Class 11 Engineering Maths - Sets

Let $A=\left \{ 1, 2, 3, 4 \right \}$
$S=\left \{ (a, b); a,b\in A, a\ \mbox{divides}\ b \right \}$ , write down $S$ explicitly

$S=\left \{ (1, 1),(1, 2),(1, 3),(1, 4),(2, 2),(2, 4),(3, 3),(4, 4) \right \}$

Let $\displaystyle A=\left \{ \left ( x, y \right ):x, y \epsilon R, x^{2}y^{2}+y^{2}=1 \right \}$ and $\displaystyle B=\left \{ \left ( x,0 \right ):x \epsilon R, -1\leq x\leq 1 \right \}.$ Then $\displaystyle A\cap B= \left \{ \left ( -a, 0 \right ), \left ( 1, b \right ) \right \}$ .Find relation between a and b

Given,

$A={(x,y):x,y \epsilon R,x^2y^2+y^2=1},B={(x,0):x \epsilon R,-1 \leq x \leq 1}$

$A \cap B={(-a,0),(1,b)}$
let the common point for A and b be

$c=(p,q)$ .

$\Rightarrow q=0$ (from condition B)

$\Rightarrow c=(p,0)$
On substituting c in A,

$p^2 \times 0^2+0^2=1$

$0=1$ which is never possible.

$\therefore$ Intersection of A and B is nullset.

If $S$ and $T$ are two sets such that $S$ has $21$ elements. T has $32$ elements and $S$ $\displaystyle \cap$ $T$ has $11$ elements, then find the number of elements in $S$ $\displaystyle \cup$ $T$ .

Given,

$\displaystyle n\left ( S \right )=21,n\left ( T \right )=32,n\left ( S\cap T \right )=11\:$

Now

$\:n\left ( S \right )+n\left ( T \right )=n\left ( S\cap T \right )+n\left ( S\cup T \right )\Rightarrow n\left ( S\cup T \right )=21+32-11=42$

$\displaystyle \left \{ a,b \right \}$ is a subset of $\left \{ b,c,a \right \}$ .If true enter 1 else 0.

$\{a,b\}$ is a subset of

$\{b,c,a\}$

Since

$\{a\},\{b\}$ are the elements in the set

$\{b,c,a\}$ .

$\displaystyle {a,e}$ is a subset of $\left \{ x:x \text{ is a vowel in the English alphabet}\right \}$ .( Enter 1 if true or 0 otherwise)

$x$ is a vowel in the English alphabet. Vowels are

$a,e,i,o$ and

$u$ .

So,

$a,e$ is a subset of

$\{x:x$ is a vowel in the English alphabet

$\}$

Let $m$ be a number in a set $A=\displaystyle \left \{ 1,2,3 \right \} \ such \ that \ A\nsubseteq \left \{ 1,3,5 \right \}$ and $m\notin \{1,3,5\}$ . Then $m$ is:

$\displaystyle \left \{ 1,2,3 \right \}\nsubseteq \left \{ 1,3,5 \right \}$ because

$2\notin \{1,3,5\}$

It is given that

$m\notin \{1,3,5\}$

Hence,

$m=2$ .

Let $A = \{1, 2, 3, 4, 5, 6\}$ , $B = \{2, 4, 6, 8\}$ . Find $A - B$ and $B - A$ .

$\displaystyle A-B=\left \{ x:x\in A\, and\, x\notin B \right \}=\left \{ 1,3,5 \right \}$ Similarly

$B - A = \{8\}$ .

Find $n(B-C)^c$ .

$(B \cup C) = \{ 4, 6\}$

$\Rightarrow(B \cup C)^C = \{ 1, 2, 3, 5, 7,8, 9\}$

$\Rightarrow n(B \cup C)^C = 7$

If $A=\left \{2, 3,4, 5, 6, 7\right \}$ and $B =\left \{3, 5, 7, 9,11, 13\right \}$ , then find $A-B$ and $B -A$

$A-B=\left \{2, 4, 6\right \}$ and

$B-A=\left \{9, 11, 13\right \}$ .

In a group of $50$ persons, $14$ drink tea but not coffee and $30$ drink tea. Find how many drink coffee but not tea?

$\displaystyle n(C-T)=n(T\cup C)-n(T)=50-30=20$

Let $A = \{2, 4, 6, 8\}$ and $B = \{6, 8, 10, 12\}$ then find $\displaystyle A\cup B$ .

$\displaystyle A\cup B$

$= \{2, 4, 6, 8, 10, 12\}$ .

Find n $(A\cup B)^c$

$A\cup B = \{ 1, 2, 3, 4, 6, 8\}$

$\Rightarrow (A\cup B)^C = U - (A\cup B) = \{ 5, 7, 9\}$

$\Rightarrow n(A\cup B)^C = 3$

$\displaystyle a \epsilon \left \{ a,b,c \right \}$ , then $\{a\}$ is a subset of $\{a,b,c\}$ .( Enter 1 if true or 0 otherwise)

$\displaystyle a\in \left \{ a,b,c \right \}$ , then

$\left \{ a \right \}\subseteq \left \{ a,b,c \right \}$

Hence,

$\left \{ a \right \}$ is a subset of

$\left \{ a,b,c \right \}$

If $A\left \{1, 2,3\right \}$ and $B =\left \{1, 3, 5, 7\right \}$ , then find $A\cup B$

$A\cup B=\left \{1, 2, 3, 5, 7\right \}$ .

If $R$ is the set of real numbers and $Q$ is the set of rational numbers, then what is $R-Q$ ?

$R$ : Set of real numbers i.e it has both rational and irrational numbers.

$Q$ : Set of rational numbers
Therefore,

$R-Q$ is a set of irrational numbers.

Let $A = \left \{1, 2, 3, 4, 5, 6\right \}$ . Insert the appropriate symbol $\in$ or $\not\in$ in the blank spaces:
(i). .A (ii) 8 . . . A (iii). .A (iv). . A (v). .A (vi). .A

If the element belongs to

$A$ we write

$\in$ ,else

$\notin$

(i)

$5\in A$ , as 5 is the 5th element in set A.
(ii)

$8\notin A$ , as 8 is not present in set A.
(iii)

$0\notin A$ , as 0 is not present in set A.
(iv)

$4\in A$ , as 4 is the 4th element in set A.
(v)

$2\in A$ , as 2 is the 2nd element in set A.
(vi)

$10 \notin A$ as 10 is not present in set A.

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

(i)

$\left \{x:x \epsilon R, -4 < x \leq 6\right \}=(4, 6]$

(ii)

$\left \{x:x \epsilon R, -12 < x < -10\right \}=(-12, -10)$

(iii)

$\left \{x:x \epsilon R, 0\leq x < 7\right \}=[0, 7)$

(iv)

$\left \{x:x \epsilon R, 3\leq x\leq 4\right \}=[3, 4]$

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) $D- C$

$A = \{3, 6, 9, 12, 15, 18, 21\}$

$B = \{ 4, 8, 12, 16, 20 \}$

$C = \{ 2, 4, 6, 8, 10, 12, 14, 16 \}$

$D = \{5, 10, 15, 20 \}$

(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)

$D - C = \{5,15,20\}$

$(P\cup Q)\cap R$

$(P\cup Q) = \left \{ a,b,c,e,d,f \right \}$ &

$R = \left \{ c,f,h,i,j \right \}$

$\therefore (P\cup Q)\cap R = \left \{ c,f \right \}\therefore \boxed {2\,elements}$

1103210_514591_ans_3d8e76e5dee34df1a18e87d6c4eb4707.jpg

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

$A=\left \{x:x\in R \text {and x satisfies}x^2-8x+12=0\right \}$
2 and 6 are the only solutions of

$x^2-8x+12=0$ .

$\therefore A=\left \{2,6\right \}$

$B=\left \{2,4,6\right \}$

$C=\left \{2,4,6,8,....\right \}$

$D=\left \{6\right \}$

Clearly,

$D\subset A\subset B\subset C$
Hence,

$A\subset B, A\subset C, B\subset C, D\subset A, D\subset B, D\subset C$

Which of the following sets are equivalent?
(i) $A=\left \{ 2,4,6,8,10 \right \}, B=\left \{ 1,3,5,7,9 \right \}$
(ii) $X=\left \{ x:x \in \mathbb{N}, 1< x< 6 \right \}, Y=$ { $x:x$ is a vowel in the English Alphabet }
(iii) $P=$ $\{$ $x:x$ is a prime number and $5< x< 23$ $\}$
$Q=\{x:x\in \mathbb{W},1<x<5\}$

Given sets in

$1,2,3$ are not equivalent sets. since the elements in both sets are not equal.

Represent the following set by using their standard notations: Set of integers

Set of integer

$=Z$

Which of the following sets are equal?
(i) $A=\left \{ 1,2,3,4 \right \}, B=\left \{ 4,3,2,1 \right \}$
(ii) $A=\left \{ 4,8,12,16 \right \}, B=\left \{ 8,4,16,18 \right \}$
(iii) $X=\left \{ 2,4,6,8 \right \}$
$Y=$ { $x:x$ is a positive even integer $0< x< 10$ }
(iv) $P=$ { $x:x$ is a multiple of $10$ , $x\in N$ }
$Q=\left \{ 10,15,20,25,30,... \right \}$

$(1)$

$A={ { \{ 1,2,3,4 } }\}$ and

$B={ { \{ 1,2,3,4 } }\}$

$A=B$

$(2)$

$A={ { \{ 4,8,12,16 } }\}$ and

$B={ { \{ 4,8,16,18 } }\}$

$A \ne B$

$(3)$

$X={ { \{2,4,6,8 } }\}$

for

$x=$ even

and

$0<x<10$

$X\ \epsilon{ { \{2,4,6,8 } }\}$

so,

$Y\ \epsilon{ { \{2,4,6,8 } }\}$

Thus

$X=Y$

$(4)$

$P$ is multiple of

$10$ and itis a subset of natural no.

so,

$P={ { \{ 10,20,30,40.... } }\}$

and

$Q={ { \{ 10,15,20,25,.... } }\}$

$P\ne Q$

Find out the set, which is defined as $\{x : x - 3 < 0\}$ ; x is any integer.

Since,

$x - 3 < 0$

$x < 3$
Set

$= \{2, 1, 0, -1, -2, -3, .....\}$

If $A = \left \{3k | k \ \epsilon \mathbb {Z}\right \}$ and $B = \mathbb {N}$ , find $A\triangle B$

Here

$A\setminus B = \left \{-3k | k\epsilon \mathbb {N}\right \}$ and

$B \setminus A = \left \{1, 2, 4, 5, 7, 8 ...\right \}$ .
Hence

$A\triangle B$ is the union of all negative integral multiple of

$3$ and all natural numbers not divisible by

$3$ . (Negative integral multiples of

$3$ means the collection

$\left \{... -12, -9, -6, -3\right \})$ . Thus

$A\triangle B = \left \{ ... -12, -9, -6, -3, 1, 2, 4, 5, 7, 8, ...\right \}$

Write the following sets in Descriptive form
(i) $A=\left \{ a,e,i,o,u \right \}$
(ii) $B=\left \{ 1,3,5,7,9,11 \right \}$
(iii) $C=\left \{ 1,4,9,16,25 \right \}$
(iv) $P=$ { $x:x$ is a letter in the word 'set theory' }
(v) $Q=$ { $x:x$ is a prime number between 10 and 20 }

(i)

$A$ is the set of all vowels in the English alphabet
(ii)

$B$ is the set of all odd natural numbers less than or equal to 11
(iii)

$C$ is the set of all square numbers less than 26.
(iv)

$P$ is the set of all letters in the word 'set theory'
(v)

$Q$ is the set of all prime numbers between 10 and 20

Find the set if it is defined as $\{x : |x^2 - 9| = 0\}$ .

Since,

$x^2 - 9 = 0$

$x^2 = 9$

$x^2 = \pm 3$
Hence, set

$= \{-3, +3\}$

Identify the following sets as finite or infinite
(i) $A= \left \{ 4,5,6,... \right \}$
(ii) $B= \left \{ 0,1,2,3,4,... 75 \right \}$
(iii) $X=$ $\{$ $x:x$ is an even natural number $\}$
(iv) $Y=$ $\{$ $x:x$ is a multiple of 6 and $x> 0$ $\}$
(v) $P=$ The set of letters in the word 'freedom'

$(1) A=\{4,5,6......\}$

infinite set

$(2) B=\{0,1,2,3,4.....75\}$

finite set

$(3) X=\{2,4,6,8,10\}$

infinite set

$(4) Y=\{6,12,18,24....\}$

infinite set

$(5) P=\{f,r,e,d,o,m\}$

finite set

Let $A=\left \{ 0,1,2,3,4,5 \right \}$ . Insert the appropriate symbol $\in$ or $\notin$ in the blank spaces
(i) $0 ---- A$
(ii) $6 ---- A$
(iii) $3 ---- A$
(iv) $4 ---- A$
(v) $7 ---- A$

(i)

$0 \in A$
(ii)

$6 \notin A$
(iii)

$3 \in A$
(iv)

$4 \in A$
(v)

$7 \in A$

Represent the following set by using their standard notations:Set of natural numbers

Set of natural number

$=N$ .

If $n\left ( A \right )=26, n\left ( B \right )=10, n\left ( A\cup B \right )=30,n\left ( {A}' \right )=17$ , find $n\left ( A\cap B \right )$ and $n\left ( \cup \right )$ .

$n(A)=26,n(B)=10,n(A\cup B)=30,n(A')=17$

$n(A\cup B) =n(A)+n(B)-n(A\cap B)$

$30=26+17-n(A\cap B)$

$30=43-n(A\cap B)$

$n(A\cap B)=43-30$

$n(A\cap B)=13$

We are given

$n(A')=17$

or,

$17=n(U)-n(A)$

or,

$n(U)=17+26$ [Since

$n(A)=26$ ]

or,

$n(U)=43$ .

Examine whether $A=$ $\{$ $x:x$ is a positive integer divisible by $3 \}$ is a subset of $B=$ $\{$ $x:x$ is a multiple of $5$ , $x\in N$ $\}$ .

$A=\{3,6,9,12,15,....\}$

$B=\{5,10,15,20.....\}$

$A$ is not a subset of

$B$ because

$B$ does not contain all elements of

$A$ .

Fill in the blanks with $\subseteq$ or $\nsubseteq$ to make each statement true.
(i) $\left \{ 3 \right \}----\left \{ 0,2,4,6 \right \}$
(ii) $\left \{ a \right \}----\left \{ a,b,c \right \}$
(iii) $\left \{ 8,18 \right \}----\left \{ 18,8 \right \}$
(iv) $\left \{ d \right \}----\left \{ a,b,c \right \}$

$(1) \{3\}\nsubseteq \{0,2,4,6\}$

$(2)\{a\}\subseteq\{a,b,c\}$

$(3)\{8,18\}\subseteq\{18,8\}$

$(4)\{d\}\nsubseteq\{a,b,c\}$

Is $\phi =\left \{ \phi \right \}$ ? Why?

No,

$\phi$ contains no element but

$\left \{ \phi \right \}$ contains one element,

If $A$ and $B$ are two sets such that $A$ has $50$ elements, $B$ has $65$ elements and $A\cup B$ has $100$ elements, how many elements does $A\cap B$ have?

$n(A)=50$

$n(B)=65$

$n(A\cup B)=100$

$n(A\cap B) = n(A)+n(B)-n(A\cup B)$

$=50+65-100$

$=115-60$

$n(A\cap B)=15$

From the sets given below, select equal sets.
$A=\left \{ 12,14,18,22 \right \}, B=\left \{ 11,12,13,14 \right \}, C=\left \{ 14,18,22,24 \right \}, D=\left \{ 13,11,12,14 \right \},$ $E=\left \{ -11,11 \right \}, F=\left \{ 10,19 \right \}, G=\left \{ 11,-11 \right \}, H=\left \{ 10,11 \right \}$ .

By the given data-

$B=D$

$E=G$

These are equal sets

If $A$ and $B$ are two sets containing $13$ and $16$ elements respectively, then find the minimum and maximum number of elements in $A\cup B$ ?

$n(A)=13$

$n(B)=16$

minimum number of elements is

$A\cup B =16$

maximum number of elements in

$A\cup B =13+16=29$

Let $X=\left \{ -3,-2,-1,0,1,2 \right \}$ and $Y=$ $\{$ $x:x$ is an integer and $-3\leq x< 2$ $\}$
(i) Is $X$ a subset of $Y$ ?
(ii) Is $Y$ a subset of $X$ ?

$X=\{-3,-2,-1,0,1,2\}$

$Y=\{-3,-2,-1,0,1\}$

$(1)$

$X$ is not a subset of

$Y$ .

$(2)$ Yes,

$Y$ is a subset of

$X$ .

If $n\left ( A\cap B \right )=5, n\left ( A\cup B \right )=35, n\left ( A \right )=13,$ find $n\left ( B \right )$ .

$n(A\cap B)=5$

$n(A\cup B)=35$

$n(A\cup B) = n(A)+n(B)-n(A\cap B)$

$35=13+n(B)-5$

$35=8+n(B)$

$n(B)=35-8$

$n(B)=27$

Which of the sets are equal sets? State the reason.
$\phi ,\left \{ 0 \right \},\left \{ \phi \right \}$

Each one is different from others.

