1   Which of the following are sets? Justify our answer.$\\$ (i) The collection of all months of a year beginning with the letter J.$\\$ (ii) The collection of ten most talented writers of India.$\\$ (iii) A team of eleven best-cricket batsmen of the world.$\\$ (iv) The collection of all boys in your class.$\\$ (v) The collection of all natural numbers less than $100$.$\\$ (vi) A collection of novels written by the writer Munshi Prem Chand.$\\$ (vii) The collection of all even integers.$\\$ (viii) The collection of questions in this chapter.$\\$ (ix) A collection of most dangerous animals of the world.

Solution :

(i) The collection of all months of a year beginning with the letter J is a well-defined collection of objects because one can definitely identity a month that belongs to this collection. Hence, this collection is a set.$\\$ (ii) The collection of ten most talented writer of India is not a well-defined collection because the criteria for determining a writer’s talent vary from person to person. Hence, this collection is not a set.$\\$ (iii) A team of eleven best cricket batsmen of the world is not a well-defined collection because the criteria for determining a batsman’s talent may vary from person to person. Hence, this collection is not a set.$\\$ (iv) The collection of all boys in your class is a well-defined collection because you can definitely identify a boy who belongs to this collection. Hence, this collection is a set.$\\$ (v) The collection of all natural numbers less than 100 is a well-defined collection because one can definitely identify a number that belongs to this collection. Hence, this collection is a set.$\\$ (vi) A collection of novels written by the writer Munshi Prem Chand is a well-defined collection because one can definitely identify a book that belongs to this collection. Hence, this collection is a set.$\\$ (vii) The collection of all even integers is a well-defined collection because one can definitely identify an even integer that belongs to this collection. Hence, this collection is a set.$\\$ (viii) The collection of questions in this chapter is a well-defined collection because one can definitely identify a question that belongs to this chapter. Hence, this collection is a set.$\\$ (ix) The collection of most dangerous animals of the world is not a well-defined collection because the criteria for determining the dangerousness of an animal can vary from person to person. Hence, this collection is not a set.$\\$

2   Let $A =\{ 1, 2,3, 4,5,6 \}$ . Insert the appropriate symbol $ \in $or$\notin $ in the blank spaces: $(i) 5...A\\ (ii) 8...A\\ (iii) 0...A\\ (iv) 4...A\\ (v) 2...A\\ (vi) 10...A$

Solution :

$(i) 5 \in A\\ (ii) 8 \notin A\\ (iii) 0 \notin A\\ (iv) 4 \in A\\ (v) 2 \in A\\ (vi) 10 \notin A$

3   Write the following sets in roster form:$\\$ (i) $A =\{ x : x$ is an integer and $- 3 < x < 7 \} .$$\\$ (ii) $B =\{ x : x$ is a natural number less than $6 \}.$$\\$ (iii) $C =\{ x : x$ is a two-digit natural number such that sum of its digitsis $8 \}$ .$\\$ (iv)$ D =\{ x : x$ is a prime number which is divisor of $60 \}.$$\\$ (v) $E$ = The set of all letters in the world TRIGONOMETRY$\\$ (vi)$ F$ = The set of all letters in the word BETTER.

Solution :

(v) E = The set of all letters in the word TRIGONOMETRY$\\$ There are $12$ letters in the word TRIGONOMETRY, out of which letters T, R and O are repeated Therefore, this set can be written in roster form as $E = \{T, R, I, G, O, N, M, E, Y\}$$\\$ (vi) F = The set of all letters in the word BETTER$\\$ There are 6 letters in the word BETTER, out of which letters E and T are repeated. Therefore, this set can be written in roster form as $F =\{ B , E , T , R \}.$

(i) $A =\{ x : x$ is an integer and $- 3 < x < 7 \} .$$\\$ The elements of this set are $-2,-1,0,1,2,3,4,5$ and $6 $ only Therefore, the given set can be written in roster form as $A=\{-2,-1,0,1,2,3,4,5,6\}$$\\$ (ii) $B =\{ x : x$ is a natural number less than $6 \}.$$\\$ The elements of this set are $1, 2, 3, 4$ and $5$ only. Therefore, the given set can be written in roster form as $B = \{1, 2, 3, 4, 5\}$$\\$ (iii) $C =\{ x : x$ is a two-digit natural number such that sum of its digitsis $8 \}$ .$\\$ The elements of this set are $17, 26, 35, 44, 53, 62, 71$ and $80$ only. Therefore, this set can be written in roster form as $C=\{17, 26, 35, 44, 53, 62, 71,80\}$$\\$ (iv)$ D =\{ x : x$ is a prime number which is divisor of $60 \}.$$\\$ $ \begin{array}{|c|c|c|c|} \hline 2 & 60 \hline 2 & 30 \hline 3 & 15 \hline & 5 \hline \end{array}$ $\\$ $\therefore 60=2*2*3*5$$\\$ The elements of this set are $2, 3 $ and $5$ only. Therefore, this set can be written in roster form as $ D=\{2,3,5 \}$

4   Write the following sets in the set-builder form:$\\$ (i) $(3, 6, 9, 12)$$\\$ (ii) $\{2, 4, 8, 16, 32\}$$\\$ (iii) $\{5, 25, 125, 625\}$$\\$ (iv) $\{2, 4, 6 ...\}$$\\$ (v) $\{1, 4, 9 ... 100\}$

Solution :

(i) A=$\{$ x : x is an odd natural number$\}=\{1,3,5,7,9,...\}$$\\$ (ii) B =$\{$ x : x is an integer;$-\dfrac{1}{2} < x < \dfrac{9}{2}\}$$\\$ It can be seen that $-\dfrac{1}{2}=-0.5 $ and $ \dfrac{9}{2}=4.5 $$\\$ $\therefore B=\{0,1,2,3,4\}$$\\$ (iii) C =$\{$x : x isan integer; x$^2 \leq 4$$\\$ It can be seen that $\\$ $(-1)^2=1 \leq 4;\\ (-2)^2 =4 \leq 4;\\ (-3)^2=9 > 4 \\ 0^2=0 \leq 4\\ 1^2=1 \leq 4\\ 2^2=4 \leq 4\\ 3^2=9> 4\\ \therefore C=\{-2,-1,0,1,2\}$$\\$ (iv) D =$\{$ x : x isa letter in the word"LOYAL" $\}=\{L,O,Y,A\}$$\\$ (v) E =$\{$ x : x isa month of a year not having $31$ days $\\$ $=\{\text{February, April, June,Septermber, November}\}$$\\$ (vi) F=$\{$ x : x isa consonant in the English alphabet which precedes k $\}$$\\$ $=\{b,c,d,f,g,h,j\}$

5   List all the elements of the following sets:$\\$ (i) A=$\{$ x : x is an odd natural number$\}$$\\$ (ii) B =$\{$ x : x is an integer;$-\dfrac{1}{2} < x < \dfrac{9}{2}\}$$\\$ (iii) C =$\{$x : x isan integer; x$^2 \leq 4$$\\$ (iv) D =$\{$ x : x isa letter in the word"LOYAL" $\}$$\\$ (v) E =$\{$ x : x isa month of a year not having $31$ days $\\$ (vi) F=$\{$ x : x isa consonant in the English alphabet which precedes k $\}$

Solution :

(i) A=$\{$ x : x is an odd natural number$\}=\{1,3,5,7,9,...\}$$\\$ (ii) B =$\{$ x : x is an integer;$-\dfrac{1}{2} < x < \dfrac{9}{2}\}$$\\$ It can be seen that $-\dfrac{1}{2}=-0.5 $ and $ \dfrac{9}{2}=4.5 $$\\$ $\therefore B=\{0,1,2,3,4\}$$\\$ (iii) C =$\{$x : x isan integer; x$^2 \leq 4$$\\$ It can be seen that $\\$ $(-1)^2=1 \leq 4;\\ (-2)^2 =4 \leq 4;\\ (-3)^2=9 > 4 \\ 0^2=0 \leq 4\\ 1^2=1 \leq 4\\ 2^2=4 \leq 4\\ 3^2=9> 4\\ \therefore C=\{-2,-1,0,1,2\}$$\\$ (iv) D =$\{$ x : x isa letter in the word"LOYAL" $\}=\{L,O,Y,A\}$$\\$ (v) E =$\{$ x : x isa month of a year not having $31$ days $\\$ $=\{\text{February, April, June,Septermber, November}\}$$\\$ (vi) F=$\{$ x : x isa consonant in the English alphabet which precedes k $\}$$\\$ $=\{b,c,d,f,g,h,j\}$

6   If$(\dfrac{x}{3}+1,y-\dfrac{2}{3})=(\dfrac{5}{3},\dfrac{1}{3}),$ find the values of x and y.

Solution :

It is given that $(\dfrac{x}{3}+1,y-\dfrac{2}{3})=(\dfrac{5}{3},\dfrac{1}{3}),$$\\$ Since the ordered pairs are equal, the corresponding elements will also be equal.$\\$ Therefore, $\dfrac{x}{3}+1=\dfrac{5}{3}$ and $ y-\dfrac{2}{3}=\dfrac{1}{3}$$\\$ $\dfrac{x}{3}+1=\dfrac{5}{3}\\ \implies \dfrac{x}{3}=\dfrac{5}{3}-1 y-\dfrac{2}{3}=\dfrac{1}{3}\\ \implies \dfrac{x}{3}=\dfrac{2}{3} \implies y=\dfrac{1}{3}+\dfrac{2}{3}\\ \implies x=2 \implies y=1\\ \therefore x=2$ and $ y=1 $

7   Match each of the set on the left in the roster form with the same set on the right described in set-builder form:$\\$ (i) $\{1, 2, 3, 6\} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ (a) $\{$ x : x is a primenumber anda divisor of $6\}$$\\$ (ii) $\{2, 3\} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ (b) $\{$ x : x is an odd natural number less than $10\}$$\\$ (iii)$ \{M, A, T, H, E, I, C, S\} \ \ $ (c) $\{$ x : x is natural number and divisor of $6 \}$$\\$ (iv) $\{1, 3, 5, 7, 9\} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ (d) $\{$ x : x is a letter of the word MATHEMATICS $\}$

Solution :

(i) All the elements of this set are natural numbers as well as the divisors of $6.$ Therefore, (i) matches with (c).$\\$ (ii) It can be seen that $2$ and $3$ are prime numbers. They are also the divisors of $6.$ Therefore, (ii) matches with (a).$\\$ (iii) All the elements of this set are letters of the word MATHEMATICS. Therefore, (iii) matches with (d).$\\$ (iv) All the elements of this set are odd natural numbers less than $10.$ Therefore, (iv) matches with (b).

