CBSE Class 11 Maths Important Questions - Chapter 1 Sets

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 $A=\{x:x\text{ is an integer},-1\le x\le 4\}$ 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, $A=\{x:x\in Z,{\displaystyle \frac{-1}{2}}\le x\le {\displaystyle \frac{11}{2}}\}$                                                                                                 

Ans: All the elements are as shown 

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

7. Write the set $B=\{3,9,27,81\}$ 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. $A=\{x:x\in N,3\le x\le 4\}$

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

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

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

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

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

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

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

Ans: $$\{x:x\in ,-5\le x\le 9\}$$

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

Ans: Clearly in interval the above set is written as

$$[-3,7)$$

15. If $A=\{1,3,5\}$ 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 $A=\{5,6\}$.                                                        

Ans: Clearly the possible values of $$A=\{5,6,\}$$is given by

$$\{\phi ,\left\{5\right\},\left\{6\right\},\{5,6\}\}$$

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

Ans: Clearly $$A\bigcup B=\{2,3,4,5,6,7\}$$

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

Ans: Clearly $$A\bigcap B=\{3,5\}$$

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

Ans: We are given with sets as shown

$$A=\{1,2,3,6\}$$

$$B=\{1,2,4,8\}$$

Hence $$B-A=\{4,8\}$$

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

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

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

Ans: The above mentioned sets are disjoint because  $$(A\bigcup B)=\phi$$.

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

Ans: We know that $$n(X\bigcup Y)$$ is given by

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

Hence we get $$n(X\bigcup Y)=44$$

23. Describe the set in Roster form.

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

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

$$\{17,26,35,44,53,62,71,80\}$$

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

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

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

Ans: We can write above mentioned sets as shown

$$A=\{F,O,L,W\}$$

$$n\left(A\right)=4$$

$$B=\{W,O,L,F\}$$

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

Hence  $$A=B$$

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

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

$$\{\varnothing ,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$$

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

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

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

27. Draw Venn diagram for ${(A\bigcap B)}^{\prime }$

Ans: We know that 

$${(A\bigcap B)}^{\prime }=U-A\bigcap B$$

Hence the region is shown in the venn diagram below

                                           Fig: $$(A\cap B{)}^{\prime }$$

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

$$A=\{T,R,I,G,N,O,M,E,T,R,Y\}$$

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

$$A=\{A,,L,O,Y\}$$

$$n\left(A\right)=4$$

Similarly, the set of letters in LOYAL is written as

$$B=\{L,O,Y,A\}$$

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

Hence $$A=B$$

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

Ans: We know that power set is written as shown

$$P(A)=\{\varnothing ,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$$

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

 (i) $$\{3,4\}\subset A$$

Ans: Clearly we can see that $$\{3,4\}\in A$$ 

Hence $$\{3,4\}\subset A$$ is incorrect

(ii) $$\{3,4\}\in A$$

Ans: Clearly we can see that $$\{3,4\}\in A$$ 

Hence $$\{3,4\}\in A$$ is correct

32. Fill in the blanks:

(i) $A\bigcup A^{\prime }$

Ans: We know that $$A\bigcup A^{\prime }=U$$ where U is the universal set

(ii) ${\left(A^{\prime }\right)}^{\prime }$

Ans: We know that $${\left(A^{\prime }\right)}^{\prime }=A$$

(iii) $A\bigcap A^{\prime }$

Ans: We know that $$A\bigcap A^{\prime }=\phi$$where $$\phi$$ is the universal set.

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

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

$$\{{\displaystyle \frac{n}{n+1}}:n\text{ is a natural number less than or equal to 6}\}$$

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

$E=\{x:\text{x is an integral positive root of the equation }{x}^2-2x-15=0\}$ 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 $\{1,\left\{2\right\}\}$.                           