$\phi$ contains no element

$\left \{ 0 \right \}$ contains one element i.e., 0.

$\left \{ \phi \right \}$ contains one element, i.e., the null set

If $A=\left \{-2,-1,0,3,4 \right \}, B=\left \{-1,3,5\right \}$ , find (i) $A-B$
(ii) $B-A$

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

$B=\{-1,3,5\}$

$(1)A-B=\{-2,0,4\}$

$(2)B-A=\{5\}$

If $A=\left \{ 3,6,9,12,15,18 \right \}, B=\left \{ 4,8,12,16,20 \right \}, C=\left \{ 2,4,6,8,10,12 \right \}$ and $D=\left \{ 5,10,15,20,25 \right \},$ find
(i) $A-B$
(ii) $B-C$
(iii) $C-D$
(iv) $D-A$
(v) $n\left ( A-C \right )$

$A=\{3,6,9,12,15,18\}$

$B=\{4,8,12,16,20\}$

$C=\{2,4,6,8,10,12\}$

$D=\{5,10,15,20,25\}$

$(1) A-B=\{3,6,9,15,18\}$

$(2)B-C=\{16,20\}$

$(3)C-D=\{2,4,6,8,12\}$

$(4)D-A=\{5,10,20,25\}$

$n(A-C)=4$

Let $A$ and $B$ be two finite sets such that $n\left ( A- B \right )=30, n\left ( A\cup B \right )=180$ . Find $n\left ( B \right )$ .

$A$ and

$B$ are finite sets

$n(A-B)=30$

Elements in set A that does not contain elements of set

$B$ .

$n(A\cup B)=180$

$n(B)=n(A\cup B) -n(A-B)$

$=180-30$

$=150$

$n(B)=150$

If $n\left ( \cup \right )=38, n\left ( A \right )=16, n\left ( A\cap B \right )=12,n\left ( {B}' \right )=20$ , find $n\left ( A\cup B \right )$ .

$n(V)=38,n(A)=16,n(A\cap B)=12 , n(B')=20\\$

$n(B)=n(U)-n(B')\\$

$=38-20\\$

$=18\\$

$n(A\cup B)= n(A)+n(B)-n(A\cap B)\\$

$n(A\cup B) = 16+18-12\\$

$n(A\cup B) = 22$

Given n(A)=285, n(B)=195, n(U)=500, $n(A \cup B)=410$ , find $n(A' \cup B')$ .

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$\Rightarrow n(A\cap B)$

$=285+195-410\\=70$

Similarly

$n(A^\prime \cup B^\prime)$

$=n(A^\prime)+n(B^\prime)-n(A^\prime \cup B^\prime)$

$=n(U)-n(A)+n(U)-n(B)-n(U$ \

$(A\cup B)$

$=500-285+500-195-(500-410)\\=430$

Given

the sets $A=\left \{4,5,6,7\right \}$ and $B=\left \{1,3,8,9\right \}$ . Find $A\cap B$

$$A=\left \{4,5,6,7

\right \}

$and$ B=\left \{1,3,8,9\right \}

$. So$ A\cap B = \varnothing$$.
Hence A

and B are disjoint sets.

Let A = {4, 6, 8, 10 } and B = { 3, 4, 5, 6, 7 }. If $f: A \rightarrow B$ is defined by $f(x)=\frac{1}{2}x+1$ then represent f(x) by a set of ordered pairs

{(4, 3), (6, 4).(8, 5), (10, 6)}

Given that $U=\left \{ 3,7,9,11,15,17,18 \right \}, M=\left \{ 3,7,9,11 \right \}$ and $N=\left \{ 7,11,15,17 \right \}$
Find
(i) $M-N$
(ii) $N-M$
(iii) ${N}'-M$
(iv) ${M}'-N$
(v) $M\cap \left ( M-N \right )$
(vi) $N\cup \left ( N-M \right )$
(vii) $n\left ( M-N \right )$

$U=\{3,7,9,11,15,17,18\}$

$M=\{3,7,9,11\}$

$N=\{7,11,15,17\}$

$(1) M-N = \{3,9\}$

$(2) N-M = \{15,17\}$

$(3)N'=\{3,9,18\}$

$M=\{3,7,9,11\}$

$N'-M=\{18\}$

$(4) M'=\{15,17,18\}$

$M'-N=\{18\}$

$(5) M\cap(M-N)=\{3,9\}$

$(6) N\cup(N-M)=\{7,11,15,17\}$

$(7) n(M-N)=2$

Given A $= \{1, 2, 3, 4\}$ , B $= \{3, 4, 5, 6\}$ and C $= \{6, 7\}$ . Verify that (A $\cap$ B) $\cap$ C = A $\cap$ (B $\cap$ C).

Given that:

$A=\{1,2,3,4\},B=\{3,4,5,6\}$ and

$C=\{6,7\}$

$A\cap B=\{3,4\}$

$L.H.S. =(A\cap B)\cap C=\{\phi\}$

Now,

$B\cap C=\{6\}$ and

$R.H.S. =A\cap (B\cap C)=\{\phi\}$

$\therefore L.H.S=R.H.S$

If A = {-1, 1}, find A $\times$ A $\times$ A.

$A=\{-1,1\}$

$A\times A=\{-1,1\}\times\{-1,1\}=\left\{(-1,-1),(-1,1),(1,-1),(1,1)\right\}$

$\Rightarrow A\times A\times A=\left\{-1,1\}\times\{(-1,-1),(-1,1),(1,-1),(1,1)\right\}\\=\left\{(-1,-1,-1),(-1,-1,1),(-1,1,-1)(-1,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1)\right\}$

For the given sets A= { 10, 0, 1, 9, 2, 4, 5} and B= { -1, -2, 5, 6, 2, 3, 4}, verify that set union is commutative. Also verify it by using Venn diagram.

Now,

$A\cup B$

$=$ { -10, 0, 1, 9, 2, 4, 5} U

$=$ { 1, 2, 5, 6, 2, 3, 4} $$ =$$ { -10, -2, -1,0, 1, 2, 3, 4, 5, 6, 9} .....(1)
Also,

B∪AB∪A.
By Venn diagram, we have
Hence, it is verified that set union is commutative.

Let A = {1, 2, 3}, B = {2, 4, 6, 8}, C = {3, 4, 5, 6} then prove the following :-
$A \, \cup \, B$ = {1, 2, 3, 4, 6, 8}

$A \, \cap \, (B \, \cup \, C) = (A \, \cap \, B) \, \cup \, (A \, \cap \, C)$ =

${2, 3}$ .

Using properties of sets, prove that $A \cup(A \cap B)=A$

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

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

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

$=A$ Common in both

$A$ and

$A\bigcup B$

$A-( A-B )=$ ...........

Ans.

$(A\cap B).$

$A-\left( A-B \right) =A-\left( A\cap B' \right) =A\cap \left( A\cap B' \right)'$

$=A\cap \left( A'\cup B \right) =\left( A\cap A' \right) \cup \left( A\cap B \right)$

$=\phi \cup \left( A\cap B \right) =(A\cap B)$

Given : A = {2, 3}, B = {4, 5}, C = {5, 6}, then prove the following :
$A \, \times \, (B \, \cap \, C)$ = {(2, 5), (3, 5)}

$B\cap C = \{ 5\}$

$A = \{ 2, 3\}$

$A \times (B\cap C) = \{ (2, 5), (3,5)\}$

Given : A = {2, 3}, B = {4, 5}, C = {5, 6}, then prove the following :
$A \, \times \, (B \, \cup \, C)$ = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

$B\cup C = \{ 4, 5, 6\}$

$A = \{ 2, 3 \}$

$A \times (B\cup C) = \{ (2, 4) , (2, 5) , (2, 6) , (3,4) , (3,5) , (3, 6)\}$

Hence proved.

Verify the following identities where A = {1, 2, 3, 4, 5}, B = {2, 3, 5, 6}, C = {4, 5, 6, 7}
$A \, \cap \, (B \, \cup \, C) \, = \, (A \, \cap \, B) \, \cup \, (A \, \cap \, C)$

${1,2,3,4,5}$ }

${2,3,5,6}$ }

${4,5,6,7}$ }

$(B\cup C)={2,3,4,5,6,7}$

$A\cap (B\cup C)={2,3,4,5}$ ...(1)

$(A\cap B)={2,3,5}$

$(A\cap C)={4,5}$

$(A\cap B)\cup (A\cap C)={2,3,4,5}$ ...(2)

From (1) and (2)

$A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$

Find the smallest set a such that a union {1}={1,2,3,5,9}

For smallest set

$\Rightarrow n(Y) = 1$

Let $A$ $=\{1,2\},$ and $B$ $=\{1,3\},$ then prove that $\left( A\times B \right) \cup \left( B\times A \right)=\left\{(1,3),(2,3),(3,1),(3,2),(1,1),(1,2),(2,1)\right\}$

$A=\left\{1,2\right\}$ and

$B=\left\{1,3\right\}$

$A\times B=\left\{(1,1)\ (1,3)\ (2,1)\ (2,3) \right\}$

$B\times A=\left\{(1,1)\ (1,2)\ (3,1)\ (3,2) \right\}$

$(A\times B)\cup (B\times A)$

$\Rightarrow \ \left\{(1,1)\ (1,3)\ (2,1)\ (2,3)\ (1,2)\ (3,1)\ (3,2) \right\}$

Given $A ={2,3}, B = {4,5}, C ={5,6},$ find
(i) $A\times \left( B\cap C \right) =\ .......$
(ii) $\left( A\times B \right) \cup \left( B\times C \right) =\ .......$

(i)

$B\cap C ={5}$

$A\times \left( B\cap C \right) =\{ (2,5),(3,5)\}$
(ii)

$\left( A\times B \right) =\{ (2,4),(2,5),(3,4),(3,5)\}$

$\left( B\times C \right) =\{ (4,5),(4,6),(5,5),(5,6)\}$

$\left( A\times B \right) \cup \left( B\times C \right) =\{ (2,4),(2,5),(3,4),(3,5),(4,5),(4,6),(5,5),(5,6)\}$

Comment on the following statements
$A-B=A\cap B'=B'-A'$

$x\in A-B\Leftrightarrow x\in A$

and

$x\notin B\Leftrightarrow x\in A$ and

$x\in B'$

$\Leftrightarrow x\in A\cap B'$

Hence

$A-B=A\cap B'.$ Again ....(1)

$x\in A\cap B'\Leftrightarrow x\in A$ and

$x\in B'$

$\Leftrightarrow \ x\notin A'$ and

$x\in B' \Leftrightarrow x\in B'$ and

$x\notin A'$

$\Leftrightarrow x\in \left( B'-A' \right)$

Hence

$A\cap B'=B'-A'$ ....(2)

Comment on the following statements
$A-\left( A-B \right) =A\cap B$

True,

the intersection

$A \cap B$ of two sets A and B is the set that contains all elements of A that also belong to B (or equivalently, all elements of B that also belong to A), but no other elements.

Let $A=\left\{1,2,3 \right\},$ $B=\left\{2,4,6,8 \right\},$ $C=\left\{2,3,5,6 \right\}$ . Then $A\cap \left( B\cup C \right).\ ...........$

$\left( B\cup C \right) =\left\{ 2,4,6,8,3,5 \right\}$

$\therefore \left\{ 1,2,3 \right\} \cap \left\{ 2,4,6,8,3,5 \right\} =\left\{ 2,3 \right\}$

Set $A=\left\{ x:x\ is\ a\ digit\ in\ the\ number\ 3591 \right\}$
$B=\left\{ x:x\in N,x<10 \right\}$ . Find $A\cup B,A\cap B,A-B$ and $B-A$ .

Ans.

$A\cup B=B,A\cap B=A,A-B=\phi$

$B-A=\left\{2,4,6,7,8\right\}$

Let A and B be two sets such that $A\times B$ consists of 6 elements. If three elements of $A\times B$ are (1,4), (2,6), (3, 6), find $A\times B$ and $B\times A$ .

From the given ordered pair of

$A\times B$ we conclude that
A= {1, 2, 3}, B={4, 6}
Since

$A\times B$ has six elements i.e.

$3\times 2$ and hence A and B are as stated above.

$\therefore A\times B$ ={(1,4), (1, 6), (2, 4), (2, 6), (3,4), (3, 6)}
By reserving the above ordered pairs, we get

$B\times A$ = { (4, 1), (6, 1), (4, 2), (6, 2), (4, 3), (6, 3)}.

State whether the following statements are true or false:
i) $1 \in \left\{ {1,2,3} \right\}$
ii) $a \subset \left\{ {b,c,a} \right\}$
iii) $\left\{ a \right\} \in \left\{ {a,b,c} \right\}$
iv) $\left\{ {a,b} \right\} = \left\{ {a,a,b,b,a} \right\}$
v) The sets $\left\{ {x:x + 8 = 8} \right\}$ is the null set.

$(1) 1\in (1,2,3)$

yes,

$1$ is an element of the set

$\{1,2,3\}$

$(2)a\subset \{b,c,a\}$

yes ,a is a subset of

$\{b,c,a\}$

$(3) a\in \{a,b,c\}$

yes,a is an element of

$\{a,b,c\}$

$(4)\{a,b\}=\{a,a,b,b,a\}$

yes, the following statement is true.

$(5) \{0\}$

The set is not a null set the statement is False.

Write the following set in roaster form:
$A=\{x\,:\,x$ is an integer and $-3\leq x\,<\,7\}$

$A=\{ x:x\quad is\quad an\quad integer\quad and\quad -3\leqq x<7\}$

$A=\{ -3,-2,-1,0,1,2,3,4,5,6\}$

List all the elements of the following sets :
$B = \left\{ {x:x = {\dfrac{1} {{2n - 1}}},1 \le n \le 5} \right\}$

$B=\{1,\cfrac{1}{3},\cfrac{1}{5},\cfrac{1}{7},\cfrac{1}{9}\}$

Describe the following sets in Roster form:
The sets of all letters in the word 'BETTER'.

The set of all letters in the word 'BETTER'

{B,E,T,R}

List all the elements of the following sets :
F = { $x$ : $x$ is a letter in the word "MISSISSIPPI"}

$F=$ {M,I,S,P}

List all the elements of the following sets :
$A = \left\{ {x:{x^2} \le 10,x \in Z} \right\}$

$A=\{x:x^2\leq10,x\epsilon Z\}$

$A=\{-3,-2,-1,0,1,2,3\}$

List all the elements of the following sets :
D = { $x$ : $x$ is a vowel in the word "EQUATION"}

$D=\{x:x$ is a vowel in the word "EQUATION"

$\}$

$D=$ {A,E,I,O,U}

For any two sets of A and B, prove that
$B' \subset A'$ if $A \subset B$

$A\subset B\Leftrightarrow A\cup B=B$

$\Rightarrow (A\cup B)'=B'$

USING DEMORGAN'S LAW

$\Rightarrow A'\cap B'=B'$

$\Leftrightarrow B'\subset A'$

Hence proved.

List all the elements of the following sets :
{ ${C = x:x}$ is an integar , ${ - {1 \over 2} < x < {9 \over 2}}$ }

$C=\{0,1,2,3,4\}$

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

Let event

$A$ be the people who like coffee.

Let event

$B$ be the people who like tea.

So, People liking one of the two

$=A\cup B$

So people liking both

$=A \cap B$

Given:

$n(A)= 37,n(B) = 52$ and

$n(A\cup B)= 70$

$\Rightarrow n(A\cap B) =n(A) +n(B) - n(A\cup B)$

or,

$n(A\cap B) = 37+52-70 = 19$

So the required people who likes both

$=19$

Let $A$ and $B$ be two sets. If $A\cup X=B\cap X=\phi$ and $A\cup X=B\cup X$ for sum set $X$ that $A=B$ .

1326502_1063161_ans_f7e83c6b72ed4a508c61abd1e9156786.jpg

If $A=\left\{6,9,11\right\}$ then find $A\cup \phi$

$A\cup \phi=A=\left\{6,9,11\right\}$

Let $U = \{1, 2, 3... 20\}$ , Let $A, B$ and $C$ be the subsets of $U$ . Let $A$ be the set of all numbers, which are perfect square, $B$ be the set of all numbers which are multiples of $5$ and $C$ be the set of all numbers, which are divisible by $2$ and $3$ ?
Consider the following statements.Prove that $A, B$ and $C$ mutually exclusive.

Given,

$U = \{1, 2, 3... 20\}$ , Let

$A, B$ .

According to the problem,

$A=\{1,4,9,16\}$ ,

$B=\{5,10,15\}$ ,

$C=\{6,12,18\}$ .

Now it is clear that

$A\cap B\cap C=\phi$ .

$A,B,C$ are mutually exclusive.

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

1326497_1063099_ans_b92b8aef87a14d7c84b676d66bede9ca.jpg

If $A=(1,2,3,4,5,6,7,8,9)$
$B=(1,3,5,,7,9,11,13)$
$C=(2,4,6,8,10,12,14)$
$D=(2,3,5,7,11,13,17)$
then find $B \cup C$

$B\cup\,C=\left\{1,3,5,7,9,11,13\right\}\cup\left\{2,4,6,8,10,12,14\right\}=\left\{1,2,3,4,5,6,7,8,9,10,11,12,13,14\right\}$

If $A$ and $B$ are two disjoint sets $n(A)=15$ and $n(B)=10$ find $n\left( {A \cup B} \right)$ and $n\left( {A \cap B} \right)$ .

