Mathematics

Set Theory and Relations

355 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

_________ determines whether a variable is a set.

  1. isset

  2. set

  3. exist

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

isset() is a PHP language construct that determines whether a variable is set and is not NULL. It returns true if the variable exists and has a value other than NULL. This is commonly used to check if form fields or array keys have been initialized before using them.

Multiple choice

Let x denote a real number. Find out the INCORRECT statement from the following, where a and b are real numbers.

  1. S = {x : x > 3} represents the set of all real numbers greater than 3.

  2. S = {x : x2 < 0} represents the empty set.
  3. S = {x : x $\in$A and x $\in$ B} represents the union of set A and set B.

  4. S = {x : a < x < b} represents the set of all real numbers between a and b.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\text{The incorrect statement is, S={$x : x\in \text{A and $x$}\in B$} represents the union of set A and set B.}\\ \text{The above symbol $(\in)$ denotes intersection of set A and set B. Therefore this statement is incorrect.}$

Multiple choice

If A = {Multiple of 3 between 1 and 20} and B = {Odd natural number less than 14}, which of the following is the union of two sets A and B?

  1. {1, 3, 5, 6, 6, 9, 11, 12, 13, 15, 18}

  2. {1, 3, 5, 6, 7, 9, 11, 12, 13, 15, 9}

  3. {1, 3, 5, 6, 7, 9, 11, 12, 13, 15, 18}

  4. {1, 3, 5, 6, 7, 9, 1, 12, 13, 15, 18}

  5. {1, 3, 5, 6, 9, 9, 11, 12, 13, 15, 18}
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As per the given information: A = {3, 6, 9, 12, 15, 18} and B = {1, 3, 5, 7, 9, 11, 13} Hence, the union of sets A and B is {1, 3, 5, 6, 7, 9, 11, 12, 13, 15, 18}.

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

If $A$ and $B$ are subsets of $U$ such that $n(U) = 700, n(A) = 200, n(B) = 300, n$$\displaystyle \left ( A\cap B \right )$ $= 100$, then find $n\displaystyle \left ( A'\cap B' \right )$

  1. $405$
  2. $305$
  3. $400$
  4. $300$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that $\displaystyle \left ( A\cup B \right )'=$ $\displaystyle A'\cap B'$ so we need to find $\displaystyle \left ( A\cup B \right ) $  first.
$n\displaystyle \left ( A\cup B \right ) =$ $n(A) + n(B) - n$$\displaystyle (A\cap B)$ $= 200 + 300 - 100 = 400$
$n\displaystyle \left ( A\cup B \right )' =$ $n(U) - n$$\displaystyle \left ( A\cup B \right )'=$$  700 - 400 = 300$
$\displaystyle \Rightarrow $ $n\displaystyle (A'\cap B')$$= 300$.

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

If A has 5 elements and B has 8 elements such that $\displaystyle A\subset B,$ then the number of elements in $\displaystyle A\cap  B,$ and $\displaystyle A\cup  B,$ are respectively :

  1. 8 , 5

  2. 3 , 3

  3. 5, 8

  4. 5, 13
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since $A\subset B$, $A\cap B$ will be all the elements of A. So, the number of elements will be 5.
Since $A\subset B$, $A\cup B$ will be all the elements of A & B. So, the number of elements will be 8.

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

With usual notations $n\left( A\cup B\cup C \right) =20,n\left( A\cap B\cap C\prime  \right) =2,n\left( B\cap C\cap A\prime  \right) =n\left( A\cap C\cap B\prime  \right) =4\quad$

$ and\quad n\left( A\cap B\cap C \right) =1$, then the number of elements belonging to exactly one of the sets is

  1. $9$
  2. $13$
  3. $14$
  4. $16$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the principle of inclusion-exclusion and Venn diagram regions: n(A only) + n(B only) + n(C only) + n(A and B only) + n(B and C only) + n(A and C only) + n(A and B and C) = 20. Given n(A and B and C) = 1, n(A and B only) = 2, n(B and C only) = 4, and n(A and C only) = 4, we solve for the sum of 'only' regions.

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

Given that the universal set,$ \xi =$ {x : 1 < x < 12 and x is an integer} and the sets P = {x : x is a prime number}, Q = {x : x is a multiple of 4} and R = {2, 3, 8, 9} the elements of the set $(Q \cup R)' \cap P$ are:

  1. {2, 3}

  2. {2, 3, 5}

  3. {5, 7, 11}

  4. {1, 5, 7, 11}
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Q={4,8};$     $P={2,3,5,7,11}$;     


$Q\cup R ={2.3.4.8.9}$

$(Q\cup R)' = {5,6,7,10,11}$

$(Q\cup R)'\cap P = {5,7,11}$

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

If among natural numbers $A={5,6,7}$ and $B={8,9,10}$ , then 

  1. $A \cap B =$ null
  2. $(A$ $\cup$ $B)' = A'\cap B'$
  3. $A$ $\cap$ $B = \{2,3,4\}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Since, no elements of A and B are the same.
Therefore, the Intersection is empty.
i.e. $A \cap B = $ null
We know, $U = $  { $ N $ }
$A = $  { $ 5,6,7 $ }
$B = $  { $ 8,9,10 $ } 
($A \cup B$) $=$ {$5,6,7$) $ \cup $ {$8,9,10$}
               $=$ {$5,6,7,8,9,10$}
Therefore,  ($A \cup B$)'$= $ All natural numbers except $ {5,6,7,8,9,10} \dots (i)$
Now, $A'= $ All Natural numbers except ${5,6,7}$
and $B'=$ All Natural numbers except ${8,9,10}$
Therefore, $A' \cap B' = $ All Natural numbers except ${5,6,7,8,9,10}\dots (ii)$
By comparing $(i) $ and $(ii)$ we get,
$(A \cup B) = A' \cap B'$

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

Given $A={x\in N :x<6} ,B={3,6,9}$ and $C={x \in N: 2x-5\le 8}$

  1. $A \cup $ (B $ \cap $C)=(A $ \cap $B) $\cap $(A $ \cap$ C)
  2. (A $\cup$B)'=A'$\cap$B'
  3. A $\cup$B=null set

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$A = { x \in N : x < 6 }$

$\therefore A = { 1,2,3,4,5 }$
$B = { 3,6,9 }$
$C = { x \in N : 2x - 5 \leq 8 }$
$\Rightarrow C = { 1,2,3,4,5,6 }$
Option A. $A \cup (B \cap C) = A \cup {3, 6 } = { 1,2,3,4,5,6 }$
RHS $= (A \cap B) \cap ( A \cup C) = { 3 } \cap {1,2,3,4,5,6} = {3}$
LHS $\neq$ RHS

Option B. $(A \cup B)' = A' \cap B'$ is always true by De Morgan's Law  . 
$A\cup B={1,2,3,4,5,6,9}$
$(A\cup B)'={7,8,10,11,...}$
$A'={x\in N:x\geq 7}$ and $B'={1,2,4,5,7,8,10,11,12...}$
$A'\cap B'={7,8,10,11,12...}$
$\therefore (A\cup B)'=A'\cap B'$

Option C. $A \cap B = { 3 } \neq$ null set
Hence, only B is correct.