Important Questions for CBSE Class 11 Maths Chapter 1 - Sets 2024-25

Very Short Questions and Answers (1 Marks Questions)

Which of the following are sets? Justify your answer.

1. The collection of all the months of a year beginning with letter  M

Ans: Set, because collection of certain and unique type of data is called a set.

2. The collection of difficult topics in Mathematics.

Ans: Not a set, because difficult topics differ person to person.

Let \[A=\{1,3,5,7,9\}\]. Insert the appropriate symbol in blank spaces:-(Question3,4) 

3. 2-A

Ans: \[\in \]

4. 5-A                                                                                                                                                 

Ans: \[\in \]

5. Write the set $A=\left\{ x:x\text{ is an integer},-1\le x \le 4 \right\}$ in roster form.

Ans: The elements in roster form is as shown

\[A=\{-1,0,1,2,3\}\]

6. List all the elements of the set, $A=\left\{ x:x \in Z,\dfrac{-1}{2}\le x\le \dfrac{11}{2} \right\}$                                                                                                 

Ans: All the elements are as shown 

\[A=\{0,1,2,3,4,5\}\]

7. Write the set $B=\left\{ 3,9,27,81 \right\}$ in set-builder form.

Ans: The above set in set builder form is as shown

\[B=\{x:x={{3}^{n}},n\in N\text{ and }1\le n\le 4\}\]

Which of the following are empty sets? Justify.

8. $A=\left\{ x:x\in N,3\le x \le 4 \right\}$

Ans: Empty set, because there is no natural number that lies between \[3\] and \[4\]

9. $B=\left\{ x:x\in N,{{x}^{2}}=x \right\}$                                                                                                            

Ans: Non-empty set, because there exist natural number which equals to square of itself. For example ${{1}^{2}}=1$ and so on.

Which of the sets are finite or infinite? Justify.

10. The set of all points on the circumference of a circle.                                     

Ans: Infinite set, because there are many points in the circumference of circle

11. $B=\left\{ x:x\in N\text{ and x is an even prime number} \right\}$

Ans: Finite set, because the only even prime number is two.

12. Are sets $A=\left\{ -2,2 \right\},B=\left\{ x:x\in R,{{x}^{2}}-4=0 \right\}$ equal? Why? 

Ans: Yes because the number of elements in A is equal to that of B 

13. Write $\left( -5,\left. 9 \right] \right.$in set-builder form

Ans: \[\left\{ x:x\in ,-5 \le x\le 9 \right\}\]

14. Write $A=\left\{ x:-3\le x \le 7 \right\}$ as interval

Ans: Clearly in interval the above set is written as

\[\left[ -3,\left. 7 \right) \right.\]

15. If $A=\left\{ 1,3,5 \right\}$ how many elements has P(A)? 

Ans: Clearly the number of elements in  \[P\left( A \right)={{2}^{3}}=8\]

16. Write all the possible subsets of $A=\left\{ 5,6 \right\}$.                                                        

Ans: Clearly the possible values of \[A=\left\{ 5,6, \right\}\]is given by

\[\left\{ \varphi ,\left\{ 5 \right\},\left\{ 6 \right\},\left\{ 5,6 \right\} \right\}\]

17. If  $A=\left\{ 2,3,4,5 \right\},B=\left\{ 3,5,6,7 \right\}$. Find $A\bigcup B$

Ans: Clearly \[A\bigcup B=\left\{ 2,3,4,5,6,7 \right\}\]

18.In above question find  $A\bigcap B$  

Ans: Clearly \[A\bigcap B=\left\{ 3,5 \right\}\]

19. If $A=\left\{ 1,2,3,6 \right\},B=\left\{ 1,2,4,8 \right\}$ find $B-A$

Ans: We are given with sets as shown

\[A=\left\{ 1,2,3,6 \right\}\]

\[B=\left\{ 1,2,4,8 \right\}\]

Hence \[B-A=\left\{ 4,8 \right\}\]

20. If $A=\left\{ p,q \right\},B=\left\{ p,q,r \right\}$, is B a superset of \[A\]? Why?

Ans: Yes, because A is a subset of B.

21. Are sets $A=\left\{ 1,2,3,4 \right\},B=\left\{ x:x\in N\text{ and 5}\le \text{x}\le \text{7} \right\}$ disjoint? Why?

Ans: The above mentioned sets are disjoint because  \[\left( A\bigcup B \right)=\varphi \].

22. If X and Y are two sets such that $n\left( X \right)=19,n\left( Y \right)=37,n\left( X\bigcap Y \right)=12$ find $n\left( X\bigcup Y \right)$.

Ans: We know that \[n\left( X\bigcup Y \right)\] is given by

\[n\left( X\bigcup Y \right)=n\left( X \right)+n\left( Y \right)-n\left( X\bigcap Y \right)\]

Hence we get \[n\left( X\bigcup Y \right)=44\]

23. Describe the set in Roster form.

$\left\{ x:\text{x is a two digit number such that the sum of its digits is 8 } \right\}$

Ans: The set in Roster form of above mentioned set is 

\[\left\{ 17,26,35,44,53,62,71,80 \right\}\]

24. Are the following pair of sets equal? Give reasons. 

$A=\left\{ x:\text{x is a letter in the word FOLLOW } \right\}$

$B=\left\{ x:\text{x is a letter in the word WOLF } \right\}$  

Ans: We can write above mentioned sets as shown

\[A=\left\{ F,O,L,W \right\}\]

\[n\left( A \right)=4\]

\[B=\left\{ W,O,L,F \right\}\]

\[n\left( B \right)=4\]

Hence  \[A=B\]

25. Write down all the subsets of the set $\left\{ 1,2,3 \right\}$

Ans: All the subsets of the set \[\{1,2,3\}\] is given below

\[\left\{ \varnothing ,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\} \right\}\]

26. Let $A=\left\{ 1,2,\left\{ 3,4 \right\},5 \right\}$ is $\left\{ \left\{ 3,4 \right\} \right\}\in A$ is incorrect. Give a reason.

Ans: Clearly \[\left\{ 3,4 \right\}\] is an element of set A, therefore \[\left\{ \left\{ 3,4 \right\} \right\}\] is a set containing element \[\left\{ 3,4 \right\}\] which belongs to A.

Hence, \[\left\{ \left\{ 3,4 \right\} \right\}\in A\] is correct.

27. Draw Venn diagram for $\left( A\bigcap B \right)'$

Ans: We know that 

\[\left( A\bigcap B \right)'=U-A\bigcap B\]

Hence the region is shown in the venn diagram below

                                           Fig: \[(A\cap B)'\]

28. Write the set in roster form A the set of letters in TRIGNOMETRY

Ans: The set of letters in TRIGNOMETRY in roster form is written as

\[A=\left\{ T,R,I,G,N,O,M,E,T,R,Y \right\}\]

29. Are the following pair of sets are equal? Give reasons    

A, the set of letters in “ALLOY” and B, the set of letters in “LOYAL”.