Given that, $A$ and $B$ are two disjoint sets

$n\left( A \right)=15$ and $n\left( B \right)=10$

Then,

$n\left( A\bigcap B \right)=0$ because $A$ and $B$ both are disjoint

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

$n\left( A\bigcup B \right)=15+10-0$

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

Hence, this is the answer.

Write all the elements of the set $S = \{ 1,\{ 3\} ,\{ 5,7\} ,9,\,10\}$ .

Consider the given set,

$S=\{1,\{3\},\{5,7\},9,10\}$

Now, All elements of the set are $1,\{3\},\{5,7\},9,10$

Let $A$ and $B$ be events such that $p\left( A \right) = \dfrac{5}{{11}},\,P\left( B \right) = \dfrac{6}{{11}}$ , and $P\left( {A \cup B} \right) = \dfrac{7}{{11}}$ and $P(B/A).$

As we know,

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

On substituting the values, we get

$\dfrac{7}{11}=\dfrac{5}{11}+\dfrac{6}{11}-P(A\cap B)$

$P(A\cap B)=\dfrac{11}{11}-\dfrac{7}{11}$

$P(A\cap B)=\dfrac{4}{11}.$

Now,

$P(A\cap B)=P(A)P(B/A).$

On substituting the values, we get

$\dfrac{4}{11}=\dfrac{5}{11}\times P(B/A)$

$P(B/A)=\dfrac{4}{5.}$

Let A $=\{a, e, i, o, u\}$ and B $=\{a, b, c, d\}$ . Is A a subset of B? No(why?). Is B a subset of A? No. (Why?)

Displaying New Doc 2019-07-16 20.44.58_29.jpg

If $f=\left\{(4,5),(5,6),(6,-4)\right\}$ and $g=\left\{(4,-4),(6,5),(8,5)\right\}$ . $|f-g|=?$

Given that:

$f=(4,5),(5,6),(6,-4), g=(4,-4),(6,5),(8,5)$

$f-g=(4-4,5-4),(5-6,6-5),(6-8,-4+5)$

$f-g=(0,1),(-1,1),(-2,1).$

$|f-g|=(0,1),(-1,1),(-2,1).$

Hence, this is the answer.

find the range of the following data 18,10,8,16,9,24,12,25,13.19.

Range of the following data:

$18,10,8,16,9,24,12,25,13,19$

Lowest Value

$=08$

Highest Value

$=25$

$\because$ Range

$=17$

If $A=(1,2,3,4,5,6,7,8,9)$
$B=(1,3,5,,7,9,11,13)$
$C=(2,4,6,8,10,12,14)$
$D=(2,3,5,7,11,13,17)$
then find $(A-D) \cup \left( {B - A} \right)$

$A-D=\left\{1,2,3,4,5,6,7,8,9\right\}-\left\{2,3,5,7,11,13,17\right\}=\left\{1,4,6,8,9\right\}$

$B-A=\left\{1,3,5,7,9,11,13\right\}-\left\{1,2,3,4,5,6,7,8,9\right\}=\left\{11,13\right\}$

$\left(A-D\right)\cup\left(B-A\right)=\left\{1,4,6,8,9\right\}\cup\left\{11,13\right\}=\left\{1,4,6,8,9,11,13\right\}$

If $A=(1,2,3,4,5,6,7,8,9)$
$B=(1,3,5,,7,9,11,13)$
$C=(2,4,6,8,10,12,14)$
$D=(2,3,5,7,11,13,17)$
then find $A-D$

$A-D=\left\{1,2,3,4,5,6,7,8,9\right\}-\left\{2,3,5,7,11,13,17\right\}=\left\{1,4,6,8,9\right\}$

$1800$ boys and $900$ girls appeared for an examination. If $42\%$ of the boys and $30\%$ of the girls passed, find
(i) number of boys passed
(ii) number of girls passed
(iii) total number of students passed
(iv) number of students failed
(v) percentage of student failed.

(i) Number of boys passed

$= 42$ % of

$1800 = \dfrac{42}{100} \times 1800 = 756$

(ii) Number of girls passed

$= 30$ % of

$900 = \dfrac{30}{100} \times 900 = 270$

(iii) Total number of students passed

$= 756 + 270 = 1026$

(iv) Total number of students failed

$= (1800 + 900) - (756 + 270) = 2700 - 1026 = 1674$

(v) Percentage of students failed

$= \dfrac{1674}{2700} \times 100 = 62$ %

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

answer is shown
1356383_1197827_ans_93f08f559a124070be17ba9365903ecd.PNG

If $A=\{ 3,4,6\} , B=\{ 1,3\}$ and $C=\{ 1,2,6\}$ then find $\left( A-B \right) \times \left( A-C \right)$

1338632_1195943_ans_1a1afd057b554a1c8ffe8f04cd824225.jpg

If $A=\left\{4,5,6\right\};B=\left\{7,8\right\}$ then show that $A\bigcup B=B\bigcup A$

$A=\{ 4,5,6\} \quad ;\quad B=\{ 7,8\} \\ A\cup B=\{ 4,5,6,7,8\} \\ B\cup A=\{ 7,8,4,5,6\}$

Both the sets contain same elements .

Hence

$A\cup B=B\cup A$

If A={1,2,3,4,5} , B={3,4,7,8} then find $A \cup B$ and $A \cap B$ .

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

$B=\left\{3, 4, 7, 8\right\}$

$A\cup B=\left\{1, 2, 3, 4, 5, 7, 8\right\}$

$A\cap B=\left\{3, 4\right\}$ .

If $A=\{6,9,11\}$ and $B=\phi$ ,find $A\cup B$ .

Given,

$A=\left \{ 6, 9, 11 \right \}, B=\phi$

Here,

$B$ is empty set.

Therefore,

$A\cup B$

$=A\cup \phi$

$=\left \{ 6, 9, 11 \right \}$

Let $O=$ Set of odd natural numbers $=\{1,3,5,7,9,...\}$ and $E=$ Set of even natural numbers $=\{2,4,6,8,10,........\}$ . Then show that $-1 \notin E$ .

$O = \{1,3,5,7,9,.....\}$

$E = \{ 2,4,6,8,10,..... \}$

So it is clear from the definition of

$E$ ,

$-1 \notin \,E\,$

Let $O=$ Set of odd natural numbers $=\{1,3,5,7,9,...\}$ and $E=$ Set of even natural numbers $=\{2,4,6,8,10,........\}$ . Then show that $3\in E$ .

$O = \{1,3,5,7,9,.....\}$

$$E = \{ {2,4,6,8,10,..... \}$$

So it is clear from the definition of

$E$ ,

$3 \in \,E\,$

Let $O=$ Set of odd natural numbers $=\{1,3,5,7,9,...\}$ and $E=$ Set of even natural numbers $=\{2,4,6,8,10,........\}$ . Then show that $0.1\notin E$ .

$O = \{1,3,5,7,9,.....\}$

$E = \{ 2,4,6,8,10,..... \}$

So it is clear from the definition of

$E$ ,

$0.1 \notin \,E\,$

Let $A=\{1,2,3,4\}$ and $B=\{2,4,6,8\}$ . Then find $A-B$ .

Given,

$A=\{1,2,3,4\}$ and

$B=\{2,4,6,8\}$ .

Then,

$A-B$

$=\{1,3\}$ .

Solve
If n(A) =15 , $n(A\cup B)=29$ , $n(A \sqcap B) = 7$ find n(B) .

$n(A)=15$

$n(A\cup B)=29$

$n(A\cap B)=7$

$\Rightarrow n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$\Rightarrow 29= 15+n(B)-7$

$\Rightarrow n(B)=29-15+7=14+7=21$

$\therefore n(B)=21$

Let $A=\left\{1, 2\right\}, B=\left\{1, 2, 3, 4\right\}$ and $C=\left\{5, 6\right\}$ . Verify that $A\times (B\cap C)=(A\times B)\cap(A\times C)$

$\begin{array}{l} A=\left\{ { 1,2 } \right\} \, \, \, B=\left\{ { 1,2,3,4 } \right\} \, \, \, \& \, C=\left\{ { 5,6 } \right\} \\ A\times \left( { B\cap C } \right) =\left\{ { 1,2 } \right\} \left\{ { 1,2,3,4,5,6 } \right\} \\ Now, \\ \left( { A\times B } \right) \cap \left( { A\times C } \right) =\left\{ { \left( { 1,1 } \right) ,\left( { 1,2 } \right) ,\left( { 1,3 } \right) ,\left( { 1,4 } \right) ,\left( { 2,1 } \right) ,\left( { 2,2 } \right) ,\left( { 2,3 } \right) ,\left( { 2,4 } \right) } \right\} \times \left\{ { \left( { 1,5 } \right) ,\left( { 1,6 } \right) ,\left( { 2,5 } \right) ,\left( { 2,6 } \right) } \right\} \\ =\left\{ { 1,2,3,4,5,6 } \right\} \\ \therefore A\times \left( { B\cap C } \right) =\left( { A\times B } \right) \cap \left( { A\times C } \right) \\ \end{array}$

Define subset of a set.

A subset B of a set 'A' is defined as a set in which all the elements of B are contained in A. i.e.,

$B\subseteq A$ .
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Let $A , B$ be two set such that $n ( A ) = 4$ and $n ( B ) = 6$ then the least possible number of elements in the power set of $( A \cup B )$ is

$\begin{array}{l} Let\, \\ A=\left\{ { 1,2,3,4 } \right\} \\ B=\left\{ { 1,2,3,4,5,6 } \right\} \\ Now, \\ A\cup B=\left\{ { 1,2,3,4,5,6 } \right\} \\ Hence,\, \, least\, possible\, number\, of\, element\, is\,4. \end{array}$

If $n(A)=7, n(B)=8$ then find the maximum and minimum number of elements of $AUB$ .

Set $A$ has $7$ elements & set $B$ has $8$ elements.

Minimum $A\cup B$ elements condition: Out of $8$ elements of set $B$ , $7$ elements are identical to that of set $A$ . In this case $A\cup B= B$ & it will have $8$ elements.

Maximum $A\cup B$ elements condition: $A$ & $B$ are disjoint sets, that no element will be common to them. Then:

$n(A\cup B)= n(A) + n(B)$ .

$A\cup B$ has $15$ elements.

U= $\left\{ {1,2,3,4,5,6,7,8,9,10} \right\}$
A= $\left\{ {2,4,6,8,10} \right\},B = \left\{ {1,3,5,7,8,10} \right\}$
Find $\left( {A \cup B} \right)$

Given, A=

$\left\{ {2,4,6,8,10} \right\},B = \left\{ {1,3,5,7,8,10} \right\}$ .

Then

$A\cup B$

$=\{1,2,3,4,5,6,7,8,10\}$ .

Write all the subsets of the sets
$\left(i\right)\left\{a\right\}$
$\left(ii\right)\phi$ .

(i)

$\left\{ a \right\} \Rightarrow$ subsets

$=\left\{ a \right\} ,\phi$

(ii)

$\phi \Rightarrow$ subsets

$=\phi$

Describe the sets of points in space whose coordinates satisfy the given inequalities or combinations of equations and inequalities.
$x^{2}+y^{2}+z^{2}\le 1$

$x^2+y^2+z^2=1$ represents a sphere of radius having centre at

$(0, 0, 0)$

$\Rightarrow x^2+y^2+z^2\leq 1$ represents all points inside the sphere.

1214454_1396072_ans_f5f70df0f84148cba40129a95a64ef0f.jpg

Let $A=\{3, 6, 12, 15, 18, 21\}, B=\{4, 8, 12, 16, 20\}, C=\{2, 4, 6, 8, 10, 12, 14, 16\}$ and $D=\{5, 10, 15, 20\}$ . Find $B-D$ .

$A=\{3, 6, 12, 15, 18, 21\},$

$B=\{4, 8, 12, 16, 20\},$

$C=\{2, 4, 6, 8, 10, 12, 14, 16\}$

$D=\{5, 10, 15, 20\}$ .

$B-D$ =

$\{4,8,12,16,20\}-\{5,10,15,20\}$

$\{4,8,12,16\}$

Let $A=\{3, 6, 12, 15, 18, 21\}, B=\{4, 8, 12, 16, 20\}, C=\{2, 4, 6, 8, 10, 12, 14, 16\}$ and $D=\{5, 10, 15, 20\}$ . Find $A-C$ .

$A=\{3, 6, 12, 15, 18, 21\},$

$B=\{4, 8, 12, 16, 20\},$

$C=\{2, 4, 6, 8, 10, 12, 14, 16\}$

$D=\{5, 10, 15, 20\}$ .

$A-C$ =

$\{3, 6, 12, 15, 18, 21\}-$

$\{2, 4, 6, 8, 10, 12, 14, 16\}$

$\{3, 15, 18, 21\}$ .

Let $A=\{3, 6, 12, 15, 18, 21\}, B=\{4, 8, 12, 16, 20\}, C=\{2, 4, 6, 8, 10, 12, 14, 16\}$ and $D=\{5, 10, 15, 20\}$ . Find $C-A$ .

$A=\{3, 6, 12, 15, 18, 21\},$

$B=\{4, 8, 12, 16, 20\},$

$C=\{2, 4, 6, 8, 10, 12, 14, 16\}$

$D=\{5, 10, 15, 20\}$ .

$C-A$ =

$\{2,4,6,8,10,12,14,16\}-\{3,6,12,15,18,21\}$

$\{2, 4, 8, 10, 14, 16\}$ .

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 A'.

$u=\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$

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

$B=\{2, 4, 6, 8\}$

$C=\{3, 4, 5, 6\}$

$A'=u-A$ =

$\{1,2,3,4,5,6,7,8,9\}-\{1,2,3,4\}$

$\{5,6,7,8,9\}$

Let $A=\{3, 6, 12, 15, 18, 21\}, B=\{4, 8, 12, 16, 20\}, C=\{2, 4, 6, 8, 10, 12, 14, 16\}$ and $D=\{5, 10, 15, 20\}$ . Find $A-D$ .

$A=\{3, 6, 12, 15, 18, 21\},$

$B=\{4, 8, 12, 16, 20\},$

$C=\{2, 4, 6, 8, 10, 12, 14, 16\}$

$D=\{5, 10, 15, 20\}$ .

$A-D$ =

$\{3, 6, 12, 15, 18, 21\}-$

$\{5, 10, 15, 20\}$ .

$\{3, 6, 12, 18, 21\}$ .

In a group of $50$ persons, $14$ drink tea but not coffee and $30$ drink tea. Find how many drink coffee but not tea.

Let A and B be sets of persons who drink tea and coffee respectively.

Then

$n(A\cup B)=50$

$n(A-B)=14$

$n(A)=30$ .

Required number

$=n(B-A)=n(B)-n(A\cap B)$ .

Now,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$\Rightarrow 50=30+n(B)-16$

$\Rightarrow n(B)=36$ .

$\therefore n(B-A)=n(B)-n(A\cap B)\Rightarrow n(B-A)=36-16=20$ .

Let $A=\{3, 6, 12, 15, 18, 21\}, B=\{4, 8, 12, 16, 20\}, C=\{2, 4, 6, 8, 10, 12, 14, 16\}$ and $D=\{5, 10, 15, 20\}$ . Find $D-A$ .

$A=\{3, 6, 12, 15, 18, 21\},$

$B=\{4, 8, 12, 16, 20\},$

$C=\{2, 4, 6, 8, 10, 12, 14, 16\}$

$D=\{5, 10, 15, 20\}$ .

$D-A$ =

$\{5,10,15,20\}-\{3,6,12,15,18,21\}$

$\{5, 10, 20\}$ .

Let $A=\{3, 6, 12, 15, 18, 21\}, B=\{4, 8, 12, 16, 20\}, C=\{2, 4, 6, 8, 10, 12, 14, 16\}$ and $D=\{5, 10, 15, 20\}$ . Find $B-C$

$A=\{3, 6, 12, 15, 18, 21\},$

$B=\{4, 8, 12, 16, 20\},$

$C=\{2, 4, 6, 8, 10, 12, 14, 16\}$

$D=\{5, 10, 15, 20\}$ .

$B-C$ =

$\{4,8,12,16,20\}-\{2,4,6,8,10,12,14,16\}$

$\{20\}$

In a group of $950$ persons, $750$ can speak Hindi and $460$ can speak English. Find how many can speak Hindi only.

Let A and B denote the sets of persons who can speak Hindi and English respectively.

Then,

$n(A\cup B)=950$

$n(A)=750$

$n(B)=460$ .

$n(A\cap B)=n(A)+n(B)-n(A\cup B)$

$750+460-950=260$

Required number

$=n(A-B)=n(A)-n(A\cap B)$ .

$=750-260=490$

490 persons can speak Hindi only.

If $A=\{1, 2, 3, 4, 5\}, B=\{4, 5, 6, 7, 8\}, C=\{7, 8, 9, 10, 11\}$ and $D=\{10, 11, 12, 13, 14\}$ . Find $B\cup C$ .