8   Which of the following are examples of the null set $\\$ (i) Set of odd natural numbers divisible by $2$$\\$ (ii) Set of even prime numbers$\\$ (iii) $\{$x : x is a natural numbers, x $ < 5$ and x $> 7\}$$\\$ (iv) $\{$y : y is a point common to any two parallel lines $\}$

Solution :

(i) A set of odd natural numbers divisible by $2$ is a null set because no odd number is divisible by $2.$$\\$ (ii) A set of even prime numbers is not a null set because $2$ is an even prime number.$\\$ (iii) $\{$x : x is a natural number, x $< 5$ and x $> 7 \}$ is a null set because a number cannot be simultaneously less than $5$ and greater than $7.$$\\$ (vi) $\{$ y : y is a point common to any two parallel lines $\}$ is a null set because parallel lines do not intersect. Hence, they have no common point.

9   Which of the following sets are finite or infinite$\\$ (i) The set of months of a year$\\$ (ii) $\{1, 2, 3....\}$$\\$ (iii) $\{1, 2, 3 ... 99, 100\}$ (iv) The set of positive integers greater than $100$$\\$ (v) The set of prime numbers less than $99$

Solution :

(i) The set of months of a year is a finite set because it has $12$ elements.$\\$ (ii) $\{1, 2, 3 ...\}$ is an infinite set as it has infinite number of natural numbers.$\\$ (iii) $\{1, 2, 3... 99, 100\}$ is a finite set because the numbers from $1$ to $100$ are finite in number.$\\$ (iv) The set of positive integers greater than $100$ is an infinite set because positive integers greater than $100$ are infinite in number.$\\$ (v) The set of prime numbers less than $99$ is a finite set because prime numbers less than $99$ are finite in number.

10   State whether each of the following set is finite or infinite:$\\$ (i) The set of lines which are parallel to the x-axis$\\$ (ii) The set of letters in the English alphabet$\\$ (iii) The set of numbers which are multiple of $5$$\\$ (iv) The set of animals living on the earth$\\$ (v) The set of circles passing through the origin $(0, 0)$

Solution :

(i) The set of lines which are parallel to the x-axis is an infinite set because lines parallel to the x-axis are infinite in number.$\\$ (ii) The set of letters in the English alphabet is a finite set because it has $26$ elements.$\\$ (iii) The set of numbers which are multiple of $5$ is an infinite set because multiple of $5$ are infinite in number.$\\$ (iv) The set of animals living on the earth is a finite set because the number of animals living on the earth is finite (although it is quite a big number).$\\$ (v) The set of circles passing through the origin $(0, 0)$ is an finite set because infinite number of circles can pass through the origin.

11   In the following, state whether A = B or not:$\\$ (i) A =$\{ a , b , c , d \}$ ; B =$\{ d , c , b , a \}$$\\$ (ii) A=$\{ 4,8,12,16 \}$ ; B =$\{ 8, 4,16,18 \}$$\\$ (iii) A=$\{ 2, 4,6,8,10 \}$ ; B =$\{$ x : x is positive even integer and x $\leq 10 \}$$\\$ (iv) A=$\{$ x : x is a multiple of $10 \}$ ; B=$\{ 10,15, 20, 25,30....... \}$

Solution :

(i) A=$\{ a , b , c , d \}$ ; B =$\{ d , c , b , a \}$$\\$ The order in which the elements of a set are listed is not significant. $\therefore A = B$$\\$ (ii) A $=\{4,8,12,16 \}$ ; B=$\{ 8, 4,16,18 \}$$\\$ It can be seen that $12 \in$ A but $12 \notin$ B .$\\$ $\therefore A \neq B$$\\$ (iii) A $=\{ 2, 4,6,8,10 \}$$\\$ B =$\{$ x : x is a positive even integer and x $\leq 10 \}$$\\$ = $\{2, 4, 6, 8, 10\}$$\\$ $\therefore A = B$$\\$ (iv) A $=\{$ x : x isa multiple of $10\}$$\\$ B =$\{10,15, 20, 25,30.......\}$$\\$ It can be seen that $15 \in B$ but $15 \notin A .$$\\$ $\therefore A \neq B$

12   Are the following pair of sets equal? Give reasons.$\\$ (i) A $=\{ 2,3 \}$ ; B =$\{$ x : x is solution of $x^ 2+ 5 x + 6 = 0\}$$\\$ (ii) A$=\{$ x : x is a letter in the word FOLLOW $\}$ ; B $=\{$ y : y is a letter in the word WOLF $\}$

Solution :

(i) A $=\{ 2,3 \}$ ; B =$\{$ x : x is a solution of $x^ 2 + 5 x + 6 = 0\}$$\\$ The equation $x ^2 + 5 x + 6 = 0$ can be solved as:$\\$ $x(x+3)+2(x+3)=0\\ (x+2)(x+3)=0\\ x=-2 \ \ 0r \ \ \ x=-3\\ \therefore A=\{2,3\};B=\{-2,-3\} \\ \therefore A\neq B$$\\$ (ii)A=$\{$x : x is a letter in the word FOLLOW $\}=\{$ F,O, L, W $\}$$\\$ B$=\{$ y : y is a letter in the word WOLF$\}=\{$ W,O, L, F $\}$$\\$ The order in which the elements of a set are listed is not significant. $\therefore A = B$

13   From the sets given below, select equal sets:$\\$ $A =\{2, 4,8,12 \} , B =\{ 1, 2,3, 4 \} ,\\ C =\{ 4,8,12,14 \} , D =\{ 3,1, 4, 2 \}\\ E =\{ -1,1 \} , F =\{ 0, a \} ,\\ G =\{ 1, -1 \} , H =\{ 0,1 \}$

Solution :

$A =\{2, 4,8,12 \} , B =\{ 1, 2,3, 4 \} ,\\ C =\{ 4,8,12,14 \} , D =\{ 3,1, 4, 2 \}\\ E =\{ -1,1 \} , F =\{ 0, a \} ,\\ G =\{ 1, -1 \} , H =\{ 0,1 \}$$\\$ It can be seen that $\\$ $8\in A,8\notin B,8\notin D,8\notin E,8\notin F,8 \notin G,8\notin H\\ \implies A\neq B,A \neq D,A\neq E,A\neq F,A \neq G ,A \neq H$$\\$ Also, $ 2\in A,2 \notin C\\ \therefore A\neq C\\ 3 \in B,3 \notin D,3\notin E,3\notin F,3 \notin G ,3 \notin H \\ \therefore B\neq C,B\neq E,B\neq F,B\neq G,B\neq H\\ 12 \in C,12\notin D,12 \notin E ,12 \notin F ,12 \notin G ,12 \notin H\\ \therefore C\neq D,C\neq E,C\neq F,C\neq G,C\neq H\\ 4 \in D ,4 \notin E ,4 \notin F ,4 \notin G ,4 \notin H\\ \therefore D \neq E , D \neq F , D \neq G , D \neq H$$\\$ Similarly,$ E \neq F , E \neq G , E \neq H\\ F \neq G , F \neq H , G \neq H $$\\$ The order in which the elements of a set are listed is not significant. $\therefore B = D and E = G$$\\$ Hence, among the given sets, B = D and E = G.