Ans: The set of letters in ALLOY is written as

\[A=\left\{ A,,L,O,Y \right\}\]

\[n\left( A \right)=4\]

Similarly, the set of letters in LOYAL is written as

\[B=\left\{ L,O,Y,A \right\}\]

\[n\left( B \right)=4\]

Hence \[A=B\]

30. Write down the power set of A, $A=\left\{ 1,2,3 \right\}$ 

Ans: We know that power set is written as shown

\[P(A)=\left\{ \varnothing ,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\} \right\}\]

31.  $A=\left\{ 1,2,\left\{ 3,4 \right\},5 \right\}$ which is incorrect and why.

 (i) \[\{3,4\}\subset A\]

Ans: Clearly we can see that \[\left\{ 3,4 \right\}\in A\] 

Hence \[\left\{ 3,4 \right\}\subset A\] is incorrect

(ii) \[\{3,4\}\in A\]

Ans: Clearly we can see that \[\left\{ 3,4 \right\}\in A\] 

Hence \[\left\{ 3,4 \right\}\in A\] is correct

32. Fill in the blanks:

(i) $A\bigcup A'$

Ans: We know that \[A\bigcup A'=U\] where U is the universal set

(ii) $\left( A' \right)'$

Ans: We know that \[\left( A' \right)'=A\]

(iii) $A\bigcap A'$

Ans: We know that \[A\bigcap A'=\varphi \]where \[\varphi \] is the universal set.

33. Write the set $\left\{\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5},\dfrac{5}{6},\dfrac{6}{7} \right\}$ in the set builder form.

Ans: The set builder form of above set is given by

\[\left\{ \dfrac{n}{n+1}:n\text{ is a natural number less than or equal to 6} \right\}\]

34. Is set $C=\left\{ x:x-5=0 \right\}$ and

$E=\left\{ x:\text{x is an integral positive root of the equation }{{x}^{2}}-2x-15=0 \right\}$ are equal?

Ans: From set C we get

\[x=5\]

Hence \[C=\left\{ 5 \right\}\]

Also on solving the equation 

\[{{x}^{2}}-2x-15=0\]

We get the positive root as shown

\[x=5\]

E={5}

So, C=E

Hence both the sets are equal

35. Write down all possible proper subsets of the set $\left\{ 1,\left\{ 2 \right\} \right\}$.                           

Ans: All possible proper subsets of the given set are

\[\varphi ,\left\{ 1 \right\},\left\{ 2 \right\},\left\{ 1,\left\{ 2 \right\} \right\}\]

36. State whether each of the following statements is true or false.

(i) $A=\left( 2,3,4,5 \right),B=\left\{ 3,6 \right\}$are disjoint sets.

Ans: Clearly we have

\[\left\{ 2,3,4,5 \right\}\bigcap \left\{ 3,6 \right\}=\left\{ 3 \right\}\ne \varphi \]

Hence the above statement is false

(ii) $A=\left( 2,6,10 \right),B=\left\{ 3,7,11 \right\}$are disjoint sets.

Ans: Clearly we have

\[\left\{ 2,6,10 \right\}\bigcap \left\{ 3,7,11 \right\}=\varphi \]

Hence the above statement is true

37. Solve the followings:

(i) $\left( A\bigcup B \right)'$

Ans: By the properties we write

\[\left( A\bigcup B \right)'=A'\bigcap B'\]

(ii) $\left( A\bigcap B \right)'$

Ans: By the properties we write

\[\left( A\bigcap B \right)'=A'\bigcup B'\]

38. Write the set of all vowels in the English alphabet which precede k in roster form.                        

Ans: The set of all vowels in the English alphabet which precede k in roster form is as shown

\[N=\left\{ a,e,i \right\}\]

39. Is pair of sets equal? Give reasons.

$A=\left( 2,3 \right),B=\left\{ x:\text{x is the solution of }{{x}^{2}}+5x+6=0 \right\}$

Ans: Given we have

\[A=\left\{ 2,3 \right\}\]

\[B=\left\{ x:x\text{ is the solution of }{{\text{x}}^{2}}+5x+6 \right\}\]

Now we can easily find the solution of  \[{{\text{x}}^{2}}+5x+6\] to be the set \[B=\left\{ -2,-3 \right\}\]

Hence  \[A\ne B\]

So the given pair of sets are not  equal

40. Write the following intervals in set builder form: $\left( -3,0 \right)$ and $\left[ \left( 6,12 \right) \right]$ 

Ans: The set builder form of above intervals is given by

\[\left\{ -3,0 \right\}\to \left\{ x:x\in R,-3\le x \le 0 \right\}\]

\[\left\{ 6,12 \right\}\to \left\{ x:x\in R,6\le x\le 12 \right\}\]  

41. If $X=\left\{ a,b,c,d \right\}$

$Y=\left\{ f,b,d,g \right\}$

Find $X-Y$ and $Y-X$

Ans: We are given with the following sets

\[X=\left\{ a,b,c,d \right\}\]

\[Y=\left\{ f,b,d,g \right\}\]

Hence \[X-Y=\left\{ a,c \right\}\]

Similarly, \[Y-X=\left\{ f,g \right\}\]

42. If A and B are two given sets, then represent the set $\left( A-B \right)'$, using the Venn diagram.

Ans: We know that 

\[\left( A-B \right)'=U-\left( A-B \right)\] and hence the venn diagram is as shown

                                    Fig-\[(A-B)'\]

43. List all the element of the set $A=\left\{ x:x\text{ is an integer},{{x}^{2}}\le 4 \right\}$ 

Ans: The elements which we will get is as shown

\[\left\{ -2,-1,0,1,2 \right\}\]

44. From the sets given below pair the equivalent sets.

$A=\left\{ 1,2,3 \right\},B=\left\{ x,y,z,t \right\},C=\left\{ a,b,c \right\},D=\left\{ 0,a \right\}$

Ans: From the data given A and C are equivalent sets because the number of elements in each is same.

45. Write the following as interval                                                                              

(i) $\left\{ x:x\in R,-4 \le x\le 6 \right\}$

Ans: The interval form of above is given as shown

\[\left( -4,6 \right]\]

(ii) $\left\{ x:x\in R,3\le x\le 4 \right\}$

Ans: The interval form of above is given as shown

\[\left[ 3,4 \right]\]

46. If $A=\left\{ 3,5,7,9,11 \right\},B=\left\{ 7,9,11,13 \right\},C=\left\{ 11,13,15 \right\}$ Find $\left( A\bigcap B \right)\bigcap \left( B\bigcup C \right)$

Ans: From the data given we have

\[A=\left\{ 3,5,7,9,11 \right\}\]

\[B=\left\{ 7,9,11,13 \right\}\]

\[C=\left\{ 11,13,15 \right\}\]

Now \[A\bigcap B=\left\{ 7,9,11 \right\}\]

\[B\bigcup C=\left\{ 7,9,11,13,15 \right\}\]

Therefore \[\left( A\bigcap B \right)\bigcap \left( B\bigcup C \right)=\left\{ 7,9,11 \right\}\]

47. Write the set $\left\{ \dfrac{1}{3},\dfrac{3}{5},\dfrac{5}{7},\dfrac{7}{9},\dfrac{9}{11},\dfrac{11}{13} \right\}$in set builder form. 