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

$B=\{4, 5, 6, 7, 8\}$

$C=\{7, 8, 9, 10, 11\}$

$D=\{10, 11, 12, 13, 14\}$

$B\cup C$ .

$=\{4,5,6,7,8\}\cup \{7,8,9,10,11\}$

$=\{4, 5, 6, 7, 8, 9, 10, 11\}$ .

If $A=\left\{a,b,c,d,e,f\right\}, B=\left\{c,e,g,h\right\}$ and $C=\left\{a,e,m,n\right\}$ , find
$A\cup C$

The union of two sets A and B is the set of elements which are in A, in B, or in both A and B.

So in,

$A=\left\{a,b,c,d,e,f\right\}, B=\left\{c,e,g,h\right\}$ and

$C=\left\{a,e,m,n\right\}$

$A\cup C$

$=\left\{a,b,c,d,e,f,m,n \right\}$

State which of the following collections are sets:
Collection of odd natural numbers less than 50.

It is a set.
If we denote the given set by A, then

$A=\left\{1, 3, 5, 7, ........., 47, 49\right\}$

In a group of $50$ persons, $14$ drink tea but not coffee and $30$ drink tea. Find how many drink tea and coffee both.

Let A and B be sets of persons who drink tea and coffee respectively.

Then

$n(A\cup B)=50$

$n(A-B)=14$

$n(A)=30$ .

$n(A-B)=14$

$\Rightarrow n(A)-n(A\cap B)=14$

$\Rightarrow n(A\cap B)$

$=n(A)-14=30-14=16$ .

16 persons drink tea and coffee both.

If $A=\left\{2,3,5,7,11 \right\}$ and $B=\phi$ , find
$A\cup B$

$A=\left\{2,3,5,7,11 \right\}$ and

$B=\phi=empty set$

The union of two sets is a new set that contains all of the elements that are in at least one of the two sets. The union is written as

$A∪B$ or

$“A \ or \ B”.$ The intersection of two sets is a new set that contains all of the elements that are in both sets.

$A\cup B$

$=\left\{2,3,5,7,11 \right\}$

State in the case whether $A\subset$ or $A\nsubseteq B$ .
$A=\left\{ x:x\in Z, x^2=1\right\}, B=\left\{ x:x\ in\ N, x^2=1\right\}$

$x^2=1 \Rightarrow x= \ 1 \ or \ -1$

$1 \in N \ and \ Z$ and

$-1 \in Z$ but

$\notin Z$

So in,

$A=\left\{ x:x\in Z, x^2=1\right\}, B=\left\{ x:x\ in\ N, x^2=1\right\}$

$A=\left\{ -1, 1\right\}$ and

$B=\{\ 1 \}$ . So,

$A\nsubseteq B$ .

If $A=\left\{3, \left\{ 4, 5\right\}, 6\right\}$ , State whether the following statement is true or not.
$\left\{ 4, 5\right\} \in A$

The given statement is false.

If $A=\{1, 2, 3, 4, 5\}, B=\{4, 5, 6, 7, 8\}, C=\{7, 8, 9, 10, 11\}$ and $D=\{10, 11, 12, 13, 14\}$ . Find $B\cup D$ .

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

$B=\{4, 5, 6, 7, 8\}$

$C=\{7, 8, 9, 10, 11\}$

$D=\{10, 11, 12, 13, 14\}$

$B\cup D$ .

$=\{4,5,6,7,8\}\cup \{10,11,12,13,14\}$

$=\{4, 5, 6, 7, 8, 10, 11,12,13,14\}$ .

If $A=\left\{1,2,3,4,5 \right\}, B=\left\{4,5,6,7,8 \right\}, C=\left\{7,8,9,10,11 \right\}$ and $D=\left\{10,11,12,13,14 \right\}$ , find :
$A\cup B$

The union of two sets

$A$ and B, is the set of elements which are in A or in B or in both. It is denoted by

$A \cup B$ , and is read "A union B".

$\therefore$

$A \cup B$ can be written as,

$A \cup B$ :

$\left\{1,2,3,4,5,6,7,8 \right\}$

State which of the given collections are sets:
Collection of all people in this world over $50$ year of age.

It is a set as there is a definite age limit.

State which of the following collections are sets:
Collection of four colours of a rainbow

It is not a set because the given collection is not well-defined-people may differ on four colours of a rainbow.

State which of the given collections are sets:
Collection of all countries of Asia.

It is set because the countries are countable.

State which of the given collections are sets:
Collection of all difficult problems in your maths book.

It is also not set because problem are different to different students.

Fill in the blanks:
A collection of ......... objects is called a set.

A collection of well defined objects is called a set.

State which of the given collections are sets:
Collection of three cities of India whose name start with the letter 'J'.

It is also not set because cities are not defined.

State which of the given collections are sets:
Collection of all poor people of Dhanbad.

It is not set because elements are not countable.

State which of the given collections are sets:
Collection of all fools.

It is not a set as the classification of a fool is based on personal understanding.

State which of the given collections are sets:
Collection of four countries of Asia.

It is not set because the elements are countable but not defined.

State which of the following collections are sets:
Collection of all clever students of your school

It is not set because the given collection is not well-defined-people may differ on whether a student is clever or not.

State which of the following collections are sets:
Collection of some multiples of $5$

It is not a set because the given collection is not well defined-people may differ on which are multiples of

$5$ .

State which of the following collections are sets:
Collection of all rich people of Bangalore

It is not a set because the given collection is not well-defined-people may differ on whether a person is rich or not.

State which of the following collections are sets:
Collection of all tall students of your class

It is not a set because the given collection is not well-defined-people may differ on whether a student is tall or not.

State which of the following collections are sets:
Collection of first three days of a week

It is set.
If we denote the given set by A, then

$A=\{\text{Sunday, Monday, Tuesday}\}$

State which of the following collections are sets:
Collection of all prime numbers

It is a set because the given collection is well defined.

State which of the following collections are sets:
Collection of all even integers which lie between $-5$ and $15$

It is a set. If we denote the given set by A, then

$A=\{-4, -2, 0, 2, 4, 6, 8, 10, 12, 14\}$

Which of the following are sets? Justify our answer.
The collection of all boys in your class.

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.

State whether the following pairs of sets are equal or not:
$A = \{\text{PUPPET}\}, B = \{P, U, E, T\}$

$A = \{PUPPET\}, B = \{P, U, E, T\}$ , then A = B because the elements in a set can be repeated or rearranged.

sets are equal

Given that $A = \{2, 5, 7, 8, 10\}, B = \{5, 7, 2, x, 10\}$ and $A= B$ , write the value of $x$ .

Given,

$A = \{2, 5, 7, 8, 10\}$

$B = \{5, 7, 2, x, 10\}$

$\because A = B$

$\therefore x = 8$

Which of the following are sets? Justify our answer.
The collection of all natural numbers less than $100$ .

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.

Which of the following are sets? Justify our answer.
A team of eleven best-cricket batsmen of the world

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.

State whether the following pairs of sets are equal or not:
$A = \{3, 5, 7, 9, 11, 13\},\quad B = \{\text{odd numbers between 2 and 14}\}$

Here we have,

$A = \{3, 5, 7, 9, 11, 13\}$

$B = \text{\{odd numbers between 2 and 14}\}$

$\Rightarrow B = \{3, 5, 7, 9,11, 13\}$

$\therefore A = B$

State which of the following collections are sets:
Collection of three healthy students of your class.

It is not a set because the given collection is not well defined -people may differ on whether a student is healthy or not. They may select different healthy students.

State which of the following collections are sets:
Collection of all good cricket players of India

It is not a set because the given collection is not well defined-people may differ on whether a cricket player of India is good or not.

Let $X=\left\{1,2,3\right\}$ and $Y=\left\{4,5\right\}$ . Find whether the following subsets of $X+Y$ are function from $X$ to $Y$ or not.
$h=\left\{(1,4),(2,5),(3,5)\right\}$ .

Given that

$x=\left\{1,2,3\right\}$ and

$y=\left\{4,5\right\}$

$X\times Y=\left\{(1,4),(1,5),(2,4),(2,5),(3,4),(3.5)\right\}$

$h=\left\{(1,4),(2,5),(3,5)\right\}$ .

It is clear that

$h$ is a function.

State which of the following collections are sets:
Collection of three youngest students of your class.

It is a set because the given collection is well defined-people can choose three youngest students of their classes.

Which of the following are sets? Justify our answer.
A collection of most dangerous animals of the world.

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.

Find the union of each of the following pairs of sets:
$A$ = { $a, e, i, o, u$ } $B$ = { $a, b, c$ }

$A$ = {

$a, e, i, o, u$ }

$B$ = {

$a, b, c$ }

$A∪ B$ = {

$a, b, c, e, i, o, u$ }

Which of the following are sets? Justify our answer.
The collection of all even integers.

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.

Which of the following are sets? Justify our answer.
A collection of novels written by the writer Munshi Prem Chand.

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.

Find the union of each of the following pairs of sets:
$X$ = { $1, 3, 5$ } $Y$ = { $1, 2, 3$ }

$X$ = {

$1, 3, 5$ } Y = {

$1, 2, 3$ }

$X ∪ Y$ = {

$1, 2, 3, 5$ }

Which of the following are sets? Justify our answer.
The collection of questions in this chapter.

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.

Find the union of each of the following pairs of sets:
$A$ = { $x: x$ is a natural number and multiple of $3$ }
$B$ = { $x: x$ is a natural number less than $6$ }

$A$ = {

$x: x$ is a natural number and multiple of

$3$ } = {

$3, 6, 9 …$ }
As,

$B$ = {

$x: x$ is a natural number less than

$6$ } = {

$1, 2, 3, 4, 5, 6$ }

$A ∪ B$ = {

$1, 2, 4, 5, 3, 6, 9, 12 …$ }

$\therefore$ ,

$A ∪ B$ = {

$x: x = 1, 2, 4, 5$ or a multiple of

$3$ }

Find the union of each of the following pairs of sets:
$A$ = { $x: x$ is a natural number and $1 < x \leq 6$ }
$B$ = { $x: x$ is a natural number and $6 < x < 10$ }

$A$ = {

$x: x$ is a natural number and

$1 < x \leq 6$ } = {

$2, 3, 4, 5, 6$ }

$B$ = {

$x: x$ is a natural number and

$6 < x < 10$ } = {

$7, 8, 9$ }

$A∪ B$ = {

$2, 3, 4, 5, 6, 7, 8, 9$ }

$\therefore$ ,

$A∪ B$ = {

$x: x ∈ N$ and

$1 < x < 10$ }.

If $A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 3, 4, 5, 6 \right \}, C = \left \{ 5, 6, 7, 8 \right \} and D = \left \{ 7, 8, 9, 10 \right \}$ ; find:
$B C$

$A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 3, 4, 5, 6 \right \}, C = \left \{ 5, 6, 7, 8 \right \}, D = \left \{ 7, 8, 9, 10 \right \}$

$B ∪ C = \left \{ 3, 4, 5, 6, 7, 8 \right \}$

If $A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 3, 4, 5, 6 \right \}, C = \left \{ 5, 6, 7, 8 \right \} and D = \left \{ 7, 8, 9, 10 \right \}$ ; find:
$A B C$

$A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 3, 4, 5, 6 \right \}, C = \left \{ 5, 6, 7, 8 \right \}, D = \left \{ 7, 8, 9, 10 \right \}$

$A ∪ B ∪ C = \left \{ 1, 2, 3, 4, 5, 6, 7, 8 \right \}$

Let $A = \left \{ a,b \right \}, B = \left \{ a,b,c \right \}$ . Is $A B$ ? What is $A B$ ?

Here,

$A = \left \{ a,b \right \}$ and

$B = \left \{ a,b,c \right \}$
Yes,

$A ⊂ B$

$A∪ B = \left \{ a,b,c \right \} = B$ .

If $A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 3, 4, 5, 6 \right \}, C = \left \{ 5, 6, 7, 8 \right \} and D = \left \{ 7, 8, 9, 10 \right \}$ ; find:
$A B$ .

$A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 3, 4, 5, 6 \right \}, C = \left \{ 5, 6, 7, 8 \right \}, D = \left \{ 7, 8, 9, 10 \right \}$

$A ∪ B = \left \{ 1, 2, 3, 4, 5, 6 \right \}$

If $A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 3, 4, 5, 6 \right \}, C = \left \{ 5, 6, 7, 8 \right \} and D = \left \{ 7, 8, 9, 10 \right \}$ ; find:
$B D$

$A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 3, 4, 5, 6 \right \}, C = \left \{ 5, 6, 7, 8 \right \}, D = \left \{ 7, 8, 9, 10 \right \}$

$B ∪ D = \left \{ 3, 4, 5, 6, 7, 8, 9, 10 \right \}$

If $A = \left \{ 3, 5, 7, 9, 11 \right \}, B = \left \{ 7, 9, 11, 13 \right \}, C = \left \{ 11, 13, 15 \right \}$ and $D = \left \{ 15, 17 \right \}$ ; find:
$A B$

$A ∩ B = \left \{ 7, 9, 11 \right \}$

If $A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 3, 4, 5, 6 \right \}, C = \left \{ 5, 6, 7, 8 \right \} and D = \left \{ 7, 8, 9, 10 \right \}$ ; find:
$A C$

$A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 3, 4, 5, 6 \right \}, C = \left \{ 5, 6, 7, 8 \right \}, D = \left \{ 7, 8, 9, 10 \right \}$

$A ∪ C = \left \{ 1, 2, 3, 4, 5, 6, 7, 8 \right \}$

If $A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 3, 4, 5, 6 \right \}, C = \left \{ 5, 6, 7, 8 \right \} and D = \left \{ 7, 8, 9, 10 \right \}$ ; find:
$B C D$

$A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 3, 4, 5, 6 \right \}, C = \left \{ 5, 6, 7, 8 \right \}, D = \left \{ 7, 8, 9, 10 \right \}$

$B ∪ C ∪ D = \left \{ 3, 4, 5, 6, 7, 8, 9, 10 \right \}$

Find the union of each of the following pairs of sets:
$A$ = { $1, 2, 3$ }, $B = \phi$

$A$ = {

$1, 2, 3$ },

$B = \phi$

$A∪ B$ = {

$1, 2, 3$ }

If $A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 3, 4, 5, 6 \right \}, C = \left \{ 5, 6, 7, 8 \right \} and D = \left \{ 7, 8, 9, 10 \right \}$ ; find:
$A B D$

$A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 3, 4, 5, 6 \right \}, C = \left \{ 5, 6, 7, 8 \right \}, D = \left \{ 7, 8, 9, 10 \right \}$

$A ∪ B ∪ D = \left \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \right \}$

If $A = \left \{ 3, 5, 7, 9, 11 \right \}, B = \left \{ 7, 9, 11, 13 \right \}, C = \left \{ 11, 13, 15 \right \}$ and $D = \left \{ 15, 17 \right \}$ ; find:
$A C D$

$A ∩ C ∩ D = \left \{ A ∩ C \right \} ∩ D$

$= \left \{ 11 \right \} ∩ \left \{ 15, 17 \right \}$

$= \phi$

If $A = \left \{ 3, 6, 9, 12, 15, 18, 21 \right \}, B = \left \{ 4, 8, 12, 16, 20 \right \}, C = \left \{ 2, 4, 6, 8, 10, 12, 14, 16 \right \}, D$ $= \left \{ 5, 10, 15, 20 \right \}$ ; find
$A – B$ .

$A – B = \left \{ 3, 6, 9, 15, 18, 21 \right \}$

If $A = \left \{ 3, 5, 7, 9, 11 \right \}, B = \left \{ 7, 9, 11, 13 \right \}, C = \left \{ 11, 13, 15 \right \}$ and $D = \left \{ 15, 17 \right \}$ ; find:
$A (B C)$

$A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)$

$= \left \{ 7, 9, 11 \right \} ∪ \left \{ 11 \right \}$

$= \left \{ 7, 9, 11 \right \}$ .

If $A = \left \{ 3, 5, 7, 9, 11 \right \}, B = \left \{ 7, 9, 11, 13 \right \}, C = \left \{ 11, 13, 15 \right \}$ and $D = \left \{ 15, 17 \right \}$ ; find:
$(A B) (B C)$

$(A ∩ B) ∩ (B ∪ C) = \left \{ 7,9,11 \right \} ∩ \left \{ 7, 9, 11, 13, 15 \right \}$

$= \left \{ 7, 9, 11 \right \}$ .

If $A = \left \{ 3, 5, 7, 9, 11 \right \}, B = \left \{ 7, 9, 11, 13 \right \}, C = \left \{ 11, 13, 15 \right \}$ and $D = \left \{ 15, 17 \right \}$ ; find:
$A C$

$A ∩ C = \left \{ 11 \right \}$

If $A = \left \{ 3, 5, 7, 9, 11 \right \}, B = \left \{ 7, 9, 11, 13 \right \}, C = \left \{ 11, 13, 15 \right \}$ and $D = \left \{ 15, 17 \right \}$ ; find:
$(A D) (B C)$

$(A ∪ D) ∩ (B ∪ C) = \left \{ 3, 5, 7, 9, 11, 15, 17 \right \} ∩ \left \{ 7, 9, 11, 13, 15 \right \}$

$= \left \{ 7, 9, 11, 15 \right \}$ .

