Tag: introduction to set

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Questions Related to introduction to set

Multiple choice maths introduction to set different sets de morgan's law de morgan's law for set theory

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice maths introduction to set different sets de morgan's law de morgan's law for set theory

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. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the total number of combatants be 100. The number of soldiers who did NOT lose an eye is 30, an ear is 20, an arm is 25, and a leg is 15. The maximum number of soldiers who lost none of these is the sum of those escaping each injury, which is 30 + 20 + 25 + 15 = 90. Therefore, at least 100 - 90 = 10 percent lost all four limbs.

Multiple choice mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

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.
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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 $a-c=\left( a-b \right) +\left( b-c \right) ={ k } _{ 1 }n+{ k } _{ 2 }n=\left( { k } _{ 1 }+{ k } _{ 2 } \right) n$ 

Multiple choice mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

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. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

Multiple choice mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

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

  1. {odd number}

  2. {composite number}

  3. {2}

  4. {whole numbers}
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

Multiple choice mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

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. $C$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
Multiple choice mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

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. $7$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 )$
$= 11 + 14 - 19 = 6$

Multiple choice mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

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. $A$
Reveal answer Fill a bubble to check yourself
D Correct answer
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, $A\cap C=A$