Ans: \[\left\{ \dfrac{2n-1}{2n+1}:n\text{ is a natural number less than 7} \right\}\]

Long Questions and Answers (4 Marks Questions)

1. In a group of $800$ people, $500$ can speak Hindi and $320$ can speak English. Find

(i) How many can speak both Hindi and English?    

Ans: We will use following notation

H-People who can speak Hindi

E-People who can speak English

It is given in the question that

\[n\left( E\bigcup H \right)=800\]

\[n\left( E \right)=320\]

\[n\left( H \right)=500\]

Also we know that

\[n\left( E\bigcup H \right)=n\left( E \right)+n\left( H \right)-n\left( E\bigcap H \right)\]

800=320+500\[-n\left( E\bigcap H \right)\]

Hence on solving above we get \[20\] people can speak both Hindi and English

(ii) How many can speak Hindi only? 

Ans: We will use following notation

H-People who can speak Hindi

E-People who can speak English

It is given in the question that

\[n\left( E\bigcup H \right)=800\]

\[n\left( E \right)=320\]

\[n\left( H \right)=500\]

Also we find that

\[n\left( E\bigcap H \right)=20\]

\[n\left( E'\bigcap H \right)=n\left( H \right)-n\left( E\bigcap H \right)\]

Hence on solving above we get \[480\] people can speak both Hindi and English

2. A survey shows that $84$ percent of Indians like grapes, whereas $45$ percent like pineapple. What percentage of Indians like both grapes and pineapple?                          

Ans: We will use following notation

A-set of Indians who like grapes

O-set of Indians who like pineapple

It is given in the question that

\[n\left( A\bigcup O \right)=100\]

\[n\left( A \right)=84\]

\[n\left( O \right)=45\]

Now we know that 

\[n\left( A\bigcup O \right)=n\left( A \right)+n\left( O \right)-n\left( A\bigcap O \right)\]

Hence on solving the above we get 

\[n\left( A\bigcap O \right)=29\]

Therefore \[29\] percent of Indians like both apples and oranges 

3. In a survey of $450$ people, it was found that $110$ play cricket, $160$ play tennis and $70$ play both cricket as well as tennis. How many plays neither cricket nor tennis? 

Ans: We will use following notation

S-set of surveyed people

A-set of people who play cricket

O- set of people who play tennis

It is given in the question that

\[n\left( A\bigcap O \right)=70\]

\[n\left( A \right)=110\]

\[n\left( O \right)=160\]

Now we know that 

\[n\left( A\bigcup O \right)=n\left( A \right)+n\left( O \right)-n\left( A\bigcap O \right)\]

\[\Rightarrow n\left( A\bigcup O \right)=110+160-70=200\]

Therefore students who like neither cricket nor tennis is given by

\[n\left( A'\bigcap O' \right)=450-200=250\]

4. In a group of students, $225$ students know French, $100$ know Spanish and $45$ know both. Each student knows either French or Spanish. How many students are there in the group? 

Ans: We will use following notation

A-set of students who know French

O- set of students who know Spanish

It is given in the question that

\[n\left( A\bigcap O \right)=45\]

\[n\left( O \right)=100\]

\[n\left( A \right)=225\]

Now we know that 

\[n\left( A\bigcup O \right)=n\left( A \right)+n\left( O \right)-n\left( A\bigcap O \right)\]

 \[\Rightarrow n\left( A\bigcup O \right)=225+100-45=280\]

Hence there are 280 students in the group.

5. If $A=\left[ \left( -3,5 \right),B=\left( 0,6 \right) \right]$  then find 

(i) $A-B$, 

Ans: Given we have 

\[A=\left( -3,5 \right)\]

\[B=\left( 0,6 \right)\]

We know that \[A-B=A\bigcap B'\]

Hence \[A-B=\left[ -3,0 \right]\]

(ii) $A\bigcup B$

Ans: Given we have 

\[A=\left( -3,5 \right)\]

\[B=\left( 0,6 \right)\]

We know that \[A\bigcup B\] means occurrence of at least one 

Hence \[A\bigcup B=\left[ -3,6 \right]\]

6.  In a survey of $400$ students in a school, $100$ were listed as taking apple juice, $150$ as taking orange juice and $75$ were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice nor orange juice.

Ans: We will use following notation

A-set of students who like apple juice

O- set of students who like orange juice

It is given in the question that

\[n\left( A\bigcap O \right)=75\]

\[n\left( A \right)=100\]

\[n\left( O \right)=150\]

Now we know that 

\[n\left( A\bigcup O \right)=n\left( A \right)+n\left( O \right)-n\left( A\bigcap O \right)\]

\[\Rightarrow n\left( A\bigcup O \right)=100+150-75=175\]

Therefore students who take neither apple nor orange juice is given by

\[n\left( A'\bigcap O' \right)=400-175=225\]

7. A survey shows that $73$ percent of Indians like apples, whereas $65$ percent like oranges. What percent of Indians like both apples and oranges?

Ans: We will use following notation

A-set of Indians who like apples

O-set of Indians who like oranges

It is given in the question that

\[n\left( A\bigcup O \right)=100\]

\[n\left( A \right)=73\]

\[n\left( O \right)=65\]

Now we know that 

\[n\left( A\bigcup O \right)=n\left( A \right)+n\left( O \right)-n\left( A\bigcap O \right)\]

Hence on solving the above we get 

\[n\left( A\bigcap O \right)=38\]

Therefore \[38\] percent of Indians like both apples and oranges 

8. In a school there are $20$ teachers who teach mathematics or physics. Of these $12$ teach mathematics and $4$ teach both physics and mathematics. How many teach physics?                                                                                                                          

Ans: We will use following notation 

P-Number of physics teachers

M- Number of mathematics teachers

We are given 

\[n\left( P\bigcup M \right)=20\]

\[n\left( M \right)=12\]

\[n\left( P\bigcap M \right)=4\]

\[n\left( P\bigcup M \right)=n\left( P \right)+n\left( M \right)-n\left( M\bigcap P \right)\]

On putting the respected values and solving we get

\[n\left( P \right)=12\]

9. Let $U=\left\{ 1,2,3,4,5,6 \right\},A=\left\{ 2,3 \right\},B=\left\{ 3,4,5 \right\}$. Find $A'\bigcap B',A\bigcup B$and hence show that $A\bigcup B=A'\bigcap B'$.               

Ans: We know that 

$A'=U-A$

$=\left\{ 1,4,5,6 \right\}$

$B'=U-B$

$=\left\{ 1,2,6 \right\}$

$A\bigcup B=\left\{ 2,3,4,5 \right\}$

$\left( A'\bigcap B' \right)=\left\{ 1,6 \right\}$

Hence proved.