If $A = \left \{ 3, 5, 7, 9, 11 \right \}, B = \left \{ 7, 9, 11, 13 \right \}, C = \left \{ 11, 13, 15 \right \}$ and $D = \left \{ 15, 17 \right \}$ ; find:
$B C$

$B ∩ C = \left \{ 11, 13 \right \}$

If $A = \left \{ 3, 5, 7, 9, 11 \right \}, B = \left \{ 7, 9, 11, 13 \right \}, C = \left \{ 11, 13, 15 \right \}$ and $D = \left \{ 15, 17 \right \}$ ; find:
$B D$

$B ∩ D = \phi$

If $A = \left \{ 3, 5, 7, 9, 11 \right \}, B = \left \{ 7, 9, 11, 13 \right \}, C = \left \{ 11, 13, 15 \right \}$ and $D = \left \{ 15, 17 \right \}$ ; find:
$A (B D)$

$A ∩ (B ∪ D) = (A ∩ B) ∪ (A ∩ D)$

$= \left \{ 7, 9, 11 \right \} ∪ \phi$

$= \left \{ 7, 9, 11 \right \}$ .

If $A = \left \{ 3, 5, 7, 9, 11 \right \}, B = \left \{ 7, 9, 11, 13 \right \}, C = \left \{ 11, 13, 15 \right \}$ and $D = \left \{ 15, 17 \right \}$ ; find:
$A D$

$A ∩ D = \phi$

If $A = \left \{ 3, 6, 9, 12, 15, 18, 21 \right \}, B = \left \{ 4, 8, 12, 16, 20 \right \}, C = \left \{ 2, 4, 6, 8, 10, 12, 14, 16 \right \}, D$ $= \left \{ 5, 10, 15, 20 \right \}$ ; find
$A – C$

$A – C = \left \{ 3, 9, 15, 18, 21 \right \}$

If $A = \left \{ 3, 6, 9, 12, 15, 18, 21 \right \}, B = \left \{ 4, 8, 12, 16, 20 \right \}, C = \left \{ 2, 4, 6, 8, 10, 12, 14, 16 \right \}, D = \left \{ 5, 10, 15, 20 \right \}$ ; find
$B C$

$B – C = \left \{ 20 \right \}$

If $A = \left \{ 3, 6, 9, 12, 15, 18, 21 \right \}, B = \left \{ 4, 8, 12, 16, 20 \right \}, C = \left \{ 2, 4, 6, 8, 10, 12, 14, 16 \right \}, D$ $= \left \{ 5, 10, 15, 20 \right \}$ ; find
$D – A$

$D – A = \left \{ 5, 10, 20 \right \}$

If $A = \left \{ 3, 6, 9, 12, 15, 18, 21 \right \}, B = \left \{ 4, 8, 12, 16, 20 \right \}, C = \left \{ 2, 4, 6, 8, 10, 12, 14, 16 \right \}, D = \left \{ 5, 10, 15, 20 \right \}$ ; find
$D B$

$D – B = \left \{ 5, 10, 15 \right \}$ .

If $A = \left \{ 3, 6, 9, 12, 15, 18, 21 \right \}, B = \left \{ 4, 8, 12, 16, 20 \right \}, C = \left \{ 2, 4, 6, 8, 10, 12, 14, 16 \right \}, D$ $= \left \{ 5, 10, 15, 20 \right \}$ ; find $C – A$

$C – A = \left \{ 2, 4, 8, 10, 14, 16 \right \}$

If $A = \left \{ 3, 6, 9, 12, 15, 18, 21 \right \}, B = \left \{ 4, 8, 12, 16, 20 \right \}, C = \left \{ 2, 4, 6, 8, 10, 12, 14, 16 \right \}, D = \left \{ 5, 10, 15, 20 \right \}$ ; find
$C D$

$C – D = \left \{ 2, 4, 6, 8, 12, 14, 16 \right \}$

If $A = \left \{ 3, 6, 9, 12, 15, 18, 21 \right \}, B = \left \{ 4, 8, 12, 16, 20 \right \}, C = \left \{ 2, 4, 6, 8, 10, 12, 14, 16 \right \}, D = \left \{ 5, 10, 15, 20 \right \}$ ; find
$C B$

$C – B = \left \{ 2, 6, 10, 14 \right \}$

If $A = \left \{ 3, 6, 9, 12, 15, 18, 21 \right \}, B = \left \{ 4, 8, 12, 16, 20 \right \}, C = \left \{ 2, 4, 6, 8, 10, 12, 14, 16 \right \}, D$ $= \left \{ 5, 10, 15, 20 \right \}$ ; find
$B – A$

$B – A = \left \{ 4, 8, 16, 20 \right \}$

If $A = \left \{ 3, 6, 9, 12, 15, 18, 21 \right \}, B = \left \{ 4, 8, 12, 16, 20 \right \}, C = \left \{ 2, 4, 6, 8, 10, 12, 14, 16 \right \}, D$ $= \left \{ 5, 10, 15, 20 \right \}$ ; find
$A – D$

$A – D = \left \{ 3, 6, 9, 12, 18, 21 \right \}$

If $A = \left \{ 3, 6, 9, 12, 15, 18, 21 \right \}, B = \left \{ 4, 8, 12, 16, 20 \right \}, C = \left \{ 2, 4, 6, 8, 10, 12, 14, 16 \right \}, D = \left \{ 5, 10, 15, 20 \right \}$ ; find
$B D$

$B – D = \left \{ 4, 8, 12, 16 \right \}$

If $R$ is the set of real numbers and $Q$ is the set of rational numbers, then what is $R Q$ ?

$R$ - Set of real numbers

$Q$ - Set of rational numbers
Therefore,

$R – Q$ is a set of irrational numbers.

If $X = \left \{ a, b, c, d \right \}$ and $Y = \left \{ f, b, d, g \right \}$ , find
$Y X$

$Y – X = \left \{ f, g \right \}$

Let $U = \left \{ 1, 2, 3; 4, 5, 6, 7, 8, 9 \right \}, A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 2, 4, 6, 8 \right \}$ and $C = \left \{ 3, 4, 5, 6 \right \}$ . Find
$A$

$U = \left \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \right \}$

$A = \left \{ 1, 2, 3, 4 \right \}$

$B = \left \{ 2, 4, 6, 8 \right \}$

$C = \left \{ 3, 4, 5, 6 \right \}$

$A’ = \left \{ 5, 6, 7, 8, 9 \right \}$

If $\xi=\{\text { natural numbers between } 10 \text { and } 40\}$
$\mathbf{A}=\{\text { multiples of } 5\}$ and $\mathrm{B}=\{\text { multiples of } 6\},$ then
(i) find $\mathrm{A} \cup \mathrm{B}$ and $\mathrm{A} \cap \mathrm{B}$
(ii) verify that $n(A \cup B)=n(A)+n(B)-n(A \cap B)$

From the question it is given that,

$\xi=\{\text { natural numbers between } 10 \text { and } 40\}$

$\xi=\{11,12,13,14,15, \ldots, 39\}$

$\xi$ is a universal set and

$A$ and

$B$ are subsets of

$\xi$ Then, the elements of A and B are to be taken only from

$\xi$

$\mathrm{A}=\{\text { multiples of } 5\}$

$A=\{15,20,25,30,35\}$

$\mathrm{B}=\{\text { multiples of } 6\}$

$B=\{12,18,24,30,36\}$
(i)

$\mathrm{A} \cup \mathrm{B}=\{15,20,25,30,35,40\} \cup\{12,18,24,30,36\}$

$A \cup B=\{15,20,25,30,35,12,18,24,36\}$

$A \cap B=\{30\}$
(ii)

$n(A \cup B)=n(A)+n(B)-n(A \cap B)$

$n(A \cup B)=5$

$n(A)=5$

$n(B)=5$

$n(A \cap B)=1$
Then,

$n(A)+n(B)-n(A \cap B)=5+5-1=9$
By comparing the results,

$9=9$
Therefore,

$n(A \cup B)=n(A)+n(B)-n(A \cap B)$

If $A = \left \{ 3, 6, 9, 12, 15, 18, 21 \right \}, B = \left \{ 4, 8, 12, 16, 20 \right \}, C = \left \{ 2, 4, 6, 8, 10, 12, 14, 16 \right \}, D = \left \{ 5, 10, 15, 20 \right \}$ ; find
$D C$

$D–C={5,15,20}$

State which of the following collections are set:
All states of India

It is a set.

If $X = \left \{ a, b, c, d \right \}$ and $Y = \left \{ f, b, d, g \right \}$ , find
$X Y$

$X – Y = \left \{ a, c \right \}$

Let $U = \left \{ 1, 2, 3; 4, 5, 6, 7, 8, 9 \right \}, A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 2, 4, 6, 8 \right \}$ and $C = \left \{ 3, 4, 5, 6 \right \}$ . Find
$(A \cup C)$

$U = \left \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \right \}$

$A = \left \{ 1, 2, 3, 4 \right \}$

$B = \left \{ 2, 4, 6, 8 \right \}$

$C = \left \{ 3, 4, 5, 6 \right \}$

$A \cup C = \left \{ 1, 2, 3, 4, 5, 6 \right \}$

$\therefore (A \cup C)’ = \left \{ 7, 8, 9 \right \}$

If $n(\xi)=40, n\left(A^{\prime}\right)=15, n(B)=12$ and $n\left((A \cap B)^{\prime}\right)=32,$ find :
(i) n(A)
(ii) $n\left(B^{\prime}\right)$
(iii) $n(A \cap B)$
(iv) $n(A \cup B)$
(v) $n(A-B)$
(vi) $n(B-A)$

From the question it is given that,

$\begin{array}{l}n(\xi)=40 \\n\left(A^{\prime}\right)=15 \\n(B)=12\end{array}$

$n\left((A \cap B)^{\prime}\right)=32$
(i)

$\mathrm{n}(\mathrm{A})$ We know that,

$n(A)=n(\xi)-n\left(A^{\prime}\right)$

$\begin{array}{l}n(A)=40-15 \\n(A)=25\end{array}$
(ii)

$n\left(B^{\prime}\right)$
We know that,

$n\left(B^{\prime}\right)=n(\xi)-n(B)$

$\begin{array}{l}n\left(B^{\prime}\right)=40-12 \\n\left(B^{\prime}\right)=28\end{array}$
(iii)

$n(A \cap B)=n(\xi)-n\left((A \cap B)^{\prime}\right)$

$=40-32$

$=8$

$\begin{aligned}(\text { iv }) \text { n }(A \cup B) &=n(A)+n(B)-n(A \cap B) \\ &=25+12-8 \\ &=37-8 \\ &=29 \end{aligned}$

$\begin{aligned}(\mathrm{v}) \mathrm{n}(\mathrm{A}-\mathrm{B})=& n(\mathrm{A})-\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \\ &=25-8 \\ &=17 \end{aligned}$

$\begin{aligned}(\mathrm{vi}) \mathrm{n}(\mathrm{B}-\mathrm{A}) &=\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \\ &=12-8 \\ &=4 \end{aligned}$

Let $X=\left\{Ram, Geeta, Akbar \right\}$ be the set of students of class $XI$ , who are in school hockey team. Let $Y= \left\{Geeta, Divid, Ashok \right\}$ be the set of students from Class $XI$ who are in school foot ball team. Find $X \cup Y$ and interpret the set.

X∪Y={Ram,Geeta,Akbar,Divid,Ashok}

This is the set of students from Class XI who are in the hockey team or the foot ball team or both

Let $A=\left\{1,2, \left\{3,4 \right\}, 5\right\}.$ Which of the following statements are incorrect and why?
$\left\{1,2,5 \right\} \subset A$

correct

∵1,2,5,∈A

If $A =\text{{5, 7, 8, 9}}$ then which of the following are subsets of $A$ ?
(i) $B = \text{{5, 8}}$
(ii) $C = \text{{0}}$
(iii) $D = \text{{7, 9, 10}}$
(iv) $E = \text{{ }}$
(v) $F =\text{{8, 7, 9, 5}}$

(i)

B={5, 8}.

Let $A=\left\{1,2, \left\{3,4 \right\}, 5\right\}.$ Which of the following statements are incorrect and why?
$\left\{1,2,3 \right\} \subset A$

incorrect ∵3∈{1,2,3} but 3∉A

Define difference of sets.

1817158_1862490_ans_d5d084f3ed984f42a576e89d42aa4af6.png

If $A=\left\{1,2,3,4 \right\}, B=\left\{3,4,5,6 \right\}, C=\left\{5,6,7,8\right\}$ and $D=\left\{7,8,9,10 \right\}$ ; find
$A \cup B$

A∪B={1,2,3,4,5,6}

Let $A =\left\{2,4,6,8 \right\}$ and $B=\left\{6,8,10,12 \right\}$ Find $A \cup B$

A∪B={2,4,6,8,10,12}

Define union of sets.

The union of two sets

$A$ and

$B$ denoted by

$A \cup B$ , is the set of all those elements which arc either in

$A$ or

$B$ or in both

$A$ and

$B$

Thus,

$AB=(x:x \in A\ or\ x \in B)$

$\therefore x \in A \cup B \ x \in A\ or\ x \in B$

$x \not \in A \cup B = x \not \in A$ and

$z \not \in B$

$\bullet$

$A \cup B= B \cap A$ (Commutative Law)

$\bullet$

$(A \cup B)\cup C= A \cup (B \cup C)$ (Associative law)

$\bullet$

$A \cup \varphi=A$

$\bullet$

$A \cup A=A$

$\bullet$

$A \cup A=U$

1817168_1862483_ans_b1824e21faef40f9b968299c170d6ea0.png

If $A=\left\{3,6,9,12,15,18,21 \right\}, B=\left\{4,8,12,16,20 \right\}, C=\left\{2,4,6,8,10,12,14,16 \right\} , D= \left\{5,10,15,20 \right\}$ find
$A-D$

A−D={3.6,9.12.18.21}

If $A=\left\{3,5,7,9,11 \right\}, B=\left\{7,9,11,13 \right\}, C=\left\{11,13,15\right\}$ and $D=\left\{15,17 \right\}$ ; find
$(A \cup D) \cap (B \cup C)$

(A∪D)∩(B∪C)={3,5,7,9,11,15,17}∩{3,5,7,9,11,13,15}(A∪D)∩(B∪C)={3,5,7,9,11,15,17}∩{3,5,7,9,11,13,15}

={7,9,11,15}

If $A=\left\{1,2,3,4 \right\}, B=\left\{3,4,5,6 \right\}, C=\left\{5,6,7,8\right\}$ and $D=\left\{7,8,9,10 \right\}$ ; find
$B \cup C$

B∪C={3,4,5,6,7,8}

If $A=\left\{3,6,9,12,15,18,21 \right\}, B=\left\{4,8,12,16,20 \right\}, C=\left\{2,4,6,8,10,12,14,16 \right\} , D= \left\{5,10,15,20 \right\}$ find
$B-D$

B−D={4,8,12,16}

If $A=\left\{3,5,7,9,11 \right\}, B=\left\{7,9,11,13 \right\}, C=\left\{11,13,15\right\}$ and $D=\left\{15,17 \right\}$ ; find
$A \cap (B \cup D)$

A∩(B∪D)={3,5,7,9,11}∩{7.9.11.13.15.17}={7,9,11}

If $A=\left\{1,2,3,4 \right\}, B=\left\{3,4,5,6 \right\}, C=\left\{5,6,7,8\right\}$ and $D=\left\{7,8,9,10 \right\}$ ; find
$B \cup D$

B∪D={3,4,5,6,7,8,9,10}

If $A=\left\{1,2,3,4 \right\}, B=\left\{3,4,5,6 \right\}, C=\left\{5,6,7,8\right\}$ and $D=\left\{7,8,9,10 \right\}$ ; find
$A \cup C$

A∪C={1,2,3,4,5,6,7,8}

If $A=\left\{3,6,9,12,15,18,21 \right\}, B=\left\{4,8,12,16,20 \right\}, C=\left\{2,4,6,8,10,12,14,16 \right\} , D= \left\{5,10,15,20 \right\}$ find
$D-A$

$D-A = \left\{5,10,20 \right\}$

If $A=\left\{3,6,9,12,15,18,21 \right\}, B=\left\{4,8,12,16,20 \right\}, C=\left\{2,4,6,8,10,12,14,16 \right\} , D= \left\{5,10,15,20 \right\}$ find
$A-C$

A−C={3.9.15.18.21}

If $A=\left\{1,2,3,4 \right\}, B=\left\{3,4,5,6 \right\}, C=\left\{5,6,7,8\right\}$ and $D=\left\{7,8,9,10 \right\}$ ; find
$B \cup C \cup D$

B∪C∪D={3,4,5,6,7,8,9,10}

If $A = \{11, 21, 31, 41\}, \ B = \{12,22,31,32\}$ , then fimd:
i) $A \cup B$
ii) $A \cap B$

1895697_1911560_ans_2ae7b9c4d5fc41f69de2e2125d9fdc04.jpeg

Which of the following are sets ? Justify your answer .
Collection of even natural less than 8.