14   Make correct statements by filling in the symbols $\subset$ or $\not\subset $ in the blank spaces:$\\$ (i) $\{ 2,3, 4 \} ... \{ 1, 2,3, 4,5 \}$$\\$ (ii) $\{ a , b , c \} ... \{ b , c , d \}$$\\$ (iii) $\{$ x : x is a student of class $XI$ of your school $\} ... \{$x : x student of your school $\}$$\\$ (iv) $\{$ x : x is a circle in the plane $\}... \{$ x : x is a circle in the same plane with radius $1$ unit $\}$$\\$ (v) $\{$ x : x is a triangle in a plane $\}... \{$ x : x is a rectangle in the plane $\}$$\\$ (vi) $\{$ x : x is an equilateral triangle in a plane $\}... \{$ x : x is a triangle in the same plane $\}$$\\$ (vii) $\{$ x : x is an even natural number $\} ... \{$ x : x is an integer \}$

Solution :

(i) $\{ 2,3, 4 \} \subset \{ 1, 2,3, 4,5 \}$$\\$ (ii) $\{ a , b , c \} \not \subset \{ b , c , d \}$$\\$ (iii) $\{$ x : x is a student of class $XI$ of your school $\} \subset \{$x : x student of your school $\}$$\\$ (iv) $\{$ x : x is a circle in the plane $\} \not \subset \{$ x : x is a circle in the same plane with radius $1$ unit $\}$$\\$ (v) $\{$ x : x is a triangle in a plane $\} \not \subset \{$ x : x is a rectangle in the plane $\}$$\\$ (vi) $\{$ x : x is an equilateral triangle in a plane $\} \subset \{$ x : x is a triangle in the same plane $\}$$\\$ (vii) $\{$ x : x is an even natural number $\} \subset \{$ x : x is an integer \}$

15   Examine whether the following statements are true or false:$\\$ (i) $\{ a , b \} \not \subset \{ b , c , a \}$$\\$ (ii) $\{ a , e \} \subset \{$ x : x is a vowel in the English alphabet $\}$$\\$ (iii) $\{ 1, 2,3\} \subset \{ 1,3,5\}$$\\$ (iv) $\{ a \} \subset \{a , b , c \}$$\\$ (v) $\{ a \}\in \{ a , b , c \}$$\\$ (vi) $\{$ x : x is an even natural number less than $6 \} \subset \{$ x : x is a natural number which divide $36 \}$

Solution :

(i) False. Each element of $\{ a , b \}$ is also an element of $\{ b , c , a \}$ .$\\$ (ii) True, a , e are two vowels of the English alphabet.$\\$ (iii) False. $2 \in 1, 2,3 \}$ ; however, $2 \notin 1,3,5 \}$$\\$ (iv) True. Each element of $\{a\}$ is also an element of $\{a, b, c\}$.$\\$ (v) False. The element of $\{a, b, c\}$ are $a, b, c.$ Therefore, $\{a \} \subset \{a , b , c \}$ (vi) True. $\{$ x : x is an even natural number less than $6 \}=\{ 2, 4 \}$$\\$ $\{$x : x is a natural number which divides $36\}=\{ 1, 2,3, 4,6,9,12,18,36 \}$

16   Let $A =\{ 1, 2, \{ 3, 4 \} ,5 \} .$ Which of the following statements are incorrect and why?$\\$ $(i) \{ 3, 4 \} \subset A\\ (ii) \{3,4\}\in A\\ (iii)\{\{3,4\}\}\subset A\\ (iv)1\notin A\\ (v)1 \subset A\\ (vi) \{1,2,5\} \subset A\\ (vii) \{1,2,5\}\in A\\ (viii) \{1,2,3\} \subset A\\ (ix) \varnothing \notin A\\ (x) \varnothing \subset A\\ (xi) \{\varnothing \} \subset A$

Solution :

A $=\{ 1, 2,\{3, 4 \} ,5\}$$\\$ (i) The statement $\{ 3, 4 \} \subset $ A is incorrect because $3 \in \{ 3, 4 \}$ ; however, $3 \notin $ A .$\\$ (ii) The statement $\{ 3, 4\} \notin $ A is correct because $\{ 3, 4 \}$ is an element of A.$\\$ (iii) The statement $\{ \{ 3, 4 \} \} \subset $ A is correct because $\{ 3, 4\}\notin \{ \{ 3, 4 \} \}$ and $\{ 3, 4 \}\in $ A .$\\$ (iv) The statement $1 \notin $ A is correct because $1$ is an element of A.$\\$ (v) The statement $1 \subset $ A is incorrect because an element of a set can never be a subset of itself.$\\$ (vi) The statement $\{ 1, 2,5\} \subset $ A is correct because each element of $\{ 1, 2,5 \}$ is also an element of A.$\\$ (vii) The statement $\{ 1, 2,5 \} \notin $ A is incorrect because $\{1, 2,5 \}$ is not an element of A. (viii) The statement $\{ 1, 2,5\} \subset $ A is incorrect because $3 \notin \{ 1, 2,3 \}$ ; however, $3 \notin A .$$\\$ (ix) The statement $\varnothing \notin $ A is incorrect because $\varnothing $ is not an element of A.$\\$ (x) The statement $\varnothing \subset $ A is correct because $\varnothing $ is a subset of every set.$\\$ (xi) The statement $\{ \varnothing \} \subset $ A is incorrect because, $\varnothing $ is a subset of A and it is not an element of A.

17   Write down all the subsets of the following sets:$\\$ (i)$\{ a \}$$\\$ (ii)$\{a,b\}$$\\$ (iii)$\{1,2,3\}$$\\$ (iv)$\varnothing $

Solution :

(i) The subsets of $\{a \}$ are $ \varnothing $ and $\{ a \} .$$\\$ (ii) The subsets $\{a , b \}$ are $\varnothing , \{ a \} , \{ b \} ,$ and $\{ a , b \}$ .$\\$ (iii) The subsets of $\{ 1, 2,3 \}$ are $\varnothing , \{ 1 \} , \{ 2 \} , \{ 3 \} , \{ 1, 2 \} , \{ 2,3 \} , \{ 1,3 \}$ and $\{ 1, 2,3 \}$ .$\\$ (iv) The only subset of $\varnothing $ is $\varnothing $ .

18   How many elements has $P ( A )$ , if $A = \varnothing ? $

Solution :

We know that if A is a set with m elements i.e., $n(A)=m,$ , then $n[p(A)]=2^m$$\\$ If A $=\varnothing $ , then $n ( A )= 0 .$$\\$ $\therefore n [ P ( A)]= 2 ^0 = 1$$\\$ Hence, $P ( A ) $ has one element.

19   Write the following as intervals:$\\$ (i) $\{ x : x \notin R , - 4 < x \leq 6 \}$$\\$ (ii) $\{ x : x \notin R , - 12 < x <- 10 \}$$\\$ (iii) $\{ x : x \notin R ,0 \leq x < 7 \}$$\\$ (iv) $\{ x : x \notin R ,3 \leq x \leq 4 \}$

Solution :

(i) $\{ x : x \notin R , - 4 < x \leq 6 \}=(-4,6)$$\\$ (ii) $\{ x : x \notin R , - 12 < x <- 10 \}=(-12,-10)$$\\$ (iii) $\{ x : x \notin R ,0 \leq x < 7 \}=(0,7)$$\\$ (iv) $\{ x : x \notin R ,3 \leq x \leq 4 \}=(3,4)$

20   Write the following intervals in set-builder form: $\\$ (i)$(-3,0)$$\\$ (ii)$[6,12)$$\\$ (iii)$(6,12)$$\\$ (iv)$(-23,5)$$\\$

Solution :

(i)$(-3,0)=\{x:x\notin R,-3 < x < 0 \}$$\\$ (ii)$[6,12)=\{x:x \notin R,6 \leq x \leq 12 \}$$\\$ (iii)$(6,12)=\{x:x\notin R,6 < x \leq 12\}$$\\$ (iv)$(-23,5)=\{ x: x \notin R,-23 \leq x < 5 \}$$\\$

21   What universal set (s) would you propose for each of the following:$\\$ (i) The set of right triangles$\\$ (ii) The set of isosceles triangles

Solution :

(i) For the set of right triangles, the universal set can be the set of triangles or the set of polygons.$\\$ (ii) For the set of isosceles triangles, the universal set can be the set of triangles or the set of polygons or the set of two-dimensional figures.

22   Given the sets $A=\{1,3,5\},B=\{2,4,6\}$ and $ C=\{0,2,4,6,8\},$which of the following may be considered as universals set (s) for all the three sets $A, B$ and $C.$$\\$ (i) $\{ 0,1, 2,3, 4,5,6 \}$$\\$ (ii) $\varnothing $$\\$ (iii) $\{ 0,1, 2,3, 4,5,6,7,8,9,10 \}$$\\$ (iv) $\{ 1, 2,3, 4,5,6,7,8 \}$

Solution :

(i) It can be seen that $A \subset \{ 0,1, 2,3, 4,5,6 \}$ $\\$ $B \subset \{ 0,1, 2,3, 4,5,6 \}$ $\\$ However, $C \not \subset \{0,1, 2,3, 4,5,6 \}$ $\\$ Therefore, the set $\{ 0,1, 2,3, 4,5,6 \}$ cannot be the universal set for the sets $A, B,$ and $C.$ $\\$ (ii) $A \not \subset \oslash , B \not \subset \oslash , C \not \subset \oslash $ $\\$ Therefore, $\oslash$ cannot be the universal set for the sets A, B and C.$\\$ (iii) $A \subset \{ 0,1, 2,3, 4,5,6,7,8,9,10 \}$ $\\$ $B \subset \{ 0,1, 2,3, 4,5,6,7,8,9,10 \}$ $\\$ $C \not \subset \{ 0,1, 2,3, 4,5,6,7,8,9,10 \}$ $\\$ Therefore, the set $\{ 0,1, 2,3, 4,5,6,7,8,9,10 \}$ is the universal set for the sets A, B, and C.$\\$ (iv) $A \subset \{ 1, 2,3, 4,5,6,7,8 \}$ $\\$ $B \subset \{1, 2,3, 4,5,6,7,8 \}$ $\\$ However, $C \not \subset \{ 1, 2,3, 4,5,6,7,8 \}$ $\\$ Therefore, the set $\{ 1, 2,3, 4,5,6,7,8 \}$ cannot be the universal set for the sets A, B and C.

