Tag: introduction to set

Questions Related to introduction to set

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 }  $ ?  

  1. $ \displaystyle \left { 2^{2},4^{2} \right } $

  2. $ \displaystyle \left { 1^{2},5^{2} \right } $

  3. $ \displaystyle \left { 1^{2},5^{2},6 ^{2} \right } $

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

  5. Answer required


Correct Option: D
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.

$|x|$ represent number of elements in region X. Now the following conditions are given
$|U|=14$, $|(A-B)^C|=12$, $|A\cup B|=9$ and $|A\Delta B|=7$, where A and B are two subsets of the universal set U and $A^C$ represents complement of set A, then?

  1. $|A|=2$

  2. $|B|=5$

  3. $|A|=4$

  4. $|B|=7$


Correct Option: A

In a battle $70\% $ of the combatants lost one eye, $80\% $ an ear, $75\% $ an arm, $85\% $ a leg and $x\% $ lost all the four limbs the minimum value of $x$ is 

  1. $10$

  2. $12$

  3. $15$

  4. $none\ of\ these$


Correct Option: B

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, $A\cap C=A$