Even natural numbers less than

$8$ are

$2, 4, 6.$

Hence

$, \{2, 4, 6 \}$ is a set.

Which of the following are sets ? Justify your answer
The collection of big cities in India.

The collection of big cities in India is not set because there is no scale to measure it.

$IV X=\left\{a,b,c,d \right\}$ and $Y=\left\{f,b,d,g \right\}$ find
$y-x$

Y−X={f,g}

List the elements of the set $P$ .

$\mu=$ {

$x : x$ is an integer,

$x\in \left[ -4,6 \right]$

$\mu = \left\{-4,-3,-2,-1,0,1,2,3,4,5,6\right\}$

$P' = \left\{-4,-3,0,1,2,4,6\right\}$

$Q' = \left\{-3,-2,1,3,6\right\}$

Since

$\mu$ in the universal set for

$P$ and

$Q$ element of set

$P$

$=\left\{-2,-1,3,5\right\}$

(There are element in universal

$\mu$ that does not contain the set

$P'$ )

If $A$ and $B$ be two sets containing $4$ and $12$ elements respectively, what can be the minimum number of elements in $A \cup B$ ? Find also, the maximum number of elements in $(A \cup B)$ .

${\textbf{Step-1: Find maximum number of elements.}}$

${\text{Given,}}$

$n(A) =4$

$n(B) =12$

${\text{Minimum number of elements in}}$

$A\cup B=12.$

${\text{A is subset of B.}}$

${\text{Maximum number of elements in}}$

$A\cup B=12+4=16.$

${\text{[All element taken in both set A and B and elements in both sets are different.]}}$

$\textbf{Hence , minimum number of elements in }\mathbf{A\cup B}\textbf{ is 12}$

${\textbf{Maximum number of elements in}}$ $\mathbf{(A \cup B)}\textbf{ is 16} .$

None of A, B and C.

Let A be the set of families which buy newspaper-A.

B be the set of families which buy newspaper-B.

and C be the set of families who buy newspaper-C.

Then,

$n (U) = 10000 \\ n (A) = 40 \% \\ n(B) = 20 \% \\ \& \, n(C) = 10 \%$

$n(A \cap B) = 5 \% , \\ n(B \cap C) = 3 \% , \\ n(A \cap C) = 4 \% \\ \& \, n(A \cap B \cap C ) = 2 \%$

Percentage of families which buy none of A, B & C

$= n(U) - n (A U B U C ) \\ \Rightarrow n(U) - [(n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n (A \cap C) + n (A \cap B \cap C)] \\ \Rightarrow 100 - [ 40 + 20 + 10 - 5 - 3 - 4 + 2 ] \\ \Rightarrow 100 - 60 = 40 \, \%$

Number of families which buy none of A, B, & C =

$10000 \times \dfrac{40}{100} = 4000$

In a town of $10,000$ families it was found that $40$ % families buy newspaper $A$ , $20$ % buy newspaper $B$ and $10$ % buy newspaper $C$ , $5$ % families buy $A$ and $B$ , $3$ % buy $B$ and $C$ and $4$ % buy $A$ and $C$ . If $2$ % families buy all three newspapers, find number of families which buy None of $A, B,C$ .

Given

$N=10,000 , n(A)=4000,n(B)=2000,n(C)=1000,n(A\cap B)=500,$

$n(B\cap C)=300,n(C\cap A)=400,n(A\cap B \cap C)=200$

$n(A'\cap B'\cap C')=n((A\cup B\cup C)')$

$=N-n(A\cup B\cup C)$

$=N-[n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)]$

$=10000-[4000+2000+1000-500-300-400+200] =4000$

Given $\mu$ = {x / x is an integer, $0 \leq x \leq 10$ }
A = {x / the remainder is 1 when x is divided by 2}
B = {x / the remainder is 2 when x is divided by 3}
Find the set $n(A \cup B)'$

$\mu = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

$A = \{1, 3, 5, 7, 9\}$

$B = \{2, 5, 8\}$

$A\cup B = \{1, 2, 3, 5, 7, 8, 9\}$

$(A\cup B)' = \{0, 4, 6, 10\}$

$\Rightarrow n(A\cup B)'=4$

A survey was carried out to find out the types of shampoo that a group of $150$ women have tried. It was found that $84$ women have used brand $A$ shampoo, $93$ have used brand $B$ , and $69$ have used brand $C$ of these women, $45$ have tried brands $A$ and $B$ , $25$ have tried brands $A$ and $C$ and $40$ have tried brand $B$ and $C$ . Determine the number of women who have tried only brand $A$ .

$n(A \cup B\cup C)=150$

$n(A)=84$

$n(B)=93$

$n(C)=69$

$n(A \cap C)=25$

$n(A \cap B)=45$

$n(B \cap C)=40$

$n(A \cap B\cap C) = n(A\cup B\cup C)-n(A)-n(B)-n(C)+n(A\cap B) +n(A\cap C)-n(B\cap C)$

$\quad \quad \quad \quad \quad \quad =150-84-93-69+45+25+40=14$

Number of women who only used brand A=

$n(A) - n(A \cap B)-n(A \cap C)=84-45-25=14$

If $A=\left \{ a, b, p, d \right \}$ , $B=\left \{ p, d , e \right \}$ , $C=\left \{ p, e, f, g \right \}$ then verify:
(i) $A\times (B\cap C)=(A\times B)\cap (A\times C)$ (ii) $A\times (B-C)=(A\times B)-(A\times C)$

$A= \left \{ a, b, p, d \right \}, B=\left \{ p, d, e \right \}, C= \left \{ p, e, f, g \right \}$
(i)

$A\times B =\left \{ (a, p) ,(a, d),(a, e),(b, p),(b, d),(b, e),(p, p),(p,d),(p, e),(d, p),(d, d),(d, e) \right \}$

$A\times C =\left \{ (a, p) ,(a, e),(a, f),(a, g),(b, p),(b, e),(b, f),(b, g),(p, p),(p,e),(p, f),(p,g),(d, p),(d, e),(d, f),(d, g) \right \}$

$B\cap C=\left \{ p, e \right \}$

$A\times (B\cap C)=\left \{ (a, p),(a, e),(b, p),(b, e),(p, p),(p, e),(d, p),(d, e) \right \}$

$(A\times B)\cap (A\times C)=\left \{ (a, p),(a, e),(b, p),(b, e),(p, p),(p, e),(d, p),(d, e) \right \}$

$\therefore A\times (B\cap C)=(A\times B)\cap (A\times C)$

(ii)

$B-C=\left \{ d \right \}$

$A\times (B-C)=\left \{ (a, d),(b, d),(p, d),(d, d) \right \}$

$(A\times B)-(A\times C)=\left \{ (a, d),(b, d),(p, d),(d, d) \right \}$

In a school with an enrolment of $950$ students, each student must join either the lions club or the country club or both. Given that $646$ students are members of the lions club and $532$ are members of the country club, calculate the number of students who are members of both clubs.

Let A and B denote the set of students who are members of lions club and country club respectively.

$n(A) = 646$

$n(B) = 532$

$n(A \cup B) = 950$

$n(A \cup B) = n(A) + n(B) - n(A \cap B)$

$\therefore n(A \cap B) = n(A) + n(B) - n(A \cup B)$

$\quad \quad \quad \quad \quad = 646+532-950 = 228$

Hence,

$228$ students are members of both the clubs.

In a certain group of $75$ students, $16$ students are taking physics, geography and english; $24$ students are taking physics and geography, $30$ students are taking physics and english; and $22$ students are taking geography and English. However, $7$ students are taking only physics, $10$ students are taking only geography and $5$ students are taking only english. How many of these students are taking physics and english, but not geography?

Let

$A$ represents numbers of students taking physics,

$B$ ,

$\rightarrow$ students taking Geography

$C\rightarrow$ students taking English

Numbers of student

$= 75$

$n(A\cap B\cap C)$

$= 16$

$n(A\cap B)$

$= 24$

$n(A\cap C)$

$= 30$

$n(B\cap C)$

$= 22$

number of students who take physics and mathematics only

$=$ number of student who take physics and English

$-$ number of students who take a three subjects

$=$

$n(A\cap C)-n(A\cap B\cap C)$

$= 30 - 16$

$= 14$

At a certain conference of $100$ people there are $29$ Indian women and $23$ Indian men. Of these Indian people $4$ are doctors and $24$ are either men or doctors. There are no foreign doctors. How many foreigners are attending the conference? How many women doctors are attending the conference?

$\textbf{Step -1: Find the total number of Indians at the conference.}$

$\text{Total Indians}=\text{Indian women}+\text{Indian men.}$

$=29+23$

$=52$

$\textbf{Step -2: Find the number of foreigners at the conference.}$

$\text{As, total number of people at the conference}=100.$

$\therefore\text{Number of foreigners}=\text{Total people}-\text{Number of total Indians.}$

$=100-52$

$=48$

$\textbf{Step -3: Find the number of women doctors at the conference.}$

$\text{Number of doctors}=4$

$\text{Given, either Indian men or doctor, i.e. }(\text{Indian men}\cup\text{ Doctor})=24$

$\mathbf{\because n(A\cup B)=n(A)+n{B}-n(A\cap B)}$

$\therefore n(\text{Indian men}\cup\text{Doctor})=n(\text{Indian men})+n(\text{Doctor})-n(\text{Indian men}\cap\text{Doctor})$

$\Rightarrow 24=23+4-n(\text{Indian men}\cap\text{Doctor})$

$\Rightarrow n(\text{Indian men}\cap\text{Doctor})=3$

$\text{As, there are no foreign doctors so, all doctors are Indian.}$

$\Rightarrow \text{Men doctors}=3$

$\text{Women doctors}=\text{Total doctors}-\text{Men doctors}$

$=4-3$

$=1$

$\textbf{Final Answer: (i) There are 48 foreigners and 1 woman doctor attending the conference.}$

In a combined test in Maths and Chemistry; $84\%$ candidates passed in Maths; $76\%$ in Chemistry and $8\%$ failed in both. If $340$ candidates passed in the test, then how many appeared ?

$\%$ of students failed in Maths

$( n(A) )= 100 - 84 = 16$

$\%$

$\%$ of students failed in Chemistry

$( n(B) ) = 100 - 76 = 24$

$\%$

$\%$ of students failed in both

$( n(C) ) = 8 \%$
Total percentage of candidates which failed can be found using the formula of union of sets.

$n(AUB) = n(A) + n (B) - n(C)$
So,

$n(AUB) = 16 + 24 - 8 = 32$

$\%$
Since

$32$

$\%$ of the total candidates failed, then

$100 - 32 = 68$ % of candidates passed which is

$= 340$

$\Rightarrow \dfrac {68}{100} \times$ total candidates

$= 340$
Therefore, total candidates are

$500$ .

In a city three daily newspaper x, y and z are published. 40% of the people of the city read x, 50% of the people read y, 30% read z, 20% read both x and y, 15% read both x and z, 10% read y and z and 24% read all of the three papers.
Can this be true ? Yes=9 No=6

Let the number of people in the city be

$100x$ (where

$x\neq 0$ , a constant).

Furher assume that

$A,B$ and

$C$ be respectively the sets of people of that city who reads the newspaper

$x,y$ and

$z$ .

And let

$S$ denote the set of people in that city.

Then we have,

$\displaystyle \left| C\cap A \right| =10x$

Assume

$\displaystyle \left| \left( A\cup B\cap C \right) ' \right| =0$

(i.e., everyone read atleast one paper)

And assume that

$\displaystyle \left| A\cap B\cap C \right| =P$

(i.e., number of people who read all the three newspapers is

$P$ ).

Now, we know that

Hence we have,

$\displaystyle \Rightarrow 40x+50x+30x-20x-15x-10x+P=100x$

$\displaystyle \Rightarrow P=25x$

i.e., people who read all newspaper is

$25%$ %.

In a combined test in English and Physics; $36\%$ candidates failed in English; $28\%$ failed in Physics and $12\%$ in both; find the total number of candidates appeared, if $208$ candidates have failed.

${\textbf{Step -1: Define the values to a variable.}}$

${\text{No}}{\text{. of students failed in English (n(A)) = 36% }}$

${\text{No}}{\text{. of students failed in Physics (n(B)) = 28% }}$

${\text{No}}{\text{. of students failed in Both (n(C)) = 12% }}$

$\textbf{Step -2: Calculate the number of candidate using formula.}$

${\text{Total percentage of candidates which failed can be found using the formula of union of sets.}}$

${\text{n(AUB) = n(A) + n(B) - n(C)}}$

${\text{So, n(AUB) = 36 + 28 - 12 = 52% }}$

${\text{Since 52% of the total candidates failed, which is = 208}}$

${\text{So,}}$

$\Rightarrow \dfrac{{52}}{{100}} \times {\text{total candidate = 208}}$

$\Rightarrow {\text{total candidate = }}\dfrac{{208 \times 100}}{{52}}$

$\Rightarrow {\text{total candidate = 400}}$

${\textbf{Hence, total number of candidate who appeared is 400}}{\text{.}}$

If $\displaystyle f_{1}\left ( x \right )$ and $f_{_{2}}\left ( x \right )$ are defined on domains $\displaystyle D_{1}$ and $D_{2}$ respectively, then $\displaystyle f_{1}\left ( x \right )+f_{2}\left ( x \right )$ is defined on $\displaystyle D_{1}\cup D_{2}.$ True or false?
If true enter 1 else 0.

$f_{1}(x)$ is defined in the domain of

$x\epsilon D_{1}$ and

$f_{2}(x)$ is defined in the domain of

$x\epsilon D_{2}$ then

$f_{1}(x)+f_{2}(x)$ is defined in the domain

$x\epsilon D_{1}\cap D_{2}$ .
Consider the possibility that it is defined in the domain

$D_{1}\cup D_{2}$
Then for any element in

$D_{1}$ ,

$f_{2}(x)$ won't be defined. Similarly for any element in

$D_{2}$ ,

$f_{1}(x)$ won't be defined. Hence the above statement is false.
Thus the correct answer is

$D_{1}\cap D_{2}$

Find total number of subsets of B = {a, b, c}

Any set

$B$ is a subset of a set

$A$ if

$A$ contains all the elements of

$B$ .

Hence, all the Subsets of

$B$ are

$\phi, \{a\} , \{b\}, \{c\}, \{a, b\} , \{a,c\}, \{ b, c \}, \{a, b, c \}$ .

Given $\displaystyle A=\left \{ 1, 2, 3 \right \}, B=\left \{ 3, 4 \right \},C=\left \{ 4, 5, 6 \right \}$ .Then find $\displaystyle \left ( A\times B \right )\cap \left ( B\times C \right )$

Given,

$\displaystyle A=\left \{ 1, 2, 3 \right \}, B=\left \{ 3, 4 \right \},C=\left \{ 4, 5, 6 \right \}$

$\displaystyle \Rightarrow A\cup \left ( B\cup C \right )=\left \{ 1, 2, 3, 4, 5, 6 \right \}$

Now

$\displaystyle A\times B=\left \{ \left ( 1, 3 \right ),\left ( 1, 4 \right ),\left ( 2,3 \right ),\left ( 2, 4 \right ),\left ( 3, 3 \right ),\left ( 3, 4 \right ) \right \}$

$\displaystyle B\times C=\left \{ \left ( 3, 4 \right ), \left ( 3, 5 \right ), \left ( 3, 6 \right ),\left ( 4, 4 \right), \left ( 4, 5 \right ),\left ( 4, 6 \right ) \right \}$

Hence

$\displaystyle \left ( A\times B \right )\cap \left ( B\times C \right )=\left \{ \left ( 3, 4 \right ) \right \}.$

Find total number of subsets of {p : p is a letter in the word 'poor'}

Set

$=$ {

$p, o ,r$ }

Any set

$B$ is a subset of a set

$A$ if

$A$ contains all the elements of

$B$ .
Hence, all the Subsets of the given set are

$\phi, \{ p\} , \{o\}, \{r\}, \{p, o\} , \{p, r\}, \{o, r\}, \{p, o, r\}$

If $A = \{$ $x$ : $x \,\,\in\,\, W$ , $3 \leq x < 6$ $\}$ , $B = \{3, 5, 7\}$ and $C = \{2, 4\}$ ; find :
$B - C$

$A=\{x:x \in W, 3\leqslant x <6\}$

$A=\{3,4,5\}$

$B=\{3,5,7\}$

$C=\{2,4\}$

$B-C =\{3,5,7\}-\{2,4\}$

$=\{3,5,7\}$

If $A = \{x : x \,\,\epsilon\,\, W, 3 \leq x < 6\}, B = \{3, 5, 7\}$ and $C = \{2, 4\}$ ;

find : $n(A - B)$ .

Given :

$A = \{x : x \,\,\epsilon\,\, W, 3 \leq x < 6\}, B = \{3, 5, 7\}$ and

$C = \{2, 4\}$

$\Rightarrow A = \{3,4,5\}$

Now,

$A - B$ means removing the elements of

$B$ which are in

$A$

$\therefore A - B =\{ 4\}$ as

$3, 5$ are common elements of

$A$ and

$B$ .

Hence,

$n(A-B)=1$ .