23   Find the union of each of the following pairs of sets: $\\$ (i) X $\{ 1,3,5 \}$ Y =$\{ 1, 2,3 \}$ $\\$ (ii) A =$\{$ a , e , i , o , u $\}$ B =$\{$ a , b , c $\}$ $\\$ (iii) A =$\{$ x : x is a natural number and multiple of $3 \}$ $\\$ B =$\{$ x : x is a natural number less than $6 \}$ $\\$ (iv) A =$\{$ x : x is a natural number and $1 < x \leq 6 \}$ $\\$ B =$\{$ x : x is a natural number and $6 < x < 10 \}$ $\\$ (v) A =$\{ 1, 2,3 \} , B = \oslash$ $\\$

Solution :

(i) X =$\{ 1,3,5 \}$ Y =$\{ 1, 2,3 \}$ $\\$ X $\cup$ Y =$\{ 1, 2,3,5 \}$ $\\$ (ii) A =$\{$ a , e , i , o , u $\}$ B =$\{$ a , b , c $\}$ $\\$ A $\cup$ B =$\{ a , b , c , e , i , o , u \}$ $\\$ (iii) A =$\{ $x : x is a natural number and multiple of $3 \}=\{ 3,6,9... \}$ $\\$ As B =$\{$ x : x isa natural number less than $6 \}=\{ 1, 2,3, 4,5,6 \}$ $\\$ A $\cup$ B=$\{ 1, 2, 4,5,3,6,9,12...\}$ $\\$ $\therefore$ A $\cup$ B =$\{$ x : x = 1, 2, 4,5 or a multiple of $3 \}$ $\\$ (iv) A =$\{$ x : x is a natural number and $1 < x \leq 6 \}=\{ 2,3, 4,5,6 \}$ $\\$ B=$\{ $ x : x is a natural number and $6 < x < 10\}=\{ 7,8,9 \}$ $\\$ A $\cup$ B =$\{2,3, 4,5,6,7,8,9 \}$ $\\$ $\therefore $ A$\cup$ B=$\{ x : x \in$ N and $1 < x < 10 \}$ $\\$ (v) A =$\{ 1, 2,3 \} , B =\oslash $ $\\$ A $\cup$ B =$\{ 1, 2,3 \}$ $\\$

24   Let A =$\{$ a , b $\} $, B =$\{$ a , b , c $\}$ . Is A $\subset$ B ? What is A $\cup$ B ?

Solution :

Here, A =$\{$ a , b $\} $, and B =$\{$ a , b , c $\} $ $\\$ Yes, A $\subset$ B$\\$ A $\cup$ B =$\{$ a , b , c $\}$= B

25   If A and B are two sets such that A $ \subset$ B , then what is A $\cup$ B ?

Solution :

If A and B are two sets such that A $ \subset$ B , then A $\cup$ B =B

26   If A =$\{$ 1, 2,3, 4 $\}$ , B =$\{$ 3, 4,5,6 $\}$ , C =$\{$ 5,6,7,8 $\}$ and D =$\{$ 7,8,9,10 $\}$ ; find $\\$ (i) A $\cup $B $\\$ (ii) A $\cup $ C $\\$ (iii) B $\cup $ C $\\$ (iv) B $\cup $ D $\\$ (v) A $\cup $ B $\cup $C $\\$ (vi) A $\cup $ B $\cup $ D $\\$ (vii) B $\cup $ C $\cup $ D $\\$

Solution :

A =$ \{$1, 2,3, 4], B =$ \{$ 3, 4,5,6 $\}$ , C=$ \{$ 5,6,7,8 $\}$ and D =$ \{$ 7,8,9,10 $\}$ $\\$ (i) A $\cup$ B =$ \{$ 1, 2,3, 4,5,6 $\}$ $\\$ (ii) A $\cup$ C =$ \{$1, 2,3, 4,5,6,7,8 $\}$ $\\$ (iii) B $\cup$ C =$ \{$ 3, 4,5,6,7,8 $\}$ $\\$ (iv) B $\cup$ D =$ \{$3, 4,5,6,7,8,9,10 $\}$ $\\$ (v) A $\cup$B $\cup$ C =$ \{$ 1, 2,3, 4,5,6,7,8 $\}$ $\\$ (vi) A $\cup$ B$\cup$ D =$ \{$ 1, 2,3, 4,5,6,7,8,9,10 $\}$ $\\$ (vii) B $\cup$ C $\cup$ D =$ \{$ 3, 4,5,6,7,8,9,10 $\}$ $\\$

27   Find the intersection of each pair of sets:$\\$ (i) X =$\{$ 1,3,5 $\}$ Y =$\{$1, 2,3 $\}$ $\\$ (ii) A =$\{$ a , e , i , o , u $\}$ B =$\{$ a , b , c $\}$ $\\$ (iii) A =$\{$ x : x isa natural number and multipleof 3$\}$ $\\$ B =$\{$ x : x isa natural number less than 6$\}$ $\\$ (iv) A =$\{$ x : x isa natural number and1 < x $\leq$ 6 $\}$ $\\$ B =$\{$ x : x isa natural number and 6 < x <10 $\}$ $\\$ (v) A =$\{$ 1, 2,3 $\}$ , B =$\oslash$ $\\$

Solution :

(i) X =$\{$ 1,3,5 $\}$ Y =$\{$1, 2,3 $\}$ $\\$ X$\cap$ Y $\{$ 1,3 $\}$ $\\$ (ii) A =$\{$ a , e , i , o , u $\}$ B =$\{$ a , b , c $\}$ $\\$ A$\cap$ B =$\{$ a $\}$ $\\$ (iii) A =$\{$ x : x isa natural number and multipleof 3$\}$ $\\$ B =$\{$ x : x isa natural number less than 6$\}$ $\\$ $\therefore$ A $\cap$B=$\{$ 3 $\}$ $\\$ ((iv) A =$\{$ x : x isa natural number and1 < x $\leq$ 6 $\}$ $\\$ B =$\{$ x : x isa natural number and 6 < x <10 $\}$ $\\$ A $\cap$B $=\oslash $$\\$ (v) A =$\{$ 1, 2,3 $\}$ , B =$\oslash$ $\\$ A$\cap$ B $=\oslash $ $\\$

28   If A =$\{$3,5,7,9,11 $\}$ , B =$\{$ 7,9,11,13 $\}$ , C=$\{$ 11,13,15 $\}$ and D =$\{$ 15,17 $\}$ ; find $\\$ (i) A $\cap $ B $\\$ (ii) B $\cap $ C $\\$ (iii) A $\cap $ C $\cap $ D$\\$ (iv) A $\cap $ C $\\$ (v) B $\cap $ D $\\$ (vi) A $\cap $ ( B $\cap $ C )$\\$ (vii) A$\cap $ D ( B$\cap $ D ) $\\$(ix) ( A$\cap $ B )( B$\cap $ C )$\\$ (x) ( A$\cap $ D )( B$\cap $ C )$\\$

Solution :

(i) A $\cap $ B =$\{$ 7,9,11 $\}$ $\\$ (ii) B $\cap $ C =$\{$ 11,13 $\}$ $\\$ (iii) A $\cap $ C $\cap $ D =$\{$ A$\cap $ C$\} \cap $ D=$\{$ 11 $\} \cap \{ $ 15,17$\} =\oslash$ $\\$ (iv) A $\cap $ C =$\{$ 11$\}$ $\\$ (v) B $\cap $ D =$\oslash$ $\\$ (vi) A $\cap $ ( B $\cap $ C ) =$\{$ (A $\cap $ B) $\cup$ ( A $\cap $C )= $\{ 7,9,11\} \cup \{11\}=\{7,9,11\}$$\\$ (vii) A$\cap $ D =$\oslash $ $\\$ (viii) A$\cap $( B$\cap $ D ) =(A$ \cap$ B)$\cup $(A$\cap$D)= $\{7,9,11\} \cup \oslash=\{7,9,11\}$ $\\$(ix) ( A$\cap $ B )( B$\cap $ C ) =$\{$ 7,9,11$\}$ =$\{$ 7,9,11,13,15$\}$ =$\{$ 7,9,11 $\}$ $\\$ (x) ( A$\cap $ D )( B$\cap $ C ) =$\{$ 3,5,7,9,11,15,17$\}$ =$\{$ 7,9,11,13,15 $\}$ =$\{$ 7,9,11,15 $\}$ $\\$

29   If A =$\{$ x : x isa natural number $\}$ , B =$\{$ x : x isan even natural number $\}$$\\$ C =$\{$ x : x isan odd natural number $\}$ and D =$\{$ x : x isa primenumber $\}$ , find$\\$ (i) A $\cap $ B$\\$ (ii) A $\cap $C$\\$ (iii) A $\cap $D$\\$ (iv) B $\cap $C$\\$ (v) B $\cap $D$\\$ (vi) C$\cap $ D$\\$

Solution :