10. For any two sets A and B prove by using properties of sets that:

$\left( A\bigcap B \right)\bigcup \left( A-B \right)=A$

Ans: We write LHS and RHS as shown

 $LHS=\left( A\bigcap B \right)\bigcup \left( A-B \right)$

$=\left( A\bigcap B \right)\bigcup \left( A\bigcap {{B}^{'}} \right)$ (since $\left( A-B \right)=\left( A\bigcap {{B}^{'}} \right)$)

$=A\bigcap \left( B\bigcup {{B}^{'}} \right)$

$=A\bigcap \left( U \right)$

$=A$

11. If A and B are two sets and $U$ is the universal set such that 

$n\left( U \right)=1000,n\left( A \right)=300,n\left( B \right)=300,n\left( A\bigcap B \right)=200$ find $n\left( {{A}^{'}}\bigcap {{B}^{'}} \right)$.

Ans: We know that 

$n\left( {{A}^{'}}\bigcap {{B}^{'}} \right)=n{{\left( A\bigcup B \right)}^{'}}$

$\Rightarrow n\left( {{A}^{'}}\bigcap {{B}^{'}} \right)=n\left( U \right)-n\left( A\bigcup B \right)$

$\Rightarrow n\left( {{A}^{'}}\bigcap {{B}^{'}} \right)=n\left( U \right)-\left[ n\left( A \right)+n\left( B \right)-n\left( A\bigcap B \right) \right]$

$\Rightarrow n\left( {{A}^{'}}\bigcap {{B}^{'}} \right)=1000-\left[ 300+300-200 \right]=600$

12. There are $210$ members in a club. $100$ of them drink tea and $65$ drink tea but not coffee, each member drinks tea or coffee. Find how many drinks coffee. How many drink coffee, but not tea.             

Ans: Let us have following notation

S-total members in the club

T-members who drink tea

C- members who drink coffee

We have

$n\left( T \right)=100$

$n\left( T-C \right)=65$

$n\left( T\bigcup C \right)=210=n\left( S \right)$(since $n\left( T\bigcap C \right)=0$

We know that

\[n\left( T-C \right)=n\left( T \right)-n\left( T\bigcap C \right)\]

\[\Rightarrow n\left( T\bigcap C \right)=35\]

Also we know that

$n\left( T\bigcup C \right)=n\left( T \right)+n\left( C \right)-n\left( T\bigcap C \right)$

$\Rightarrow n\left( C \right)=145$

Therefore $n\left( C-T \right)=110$

13. If $P\left( A \right)=P\left( B \right)$, Show that $A=B$

Ans: For every $a\in A$ 

$\left\{ a \right\}\subset A$

$\Rightarrow \left\{ a \right\}\in P\left( A \right)$

$\Rightarrow \left\{ a \right\}\in P\left( B \right)$ (since $P\left( A \right)=P\left( B \right)$)

$\Rightarrow \left\{ a \right\}\in B$

$\left\{ a \right\}\subset B$

$\Rightarrow A\subset B$

Similarly we can easily say $B\subset A$

Therefore $B=A$

14. In a class of $25$ students, $12$ have taken mathematics, $8$ have taken mathematics but not biology. Find the no. of students who have taken both mathematics and biology and the no. of those who have taken biology but not mathematics each student has taken either mathematics or biology or both.

Ans: Let us have following notation

T-total number of students

M- number of students who have taken mathematics

B- number of students who have taken biology

$n\left( M \right)=12$

$n\left( M-B \right)=8$

$n\left( M\bigcup B \right)=25$

Now we know that 

$n\left( M\bigcup B \right)=n\left( M \right)+n\left( B-M \right)$

$\Rightarrow 25=12+n\left( B-M \right)$

$\Rightarrow n\left( B-M \right)=13$

\[n\left( M\bigcup B \right)=n\left( M-B \right)+n\left( B-M \right)+n\left( M\bigcap B \right)\]

Hence we get \[n\left( M\bigcap B \right)=4\]  

15. A and B are two sets such that $n\left( A-B \right)=14+x,n\left( B-A \right)=3x,n\left( A\bigcap B \right)=x$. Draw a Venn diagram to illustrate this information. If $n\left( A \right)=n\left( B \right)$, Find 

(i) the value of $x$ 

Ans: It is given in the question 

 $n\left( A-B \right)=14+x$

$n\left( B-A \right)=3x$

$n\left( A\bigcap B \right)=x$

The venn diagram is as shown

We have

$n\left( A \right)=n\left( A-B \right)+n\left( A\bigcap B \right)$

$\Rightarrow n\left( A \right)=14+2x$

$n\left( A \right)=n\left( B-A \right)+n\left( A\bigcap B \right)$

$\Rightarrow n\left( B \right)=4x$

Also it is given that $n\left( B \right)=n\left( A \right)$

Hence $14+2x=4x$

$\Rightarrow x=7$

(ii) $n\left( A\bigcup B \right)$         

Ans: From the above data we have

$n\left( A\bigcup B \right)=n\left( A-B \right)+n\left( B-A \right)+n\left( A\bigcap B \right)$

$\Rightarrow n\left( A\bigcup B \right)=14+x+3x+x=14+5x$

Hence  $n\left( A\bigcup B \right)=49$ (since $x=7$)

16. If A and B are two sets such that $A\bigcup B=A\bigcap B$ , then prove that $A=B$. 

Ans: Let us have $a\in A\Rightarrow a\in A\bigcap B$ 

It is given that $A\bigcup B=A\bigcap B$

Since we have $a\in A\bigcap B$ 

Therefore $A\subset B$

And similarly $B\subset A$

Therefore $A=B$ proved

17. Prove that if $A\bigcup B=C$ and $A\bigcap B=\varphi $ then $A=C-B$ 

Ans: Given $\left( A\bigcup B \right)=C$and $\left( A\bigcap B \right)=\varphi $

Now 

$\left( A\bigcup B \right)-B=\left( A\bigcup B \right)\bigcap {{B}^{'}}$

$=\left( {{B}^{'}}\bigcap A \right)\bigcup \left( {{B}^{'}}\bigcap B \right)$

$=\left( {{B}^{'}}\bigcap A \right)$

$=A-B$

$=A$(since $\left( A\bigcap B \right)=\varphi $)

Hence proved

18. In a group of $65$ people, $40$ like cricket, $10$ like both cricket and tennis. How many like tennis only and not cricket? How many like tennis? 

