Tag: cardinal number of a finite set

Questions Related to cardinal number of a finite set

A market research group conducted a survey of $2500$ consumers and reported that $1620$ consumers like product $p _{1}$ and $1500$ consumers like product $p _{2}$ then (Note $A$ and $B$ denotes the set of products $p _{1}$ and $p _{2}$ respectively)

maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

  1. $\displaystyle n\left ( A \cup B \right )\geq 620$

  2. $\displaystyle n\left ( A \cap B \right )\leq 1500$

  3. $\displaystyle 620 \leq n\left ( A \cap B \right )\leq 1500$

  4. All of these

Show answer
Correct Option: D
Explanation:

$n(A)  =1620, n(B) = 1500$

$\Rightarrow n(A\cap B) \leq min{n(A), n(B)} $

$\Rightarrow n(A\cap B) \leq 1500$

Also $n(A\cup B) \geq  n(A)+n(B) -n(U) =620$

 $n(A\cup B) \geq620$

Hence all options are correct.

In a community it is found that $52$% people like coffee and $73$% like tea. If $x\%$ like both coffee and tea then

maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

  1. $\displaystyle x\geq 25$

  2. $\displaystyle x\leq 52$

  3. $\displaystyle 25\leq x\leq 52 $

  4. all of these

Show answer
Correct Option: C
Explanation:

Let $A=$ number of people like coffee, $B=$ number of people like tea.


$\therefore \ n(A)=52$%  $ \ n(B) = 73$%


$\displaystyle n\left ( A\cap B \right ) =x$%

Let total people in the community =$100 \displaystyle = n\left ( U \right )$
 
$\displaystyle \therefore n\left ( A\cup B \right )\leq 100$ 

$\displaystyle n\left ( A \right )+n\left ( B \right )-n\left ( A\cap B \right )\leq 100$ 

$\displaystyle 52+73-x\leq 100$ 

$\Rightarrow \displaystyle 125-x\leq 100$ 

$\displaystyle\Rightarrow  \therefore x\geq 25$...(i)

Again $\displaystyle A\cap B\subseteq A$

$\displaystyle \Rightarrow n\left ( A\cap B \right )\leq n\left ( A \right )$ 

$\displaystyle x \leq 52$ ...(ii)

$\displaystyle

\therefore $ By (i) and (ii)$\displaystyle 25\leq x\leq 52$

Let $\displaystyle n\left ( u \right )=700,n\left ( A \right )=200, n\left ( B \right )=300, n\left (A\cap B \right )=100$, then $n\left ( A'\cap B' \right )=$

maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

  1. $400$

  2. $600$

  3. $300$

  4. None of these

Show answer
Correct Option: C
Explanation:

$\displaystyle n \left ( A' \cap B'\right )=n\left ( A\cup  B\right

)'$ $\displaystyle =n\left ( u \right )-n\left ( A\cup B \right

)$ $\displaystyle =n\left ( u \right )-\left {n \left ( A \right

)+n\left ( B \right )-n\left ( A\cap B \right ) \right

}$ $\displaystyle =700-\left { 200+300-100 \right }=300$

Let $A$ and $B$ be two sets such that $\displaystyle n\left( A \right) =70$ and $\displaystyle n\left( B \right) =60$ and $\displaystyle n\left( A \cup B \right) =110 $. Then $\displaystyle n\left( A \cap B \right) $ is equal to

maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

  1. $240$

  2. $20$

  3. $100$

  4. $120$

Show answer
Correct Option: B
Explanation:

Given $\displaystyle n\left( A \right) =70$ and $\displaystyle n\left( B \right) =60$ and $\displaystyle n\left( A\quad \cup \quad B \right) =110 $.

$\displaystyle n\left( A\quad \cup \quad B \right)=\displaystyle n\left( A \right)+n\left( B \right)-n\left( A\quad \cap \quad B \right)$

$\Rightarrow 110=70+60-\displaystyle n\left( A\quad \cap \quad B \right)$

$\therefore \displaystyle n\left( A\quad \cap \quad B \right)=20$

Hence, option B. 