A=$\{$ x : x isa natural number $\}=\{$1, 2,3, 4,5... $\}$$\\$ B =$\{$ x : x isan even natural number $\}=\{$ 2, 4,6,8...$\}$$\\$ C =$\{$ x : x isan odd natural number $\}=\{$ 1,3,5,7,9... $\}$$\\$ D =$\{$ x : x isa primenumber $\}=\{$2,3,5,7... $\}$$\\$ i) A$\cap $ B =$\{$ x : x isa even natural number $\}=$ B$\\$ (ii) A$\cap $ C=$\{$ x : x isan odd natural number $\}=$C$\\$ (iii) A $\cap $D =$\{$ x : x isa prime number $\}=$D$\\$ (iv) B$\cap $ C= (v) B $\cap $D =$\{$ 2 $\}$$\\$ (vi) C $\cap $D =$\{$ x : x isodd primenumber $\}$

30   Which of the following pairs of sets are disjoint $\\$ (i) $\{$ 1, 2,3, 4 $\}$and $\{$ x : x isa natural number and 4 $\leq $x $\leq $ 6$\}$$\\$ (ii) $\{$ a , e , i , o , u $\}$ and $\{$ c , d , e , f $\}$$\\$ (iii) $\{$ x : x isan eveninteger $\}$ and $\{$ x : x isan oddinteger $\}$

Solution :

(i) $\{$ 1, 2,3, 4 $\}$ $\\$ $\{$ x : x isa natural number and 4 $\leq$ x $\leq$6 $\}=\{$4,5,6 $\}$ $\\$ Now, $\{$ 1, 2,3, 4 $\}$ $\\$ $\{$ 4,5,6 $\}=\{$4 $\}$ $\\$ Therefore, this pair of sets is not disjoint. $\\$ (ii) $\{$ a , e , i , o , u $\}$ $\cap$ $\{$ c , d , e , f $\}=\{$e$\}$ $\\$ Therefore,$\{$ a , e , i , o , u $\}$ and $\{$c , d , e , f $\}$ are not disjoint. $\\$ (iii) $\{$ x : x isan eveninteger $\}$$\cap$ $\\$ $\{$ x : x isan oddinteger $\}=\oslash $ $\\$ Therefore, this pair of sets is disjoint. $\\$

31   If A =$\{$3,6,9,12,15,18, 21 $\}$ , B =$\{$ 4,8,12,16, 20 $\}$ ,$\\$ C =$\{$2, 4,6,8,10,12,14,16 $\}$ , D =$\{$ 5,10,15, 20 $\}$; find$\\$ (i) A - B$\\$ (ii) A - C$\\$ (iii) A - D$\\$ (iv) B - A$\\$ (v) C - A$\\$ (vi) D - A$\\$ (vii) B -C$\\$ (viii) B - D$\\$ (ix) C - B$\\$ (x) D - B$\\$ (xi) C - D$\\$ (xii) D - C$\\$

Solution :

(i) A - B=$\{$ 3,6,9,15,18, 21 $\}$$\\$ (ii) A - C=$\{$3,9,15,18, 21$\}$$\\$ (iii) A - D=$\{$ 3,6,9,12,18, 21 $\}$$\\$ (iv) B - A=$\{$ 4,8,16, 20 $\}$$\\$ (v) C - A=$\{$ 2, 4,8,10,14,16 $\}$$\\$ (vi) D - A=$\{$ 5,10, 20 $\}$$\\$ (vii) B -C=$\{$ 20 $\}$$\\$ (viii) B - D=$\{$ 4,8,12,16 $\}$$\\$ (ix) C - B=$\{$ 2,6,10,14 $\}$$\\$ (x) D - B=$\{$ 5,10,15 $\}$$\\$ (xi) C - D=$\{$ 2, 4,6,8,12,14,16 $\}$$\\$ (xii) D - C=$\{$5,15, 20 $\}$$\\$

32   If X =$\{$ a , b , c , d $\}$ and Y =$\{$ f , b , d , g $\}$ , find $\\$ (i) X - Y $\\$ (ii) Y - X $\\$ (iii) X $\cap$ Y $\\$

Solution :

(i) X - Y =$\{$ a , c $\}$$\\$ (ii) Y - X =$\{$ f , g $\}$$\\$ (iii) X $\cap$ Y=$\{$ b , d $\}$ $\\$

33   IF R is the set real numbers and Q is the set of rational numbers, then what is R – Q?

Solution :

R: Set of real numbers $\\$ Q: Set of rational numbers $\\$ Therefore, R – Q is a set of irrational number.

34   State whether each of the following statement is true or false. Justify you answer.$\\$ (i) $\{$2,3, 4,5 $\}$ and $\{$ 3,6 $\}$ are disjoint sets.$\\$ (ii) $\{$ a , e , i , o , u $\}$ and$\{$ a , b , c , d $\}$ are disjoint sets.$\\$ (iii) $\{$ 2,6,10,14 $\}$ and $\{$ 3,7,11,15$\}$are disjoint sets.$\\$ (iv) $\{$ 2,6,10 $\}$ and $\{$ 3,7,11 $\}$ are disjoint sets.

Solution :

(i) False $\\$ As 3 $\in \{$ 2,3, 4,5 $\}$ ,3 $\in \{$ 3,6 $\}$ $\\$ $\implies \{$ 2,3, 4,5 $\}\cap\{ $ 3,6 $\}=\{$ 3 $\}$ $\\$ (ii) False As a $\in \{$ a , e , i , o , u $\}$ , a $\in \{$ a , b , c , d $\}$ $\\$ $\implies \{$ a , e , i , o , u $\}\cap \{$ a , b , c , d $\}=\{$ a $\}$ $\\$ (iii) True$\\$ As $\{$ 2,6,10,14 $\}\cap\{$ 3,7,11,15 $\}=\oslash$ $\\$ (iv) True$\\$ As $\{$ 2,6,10 $\}\cap\{$ 3,7,11 $\}=\oslash$

35   Let U $=\{$ 1, 2,3, 4,5,6,7,8,9 $\}$ , A $=\{$1, 2,3, 4 $\}$ , B $=\{$ 2, 4,6,8$\}$ and C $=\{$3, 4,5,6 $\}$ . Find $\\$ (i) A '$\\$ (ii) B '$\\$ (iii) ( A $\cup$ C ) '$\\$ (iv) ( A $\cup$ B ) '$\\$ (v) (A ' ) '$\\$ (vi) ( B - C ) '$\\$

Solution :

U =$\{$ 1, 2,3, 4,5,6,7,8,9 $\}$ $\\$ A =$\{$ 1, 2,3, 4 $\}$ $\\$ B =$\{$ 2, 4,6,8 $\}$ $\\$ C =$\{$ 3, 4,5,6 $\}$ $\\$ (i) A ' =$\{$ 5,6,7,8,9 $\}$ $\\$ (ii) B ' =$\{$ 1,3,5,7,9 $\}$ $\\$ (iii) A $\cup$ C =$\{$ 1, 2,3, 4,5,6 $\}$ $\\$ $\therefore$( A $\cup$ C ) ' =$\{$ 7,8,9 $\}$ $\\$ (iv) A $\cup$ B $\{$ 1, 2,3, 4,6,8 $\}$ $\\$ ( A $\cup$ B ) ' =$\{$ 5,7,9 $\}$ $\\$ (v) ( A ' ) ' = A =$\{$ 1, 2,3, 4 $\}$ $\\$ (vi) B - C =$\{$ 2,8 $\}$ $\\$ $\therefore$( B - C) ' =$\{$ 1,3, 4,5,6,7,9 $\}$ $\\$

36   If U =$\{$ a , b , c , d , e , f , g , h $\}$ , find the complements of the following sets: $\\$ (i) A =$\{$ a , b , c $\}$ $\\$ (ii) B =$\{$d , e , f , g $\}$ $\\$ (iii) C =$\{$ a , c , e , g $\}$ $\\$ (iv) D =$\{$ f , g , h , a $\}$ $\\$

Solution :

U=$\{$ a , b , c , d , e , f , g , h $\}$ , $\\$ (i) A =$\{$ a , b , c $\}$$\\$ A ' =$\{$ d , e , f , g , h $\}$$\\$ (ii) B =$\{$ d , e , f , g $\}$$\\$ $\therefore$ B ' =$\{$ a , b , c , h $\}$$\\$ (iii) C =$\{$ a , c , e , g $\}$$\\$ $\therefore$ C ' =$\{$ b , d , f , h $\}$$\\$ (iv) D =$\{$ f , g , h , a $\}$$\\$ $\therefore$ D ' =$\{$ b , c , d , e $\}$$\\$

37   Taking the set of natural numbers as the universal set, write down the complements of the following sets: $\\$ (i) $\{$ x : x isan even natural number $\}$ $\\$ (ii) $\{$ x : x isan odd natural number $\}$ $\\$ (iii) $\{$ x : x isa positive multipleof 3 $\}$ $\\$ (iv) $\{$ x : x isa prime number $\}$ $\\$ (v) $\{$ x : x isa natural number divisible by 3 and 5 $\}$ $\\$ (vi) $\{$ x : x isa perfect square $\}$ $\\$ (vii)$\{$ x : x isa perfect cube $\}$ $\\$ (viii) $\{$ x : x +5=8 $\}$ $\\$ (ix) $\{$ x : 2 x + 5 = 9 $\}$ $\\$ (x) $\{$ x : x $\geq$ 7 $\}$ $\\$ (xi) $\{$x : x $\in$ N and 2 x + 1 > 10 $\}$ $\\$