Ans: Let us have following denotion

C-the set of people who like cricket

T-the set of people who like tennis

$n\left( C\bigcup T \right)=65$

$n\left( C \right)=40$

$n\left( C\bigcap T \right)=10$

We know that 

$n\left( C\bigcup T \right)=n\left( C \right)+n\left( T \right)-n\left( C\bigcap T \right)$

$\Rightarrow 65=40+n\left( T \right)-10$

Hence we get people who like tennis as $n\left( T \right)=35$

Now people who like tennis only not cricket is given by

$n\left( T-C \right)=n\left( T \right)-n\left( C\bigcap T \right)$

$\Rightarrow n\left( T-C \right)=35-10=25$

19. Let A,B and C be three sets $A\bigcup B=A\bigcup C$ and $A\bigcap B=A\bigcap C$ show that $B=C$                                                                                                                                  

Ans: Let us have $b\in B\Rightarrow b\in A\bigcup B$

Also it is given $A\bigcup B=A\bigcup C$

Therefore $b\in A\bigcup C$

Hence we get $b\in A\text{ or }b\in C$

In both cases B is subset of C

Similarly in both cases C is subset of B

Therefore $B=C$

20. If $U=\left\{ a,e,i,o,u \right\}$

$A=\left\{ a,e,i \right\}$ and $B=\left\{ e,o,u \right\}$, $C=\left\{ a,e,i \right\}$

Then verify that \[A\bigcap \left( B-C \right)=A\bigcap B-A\bigcap C\]\[A\cap (B-C)=(A\cap B)-(A\cap C)\]

Ans: From the above data we have

$B-C=\left\{ e,o \right\}$

$A\bigcap \left( B-C \right)=\left\{ e \right\}$

Also 

$A\bigcap B=\left\{ e,o \right\}$and

$A\bigcap C=\left\{ a \right\}$

Hence proved 

$A\bigcap \left( B-C \right)=\left( A\bigcap B \right)-\left( A\bigcap C \right)$

Very Long Questions and Answers (6 Marks Questions)

1. In a survey it is found that $21$ people like product A, $26$ people like product B   and $29$ like product C. If $14$ people like product A and B, $15$ people like product and C, $12$ people like product C and A, and $8$ people like all the three products. Find 

(i) How many people are surveyed in all? 

Ans: Let us have A, B, C denote respectively the set of people who like the products A, B, C.

Then we can have a venn diagram as shown

From the above diagram 

Total number of surveyed people is given by

$a+b+c+d+e+f+g$

It is given in the question that

$a=21,e=26,g=29,d=12,b=14,f=15,c=8$

Therefore total number of surveyed people is given by

$21+14+8+12+26+15+29=125$

(ii) How many like product C only?

Ans: The number of people who like product C only is $29$                                                                 

2. A college awarded $38$ medals in football, $15$ in basketball and $20$ in cricket. If these medals went to a total of $50$ men and only five men got medals in all the three sports, how many received medals in exactly two of the three sports?

Ans: Let us have a notation F, B, and C for medals in football, basketball, and cricket respectively

C is intersection of all A,B,C and a,e,g are intersections of A and not B, B and not C, A and not C respectively.

From the above venn diagram      

\[f=5\]                ……(a)

\[a+b+e+f=38\]……(b)

\[b+c+d+f=15\]……(c)

\[e+d+f+g=20\]……(d)

$a+b+c+d+e+f+g=50$ ……(e)

From equations (d), (e) we get as shown

$a+b+c=30$……(f)

Now from equation (b) and (f) we get as shown

$e-3=c$       …….(g)

put value of c in the equation € as shown

$a+e+g+b+e+d=50-5+3$

Also from equation (d) and (e) we get

$a+e+g=35$

Therefore the medals received in exactly 2 of three sports is given by solving above equations as shown

 \[b+e+d=13\]

3. There are 200 individuals with a skin disorder, $120$ had been exposed to the chemical ${{C}_{1}}$, 50 to chemical ${{C}_{2}}$, and 30 to both the chemicals ${{C}_{1}}$ and ${{C}_{2}}$. Find the number of individuals exposed to 

(i). Chemical ${{C}_{1}}$ but not chemical ${{C}_{2}}$

Ans: Let us have a following notation

A- Denote the set of individuals exposed to the chemical \[{{C}_{1}}\]

B- Denote the set of individuals exposed to the chemical \[{{C}_{2}}\]

Given 

\[n\left( S \right)=200\] 

\[n\left( A \right)=120\]

\[n\left( B \right)=50\]

\[n\left( A\bigcap B \right)=30\]

\[\therefore n\left( A\bigcap \overline{B} \right)=n\left( A \right)-n\left( A\bigcap B \right)\]

\[\Rightarrow n\left( A\bigcap \overline{B} \right)=120-30=90\]

Hence the number of individuals exposed to chemical \[{{C}_{1}}\] but not to \[{{C}_{2}}\] is \[90\]

(ii). Chemical ${{C}_{2}}$ but not chemical ${{C}_{1}}$

Ans: Let us have a following notation

A- Denote the set of individuals exposed to the chemical \[{{C}_{1}}\]

B- Denote the set of individuals exposed to the chemical \[{{C}_{2}}\]

Given 

\[n\left( S \right)=200\] 

\[n\left( A \right)=120\]

\[n\left( B \right)=50\]

\[n\left( A\bigcap B \right)=30\]

\[\therefore n\left( \overline{A}\bigcap B \right)=n\left( B \right)-n\left( A\bigcap B \right)\]

\[\Rightarrow n\left( \overline{A}\bigcap B \right)=50-30=20\]

Hence the number of individuals exposed to chemical \[{{C}_{2}}\] but not to \[{{C}_{1}}\]  is \[20\]

(iii). Chemical ${{C}_{1}}$ or chemical ${{C}_{2}}$

Ans: Let us have a following notation

A- Denote the set of individuals exposed to the chemical \[{{C}_{1}}\]

B- Denote the set of individuals exposed to the chemical \[{{C}_{2}}\]

Given 

\[n\left( S \right)=200\] 

\[n\left( A \right)=120\]

\[n\left( B \right)=50\]

\[n\left( A\bigcap B \right)=30\]

\[\therefore n\left( A\bigcup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\bigcap B \right)\]

\[\Rightarrow n\left( A\bigcup B \right)=120+50-30=140\]

Hence the number of individuals exposed to chemical \[{{C}_{2}}\]or \[{{C}_{1}}\]  is \[140\]

4. In a survey it was found that $21$ people liked product A, $26$ liked product B and $29$ liked product C. If $14$ people liked products A and B, $12$ people like C and A, $15$ people like B and C and $8$ liked all the three products. Find now many liked product C only.

Ans: Let us have a venn diagram of above information as shown

The followings are given in the question

$a+b+c+d=21$

$b+c+e+f=26$

$c+d+f+g=29$

Also it is given in the question 

$b+c=14$

$c+f=15$

$c+d=12$

$c=8$

$\therefore d=4$

$\therefore f=7$

Hence the number of people who like product C only is $g=10$ 

5. A college awarded $38$ medals in football, $15$ in basketball and $20$ in cricket. If these medals went to a total of $58$ men and only three men got medals in all the three sports, how many received medals in exactly two of the three sports?

Ans: Let us denote A, B and C as the sets of men who received medals in football, basketball and cricket respectively.

\[n\left( A \right)=38\]

\[n\left( B \right)=15\]

\[n\left( C \right)=20\]

\[n\left( A\bigcup B\bigcup C \right)=58\]

\[n\left( A\bigcap B\bigcap C \right)=3\]

Now we know that 

\[\left( A\bigcup B\bigcup C \right)=n\left( A \right)+n\left( B \right)+n\left( C \right)-\left[ n\left( A\bigcap B \right)+n\left( B\bigcap C \right)+n\left( C\bigcap A \right) \right]+n\left( A\bigcap B\bigcap C \right)\]

\[\Rightarrow 58=38+15+20-\left( a+d \right)-\left( d+c \right)-\left( b+d \right)+3\]

\[\Rightarrow 18=a+b+c+3d\]

Hence we get \[a+b+c=9\]

6. In a survey of $60$ people, it was found that $25$ people read newspaper H, $26$ read newspaper T, $26$ read newspaper I, $9$ read both H and I, $11$ read both H and T, $8$ read both T and I, $3$ read all three newspapers. Find 

i) The no. of people who read at least one of the newspapers. 