Find the total number of subsets of each of the following set:
$C=\{x \mid x \in W, x \leq 2\}$

Given :

$C=\{x:x\in W, x\le 2\}$

$C =\{ 0,1,2 \}$
Any set

$B$ is a subset of a set

$A$ if

$A$ contains all the elements of

$B$ .
So, all the Subsets of

$C$ are

$\phi, \{0\} , \{1\}, \{2\}, \{0, 1 \} , \{0, 2 \}, \{ 1,2 \}, \{0, 1,2 \}$ .

Hence, the total number of subsets of

$C$ is

$8$ .

If $C$ is the set of letters in the word 'cooler' find number of its subsets

Set

$C =\{ c,o,l,e,r\}$
Elements in the set are unique, so we do not repeat the letters.

Thus,

$o$ is written only once.
Now,

$n(C) = 5$ as there are

$5$ elements in the set.

We can form

${2}^{n}$ subsets of a set containing

$n$ elements.
Hence,

${2}^{5} = 32$ subsets can be formed for set

$C$ .

Find total number of subsets of A = {5, 7}

Subsets of a set can be formed by choosing no elements to one or more elements from the set.

If a set contains

$‘n’$ elements, then the number of proper subsets of the set is

$2^n - 1.$

$A = \{p, q\}$ the proper subsets of

$A$ are

$[\{ \}, \{p\}, \{q\}]$

⇒ Number of proper subsets of A are

$3 = 2^2 - 1 = 4 - 1$

Hence, all the Subsets of

$A = \phi, \{5\}, \{7\}, \{5,7\}$ .

If $T = \{ \text {x : x is a letter in the word 'TEETH'} \}$ , find all its subsets.

Set

$T =$ {

$T, E, H$ }

We can form subsets of a set by taking no element or one or more elements from it.
So, the subsets of set

$T$ are

$\phi, \{t\}, \{e \}, \{h \}, \{t, e\} , \{ t, h\} , \{e,h\}, \{ t,e,h\}$ .

Given, $A = \{$ Quadrilaterals $\}$ , $B = \{$ Rectangles $\}$ , $C = \{$ Squares $\}$ , $D = \{$ Rhombuses $\}$ . State whether the following statement is correct or incorrect. Give reasons.
$D$ $\subset$ $A$

$A = \{$ Quadrilaterals

$\}$

$B = \{$ Rectangles

$\}$

$C = \{$ Squares

$\}$

$D = \{$ Rhombuses

$\}$

Yes,

$D\subset A$ [

$\because$ Rhombuses are Quadrilaterals but all quadrilaterals are not Rhombuses..]

If $P =$ {factors of 36} and $Q =$ {factors of 48}, find $Q - P$

Factors of

$36$ are

$1,2,3,4,6,9,12,18, 36$
Factors of

$48$ are

$1,2,3,4,6,8, 12,16, 24, 48$

So, P

$=\{1,2,3,4,6,9,12,18, 36 \}$

And Q

$= \{ 1,2,3,4,6,8, 12,16, 24, 48 \}$

$Q - P$ will have the elements of

$Q$ which are not in

$PB$

So,

$Q - P = \{8,16,24,48 \}$ .

An investigator interviewed 100 students to determine their preferences for the three drinks milk coffee and tea. He reported the following, 10 students had all the three drinks, 20 had milk and coffee, 30 had coffee and tea , 25 had milk and tea, 12 had milk only. 5 had coffee only, 8 had tea only. The number of students that did not take any of the three drinks is.

According to given information, the venn diagram is below.
Now,

$n(M\cup C\cup T)=12+5+8+10+15+20+10=80$

Now, number of people who did not took any of the drinks is

$n(M'\cap C' \cap T')= n(M\cup C \cup T)'$

$\Rightarrow n(M\cup C \cup T)'=N-n(M\cup C \cup T)$

$\Rightarrow n(M\cup C\cup T)'=100-80=20$

Given A = { $x$ : $x \,\,\in\,\, N$ and $3 < x \leq 6$ } and B = { $x$ : $x \,\,\in\,\, W$ and $x < 4$ }, then find : B - A.

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

$x$ is greater than

$3$ but less than or equal to

$6$

And B

$= \{ 0,1,2,3 \}$ as

$x$ is a whole number less than

$4$

$B - A$ will have the elements of

$B$ which are not in

$A$

So,

$B - A =\{ 0,1,2,3\}$ .

Let $\displaystyle A=\left \{ \theta :2\cos ^{2}+\theta +\sin \theta \leq 2 \right \}$ and $\displaystyle B= \left \{ \theta :\pi /2\leq \theta \leq 3\pi /2 \right \}.$ Then $\displaystyle A\cap B=\left \{ \theta :\pi /2\leq \theta \leq 5\pi /6 or \pi \leq \theta \leq 3\pi /4 \right \}$
If true enter 1 else 0

$2\cos^{2}\theta+\sin \theta\leq 2$

$2\sin^{2}\theta-\sin \theta\geq 0$

$\sin \theta \in [-1,0]\cup [\dfrac{1}{2},1]$

$\theta\in [\dfrac{3\pi}{2},2\pi]\cup [\dfrac{\pi}{6},\dfrac{5\pi}{6}]$

$A\cap B\in [\dfrac{\pi}{2},\dfrac{5\pi}{6}]$

The correct answer is

$0$

Given, $A = \{$ Triangles $\}, B = \{$ Isosceles triangles $\},$
$C =\{$ Equilateral triangles $\}.$ State whether the following statements are correct or incorrect. Give reasons.
A $\subset$ B

$A = \{$ Triangles

$\}$

$B = \{$ Isosceles triangle

$\}$

$C = \{$ Equilateral triangle

$\}$

No,

$A$

$\not\subset$

$B$ but

$B$

$\subset$ A

[

$\because$ all triangle contain every isosceles triangle.]

Given, $A = \{$ Triangles $\}, B = \{$ Isosceles triangles $\},$
$C =\{$ Equilateral triangles $\}.$ State whether the following statement are correct or incorrect. Give reasons.
$C$ $\subset$ $A$

$A = \{$ Triangle

$\}$

$B = \{$ Isosceles triangle

$\}$

$C = \{$ Equilateral

$\triangle$

$\}$

Yes

$C$

$\subset$

$A$ [

$\because$ Equilateral

$\triangle s$ come under

$\triangle$ Family. Again there are some triangles belonging to the triangle family which are not Equilateral so the subset is proper.]

In a group of 1000 people , there are 750 who can speak Hindi and 400 who can speak Bengali. How many can speak Bengali ?

${\textbf{Step-1: Finding total number of People who speak both Hindi and Bengali.}}$

${\text{Let H and B be the who can speck Hindi and Bengali.}}$

${\text{So,}}$

$n(H \cup B) = 1000, n(H) = 750, n(B) = 400$

$n(H \cup B) = n(H) + n(B) - n(H \cap B)$

$\Rightarrow 1000 = 750 + 400 - n(H \cap B)$

$\Rightarrow n(H \cap B) = 1150 - 1000$

$\Rightarrow n(H \cap B) = 150$

${\textbf{Step-2: Finding only Bengali speakers.}}$

${\text{Now we have to find the number of people who speak only Bengali.}}$

${\text{So, for that we will subtract the number of people who can speak both the languages}}$

${\text{from the number of people who can speak Bengali.}}$

${\text{So, we get,}}$

$n(\text{only Bengali speakers})$

$= n(B) - n(H \cap B)$

$= 400 - 150$

$= 250$

${\textbf{Hence,250 people only speak Bengali.}}$

If $A = (6, 7, 8, 9), B = (4, 6, 8, 10)$ and $C = \{x : x \,\,\epsilon\,\,N : 2 < x \leq 7\}$ ; find : $B - (A \cap C)$
The sum of the elements in the above set is?

$= \{3,4,5,6,7 \}$ as

$x$ is greater than

$2$ and less than equal to

$7$

So,

$A \cap C =\{ 6,7 \}$

$B - (A \cap C)$ will have the elements of

$B$ which are not in

$(A \cap C)$
So,

$B - (A \cap C) = \{ 4,8,10 \}$

Thus the sum of the elements in the above set :

$22$

If total no. of elements in set A is $a$ and in set B is $b$ , then $\cfrac{a+b}{4} =$

Whole numbers which are Square numbers less than or equal to

$16$ are

$0, 1, 4, 9, 16$
So, Set

$A =$ {

$0, 1, 4, 9, 16$ }

$\Rightarrow a = 5$
And

$\dfrac {x}{3} - 2 < 3$

$\Rightarrow \dfrac {x}{3} < 5$

$\Rightarrow x < 15$
Whole numbers which are less than

$15$ are

$0, 1, 2,3,4,5,6.... 14$ .
So, Set

$B =$ {

$0, 1, 2,3,4,5,6.... 14$ }

$\Rightarrow b = 15$

If universal set $\xi = \{a, b, c, d, e, f, g, h\}, A = \{b, c, d, e, f\}, B =\{a, b, c, g, h\}$ and $C = \{c, d, e, f, g\}$ find : $C - (B \cap A)$
Thus find the number of elements in the above set.

Intersection of two sets has the common elements of both the sets.

So,

$B \cap A =$ {

$b,c$ }

$C - (B \cap A)$ will have elements of C which are not in

$(B \cap A)$

So,

$C - (B \cap A) =$ {

$d,e,f,g$ }

There are thus

$4$ elements in the above set.

If $A = (6, 7, 8, 9), B = (4, 6, 8, 10)$ and $C = \{x : x \,\,\epsilon\,\,N : 2 < x \leq 7\}$ ; find : $n(B - (A - C))$

$A=\left\{ 6,7,8,9 \right\}$

$B=\left\{ 4,6,8,10 \right\}$

$C=\left\{ x:x\in N\quad and\quad 2<x\le 7 \right\}$

$\therefore C=\left\{ 3,4,5,6,7 \right\}$

$A-C=$ Contains elements present only in

$A$ and NOT in

$C$

$A-C=\left\{ 6,7,8,9 \right\} -\left\{ 3,4,5,6,7 \right\}$

$A-C=\left\{ 8,9 \right\}$

$B-\left\{ A-C \right\} =\left\{ 4,6,8,10 \right\} -\left\{ 8,9 \right\}$

$=\left\{ 4,6,10 \right\}$

Hence,

$n(B-(A-C))=3$ .

If $A = (6, 7, 8, 9), B = (4, 6, 8, 10)$ and $C = \{x : x \,\,\epsilon\,\,N : 2 < x \leq 7\}$ ; find : $A - (B \cup C)$

$= \{3,4,5,6,7 \}$ as

$x$ is greater than

$2$ and less than equal to

$7$

$A - C$ will have the elements of

$A$ which are not in

$C$

Union of two sets has all the elements of both the sets

So,

$B \cup C = \{ 3,4,5,6,7, 8,10 \}$

$A - (B \cup C)$ will have the elements of

$A$ which are not in

$(B \cup C)$
So,

$A - (B \cup C) = \{ 9 \}$

$n(A$ $'$ ).

Given,

$n($

$\xi$

$) = 30$

$n(A) = 22$

$n(B) = 15$

$n(A$

$\cup$

$B) = 25$

$n($

${ A'}$

$) =$

$n(\xi$ )

$- n(A)$

$= 30 - 22$

$= 8$

$n(A$ $\cup$ $B)$

Given,

$n(A)=40, n(B)=27$ and

$n(A\cap B)=15$

Using the formula,

$n (A \cup B ) = n(A) + n(B) - n(A \cap B)$
So,

$n (A \cup B ) = 40 + 27 - 15 =52$

$n(A) + n(B) = n(A$ $\cup$ $B) + n(A$ $\cap$ $B$ ) .If true enter 1 else 0.

From the given sets, we have

$\varepsilon=[26,27,28,29,30,31,32,33,34,35,35,37,38,39,40,41,42,43,44]$

$A=[26,28,30,32,24,26,38,40,42,44]$

$B=[27,30,33,36,39,42]$

$n(A)=10$ and

$n(B)=6$

$\therefore n(A)+n(B)=10+6=16$

We know that

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$\Rightarrow n(A\cup B)+n(A\cap B)=n(A)+n(B)$

$\Rightarrow n(A\cup B)+n(A\cap B)=16$

Therefore,

$n(A)+n(B)=n(A\cup B)+n(A\cap B)$

$n(A$ $\cap$ $B$ $^{'}$ $)$

Using the formula,

$n (A \cup B ) = n(A) + n(B) - n(A \cap B)$

So,

$n (A \cap B ) = n(A) + n(B) - n (A \cup B ) = 22 + 15 - 25 = 12$

And

$n ({A \cap B} ^{'} ) = n (\xi) - n (A \cap B ) = 30 - 12 = 18$

$n(B$ $'$ $)$ .

$n(\xi ) =30$

$n(A)=22$

$n(B)=15$

$n(A\cup B) =25$

$n(B')=n(\xi)-n(B)$

$=30-15$

$=15$

$n(B')=15$

A and B are two overlapping sets such that n(A $\cap$ B) = x + 4, n(A - B) = 4x - 8 and n(B) = 3x +Find x, if n(A $\cup$ B) = 70.

Given,

$n(A\cap B)=x+4, n(A-B)=4x-8, n(B)=3x+8$

Also given

$n(A\cup B)=70$

We know that

$n(A-B) = n(A \cup B) - n(B)$
Substituting the given values, we get

$4x-8 = 70 - 3x - 8$

$\Rightarrow 7x = 70$

$\Rightarrow x = 10$

$n[(A$ $\cup$ $B) - A]$

Given,

$n(A)=40, n(B)=27$ and

$n(A\cup B)=15$

From the figure,

$(A \cup B) - A$ is the region of " ONLY B "
Hence,

$n ((A \cup B) - A) = n (only B)$

$= n(B) - n(A \cap B)$

$= 27 - 15$

$= 12$

If $n(A) = 40, n(B) = 27$ and $n(A$ $\cap$ $B) = 15$ ; find :
$n(\mbox{only}\ A)$

$n(A) = 40$

$n(B) = 27$

$n(A\cap B)=15$ .

$n(\mbox{only}\ A) = n(A) - n(A$

$\cap$

$B)$

$= 40 - 15$

$= 25$

$n(B - A)$

Given,

$n(A)=40, n(B=27, n(A \cap B=15)$

Using the formula,

$n (B -A ) = n(B) - n(A \cap B)$
So,

$n (B -A ) = 27 - 15 =12$

$n(A - B)$ = $n(A'$ $\cap$ $B)$ .If true enter 1 else 0.

From the question, we have

$\varepsilon=[26,27,28,29,30,31,32,33,34,35,35,37,38,39,40,41,42,43,4]$

$A=[26,28,30,32,34,36,38,40,42,44]$

$B=[27,30,33,36,39,42]$

$n(A-B)=[26,28,30,32,24,26,28,40,42,44]-[27,30,33,36,39,42]$

$= [26,28,32,34,38,40,44]$

Therefore,

$n(A-B)=7$

$A'=[27,29,31,33,35,37,39,41,43]$

$n(A'\cap B)=[27,29,31,33,35,37,39,41,43]\cap [27,30,33,36,39,42]$

$\Rightarrow [27,33,39]$

$\Rightarrow n(A' \cap B)=3$

Thus

$n(A-B)\neq n(A'\cap B)$

In a community of $175$ persons, $40$ read TOI, $50$ read the Samachar Patrika and $100$ do not read any. How many persons read both the papers.

Let

$x$ no. of people read both the newspapers. Since

$100$ do not read any, no. of people reading both or either of newspapers

$=175-100=75$

$\displaystyle \therefore \left ( 40-x \right )+x+\left ( 50-x \right )=75\Rightarrow x=15$

Let $n$ be a positive integer. Call a non-empty subset $S$ of $\left \{1, 2, . . . , n\right \}$ good, if the arithmetic mean of the elements of $S$ , is also an integer. Further let $t_n$ denote the number of good subsets of $\left \{1, 2, . . . , n\right \}$ . Prove that $t_n$ and $n$ are both odd or both even.

We show that

$T_n,n$ is even.
Note that the subsets

$\left \{1\right \}, \left \{2\right \}, , \left \{n\right \}$ are good.
Among the other good subsets, let

$A$ be the collection of subsets with an integer average which belongs to the subset, and let

$B$ be the collection of subsets with an integer average which is not a member of the subset.
Then there is a bijection between

$A$ and

$B$ , because removing the average takes a member of

$A$ to a member of

$B$ ; and including the average in a member of

$B$ takes it to its inverse.
So

$T_n n = |A| + |B|$ is even.

Alternate solution:
Let

$S = \left \{1, 2, . . . , n\right \}$ .
For a subset

$\overline A$ of

$S$ , let

$A =\left \{n + 1 a|a \epsilon A\right \}$ .
We call a subset

$A$ symmetric if

$\overline A = A$ . Note that the arithmetic mean of a symmetric subset is

$(n+1)/2$ .
Therefore, if

$n$ is even, then there are no symmetric good subsets, while if

$n$ is odd then every symmetric subset is good.
If

$A$ is a proper good subset of

$S$ , then so is

$\overline A$ .
Therefore, all the good subsets that are not symmetric can be paired. If

$n$ is even then this proves that tn is even.
If

$n$ is odd, we have to show that there are odd number of symmetric subsets. For this, we note that a symmetric subset contains the element

$(n + 1)/2$ if and only if it has odd number of elements.
Therefore, for any natural number k, the number of symmetric subsets of size

$2k$ equals the number of symmetric subsets of size

$2k +1$ .
The result now follows since there is exactly one symmetric subset with only one element.

