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

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

  1. $400$
  2. $600$
  3. $300$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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

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

  1. $240$
  2. $20$
  3. $100$
  4. $120$
Reveal answer Fill a bubble to check yourself
B Correct answer
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. 

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

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

  1. $8$
  2. $20$
  3. $32$
  4. $50$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $=100-68=32$

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

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-

  1. $240$
  2. $20$
  3. $100$
  4. $120$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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


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

                      $=70+60-110$

                      $=20$

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

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 $S$?

  1. $21$
  2. $23$
  3. $32$
  4. $36$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
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, $n = 4$

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

Americans like at least one of cheese or apples. A survey shows that $63$% of the Americans like cheese while $76$% like apples. If $x$ % of the Americans like both cheese and apples, then

  1. $x = 39$
  2. $x= 63$
  3. $3 \leq x \leq 63$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given $A$ be the percent of americans like cheese
           $B$ be the percent of americans like apples

$p(A)=\dfrac{63}{100}=0.63$

$p(B)=\dfrac{76}{100}=0.76$

Let $p(A\cap B)=x$

$p(A\cup B)=1$ as every american likes either cheese or apples

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

$1=0.63+0.76-x$

$x=1.39-1$

$x=0.39\Rightarrow x=0.39\times 100=39\%$
Multiple choice maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

Let the sets $A={2,4,6, 8, ...}$ and $B={3, 6, 9, 12, ...}$, and $n(A)=200, n(B)=250$. Then

  1. $n\left ( A\cap B \right )=67$
  2. $n\left ( A\cup B \right )=450$
  3. $n\left ( A\cap B \right )=66$
  4. $n\left ( A\cup B \right )=384$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

In A, last term will be $400$.

In B, the terms are also in A.P having a common difference of $3$.

Hence 

$a _{n}=a _1+(n-1)d$.

Now $n=250$ for the last term.

Hence

$a _{250}=3+(250-1).3$
$=3(1+250-1)$
$=750.$

Now $A\cap B$ will have elements which are multiples of $6$.

Last term will be $400-4=396$.

Hence
$a _{n}=a+(n-1).d$
$d=6,n=?,a=6$ and $a _{n}=396$

Hence
$396=6+(n-1).6$
Or 
$66=n$.

Hence
$n(A\cap B)=66$.

Now 
$n(A \cup B)=n(A)+n(B)-n(A\cap B)$
$=200+250-66$
$=384$.