Solution :

U = N set of natural numbers $\\$ (i) $\{$ x : x isan even natural number $\} ' =\{$ x : x isan odd natural number $\}$ $\\$ (ii) $\{$ x : x isan odd natural number $\} ' =\{$ x : x isan even natural number $\}$ $\\$ (iii) $\{$ x : x isa positivemultipleof 3 $\} ' =\{$ x : x $\in$ N and x is not a multipleof 3 $\}$ $\\$ (iv) $\{$ x : x isa primenumber $\} ' =\{$ x : x isa positivecompositenumber and x = 1 $\}$ $\\$ (v) $\{$ x : x isa natural number divisibleby3and5 $\} ' =\{$ x : x isa natural number that is not divisibleby3or 5 $\}$ $\\$ (vi) $\{$ x : x isa perfect square $\} ' =\{$ x : x $\in$ N and x is not a perfect square $\}$ $\\$ (vii) $\{$ x : x isa perfect cube $\} ' =\{$ x : x$\in$ N and x is not a perfect cube $\}$ $\\$ (viii) $\{$ x : x + 5 = 8 $\} ' =\{$ x : x $\in$ N and x $\neq$ 3 $\}$ $\\$ (ix)$\{$ x : 2 x + 5 = 9$\} ' =\{$ x : x $\in$ N and x $\neq$ 2 $\}$ $\\$ (x) $\{$ x : x $\geq$ 7 $\} ' =\{$ x : x $\in$ N and x < 7 $\}$ $\\$ (xi) $\{$ x : x $\in$ N and 2 x + 1 = 10 $\} ' =\{$x : x $\in$N and x $\leq$ 9 / 2 $\}$ $\\$

38   If U =$\{$1, 2,3, 4,5,6,7,8,9 $\}$ , A =$\{$ 2, 4,6,8 $\}$ and B =$\{$ 2,3,5,7 $\}$ . Verify that $\\$ (i) ( A $\cup$ B ) ' = A ' $\cap$ B '$\\$ (ii) ( A$\cap$B ) '= A ' $\cup$ B '

Solution :

U =$\{$ 1, 2,3, 4,5,6,7,8,9 $\}$ $\\$ A =$\{$ 2, 4,6,8 $\}$ , B =$\{$ 2,3,5,7 $\}$ $\\$ (i) ( A $\cup$ B ) ' =$\{$ 2,3, 4,5,6,7,8$\}$ ' =$\{$ 1,9 $\}$$\\$ A ' $\cap$ B ' =$\{$ 1,3,5,7,9 $\}\cap\{$ 1, 4,6,8,9 $\}=\{$ 1,9$\}$$\\$ $\therefore$ ( A $\cup$ B ) ' = A ' $\cap$ B '$\\$ (ii) ( A $\cap$ B ) ' =$\{$ 2$\}$ ' =$\{$ 1,3, 4,5,6,7,8,9 $\}$$\\$ A ' $\cup$B ' =$\{$ 1,3,5,7,9 $\}\cup\{$1, 4,6,8,9 $\}=\{$ 1,3, 4,5,6,7,8,9 $\}$ $\\$ $\therefore$ ( A $\cap$B ) ' = A ' $\cup$ B '$\\$

39   Draw appropriate Venn diagram for each of the following:$\\$ (i) ( A $\cup$ B ) '$\\$ (ii) A ' $\cap$ B '$\\$ (iii) ( A $\cap$ B ) '$\\$ (iv) A ' $\cup$ B '

Solution :

(i) ( A $\cup$ B ) '$\\$

(ii) A ' $\cap$ B '$\\$

(iii) ( A $\cap$ B ) '$\\$

(iv) A ' $\cup$ B '

40   Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60$^o$ , what is A ' ?

Solution :

A ' is the set of all equilateral triangles.

41   Fill in the blanks to make each of the following a true statement:$\\$ (i) A $\cup$ A ' = ... $\\$ (ii)$\oslash$ ' A$\\$ (iii) A $\cap$ A ' = ...$\\$ (iv) U ' $\cap$ A = ...$\\$

Solution :

(i) A $\cup$ A ' = U $\\$ (ii)$\oslash$ ' A=U$\cup$A=A$\\$ $\therefore \oslash'$A=A$\\$ (iii) A $\cap$ A ' = $\oslash$$\\$ (iv) U ' $\cap$ A = $\oslash \cap $A=$\oslash$$\\$ $\therefore $U' $\cap$ A=$\oslash$

42   If X and Y are two sets such that n ( X )= 17, n ( Y )= 23 and n ( X $\cap$ Y )= 38 , find n ( X $\cap$ Y ) .

Solution :

It is given that:$\\$ n ( X )= 17, n ( Y )= 23, n ( X $\cup$ Y)= 38 We know that:$\\$ n ( X $\cup$Y )= n ( X )+ n ( Y )- n ( X $\cap$Y )$\\$ $\therefore$ 38 = 17 + 23 = n ( X $\cap$Y )$\\$ $\implies$ n ( X $\cap$ Y )= 40 - 38= 2$\\$ $\therefore$ n ( X $\cap$ Y )= 2

43   If X and Y are two sets such that X $\cup$ Y has 18 elements, X has 8 elements and Y has 15 elements; how many elements does X $\cap Y have?

Solution :

It is given that:$\\$ n ( X $\cup$ Y )= 18, n ( X )= 8, n ( Y )= 15$\\$ n ( X $\cap$ Y )= ?$\\$ We know that:$\\$ n ( X $\cup$ Y )= n (X )+ n ( Y )- n ( X $\cap$ Y )$\\$ $\therefore$ 18 = 8 + 15 - n (X $\cap$ Y )$\\$ $\implies$ n ( X $\cap$ Y )= 23 - 18 = 5$\\$ $\therefore$ n ( X $\cap$ Y )= 5

44   In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?

Solution :

Let H be the set of people who speak Hindi, and E be the set of people who speak English $\\$ $\therefore$ n ( H $\cup$ E )= 400, n ( H )= 250, n ( E )= 200$\\$ n ( H $\cap$ E )= ?$\\$ We know that:$\\$ n ( H $\cup$ E )= n ( H )= n ( E )= n ( H $\cap$ E )$\\$ $\therefore$ 400 = 250 + 200 - n ( H $\cap$ E )$\\$ $\implies$ 400 = 450 -n ( H $\cup$ E )$\\$ $\implies$ n ( H $\cup$ E )= 450 - 400$\\$ $\therefore$ n ( H $\cup$ E)= 50$\\$ Thus, 50 people can speak both Hindi and English.

45   If S and T are two sets such that S has 21 elements, T has 32 elements, and S $\cap$T has 11 elements, how many elements does S $\cup$ T have?

Solution :

It is given that:$\\$ n ( S )= 21, n ( T )= 32, n ( S $\cap$ T )= 11$\\$ We know that:$\\$ n ( S $\cup$ T )= n ( S )= n ( T )- n( S $\cap$ T )$\\$ $\therefore$ n ( S $\cup$ T )= 21 + 32 - 11 = 42$\\$ Thus, the set ( S $\cup$ T ) has 42 elements.$\\$

46   If X and Y are two sets such that X has 40 elements, X $\cup$ Y has 60 elements and X $\cap$Y has 10 elements, how many elements does Y have?

Solution :

It is given that:$\\$ n ( X )= 40, n ( X $\cup$ Y )=60, n ( X $\cap$ Y )= 10$\\$ We know that:$\\$ n ( X $\cup$Y )= n (X )+ n ( Y )- n ( X $\cap$ Y )$\\$ $\therefore$ 60 = 40 + n ( Y )- 10$\\$ $\therefore$ n ( Y )= 60 - ( 40 - 10 )= 30$\\$ Thus, the set Y has 30 elements.$\\$

47   In a group of 70 people, 37 like coffee, 52 like tea, and each person likes at least one of the two drinks. How many people like both coffee and tea?

Solution :

Let C denote the set of people who like coffee, and T denote the set of people who like tea$\\$ n ( C $\cup$ T )= 70, n ( C )= 37, n ( T )= 52$\\$ We know that:$\\$ n ( C $\cup$ T )= n ( C)+ n ( T )- n ( C $\cap$ T )$\\$ $\therefore$ 70 = 37 + 52 - n ( C $\cup$ T )$\\$ $\implies$ 70 = 89 - n ( C $\cup$ T)$\\$ $\implies$ ( C $\cup$ T )= 89 - 70 = 19$\\$ Thus, 19 people like both coffee and tea.

48   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?

Solution :

Let C denote the set of people who like cricket, and T denote the set of people who like tennis $\\$ $\therefore$ n ( C $\cup$ T )= 65, n (C )= 40, n ( C $\cap$ T )= 10 $\\$ We know that: $\\$ n ( C$\cup$ T )= n ( C )+ n ( T )- n (C $\cap$ T ) $\\$ $\therefore$ 65 = 40 + n ( T )- 10 $\\$ $\implies$ 65 = 30 + n ( T ) $\\$ $\implies$ n ( T )= 65 - 30 =35 $\\$ Therefore, 35 people like tennis. $\\$ Now, $\\$ ( T - C )$\cup$( T $\cap$ C )= T $\\$ Also, $\\$ ( T - C )$\cap$( T $\cap$ C )= $\oslash$ $\\$ $\therefore$ n ( T )= n ( T -C )+ n ( T $\cap$ C ) $\\$ $\implies$ 35 = n ( T - C )+ 10 $\\$ $\implies$ n ( T - C )= 35 - 10 = 25 $\\$ Thus, 25 people like only tennis.