Ans: Let us have a venn diagram as shown

We are given with the following data

\[a+b+c+d=25\]

\[b+c+e+f=26\]

\[d+c+g+f=26\]

And also it is given 

\[c+d=9\]

\[c+b=11\]

\[c+f=8\]

\[c=3\]

\[\therefore f=5\]

\[\therefore b=8\]

\[\therefore d=6\]

\[\therefore g=12\]

\[\therefore e=10\]

\[\therefore a=8\]

The no. of people who read at least one of the newspapers is \[a+b+c+d+e+f+g=52\]

ii) The no. of people who read exactly one newspaper

Ans: Let us have a venn diagram as shown

We are given with the following data

\[a+b+c+d=25\]

\[b+c+e+f=26\]

\[d+c+g+f=26\]

And also it is given 

\[c+d=9\]

\[c+b=11\]

\[c+f=8\]

\[c=3\]

\[\therefore f=5\]

\[\therefore b=8\]

\[\therefore d=6\]

\[\therefore g=12\]

\[\therefore e=10\]

\[\therefore a=8\]

The no. of people who read exactly one newspaper is \[a+e+g=30\]

7. These are $20$ students in a chemistry class and $30$ students in a physics class. Find the number of students which are either in physics class or chemistry class in the following cases.

(i) Two classes meet at the same hour.

Ans: Let \[C\] be the set of students in chemistry class and \[P\] be the set of students in physics class.

\[n\left( C \right)=20\]

\[n\left( P \right)=30\]

Now it is given that two classes meet at the same hour and hence

\[n\left( C\bigcap P \right)=0\]

\[\therefore n\left( C\bigcup P \right)=n\left( C \right)+n\left( P \right)-0\]

\[\Rightarrow n\left( C\bigcup P \right)=20+30=50\]

Hence the number of students which are either in physics class or chemistry class when classes are at the same time is \[50\].

(ii) The two classes met at different hours and ten students are enrolled in both the courses.

Ans: Let \[C\] be the set of students in chemistry class and \[P\] be the set of students in physics class.

\[n\left( C \right)=20\]

\[n\left( P \right)=30\]

Now it is given that two classes meet at the same hour and hence

\[n\left( C\bigcap P \right)=10\]

\[\therefore n\left( C\bigcup P \right)=n\left( C \right)+n\left( P \right)-10\]

\[\Rightarrow n\left( C\bigcup P \right)=20+30-10=40\]

The number of students which are either in physics class or chemistry class when the two classes met at different hours and ten students are enrolled in both the courses is.

8. In a survey of $25$ students, it was found that $15$ had taken mathematics, $12$ had taken physics and $11$ had taken chemistry, $5$ had taken mathematics and chemistry, $9$ had taken mathematics and physics, $4$ had taken physics and chemistry and $3$ had taken all three subjects. 

Find the no. of students that had taken 

(i). only chemistry 

Ans: Let us have a venn diagram of above information as shown

\[n\left( M \right)=a+b+d+e=15\]

\[n\left( P \right)=b+c+f+e=12\]

\[n\left( C \right)=d+e+f+g=11\]

\[n\left( M\bigcap P \right)=b+e=9\]

\[n\left( M\bigcap C \right)=d+e=5\]

\[n\left( P\bigcap C \right)=f+e=4\]

Also it is given that \[e=3\]

\[\therefore b=6,\therefore d=2,\therefore f=1\]

Also \[\therefore a=4,\therefore g=5,\therefore c=2\]

Therefore the number of students who had taken only chemistry is \[g=5\]

(ii). only mathematics 

Ans: Let us have a venn diagram of above information as shown

\[n\left( M \right)=a+b+d+e=15\]

\[n\left( P \right)=b+c+f+e=12\]

\[n\left( C \right)=d+e+f+g=11\]

\[n\left( M\bigcap P \right)=b+e=9\]

\[n\left( M\bigcap C \right)=d+e=5\]

\[n\left( P\bigcap C \right)=f+e=4\]

Also it is given that \[e=3\]

\[\therefore b=6,\therefore d=2,\therefore f=1\]

Also \[\therefore a=4,\therefore g=5,\therefore c=2\]

Therefore the number of students who had taken only mathematics is \[a=4\]

(iii). only physics 

Ans: Let us have a venn diagram of above information as shown

\[n\left( M \right)=a+b+d+e=15\]

\[n\left( P \right)=b+c+f+e=12\]

\[n\left( C \right)=d+e+f+g=11\]

\[n\left( M\bigcap P \right)=b+e=9\]

\[n\left( M\bigcap C \right)=d+e=5\]

\[n\left( P\bigcap C \right)=f+e=4\]

Also it is given that \[e=3\]

\[\therefore b=6,\therefore d=2,\therefore f=1\]

Also \[\therefore a=4,\therefore g=5,\therefore c=2\]

Therefore the number of students who had taken only physics is \[c=2\]

(iv). physics and chemistry but not  mathematics 

Ans: Let us have a venn diagram of above information as shown

\[n\left( M \right)=a+b+d+e=15\]

\[n\left( P \right)=b+c+f+e=12\]

\[n\left( C \right)=d+e+f+g=11\]

\[n\left( M\bigcap P \right)=b+e=9\]

\[n\left( M\bigcap C \right)=d+e=5\]

\[n\left( P\bigcap C \right)=f+e=4\]

Also it is given that \[e=3\]

\[\therefore b=6,\therefore d=2,\therefore f=1\]

Also \[\therefore a=4,\therefore g=5,\therefore c=2\]

Therefore the number of students who had taken physics and chemistry but not mathematics is \[f=1\]

(v). mathematics and physics but not chemistry 

Ans: Let us have a venn diagram of above information as shown

\[n\left( M \right)=a+b+d+e=15\]

\[n\left( P \right)=b+c+f+e=12\]

\[n\left( C \right)=d+e+f+g=11\]

\[n\left( M\bigcap P \right)=b+e=9\]

\[n\left( M\bigcap C \right)=d+e=5\]

\[n\left( P\bigcap C \right)=f+e=4\]

Also it is given that \[e=3\]

\[\therefore b=6,\therefore d=2,\therefore f=1\]

Also \[\therefore a=4,\therefore g=5,\therefore c=2\]

Therefore the number of students who had taken physics and mathematics but not chemistry is \[b=6\]

(vi). only one of the subjects 

Ans: Let us have a venn diagram of above information as shown

\[n\left( M \right)=a+b+d+e=15\]

\[n\left( P \right)=b+c+f+e=12\]

\[n\left( C \right)=d+e+f+g=11\]

\[n\left( M\bigcap P \right)=b+e=9\]

\[n\left( M\bigcap C \right)=d+e=5\]

\[n\left( P\bigcap C \right)=f+e=4\]

Also it is given that \[e=3\]

\[\therefore b=6,\therefore d=2,\therefore f=1\]

Also \[\therefore a=4,\therefore g=5,\therefore c=2\]

Therefore the number of students who had taken only one of the subjects is \[\therefore a+g+c=11\]

(vii). at least one of three subjects 

Ans: Let us have a venn diagram of above information as shown

\[n\left( M \right)=a+b+d+e=15\]

\[n\left( P \right)=b+c+f+e=12\]

\[n\left( C \right)=d+e+f+g=11\]

\[n\left( M\bigcap P \right)=b+e=9\]

\[n\left( M\bigcap C \right)=d+e=5\]

\[n\left( P\bigcap C \right)=f+e=4\]

Also it is given that \[e=3\]

\[\therefore b=6,\therefore d=2,\therefore f=1\]

Also \[\therefore a=4,\therefore g=5,\therefore c=2\]

Therefore the number of students who had taken atleast one of the subjects is \[a+b+c+d+e+f+g=23\]

(viii). None of three subjects.

