Mathematics

Set Theory and Relations

368 Questions

Set theory involves the study of collections of objects and includes operations like union, intersection, and finding complements. Questions cover power sets, Cartesian products, and properties of empty sets. This foundational mathematical topic frequently appears in various competitive exams and university entrance tests.

set operationspower setscartesian productsset complementsproperties of empty setsset partitions

Set Theory and Relations Questions

Multiple choice mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

The set of integers $Z$ with the binary operation $*$ defined as $a * b = a + b+ 1$ for $a, b, Z$ is a group. The identity element of this group is

  1. $0$
  2. $1$
  3. $-1$
  4. $15$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$a\ast b=a+b+1$   (a,b,z is a group)

at $a=-1 \Rightarrow a\ast b=-1+b+1=b$
at $b=-1  \Rightarrow a\ast b=a-1+1=a$
$\Rightarrow a\ast 0=a+0+1$
$\Rightarrow$ identity element is $-1$.

Multiple choice mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

If * is defined on the set R of all real numbers by $a*b=\sqrt{a^2+b^2}$, find the identity element in R with respect to *.

  1. 0

  2. 1

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let e be the identity element in R with respect to . Then,

$a*e=a=e*a$ for all $a\in R$
$a*e=a$ and $e*a=a$ for all $a\in R$
$\sqrt{a^2+e^2}=a$ and $\sqrt{e^2+a^2}=a$ for all $a\in R$
$a^2+e^2=a^2$ and $e^2+a^2=a^2$ for all $a\in R$
$e=0$
Hence, 0 is the identity element in R with respect to $$.

Multiple choice maths real number fundemental theorem of arithmetic real numbers on number line fundamental theorem of arithmetic

The ........... when multiplied always give a new unique natural number.

  1. decimal numbers

  2. fractions

  3. irrational numbers

  4. prime numbers
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For example: $24$ is made by multiplying the prime numbers $2, 2, 2$ and $3$ together. $24 = 2 \times 2 \times 2 \times 3$
It makes a unique number using a unique combination of $2, 2, 2$ and $3.$
Therefore, $D$ is the correct answer.

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

If $A\subset B$, then $n[P(A)]$ ______ $n[P(B)]$

  1. $=$
  2. $<$
  3. $\leq $
  4. $>$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Assume $ A \subset B$ is true. 

Then, every element of $A$ i.e. $a _1,a _2, ... , a _n$ in A are also in B.

So, number of elements in $B$ will always be greater than no. of elements in $A$

And $P(A)$ will contain less number of subsets than $P(B)$

Hence, $n[P(A)] <  n[P(B)]$

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

Let $U$ be the universal set for sets $A$ and $B$ such that $n(A)=200 , n(B)=300$ and $n(A\cap B)=100$, then $n(A'\cap B')$ is equal to $300$ provided that $n(U)$ is equal to

  1. $600$
  2. $700$
  3. $800$
  4. $900$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$
$=200+300-100$
$=400$
$n(A'\cap B')=n(A\cup B)'$
                    $=n(U) - n(A\cup B)$
$300=n(U)-400$
$n(U)=700$

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

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

Multiple choice maths set concepts finite and infinite sets types of sets set language

The set of all animals on the earth is a 

  1. Finite set

  2. Singleton set

  3. Null set

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

Since the numbers of animals on the earth are countable(limited).
Therefore, set of all the animals on the earth is "finite" set.
Option A is correct.