49   In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?

Solution :

Let F be the set of people in the committee who speak French, and S be the set of people in the committee who speak Spanish $\\$ $\therefore$ n ( F )= 50, n ( S )= 20, n ( S $\cap$ F )= 10$\\$ We know that:$\\$ n ( S $\cup$ F )= n ( S )+ n ( F )- n (S $\cap$F )$\\$ = 20 + 50- 10$\\$ = 70 - 10 = 60$\\$ Thus, 60 people in the committee speak at least one of the two languages.

50   Decide, among the following sets, which sets are subsets of one and another: $\\$ A =$\{$ x : x $\in$ R and x satisfy x $^2 $- 8 x + 12 = 0 $\}$ ,$\\$ B =$\{$ 2, 4,6 $\}$ , C =$\{$ 2, 4,6,8... $\}$ , D =$\{$ 6 $\}$ .

Solution :

A =$\{$ x : x $\in$ R and x satisfy x $^2 $- 8 x + 12 = 0 $\}$ ,$\\$ 2 and 6 are the only solutions of x $^2 -$ 8 x + 12 = 0 .$\\$ $\therefore$ A =$\{$ 2,6 $\}$ $\\$ B=$\{$ 2, 4,6 $\}$ , C =$\{$ 2, 4,6,8... $\}$ , D =$\{$ 6 $\}$ $\\$ $\therefore$ D $\subset$ A $\subset$ B $\subset$ C$\\$ Hence, A$\subset$ B , A $\subset$ C , B $\subset$ C , D $\subset$ A , D $\subset$ B , D $\subset$ C

51   In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example. $\\$ (i) If x $\in$A and A $\in$ B , then x $\in$ B $\\$ (ii) If A $\subset$ B and B $\in$ C , then A $\in$ C $\\$ (iii) If A $\subset$ B and B $\subset$ C , then A $\subset$ C $\\$ (iv) If A $\not \subset$ B and B $\not \subset$ C , then A $\not \subset$ C $\\$ (v) If x $\in$ A and A $\not \subset$ B , then x $\in$ B $\\$ (vi) If A $\subset$ B and x $\not \in$B , then x $\not \in$ A $\\$

Solution :

(i) False $\\$ Let A $=\{$ 1, 2 $\}$ and B $=\{$ 1, $\{$1, 2 $\}$ , c3 $\}$ $\}$$\\$ Now, 2 $\in\{$ 1, 2 $\}$ and $\{$ 1, 2 $\}\in\{\{$ 3 $\}$ ,1, 1, 2 $\}\}$$\\$ $\therefore$ A $\in$ B$\\$ However, 2 $\in\{\{$3 $\}$ ,1, $\{$ 1, 2 $\}\}$$\\$ (ii) False$\\$ Let A$=\{$2 $\}$ , B $=\{$ 0, 2 $\}$ ,and C $=\{$ 1, $\{$ 0, 2 $\}$ ,3 $\}$$\\$ As A $\subset$ B$\\$ B $\in$ C$\\$ (iii) True$\\$ Let A $\subset$ B and B $\subset$ C .$\\$ Let x $\in$ A$\\$ $\implies$ x $\in$ B $\qquad$ [A $\subset$ B ]$\\$ $\implies$ x $\in$ C $\qquad$ [B $\subset$ C ]$\\$ (iv) False$\\$ Let A $=\{$ 1, 2 $\}$ , B $=\{$ 0,6,8 $\}$ , and C $=\{$ 0,1, 2,6,9 $\}$$\\$ Accordingly, A $\not \subset$ B and B $\not \subset$ C .$\\$ However, A $\subset$ C$\\$ (v) False$\\$ Let A $=\{$ 3,5,7 $\}$ and B $=\{$ 3, 4,6 $\}$$\\$ Now, 5 $\in$ A and A $\not \subset$ B$\\$ However, 5 $\not \in$ B$\\$ (vi) True$\\$ Let A$\subset$ B and x $\not \in$ B .$\\$ To show: x $\not \in$ A$\\$ If possible, suppose x $\in$ A .$\\$ Then, x $\in$ B , which is a contradiction as x $\not \in$ B$\\$ $\therefore$ x $\not \in$ A

52   Let A, B and C be the set such that A $\cup$ B = A $\cup$ C and A $\cap$ B = A $\cap$ C . Show that B = C.

Solution :

Let, A, B and C be the sets such that A $\cup$ B = A $\cup$ C and A $\cap$ B = A $\cap$ C . $\\$ To show: B = C $\\$ Let x $\in$ B $\\$ $\implies$ x $\in$ A $\cup$ B $\qquad$[ B $\subset$ A $\cup$ B ] $\\$ $\implies$ x $\in$ A $\cup$ C $\qquad$[ A $\cup$ B = A $\cup$ C ] $\\$ $\implies$ x $\in$ A or x $\in$ C $\\$ Case I $\\$ Also, x $\in$ B $\\$ $\therefore$ x$\in$ A$\cap$ B $\\$ $\implies$ x $\in$ A $\cap$ C $\qquad$[ A $\cap$ B = A$\cap$ C ] $\\$ $\therefore$ x $\in$ A and x $\in$ C $\\$ $\therefore$ x $\in$ C $\\$ $\therefore$ B $\subset$ C $\\$ Similarly, we can show that C $\subset$B . $\\$ $\therefore$ B = C

53   Show that the following four conditions are equivalent: $\\$ (i) A $\subset$ B $\\$ (ii) A - B =$\oslash$ $\\$ (iii) A $\cup$ B = B $\\$ (iv) A $\cap$ B = A $\\$

Solution :

First, we have to show that $(i)\iff(ii)$.$\\$ Let A $\subset$ B$\\$ To show: A - B $\neq \oslash$$\\$ If possible, suppose A - B$\neq \oslash$$\\$ This means that there exists x $\in$ A , x $\neq$ B , which is not possible as A $\subset$ B .$\\$ $\therefore$ A - B$=\oslash $ $\\$ $\therefore$ A $\subset$ B $\implies$ A - B $=\oslash $ $\\$ Let A - B $=\oslash $ $\\$ To show: A $\subset$ B$\\$ Let x $\in$ A$\\$ Clearly, x $\in$ B because if x $\not \in$ B , then A - B $\neq \oslash$$\\$ $\therefore$ A - B $=\oslash $ $\implies$ A $\subset$ B$\\$ $\therefore$ $(i)\iff(iii)$$\\$ Let A $\subset$ B$\\$ To show: A $\cup$ B = B$\\$ Clearly, B $\subset$ A $\cup$ B Let x $\in$ A $\cup$ B $\implies$ x $\in$ A or x $\in$ B$\\$ Case I: x $\in$ A$\\$ $\implies$x $\in$ B $\qquad$ [$\therefore$A $\subset$ B ]$\\$ $\therefore$ A $\cup$ B $\subset$ B$\\$ Case II: x $\in$ B$\\$ Then, A $\cup$ B = B$\\$ Conversely, let A $\cup$ B = B$\\$ Let x $\in$ A$\\$ $\implies$ x $\in$ A $\cup$ B $\qquad$[$\therefore$ A $\subset$A $\cup$ B ]$\\$ $\implies$ x $\in$ B$\qquad$ [$\therefore$A $\cup$ B = B ]$\\$ $\therefore$ A $\subset$ B$\\$ Hence, $(i)\iff(iii)$$\\$$\\$ Now, we have to show that $(i)\iff(iv)$$\\$ .$\\$ Let A $\subset$ B$\\$ Clearly A $\cap$ B $\subset$ A$\\$ Let x $\in$ A$\\$ We have to show that x $\in$ A $\cap$ B$\\$ As A $\subset$ B , x $\in$ B $\therefore$ x $\in$ A $\cap$ B $\therefore$ A $\subset$ A $\in$ B Hence, A = A$\cap$ B$\\$ Conversely, suppose A $\cap$B = A$\\$ Let x $\in$ A$\\$ $\implies$ x$\in$ A $\cap$ B$\\$ $\implies$ x $\in$ A and x $\in$ B$\\$ $\implies$ x $\in$ B$\\$ $\therefore$ A $\subset$ B$\\$ Hence, $(i)\iff(iv)$$\\$ .$\\$

54   Show that if A$\subset$ B , then C - B $\subset$ C - A

Solution :

Let A $\subset$ B $\\$ To show: C - B $\subset$ C- A $\\$ Let x $\in$ C - B $\\$ $\implies$ x $\in$ C and x $\not \in$ B $\\$ $\implies$ x $\in$ C and x $\not \in$ A [ A $\subset$ B ] $\\$ $\implies$ x $\in$ C - A $\\$ $\therefore$ C - B $\subset$ C - A $\\$

55   Assume that P ( A )= P ( B ) . Show that A = B .

Solution :