Ans: Let us have a venn diagram of above information as shown

\[n\left( M \right)=a+b+d+e=15\]

\[n\left( P \right)=b+c+f+e=12\]

\[n\left( C \right)=d+e+f+g=11\]

\[n\left( M\bigcap P \right)=b+e=9\]

\[n\left( M\bigcap C \right)=d+e=5\]

\[n\left( P\bigcap C \right)=f+e=4\]

Also it is given that \[e=3\]

\[\therefore b=6,\therefore d=2,\therefore f=1\]

Also \[\therefore a=4,\therefore g=5,\therefore c=2\]

Therefore the number of students who had taken none of the subjects is \[25-\left( a+b+c+d+e+f+g \right)=2\]

9. In a survey of $100$ students, the no. of students studying the various languages were found to be English only $18$, English but not Hindi $23$, English and Sanskrit $8$, English $26$, Sanskrit $48$, Sanskrit and Hindi $8$, no language $24$. Find 

(i) How many students were studying Hindi? 

Ans: Let the total number of students be 

\[S=100\]

Let us have the venn diagram as shown

\[a=18\]

\[a+e=23\]

\[g+e=8\]

\[a+e+g+d=26\]

\[g+e+f+c=48\]

\[g+f=8\]

So we get 

\[e=5,g=3,d=0,f=5,c=35\]

Therefore the number of students studying Hindi is \[f+b+g+d=18\]

(ii) How many students were studying English and Hindi?

Ans: Let the total number of students be 

\[S=100\]

Let us have the venn diagram as shown

\[a=18\]

\[a+e=23\]

\[g+e=8\]

\[a+e+g+d=26\]

\[g+e+f+c=48\]

\[g+f=8\]

So we get 

\[e=5,g=3,d=0,f=5,c=35\]

Therefore the number of students studying Hindi and English is \[g+d=3\] 

10. In a class of $50$ students, $30$ students like Hindi, $25$ like science and $16$ like both. Find the no. of students who like 

(i) Either Hindi or Science

Ans: Let the total number of students be

\[T=50\]

Let us denote number of students who like Hindi with H and who like science with S

\[n\left( H\bigcup S \right)=n\left( H \right)+n\left( S \right)-n\left( H\bigcap S \right)\]

\[\Rightarrow n\left( H\bigcup S \right)=30+25-16=39\]

Therefore the number of students who like either Hindi or Science is \[39\]

(ii) Neither Hindi nor Science.

Ans: Let the total number of students be

\[T=50\]

Let us denote number of students who like Hindi with H and who like science with S

\[n\left( {{H}^{'}}\bigcap {{S}^{'}} \right)=T-n\left( H\bigcup S \right)\]

\[\Rightarrow n\left( {{H}^{'}}\bigcap {{S}^{'}} \right)=50-39=11\]

\[\Rightarrow n\left( H\bigcup S \right)=30+25-16=39\]

Therefore the number of students who like either Hindi or Science is \[39\]  

11. In a town of 10,000 families, it was found that 40% of families buy newspaper A, 20% families buy newspaper B, and 10% of families buy newspaper C. 5% of families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three papers. Find the no. of families which buy

(i) A only 

Ans: Let the total number of families be 

\[T=10,000\]

Let us have the venn diagram of above information as shown

It is given in the question that 

\[x+a+c+d=4000\]

\[y+a+b+d=2000\]

\[z+b+c+d=1000\]

\[a+d=500\]

\[b+d=300\]

\[c+d=400\]

\[d=200\]

Hence on solving we get

\[a=300,b=100,c=200\]

Therefore the number of families who buy newspaper A only is \[x=4000-300-200-20=3380\]

(ii) B only

Ans: Let the total number of families be 

\[T=10,000\]

Let us have the venn diagram of above informations as shown

It is given in the question that 

\[x+a+c+d=4000\]

\[y+a+b+d=2000\]

\[z+b+c+d=1000\]

\[a+d=500\]

\[b+d=300\]

\[c+d=400\]

\[d=200\]

Hence on solving we get

\[a=300,b=100,c=200\]

Therefore the number of families who buy newspaper B only is \[y=2000-300-200-100=1400\]

(iii) none of A, B, and C. 

Ans: Let the total number of families be 

\[T=10,000\]

Let us have the venn diagram of above informations as shown

It is given in the question that 

\[x+a+c+d=4000\]

\[y+a+b+d=2000\]

\[z+b+c+d=1000\]

\[a+d=500\]

\[b+d=300\]

\[c+d=400\]

\[d=200\]

Hence on solving we get

\[a=300,b=100,c=200\]

From the above we get

\[z=1000-100-200-200=500\]

Therefore the number of families who buy newspaper none of A, B or C is 

\[10000-\left[ 3300+1400+500+300+100+200+200 \right]=5000\]

12. Two finite sets have m and n elements. The total no. of subsets of the first set is 56 more than the total no. of subsets of the second set. Find the value of m and n.      

Ans: Assume A and B be two sets having m and n elements respectively

Hence we know that number of subsets will be given as shown

Number of subsets of A is \[{{2}^{m}}\]

Number of subsets of B is \[{{2}^{n}}\]

According to the question 

\[{{2}^{m}}=56+{{2}^{n}}\]

\[\Rightarrow {{2}^{m}}-{{2}^{n}}=56\]

\[\Rightarrow {{2}^{n}}\left( {{2}^{m-n}}-1 \right)=56\]

On comparing we get

\[n=3,m-n=3\]
\[\Rightarrow m=6\]

Sets Important Question PDF For Download

There is no doubt that we need help when we are solving something for the first time. The same goes for the important questions for class 11 maths chapter 1. Vedantu has provided its students with some tips in the pdf which can make their learning of sets in class 11 extra questions a bit less complicated and fun.

The chapter first of the 11th notebook is easy and has all the essential questions which make students test their formula-solving skills for sets. You can quickly check out the step-by-step solutions of this chapter's important questions in the pdf format and download it offline, so it can be viewed anytime even when the person is offline. 

Important Concepts Class 11 Maths Chapter 1 Related to Sets

Given below, we have breakdown of important concepts you will study in class 11 maths chapter 1. These will help you get a better grip of the formulas and the theorems which you need to use to solve the questions. 

