\cap A$ is

  1. $0$

  2. $1$

  3. $2$

  4. $3$


Correct Option: D
Explanation:

$A-B={a,b,d}$
$(A-B)\cap A={a,b,d}$

In a group of $760$ persons, $510$ can speak Hindi and $360$ can speak English. Find how many can speak Hindi only.

  1. $250$

  2. $400$

  3. $1270$

  4. $150$


Correct Option: B
Explanation:
People who can speak both Hindi and English $= n (H ∩ E)$

$n (P) = n(E) + n(H) – n (H ∩ E)$

$n(E\cap H)=510+360-760=110$

We can see that, $H$ is disjoint union of $n(H–E)$ and $n (H ∩ E).$


(If $A$ and $B$ are disjoint then $n (A ∪ B) = n(A) + n(B))$

$∴ H = n(H–E) ∪ n (H ∩ E).$

$⇒ n(H) = n(H–E) + n (H ∩ E).$

$⇒ 510 = n (H – E)+ 110$

$⇒ n(H–E) = 400.$

There are $25$ trays on a table in the cafeteria. Each tray contains a cup only, a plate only, or both a cup and a plate. If $15$ of the trays contain cups and $21$ of the trays contain plates, how many contain both a cup and a plate?

  1. $10$

  2. $11$

  3. $12$

  4. $13$


Correct Option: B
Explanation:
Let $'A'$ denote the set with a cup
$'B'$ denote the set with a plate
$'A$ $\cap$ $B'$ denote the set with both a cup and a plate
$'A$ $\cup$ $B'$ denote the set with the trays either a cup or a plate
$n($ $X)$ denote the number of elements in the set $'X'$

Given, total number of trays $n(A$ $\cup$ $B)$ $=$ $25$
Number of trays that contain cups $n($$A)$ $=$ $15$
Number of trays that contain cups $n($$B)$ $=$ $21$

To find the trays with both a cup and a plate $n(A$ $\cap$ $B)$,

We know that
$n(A$ $\cup$ $B)$ $=$ $n($$A)$$+$ $n($$B)$ $-$ $n(A$ $\cap$ $B)$
Rearranging the terms, we get
$n(A$ $\cap$ $B)$ $=$ $n($$A)$$+$ $n($$B)$ $-$ $n(A$ $\cup$ $B)$
From the above,
$n(A$ $\cap$ $B)$ $=$ $15$ $+$ $21$ $-$ $25$
$=$ $11$
$n(A$ $\cap$ $B)$ $=$ $11$

Therefore, number of trays with both a cup and a plate is $'11'$.

Let A = {x : x is a square of a natural number and x is less than 100} and B is a set of even natural numbers. What is the cardinality of $ A \cap B$ ?

  1. 4

  2. 5

  3. 9

  4. None of the above


Correct Option: A
Explanation:

$A = {1, 4. 9. 16, 25, 36, 49, 64, 81 }$

$B = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 24, 26, 28, 30, 32, 34, 36, 38,...}$
$A \cap B = {4, 16, 36, 64}$
Hence, $n(a \cap B) = 4$.

If $A = ${1,4,6}, $B = ${3,6}, then find $(A \cap B)$

  1. $({1,3,4,6})$

  2. $\phi$

  3. ${6}$

  4. none of these


Correct Option: C
Explanation:
Given,
$A=(1,4,6)$, $B=(3,6)$

$(A\cap B)=common \  of \  A \  and \ B=6  $

$\therefore (A \cap B)=6$

$If\ A = {1, 2, 3}, B = {4, 5}, then\ find\ A \cap B$

  1. ${1, 2, 3, 4, 5}$

  2. $\phi $

  3. ${4,5}$

  4. ${1, 2, 3}$


Correct Option: B

A survey shows that $63\%$ of Indians like mangoes whereas $76\%$ like apple. If $x%$ of the Indians like both mangoes and apples, then

  1. $x=39$

  2. $x=63$

  3. $39\le x\le 63$

  4. $None\ of\ these$


Correct Option: C
Explanation:
Let $100$ denote the population of India, then
$n(A)=63$, $n(B)=76$, $n(A\cap B)=x$
$n(A\cup B)=n(A)+n(B)-n(A\cap B)$
$\therefore x=63+76-n(A\cap B)$ ..$(1)$
If should be noted that $n(A\cup B)\neq 100$ but $n(A\cup B)\le 100$ because there are Indians who may like other fruits besides mangoes and apples.
Hence from $(1)$, $x=139-n(A\cup B)$
$\therefore x\ge 39$ or $39\le x$
Again $A\cap B\subset A$, $A\cap B\subset B$
$\therefore n(A\cap B)\le n(A)=63$, $n(A\cap B)\le n(B)=76$
$\therefore x\le 63$.
Hence $39\le x\le 63$.

