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 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

A survey on a sample of $25$ new cars being sold at a local auto dealer was conducted to see which of the three popular options - air-conditioning, radio and  power windows - were already installed.
The survey found:
$15$ had air-conditioning
$2$ had air-conditioning and power windows but no radios.
$12$ had power windows
$6$ had air-conditioning and radio but no power windows.
$11$ had radio.
$4$ had radio and power windows.
$3$ had all three options.
What is the number of cars that had none of the options?

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

$n(A\cup R\cup P)= n(A) +n(R) +n(P)-n(A\cap R)-n(R\cap P) -n(P\cap A) + n(A\cap R \cap P)$


$2= n(A\cap P)-n(A\cap P\cap R)$

$\Rightarrow n(A\cap P) = 2+3=5$

$6= n(A\cap R)-n(A\cap P\cap R)$

$\Rightarrow n(A\cap R) = 6+3=9$

$n(A\cup R\cup P)= n(A) +n(R) +n(P)-n(A\cap R)-n(R\cap P) -n(P\cap A) + n(A\cap R \cap P)$

$=15+11+12-9-5-4+3$

So answer $= 25-23 = 2$

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.

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

For any two sets A and B, $\left { (A\setminus B)\cup (B\setminus A) \right }\cap (A\cap B)$ is:

  1. $\phi $
  2. $ A \cup B$
  3. $A \cap B$
  4. $A' \cap B'$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$(A$ \ $B)\cup(B$ \ $A)$ contains that elements form set $A$ and $B$ that are not contained in the other set.

$(A$ \ $B)\cup(B$ \ $A)=(A\cap B)$ \ $(A\cap B)$
$\therefore (A$ \ $B)\cup(B$ \ $A)$ does not contain elements of $A\cap B$
Hence {$(A$ \ $B)\cup(B$ \ $A)$}$\cap (A\cap B)=${$\phi$}

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

Let the universal set, $\xi$ = {$x : 1 \leq  x \leq  15$ and x is an  integer} set H = {x : x is a multiple of 3} and set K = {x : x is an even number}. Find $n(H' \cap K)$.

  1. $2$
  2. $5$
  3. $7$
  4. $13$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

H' includes all numbers from 1 to 15 that are not multiples of 3.


$H'\cap K ={2,4,8,10,14}$

$n(H'\cap K) =5$

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

Consider the following statements for non empty sets A, B and C
1 $\displaystyle A-\left ( B-C \right )=\left ( A-B \right )\cup C $
2 $\displaystyle A-\left ( B\cup C \right )=\left ( A-B \right )- C $
which of the statements given above is/are correct?

  1. 1 only

  2. 2 only

  3. Both 1 and 2

  4. Neither 1 nor 2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

1. $A- (B - C) = \displaystyle A-(B\cap C')$
$\displaystyle =A\cap (B\cap C')'$
$\displaystyle =A\cap (B'\cup (C'))$
$\displaystyle = A\cap (B'\cup C)$
Thus, $\displaystyle A-(B-C)\neq (A-B)\cup C$


2. A- $\displaystyle (B\cup C)=A\cap (B\cap C)'$
$\displaystyle = A\cap (B'\cap C)'$
$\displaystyle (A-B)-C=(A\cap B')-C$
$\displaystyle =A\cap B'\cap C'$
$\displaystyle \Rightarrow A-(B\cup C)=(A-B)-C$
Associative property.