Equal Sets 

For sets in class 11 important questions, one needs to know about sets as they are defined as a collection of well-defined, distinct objects. On the other hand, items which come together to form a set are called elements. The condition of two sets to become equal can happen when each set's element is also a part of the other set. Likewise, if both the sets are subsets of each other, you can even say these two sets are equal. 

Venn Diagram 

It is a diagram which is used by students and mathematicians to represent sets and their relation from each other. By seeing a Venn diagram, you can determine which operation has been done on the given two sets such as the intersection of the sets and their difference. Likewise, one can easily show the subsets of a given set using these diagrams. 

Union & Intersection of Sets

In class 11 sets important questions students will learn about the concept of a cardinal number of a set which is several distinct elements or members in a finite set. With the cardinality's help, we can define the size of a set if you want to denote the cardinal number of a set A you need to write it down like this n(A).

There are three properties of which you need to remember for the cardinal numbers and these are:

• If A ∩ B = ∅, then, n(A ∪ B) = n(A) + n(B) this is a Union of disjoint sets.

• If A and B are two finite sets, then n(A ∪ B) = n(A) + n(B) – n(A ∩ B) which is said to be a union of two sets.

• If A, B and C are three finite sets, then; n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) this shows the union of three sets. 

Sets Class 11 Extra Questions

Well, if you are preparing to give the exams this year or next year, one thing is sure you need to prepare for the additional questions which are a bit tricky, and you can't find them in your textbook as well. Students studying in 11th can prepare their academic and competitive exams by solving these additional questions from Sets class 11. 

Q.1 Which of the following sets. Explain your answer.

(a). A collection of all days which are present in a single week and starts with an alphabet S.

(b).  The collection of the ten most famous singers of India.

(c).  A group of best football strikers in the world.

(d).  The collection of all boys in your school.

(e).  The collection of all the possible odd numbers below 100.

(f).  A collection of poems which are penned down by the famous poet Shakespeare.

(g).  The collection of all prime numbers.

(h).  The collection of questions in a science book.

(i).  A collection of most dangerous reptiles in India.

Q.2: Let P = {2, 3, 4, 5, 6, 7}. Insert the correct symbol  inside the given blank spaces below:

(a).  2 . . . . . . . . . . P

(b).  9 . . . . . . . . . P

(c).  11 . . . . . . . . P

(d).  4 . . . . . . . . P

(e).  0 . . . . . . . . P

(f).  7 . . . . . . . . P

Q.3: List all the elements from the given set P = {y: y is even natural number}

Q.4: If A = {(x,y) : x² + y²= 25  where x, y ∈ W } write a set of all possible ordered pairs.

Q.5: If P(X) = P(Y) show that A = B.

Q.6: Let N  and M be sets ; if N∩M = M∩X = ∅ and N∪X = M∪X for some set X.Show that N=M.

Q.7: If X ={1,2,3,4,5}, then solve the question to find out the proper subsets of A.

Q.8: For this question Write a Roster form of the given set A={x: x ∈ R, 2x+10 =12}

Q.9 Let X and Y are the two sets which have 3 and 6 elements present in them respectively. Find the maximum and the minimum number of elements in X ∪ Y.

Q.10:  If X = {(a,b) : a² + b²= 25  where a, b ∈ W } write a set of all possible ordered pairs.

Practice Questions

• Write the set {-2,7} in the set builder form.

• If the set N = { 1,3,7), then how many elements have set P(N).

• If the universal set (U) = { 1,2,3,4,5,6,7}, A = {2,4,6} , B= { 3,5} and C = { 1,2,4,7}, Find: A′ ∪ (B ∩ C′), and  (B – A) ∪ (A – C)

• If X, Y, and Z are three sets, then X – (Y ∪ Z) is equal to.

• If A = {x, y} and B = { x, y, z). Is Z a superset of Y? Why?

Chapterwise Links for CBSE Class 11 Maths

• Chapter 2 - Relations and Functions

• Chapter 3 - Trigonometric Functions

• Chapter 4 - Principle of Mathematical Induction

• Chapter 5 - Complex Numbers and Quadratic Equations

• Chapter 6 - Linear Inequalities

• Chapter 7 - Permutations and Combinations

• Chapter 8 - Binomial Theorem

• Chapter 9 - Sequences and Series

• Chapter 10 - Straight Lines

• Chapter 11 - Conic Sections

• Chapter 12 - Introduction to Three Dimensional Geometry

• Chapter 13 - Limits and Derivatives

• Chapter 14 - Mathematical Reasoning

• Chapter 15 - Statistics

• Chapter 16 - Probability

Benefits of Solving Important Questions For Class 11 Maths Chapter 1

Let's try to find out why solving the important questions for class 11th maths chapters are pretty essential and need to be done as much as possible. 

• Understanding the Formulae: many times, students might skip the derivation and keep on mugging the formula all along. Knowing your formulas is a good thing, but when you don't know which one to use to solve a question is when the problem comes. With our PDF of solved sets examples, you will be able to understand which formula will be suitable for solving the problem.

• Makes Your Problem Solving Efficient: Once you get a good grip on how to solve the problem, you can easily find out which problem will take more time and start writing it before anything else. 

• Covers Important Topics: With this Pdf designed by Vedantu students get to know about all the main concepts of Venn Diagrams and how to use them along with Complement's properties. As a result, students will understand every topic they need to learn for their upcoming exams. 

• Confidence Booster: When you start solving a question, and it comes out that you managed to get the right answer, you feel uplifted as it boosts your morale. If you have an issue with the answer, you can find out the step by step solving of the union and intersection sets answers. 

• Gives More Questions for Practise: A student needs to be solving different types of problems to sharpen their mind and test their knowledge of the subject, and these important questions do the same thing. 

Conclusion

There you have it, these are some of the basic concepts you are going to study in the class 11th maths chapter 1 based on Sets. The important questions are solved and were written down so that it will be easier for you to understand their language. You need to keep on practising even if you think you are done with the chapter and have enough understanding. Always revise the chapter by doing some questions before you finally appear in exams. For the important questions of class 11th maths chapter 1, you need to put both your mind and heart to study its concepts and get them memorized.

Important Related Links for CBSE Class 11 

Conclusion

Important Questions for Class 11 Maths Chapter 1 - Sets offered by Vedantu is an excellent resource for students who want to excel in their mathematical studies. The questions cover all the important topics in the chapter, including the definition of sets, types of sets, operations on sets, and Venn diagrams, making it easier for students to understand and improve their mathematical skills. The questions are designed by subject matter experts according to the CBSE syllabus for Class 11 students and provide a comprehensive and detailed explanation of the concepts. Additionally, the questions offer practice exercises that help students test their understanding of the chapter and prepare for their exams. Vedantu also provides interactive live classes and doubt-solving sessions to help students clarify their doubts and improve their understanding of the chapter. Overall, the Important Questions for Class 11 Maths Chapter 1 - Sets offered by Vedantu are an essential resource for students who want to improve their mathematical skills and score well in their exams.