Ans: All possible proper subsets of the given set are

$$\phi ,\left\{1\right\},\left\{2\right\},\{1,\left\{2\right\}\}$$

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

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

Ans: Clearly we have

$$\{2,3,4,5\}\bigcap \{3,6\}=\left\{3\right\}\ne \phi$$

Hence the above statement is false

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

Ans: Clearly we have

$$\{2,6,10\}\bigcap \{3,7,11\}=\phi$$

Hence the above statement is true

37. Solve the followings:

(i) ${(A\bigcup B)}^{\prime }$

Ans: By the properties we write

$${(A\bigcup B)}^{\prime }=A^{\prime }\bigcap B^{\prime }$$

(ii) ${(A\bigcap B)}^{\prime }$

Ans: By the properties we write

$${(A\bigcap B)}^{\prime }=A^{\prime }\bigcup B^{\prime }$$

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

$$N=\{a,e,i\}$$

39. Is pair of sets equal? Give reasons.

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

Ans: Given we have

$$A=\{2,3\}$$

$$B=\{x:x\text{ is the solution of }{\text{x}}^2+5x+6\}$$

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

Hence  $$A\ne B$$

So the given pair of sets are not  equal

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

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

$$\{-3,0\}\to \{x:x\in R,-3\le x\le 0\}$$

$$\{6,12\}\to \{x:x\in R,6\le x\le 12\}$$  

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

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

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

Ans: We are given with the following sets

$$X=\{a,b,c,d\}$$

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

Hence $$X-Y=\{a,c\}$$

Similarly, $$Y-X=\{f,g\}$$

42. If A and B are two given sets, then represent the set ${(A-B)}^{\prime }$, using the Venn diagram.

Ans: We know that 

$${(A-B)}^{\prime }=U-(A-B)$$ and hence the venn diagram is as shown

                                    Fig-$$(A-B{)}^{\prime }$$

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

Ans: The elements which we will get is as shown

$$\{-2,-1,0,1,2\}$$

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

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

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) $\{x:x\in R,-4\le x\le 6\}$

Ans: The interval form of above is given as shown

$$(-4,6]$$

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

Ans: The interval form of above is given as shown

$$[3,4]$$

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

Ans: From the data given we have

$$A=\{3,5,7,9,11\}$$

$$B=\{7,9,11,13\}$$

$$C=\{11,13,15\}$$

Now $$A\bigcap B=\{7,9,11\}$$

$$B\bigcup C=\{7,9,11,13,15\}$$

Therefore $$(A\bigcap B)\bigcap (B\bigcup C)=\{7,9,11\}$$

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

Ans: $$\{{\displaystyle \frac{2n-1}{2n+1}}:n\text{ is a natural number less than 7}\}$$

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

$$n(E\bigcup H)=800$$

$$n\left(E\right)=320$$

$$n\left(H\right)=500$$

Also we know that

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

800=320+500$$-n(E\bigcap H)$$

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

$$n(E\bigcup H)=800$$

$$n\left(E\right)=320$$

$$n\left(H\right)=500$$

Also we find that

$$n(E\bigcap H)=20$$

$$n(E^{\prime }\bigcap H)=n\left(H\right)-n(E\bigcap H)$$

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

$$n(A\bigcup O)=100$$

$$n\left(A\right)=84$$

$$n\left(O\right)=45$$

Now we know that 

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

Hence on solving the above we get 

$$n(A\bigcap O)=29$$

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

$$n(A\bigcap O)=70$$

$$n\left(A\right)=110$$

$$n\left(O\right)=160$$

Now we know that 

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

$$\Rightarrow n(A\bigcup O)=110+160-70=200$$

Therefore students who like neither cricket nor tennis is given by

$$n(A^{\prime }\bigcap O^{\prime })=450-200=250$$

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

$$n(A\bigcap O)=45$$

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

$$n\left(A\right)=225$$

Now we know that 

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

 $$\Rightarrow n(A\bigcup O)=225+100-45=280$$

Hence there are 280 students in the group.

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

(i) $A-B$, 

Ans: Given we have 

$$A=(-3,5)$$

$$B=(0,6)$$

We know that $$A-B=A\bigcap B^{\prime }$$

Hence $$A-B=[-3,0]$$

(ii) $A\bigcup B$

Ans: Given we have 

$$A=(-3,5)$$

$$B=(0,6)$$

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

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

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

$$n(A\bigcap O)=75$$

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

$$n\left(O\right)=150$$

Now we know that 

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

$$\Rightarrow n(A\bigcup O)=100+150-75=175$$

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

$$n(A^{\prime }\bigcap O^{\prime })=400-175=225$$

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

$$n(A\bigcup O)=100$$

$$n\left(A\right)=73$$

$$n\left(O\right)=65$$

Now we know that 

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

Hence on solving the above we get 

$$n(A\bigcap O)=38$$

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 

$$n(P\bigcup M)=20$$

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

$$n(P\bigcap M)=4$$

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

On putting the respected values and solving we get

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

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

Ans: We know that 

$A^{\prime }=U-A$

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

$B^{\prime }=U-B$

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

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

$(A^{\prime }\bigcap B^{\prime })=\{1,6\}$

Hence proved.