Out of 100 students, 50 fail in English and 30 in Mathematics. It 12 students fail in both English and Mathematics, the number of students passing both these subjects is

maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

  1. $8$

  2. $20$

  3. $32$

  4. $50$

Show answer
Correct Option: C
Explanation:

$n(A)=50,n(B)=30,n(A\cap B)=12,n(A\cup B)'=?$
No. of students failed in both subjects
$n(A\cup B)=(50-12)+(30-12)+12$
$=68$
$\therefore$ Req. No. of students passed in both subjects $=100-68=32$

Let $A$ and $B$ be two sets such that $n(A)=70, n(B)=60$ and $n(A\cup B)=110$. Then $n(A\cap B)$ is equal to-

maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

  1. $240$

  2. $20$

  3. $100$

  4. $120$

Show answer
Correct Option: B
Explanation:

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


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

                      $=70+60-110$

                      $=20$

If sets $A$ and $B$ are not disjoint, then $n(A\cup B)$ is equal to

maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

  1. $n(A)+n(B)$

  2. $n(A)+n(B)-n(A\cap B)$

  3. $n(A)+n(B)+n(A\cap B)$

  4. $n(A)$, $n(B)$

Show answer
Correct Option: B
Explanation:

In all cases(even if sets A and B are disjoint)


$n(A\cup B) = n(A) +n(B)-n(A\cap B)$, only that $n(A\cap B)=0$ if they are disjoint.

$A = {$ An integer whose square is a negative value$}$ is 

maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

  1. singleton set

  2. null set

  3. infinite set

  4. disjoint set

Show answer
Correct Option: B
Explanation:

The square of every integer is a non-negative number
$\displaystyle \therefore A = \phi   $ i.e. the null set.

Let $S$ be a set of all distinct numbers of the form $\dfrac{p}{q}$, where $p, q$ $\in [1, 2, 3, 4, 5, 6]$. What is the caardinality of the set $S$?

maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

  1. $21$

  2. $23$

  3. $32$

  4. $36$

Show answer
Correct Option: B
Explanation:

Total possible numbers of form $\dfrac { p }{ q } $ when $p\neq q$ is $={}^{ 6 }C _{ 2 }=30$

Numbers when $p=q$ is $={}^{ 6 }C _{ 1 }=6$

Therefore total numbers $30+6=36$

$\dfrac { 1 }{ 1 } =\dfrac { 2 }{ 2 } =\dfrac { 3 }{ 3 } =\dfrac { 4 }{ 4 } =\dfrac { 5 }{ 5 }= \dfrac { 6 }{ 6 } $      (five numbers deducted from caardinality of set) 

$\dfrac { 1 }{ 2 } =\dfrac { 2 }{ 4 } =\dfrac { 3 }{ 6 } $    (two numbers deducted from caardinality of set) 

$\dfrac { 2 }{ 1 } =\dfrac { 4 }{ 2 } =\dfrac { 6 }{ 3 } $     (two more  numbers deducted from caardinality of set) 

$\dfrac { 1 }{ 3 } =\dfrac { 2 }{ 6 } $          (one number deducted from caardinality of set) 

$\dfrac { 3 }{ 1 } =\dfrac { 6 }{ 2 } $          (one more number deducted from caardinality of set) 

$\dfrac { 2 }{ 3 } =\dfrac { 4 }{ 6 } $          (one number deducted from caardinality of set) 

$\dfrac { 3 }{ 2 } =\dfrac { 6 }{ 4 } $          (one more number deducted from caardinality of set) 

So, the caardiality of set $=36-5-2-2-1-1-1-1=23$.

So, option B is correct.

$n[P(A)] = 16$, then $n(A) =$ ________

maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: D
Explanation:

If $A$ is a set. Then, $n(A)$ is called cardinal number of the set A = number of elements in set A.

$P(A)$ is called power set of $A$. Power set $= P(A) =$ set of all subsets of set $A$

We know that if number of elements in set $A$ is $n$

$n(A) = n$

Number of subsets of power set $= 2 ^n$

Given,

$n[ P(A)] = 16$

$2^n = 16$

$2 ^n = 2 ^4$

Therefore, $n = 4$