Two set A and B are as under 
A = {(a,b) $\epsilon$ R $\times$ R : $\mid a - 5\mid$ < $1$ and  $\mid b - 5\mid$ < $1$};
B = {(a,b) $\epsilon$ R $\times$ R : $4(a-6)^2 + 9(b-5)^2$ $\leq 36$. Then, 

  1. B $\subset$ A

  2. A $\subset$ B

  3. A $\bigcap$ B = $\phi$ (an empty set)

  4. nither A $\subset$ B nor B $\subset$ A


Correct Option: B

$R$ is the set of all positive odd integers less than $20$; $S$ is the set of all multiples of $3$ that are less than $20$. How many elements are in the set $R$ $\cap$ $S$?

  1. 0

  2. 1

  3. 2

  4. 3

  5. 4


Correct Option: D
Explanation:

Given, $R=$ {$1,3,5,7,9,11,13,15,17,19$} , $S=$ {$3,6,9,12,15,18$}
Therefore the intersection of $S$ and $R$ is {$3,9,15$}.
So the number of elements which are common to both is $3$.

Let $Z$ denotes the set of all integers and $A=\left{ \left( a,b \right) :{ a }^{ 2 }+3{ b }^{ 2 }=28,a,b\in Z \right} $ and $B=\left{ \left( a,b \right) :a < b,a,b\in Z \right} $. Then, the number of elements in $A\cap B$ is

  1. $2$

  2. $4$

  3. $6$

  4. $5$


Correct Option: C
Explanation:

$\because A=\left{ \left( a,b \right) :{ a }^{ 2 }+3{ b }^{ 2 }=28,a,b\in Z \right} $
   $=\left{ \left( 5,1 \right) ,\left( -5,-1 \right) ,\left( 5,-1 \right) ,\left( -5,1 \right) ,\left( 4,2 \right) ,\left( -4,-2 \right) ,\left( 4,-2 \right) ,\left( -4,2 \right) ,\left( 1,3 \right) ,\left( -1,-3 \right) ,\left( 1,-3 \right) ,\left( -1,3 \right)  \right}$
and $B=\left{ \left( a,b \right) :a<b,a,b\in Z \right} $
$\therefore A\cap B=\left{ \left( 1,3 \right) ,\left( -1,3 \right) ,\left( -4,-2 \right) ,\left( -4,2 \right) ,\left( -5,-1 \right) ,\left( -5,1 \right)  \right} $
$\therefore $ The number of elements in $A\cap B$ is $6$.

Let $A,B$ be two sets such that $A\cap B=\phi$, then $A=\phi$ and $B=\phi$?

  1. True

  2. False


Correct Option: B
Explanation:

For two sets $A,B$ if $A\cap B=\phi$ then it is not necessary that $A=\phi$ and $B=\phi$.

For example,
Let $A={1,2,3}$ and $B={4,5,6}$.
Now $A\cap B=\phi$ still $A\ne \phi $ and $B\ne \phi$.

If $A={\theta :\tan \theta -\tan^2\theta > 0}, B={\theta :|\sin \theta | < 1/2}$ find $A\cap B$.

  1. $\left(0 , \cfrac{7\pi}{6}\right)$

  2. $\left(0 , \cfrac{\pi}{6}\right)$

  3. $\left(0 , -\cfrac{\pi}{6}\right)$

  4. None of these


Correct Option: B
Explanation:

$\tan \theta - \tan ^2 \theta > 0$ is only true when 

$0<\tan \theta < 1$
or,$ 0< \theta < \cfrac{\pi}{4}$
$A = (0,\cfrac{\pi}{4})$
$|\sin \theta| < \cfrac{1}{2}$
$-\cfrac{1}{2} < \sin \theta < \cfrac{1}{2}$
$B = (-\cfrac{\pi}{6} , \cfrac{\pi}{6})$
$A \cap B = (0 , \cfrac{\pi}{6})$

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