Let P ( A )= P ( B ) $\\$ To show: A = B $\\$ Let x $\in$ A $\\$ A $\in$ P ( A )= P ( B ) $\\$ $\therefore$ x $\in$ C , for some C $\in$ P ( B ) $\\$ Now, C $\subset$ B $\\$ $\therefore$ x $\in$ B $\\$ $\therefore$ A $\subset$ B $\\$ Similarly, B $\subset$ A $\\$ $\therefore$ A = B $\\$

56   Is it true for any sets A and B, P ( A ) $\cup$P ( B )= P ( A $\cup$ B ) ? Justify your answer.

Solution :

False $\\$ Let $A=\{0,1\}$ and $B=\{1,2\}$ $\\$ $\therefore A\cup B=\{0,1,2\}$ $\\$ $p(A)=\{\oslash,\{0\},\{1\},\{0,1\}\}$ $\\$ $p(B)=\{\oslash,\{1\},\{2\},\{1,2\}\}$ $\\$ $p(A\cup B)=\{\oslash,\{1\},\{2\},\{0,1\},\{1,2\},\{0,2\},\{0,1,2\}\}$ $\\$ $p(A)\cup p(B)=\{\oslash,\{1\},\{0,1\},\{2\},\{1,2\}\}$$\\$ $p(A)\cup p(B)=\{\oslash,\{1\},\{0,1\},\{2\},\{1,2\}\}$ $\\$ $\therefore p(A)\cup p(B)\neq p(A\cup B)$ $\\$

57   Show that for any sets A and B, $A =( A\cap B )\cup( A - B )$ and $A \cup( B - A )=( A \cup B )$

Solution :

To show: A =( A $cap$B )$\cup$(A - B ) $\\$ Let x $\in$ A $\\$ We have to show that x$\in$( A B)$\cup$( A - B ) $\\$ Case I x $\in$ A$cap$ B $\\$ Then, x $\in$(A$\cap$ B )$\subset$( A $\cup$ B )$\cup$ (A - B ) $\\$ Case II $\\$ x $\not \in$ A $cap$B $\\$ x $\not \in$ A or x $\not \in$ B $\\$ $\therefore$ x $\not \in$ B [ x $\not \in$ A ] $\\$ $\therefore$ x $\not \in$ A - B $\subset$ (A $\cup$ B ) $\cup$ ( A - B ) $\\$ $\therefore$ A $\subset$ (A$cap$ B )$\cup $( A - B ) ....... 1 It is clear that

58   Using properties of sets show that $\\$ (i) A $\cup$( A $\cap $ B )= A $\\$ (ii) A $\cap$ (A $\cup$ B )= A

Solution :

(i) To show: A $\cup$( A $\cap $ B )= A $\\$ We know that $\\$ A$\subset$ A $\\$ A B $\subset$A $\\$ $\therefore$ A $\cup$( A B )$\subset$ A ..... ( 1 ) $\\$ Also, A $\subset$ A $\cup$( A$\cap $ B ) ...... ( 2 ) $\\$ $\therefore$ From ( 1 ) and ( 2 ) , A $\cup$( A $\cap $B )= A $\\$ (ii) To show: A $\cap $ A ( A $\cup$ B )= A $\\$ A$\cap $( A B )= ( A $\cap $A)( A $\cap $B ) $\\$ =A$\cup$ (A$\cap $B) $\\$ = A$\{$ from ( 1 ) $\}$ $\\$

59   Let A and B be sets. If A $\cap$ X = B$\cap$ X =$\oslash$ and A $\cup$ X = B $\cup$ X for some set X, show that A = B.$\\$ (Hints A = A $\cap$( A $\cup$ X ) , B = B $\cap$( B $\cup$ X ) and use distributive law)

Solution :

Let A and B be two sets such that A $\cap$ X = B$\cap$x = f and A $\cup$ X = B $\cup$ X for some set X. $\\$ To show: A = B $\\$ It can be seen that $\\$ A = A $\cap$( A $\cup$ X )= A $\cap$ (B $\cup$ X )[ A $\cup$ X = B $\cup$ X ] $\\$ =( A B ) $\cup$( A X ) [Distributive law] $\\$ =( A$\cap$ B )$\cup$=$\oslash$ [ A $\cap$X =$\oslash$ ] $\\$ = A$\cap$ B ..... ( 1 ) $\\$ Now, B = B $\cap$( B $\cup$ X ) $\\$ =B $\cap$( A $\cup$ X ) [A $\cup$ X = B $\cup$ X ] $\\$ =( B A )$\cup$( B X)[Distributive law] $\\$ =( B A )$\cup \oslash$[ B $\cap$ X=$\oslash$] $\\$ =B $\cap$ A $\\$ =A$\cap$ B....(2) $\\$ Hence, from (1) and (2), we obtain A = B.

60   Find sets A, B and C such that A$\cap$ B , B$\cap$ C and A$\cap$ C are non-empty sets and A $\cap$ B $\cap$ C =$\oslash$

Solution :

Let A =$\{$ 0,1 $\}$ , B =$\{$ 1, 2 $\}$ , and C =$\{$ 2,0 $\}$ . $\\$ Accordingly, A$\cap$ B =$\{$ 1 $\}$ , B$\cap$ C =$\{$ 2 $\}$ ,and A $\cap$C =$\{$ 0 $\}$ . $\\$ A$\cap$ B , B$\cap$ C , and A$\cap$ C are non-empty. $\\$ However, A $\cap$B$\cap$ C =$\oslash$ $\\$

61   In a survey of $600$ students in a school, $150 $ students were found to be taking tea and $225$ taking coffee, $100$ were taking both tea and coffee. Find how many students were taking neither tea nor coffee?

Solution :

Let U be the set of all students who took part in the survey. $\\$ Let T be the set of students taking tea. $\\$ Let C be the set of students taking coffee. $\\$ Accordingly, n ( U )= 600, n ( T )= 150, n ( C )= 225, n ( T$\cap$ C )= 100 $\\$ To find: Number of student taking neither tea nor coffee i.e., $\\$we have to find n (T $\cap$ C )'. $\\$ n (T '$\cap$ C ' )= n (T '$\cap$ C ' ) $\\$ = n ( U) -n (T $\cap$ C ) $\\$ = n ( U )-[n (T )+ n ( C )- n ( T$\cap$ C )] $\\$ = 600 -[ 150 + 225 - 100 ] $\\$ = 600 - 275 $\\$ = 325 $\\$ Hence, 325 students were taking neither tea nor coffee.

62   In a group of students 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?

Solution :

Let U be the set of all students in the group. $\\$ Let E be the set of all students who know English. $\\$ Let H be the set of all students who know Hindi. $\\$ $\therefore$ H $\cup$ E = U $\\$ Accordingly, n ( H )= 100 and n ( E )= 50 $\\$ n ( H $\cap$ E )= 25 $\\$ n ( U )= n ( H )+ n ( E )- n ( H$\cap$ E ) $\\$ = 100 + 50 - 25 $\\$ =125 $\\$ Hence, there are 125 students in the group.

63   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 number of people who read at least one of the newspapers. $\\$ (ii) The number of people who read exactly one newspaper. $\\$

Solution :

Let A be the set of people who read newspaper H. $\\$ Let B be the of people who read newspaper T. $\\$ Let C be the set of people who read newspaper I. $\\$ Accordingly, n ( A )= 25, n ( B )= 26, and n ( C )= 26 $\\$ n ( A $\cap$ C )= 9, n ( A$\cap$ B )= 11, and n ( B $\cap$ C )= 8 n ( A $\cap$B$\cap$ C )= 3 $\\$ Let U be the set of people who took part in the survey. $\\$ (i) Accordingly, $\\$ n ( A $\cup$ B $\cup$ C )= n ( A )+ n ( B )+ n ( C )- n ( A $\cap$ B )- n ( B $\cap$ C ) n (C$\cap$ A )+ n ( A $\cap$B$\cap$ C ) $\\$ = 25 + 26 + 26 - 11 - 8 - 9 + 3 $\\$ = 52 $\\$ Hence, 52 people read at least one of the newspapers. $\\$ (ii) Let a be the number of people who read newspapers H and T only. $\\$

Let b denote the number of people who read newspapers I and H only. $\\$ Let c denote the number of people who read newspapers T and I only. $\\$ Let d denote the number of people who read all three newspapers. $\\$ Accordingly, d = n ( A $\cap $ B$\cap $ C )= 3 $\\$ Now, n ( A$\cap $ B )= a + d $\\$ n ( B$\cap $ C )= c + d $\\$ n ( C$\cap $ A )= b + d $\\$ a + d + c + d + b + d = 11 + 8 + 9 = 28 $\\$ a + d + c + d= 28 - 2 d = 28 - 6 = 22 $\\$ Hence, ( 52 - 22 )= 30 people read exactly one newspaper.

64   In 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 liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.

Solution :

Let A, B and C be the set of people who like product A, product B, and product C respectively. $\\$ Accordingly, n ( A )= 21, n ( B )= 26, n ( C )= 29, n ( A$\cap$ B )= 14, n ( C$\cap$ A )= 12 $\\$ n ( B$\cap$ C )= 14, n ( A$\cap$ B$\cap$ C )= 8 $\\$ The Venn diagram for the given problem can be drawn as $\\$

It can be seen that number of people who like product C only is $\{ 29 -( 4 + 8 + 6 )\}11 .$