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

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

Ans: We write LHS and RHS as shown

 $LHS=(A\bigcap B)\bigcup (A-B)$

$=(A\bigcap B)\bigcup (A\bigcap B^{{}^{\prime }})$ (since $(A-B)=(A\bigcap B^{{}^{\prime }})$)

$=A\bigcap (B\bigcup B^{{}^{\prime }})$

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

$=A$

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

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

Ans: We know that 

$n(A^{{}^{\prime }}\bigcap B^{{}^{\prime }})=n{(A\bigcup B)}^{{}^{\prime }}$

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

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

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

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

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

$n(T-C)=65$

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

We know that

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

$$\Rightarrow n(T\bigcap C)=35$$

Also we know that

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

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

Therefore $n(C-T)=110$

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

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

$n(M-B)=8$

$n(M\bigcup B)=25$

Now we know that 

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

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

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

$$n(M\bigcup B)=n(M-B)+n(B-M)+n(M\bigcap B)$$

Hence we get $$n(M\bigcap B)=4$$  

15. A and B are two sets such that $n(A-B)=14+x,n(B-A)=3x,n(A\bigcap B)=x$. 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 

 $n(A-B)=14+x$

$n(B-A)=3x$

$n(A\bigcap B)=x$

The venn diagram is as shown

We have

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

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

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

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

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

Hence $14+2x=4x$

$\Rightarrow x=7$

(ii) $n(A\bigcup B)$         

Ans: From the above data we have

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

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

Hence  $n(A\bigcup B)=49$ (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 $A\bigcap B=\phi$ then $A=C-B$ 

Ans: Given $(A\bigcup B)=C$and $(A\bigcap B)=\phi$

Now 

$(A\bigcup B)-B=(A\bigcup B)\bigcap B^{{}^{\prime }}$

$=(B^{{}^{\prime }}\bigcap A)\bigcup (B^{{}^{\prime }}\bigcap B)$

$=(B^{{}^{\prime }}\bigcap A)$

$=A-B$

$=A$(since $(A\bigcap B)=\phi$)

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

$n(C\bigcup T)=65$

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

$n(C\bigcap T)=10$

We know that 

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

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

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

Now people who like tennis only not cricket is given by

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

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

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 $U=\{a,e,i,o,u\}$

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

Then verify that $$A\bigcap (B-C)=A\bigcap B-A\bigcap C$$$$A\cap (B-C)=(A\cap B)-(A\cap C)$$

Ans: From the above data we have

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

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

Also 

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

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

Hence proved 

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

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 

$$n\left(S\right)=200$$ 

$$n\left(A\right)=120$$

$$n\left(B\right)=50$$

$$n(A\bigcap B)=30$$

$$\therefore n(A\bigcap \bar{B})=n\left(A\right)-n(A\bigcap B)$$

$$\Rightarrow n(A\bigcap \bar{B})=120-30=90$$

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 

$$n\left(S\right)=200$$ 

$$n\left(A\right)=120$$

$$n\left(B\right)=50$$

$$n(A\bigcap B)=30$$

$$\therefore n(\bar{A}\bigcap B)=n\left(B\right)-n(A\bigcap B)$$

$$\Rightarrow n(\bar{A}\bigcap B)=50-30=20$$

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 

$$n\left(S\right)=200$$ 

$$n\left(A\right)=120$$

$$n\left(B\right)=50$$

$$n(A\bigcap B)=30$$

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

$$\Rightarrow n(A\bigcup B)=120+50-30=140$$

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.

$$n\left(A\right)=38$$

$$n\left(B\right)=15$$

$$n\left(C\right)=20$$

$$n(A\bigcup B\bigcup C)=58$$

$$n(A\bigcap B\bigcap C)=3$$

Now we know that 

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

$$\Rightarrow 58=38+15+20-(a+d)-(d+c)-(b+d)+3$$

$$\Rightarrow 18=a+b+c+3d$$

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 

$$c+d=9$$

$$c+b=11$$

$$c+f=8$$

$$c=3$$

$$\therefore f=5$$

$$\therefore b=8$$