Find $n({A\cap C}^c)$ .

Let

$U=\left \{ 1,2,3,4,5,6,7,8,9 \right \}$

$A=\left \{ 1,2,3,4 \right \}$

$B=\left \{ 2,4,6,8 \right \}$

$C=\left \{ 3,4,5,6 \right \}$

solution

$C=\left \{ 3,4,5,6 \right \}$

$\therefore C^{c}=\left \{ 1,2,7,8,9 \right \}$

$A=\left \{ 1,2,3,4 \right \}$

$\therefore A\cap C^{c}=\left \{ 1,2 \right \}$

$n(A\cap C^{c})=2$

No. of element

In a group of $50$ persons, $14$ drink tea but not coffee and $30$ drink tea, then find how many drink tea and coffee both?

T : People drinking Tea

C : People drinking coffee

$\displaystyle n(T)=n(T-C)+n(T\cap C)\Rightarrow 30=14+n(T\cap C)\Rightarrow n(T\cap C)=16$

Let $n$ be a natural number and $X =\left \{1, 2,......, n\right \}$ . For subsets $A$ and $B$ of $X,$ we denote $A\Delta B$ to be the set of all those elements of $X$ which belong to exactly one of $A$ and $B.$ Let $F$ be a collection of subsets of $X$ such that for any two distinct elements $A$ and $B$ in $F$ , the set $A\Delta B$ has at least two elements. Show that $F$ has at most $2^{n-1}$ elements. Find all such collections $F$ with $2^{n-1}$ elements.

For each subset

$A$ of

$\left \{1, 2,....., n-1\right \}$ , we pair it with

$A\cup \left \{n\right \}$ .
Note that for any such pair

$(A, B)$ not both

$A$ and

$B$ can be in

$F$ .
Since there are

$2^{n-1}$ such pairs it follows that F can have at most

$2^{n-1}$ elements.
By induction

$|F| = 2^{n-1}$ , then F contains either all the subsets with odd number of elements or all the subsets with even number of elements.
Suppose that the result is true for

$n = m - 1$ .
Now consider

$n = m$ .
Let

$F_1$ be the set of those elements in F which contain

$m$ and

$F_2$ be the set of those elements which do not contain

$m$ .
By induction,

$F_2$ can have at most

$2^{m-2}$ elements.
Further, for each element of

$A$

$F_1$ we consider

$A/\left \{m\right \}$ .
This new collection also satisfies the required property,
So

$F_1$ has at most

$2^{m-2}$ elements.
Thus, if

$|F| = 2^{m-1}$ then it follows that

$|F_1| = |F_2| = 2^{m-2}$ .
Further, by induction hypothesis,

$F_2$ contains all those subsets of

$\left \{1, 2, ....., m - 1\right \}$ with (say) even number of elements.
It then follows that

$F_1$ contains all those subsets of

$\left \{1, 2, ....., m\right \}$ which contain the element

$m$ and which contains an even number of elements.
This proves that

$F$ contains either all the subsets with odd number of elements or all the subsets by even number of elements.

There are 40 students in a chemistry class and 60 students in a physics class. Find the number of students which are either in physics class or chemistry class in the following case:
The two classes meet at different hours and 20 students are enrolled in both the subjects

Let A be the set of students in chemistry class and B be the set of students in physics class. It is given that

$n (A) = 40$ and

$n(B) = 60$ .
If two classes meet at different timings then there can be some student sitting in both the classes.
Therefore,

$n (A \cap B) = 20$

$n(A\cup B) =n(A) +n(B) -n(A \cap B)$

$=40+60-20=80$ .

There are 40 students in a chemistry class and 60 students in a physics class. Find the number of students which are either in physics class or chemistry class in the following case:
The two classes meet at the same hour

Let A be the set of students in chemistry class and B be the set of students in physics class. It is given that

$n (A) = 40$ and

$n(B) = 60$ .
If two classes meet at the same hour, then there will not be a common student sitting in both the classes.
Therefore,

$n (A \cap B) = 0$

$\therefore n(A\cup B)=n(A) +n(B)-n(A\cap B)$

$\Rightarrow n(A\cup B) =40+60-0 =100$

Let A and B be sets, If $A\cap X=B\cap X= \phi$ and $A\cup X= B \cup X$ for some set X, prove that $A= B$ .

$A\cup X=B\cup X$ for some set X

$\Rightarrow A\cap (A\cup X)=A\cap (B\cup X)$

$\Rightarrow (A\cap A)\cup (A\cap X)=A\cap (B\cup X)$

$\Rightarrow A=(A\cap B)\cup (A\cap X)$

$\Rightarrow A=(A\cap B)\cup \phi$ [

$\because A\cap X=\phi$ (given)]

$\Rightarrow A=A\cap B$

$\Rightarrow A\subset B$ .......(i)
Again,

$A\cup X=B\cup X$

$\Rightarrow B\cap (A\cup X)=B\cap (B\cup X)$

$\Rightarrow (B\cap A)\cup (B\cap X)=(B\cap B)\cup (B\cap X)$

$\Rightarrow (B\cap A)\cup (B\cap X)=B$

$\Rightarrow (B\cap A)\cup \phi=B$ [

$\because B\cap X=\phi$ (given)]

$\Rightarrow B\cap A=B$

$\Rightarrow B\subset A$ .........(ii)
From (i) and (ii)

$A=B$ .

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) $\left \{ x : x \text { is a natural numbers, $$x < 5$$ and $$x > 7}\right \}$
(iv) $\left \{ y : y \text { is a point common to any two parallel lines}\right \}$

(i) A set of odd natural numbers divisible by

$2$ is a null set because no odd number is divisible by

$2$ .

(ii)

$A=\{2\}$ .
A set of even prime numbers is not a null set because

$2$ is an even prime number.

(iii)

$\left \{x:x \text { is a natural number}, x < 5 \ \text{and} x > 7\right \}$ is a null set because a number cannot be simultaneously less than

$5$ and greater than

$7$ .

(iv)

$\left \{y:y\text { is a point common to any two parallel lines}\right \}$ is a null set because parallel lines do not intersect. Hence, they have no common point.

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

(i) The set of months of a year is a finite set because it has

$12$ elements.

(ii)

$\left \{1,2,3....\right \}$ is an infinite set as it has infinite number of natural numbers.

(iii)

$\left \{1,2,3,...99,100\right \}$ 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.

Make correct statements by filling in the symbols $\subset$ or $\not\subset$ in blank spaces:
(i) $\left \{2, 3, 4\right \} ...... \left \{1, 2, 3, 4,5\right \}$
(ii) $\left \{ a, b, c \right \} . . . \left \{ b, c, d \right \}$
(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}

(i)

$\left \{2,3,4\right \}\subset \left \{1,2,3,4,5\right \}$

(ii)

$\left \{a,b,c\right \}\not\subset \left \{b,c,d\right \}$

(iii)

$\{x:x \text{ is a student of class XI of your school} \} \subset \{x:x \text { is student of your school}\}$

(iv)

$\{x:x \text{ is a circle in the plane}\}\not\subset \{x:x \text{ is a circle in the same plane with radius 1 unit} \}$

(v)

$\{x:x \text{ is a triangle in a plane}\} \not\subset \{x:x \text{ is a rectangle in the plane}\}$

(vi)

$\{x:x \text{ is an equilateral triangle in a plane}\} \subset \{x:x \text{ is a triangle in the same plane}\}$

(vii)

$\{x:x \text { is an even natural number}\} \subset \{x:x \text{ is an integer}\}$

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'?

$U$ - Set of all triangles in a plane

$A$ - Set of triangles with at least one angle different from

$60^{0}$ i.e. the set cannot have equilateral triangles

$A'=U-A$
So,

$A'$ is the set of all equilateral triangles.

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

(i)

$A=\left \{x:x\text { is an odd natural number}\right \}=\left \{1,3,5,7,9 ..... \right \}$

(ii)

$B=\left \{x:x \text { is an integer}, -\frac {1}{2} < x < \frac {9}{2}\right \}$
Since,

$-\dfrac {1}{2}=-0.5$ and

$\dfrac {9}{2}=4.5$

$\therefore B=\left \{0,1,2,3,4\right \}$

(iii)

$C=\left \{x:x\text { is an integer}; x^2\leq 4\right \}$
It can be seen that

$(-1)^2=1\leq 4$

$; (-2)^2=4\leq 4$

$; (-3)^2=9 > 4$

$0^2=0\leq 4$

$1^2=1\leq 4$

$2^2=4\leq 4$

$3^2=9 > 4$

$\therefore C=\left \{-2, -1, 0, 1, 2\right \}$

(iv)

$D= \{x:x \text { is a letter in the word LOYAL}\} =\left \{L,O,Y,A\right \}$

(v)

$E=\left \{x:x\text { is a month of a year not having 31 days}\right \}$

$=\left \{\text {February, April, June, September, November}\right \}$

(vi)

$F=\left \{x:x \text { is a consonant in the English alphabet which precedes k}\right \}=\left \{b,c,d,f,g,h,j \right \}$

Which of the following are sets ? Justify your answer.
(i) The collection of all the 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

(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 identify a month that belongs to this collection.
Hence, this collection is a set.

(ii) The collection of ten most talented writers of India is not a well-defined collection because the criteria for determining a writer's talent may 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 that criteria for determining the dangerousness of an animal can vary from animal to animal.
Hence, this collection is not a set.

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 $A C$
(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 $x \notin B$ , then $x \notin A$

(i) False

Let

$A=\left \{1,2\right \}$ and

$B=\left \{1, \left \{1, 2\right \}, \left \{3\right \}\right \}$

$2\in \left \{1,2\right \}$ and

$\left \{1,2\right \}\in \left \{\left \{3\right \},1,\left \{1,2\right \}\right \}$

Now,

$\therefore A\in B$

However,

$2\not {\in}\left \{\left \{3\right \},1,\left \{1,2\right \}\right \}$

(ii) False.

$A\subset B$ ,

$B\in C$

Let

$A=\left \{2\right \}, B=\left \{0,2\right \}$ , and

$C=\left \{1, \left \{0, 2\right \},3\right \}$

However,

$A\not {\in}C$

(iii) True

Let

$A\subset B$ and

$B\subset C$ .

Let

$x\in A$

$\Rightarrow x\in B$

$[\because A\subset B]$

$\Rightarrow x\in C$

$[\because B\subset C]$

$\therefore A\subset C$

(iv) False

As,

$A\not {\subset}B$ and

$B\not {\subset}C$

Let

$A=\left \{1, 2\right \}, B=\left \{0,6,8\right \}$ , and

$C=\left \{0,1,2,6,9\right \}$

However,

$A\subset C$

(v) False

Let

$A=\left \{3,5,7\right \}$ and

$B=\left \{3,4,6\right \}$

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

$x\not {\in}B$

$\therefore x\not {\in}A$

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

We know that number of subsets of any set is

$2^n$ where

$n$ is the number of elements in that set.
Every non-empty set has two subsets

$\phi$ and the set itself.

(i) The subsets of

$\{a\}$ are

$\phi$ and

$\{a\}$ .

(ii) The subsets of

$\{a, b\}$ are

$\phi, \{a\}, \{b\}, and \{a, b\}$ .

(iii) The subsets of

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

$\phi, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{2, 3\}, \{1, 3\}, and \{1, 2, 3\}$

(iv) The only subset of

$\phi$ is

$\phi$ .

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)

(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 multiples 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 infinite set because infinite number of circles can pass through the origin.

In a town of $10,000$ families it was found that $40$ % families buy newspaper A, $20$ % families buy newspaper B and $10$ % families buy newspaper C. $5$ % families buy A and B, $3$ % buy B and C and $4$ % buy A and C. If $2$ % families buy all the three newspaper, the member of families which buy A only is ___.

Let

$n(A),n(B),n(C)$ be the families that buy newspaper

$A,B$ and

$C.$

$n(A)=40\%$ of

$10,000=\dfrac{40}{100}\times 10,000=4,000$

$n(B)=20\%$ of

$10,000=\dfrac{20}{100}\times 10,000=2,000$

$n(C)=10\%$ of

$10,000=\dfrac{10}{100}\times 10,000=1,000$

$n(A\cap B)=5\%$ of

$10,000=\dfrac{5}{100}\times 10,000=500$

$n(B\cap C)=3\%$ of

$10,000=\dfrac{3}{100}\times 10,000=300$

$n(C\cap A)=4\%$ of

$10,000=\dfrac{4}{100}\times 10,000=400$

$n(A\cap B\cap C)=2\%$ of

$10,000=\dfrac{2}{100}\times 10,000=200$

Number of families buys only newspaper

$A$

$=n(A)-[n(A\cap B)+n(A\cap C)]+n(A\cap B\cap C)]$

$=4000-[500+400]+200$

$=4000-900+200$

$=3300$

If $A, B$ and $C$ are any three sets, then $A\cap (B\triangle C) = (A\cap B)\Delta (A\cap C)$ .If true enter 1 else 0.

$A\cap (B\Delta C) = A\cap \left [ (B\cup C)-(B\cap C) \right ]$

$\Rightarrow A\cap (B\cup C)-A\cap (B\cap C)$

$\Rightarrow (A\cap B)\cup (A\cap C)-(A\cap B)\cap (A\cap C)$

(using distributive low)

$\Rightarrow \boxed {(A\cap B)\Delta (A\cap C)}$

1125767_514860_ans_118b7814b41c466e90efc8e05f1e6dff.jpg

Write the following sets in Set - Builder form
(i) The set of all positive even numbers
(ii) The set of all whole numbers less than 20
(iii) The set of all positive integers which are multiples of 3
(iv) The set of all odd natural numbers less than 15
(v) The set of all letters in the word 'computer'

(i) {

$x:x$ is a positive even number }
(ii) {

$x:x$ is a whole number and

$x< 20$ }
(iii) {

$x:x$ is a positive integer and multiple of 3 }
(iv) {

$x:x$ is an odd natural number and

$x< 15$ }
(v) {

$x:x$ is a letter in the word 'computer' }

List all of the subsets of the set $\left\{ a,b,c,d \right\}$

The list of all subsets of ${a,b,c,d}$ is $\phi$ ,{ $a$ },{ $b$ },{ $c$ },{ $d$ },{ $a,b$ },{ $a,c$ },{ $a,d$ },{ $b,c$ },{ $b,d$ },{ $c,d$ },{ $a,b,c$ },{ $a,b,d$ },{ $a,c,d$ },{ $b,c,d$ },{ $a,b,c,d$ }

$A$ and $B$ are two sets such that $n\left ( A-B \right )=32+x, n\left ( B-A \right )=5x$ and $n\left (A\cap B \right )=x$ Illustrate the information by means of a Venn diagram. Given that $n\left ( A \right )=n\left ( B \right )$ .
Calculate (i) The value of $x$ (ii) $n\left ( A\cup B \right )$ .

$n(A-B)=32+x$

$n(B-A)=5x$

$n(A\cap B)=x$

$n(A)=n(B)$

$(1) n(A)=n(B)$

$32+x+x=5x+x$

$32+2x=6x$

$6x-2x=32$

$4x=32$

$x=8$

$(2)n(A)=32+x+x$

$=32+2x$

$=32+2\times 8$

$x=48$

$n(B) = 5x+x$

$=6x$

$=6\times 8$

$=48$

$n(A\cup B) = n(A)+n(B)-n(A\cap B)$

$=48+48-8$

$=96-8$

$=88$

so, the value of

$x=8$ and

$n(A\cup B) =88$

Let $A=$ { $x:x$ is a natural number $< 11$ }
$B=$ { $x:x$ is an even number and $1< x< 21$ }
$C=$ { $x:x$ is an integer and $15\leq x\leq 25$ }
(i) List the element of $A, B, C$
(ii) Find $n(A), n(B), n(C)$ .
(iii) State whether the following are True (T) or False (F)
(a) $7\in B$
(b) $16\notin A$
(c) $\left \{ 15,20,25 \right \}\subset C$
(d) $\left \{ 10,12 \right \}\subset B$

(i)

$A=\left \{ 1,2,3,4,5,6,7,8,9,10 \right \}, B=\left \{ 2,4,6,8,10,12,14,16,18,20 \right \}, C=\left \{ 15,16,17,18,19,20,21,22,23,24,25 \right \}$
(ii)

$n(A)=10, n(B)=10, n(C)=11$
(iii) a) F b) T c) T d) T

Let A={-5, -3, -2, -1}, B={-2, -1, 0} and C={-6, -4, -2}. Find A ( B \ C) and (A \ B) \ C. What can we conclude about set difference operation?

$B$ \

$C=$ {

$-1,0$ }

$\Rightarrow A$ \

$(B$ \

$C)=$ {

$-5,-3,-2$ }

$A$ \

$B=$ {

$-5,-3$ }

$\Rightarrow (A$ \

$B)$ \

$C=$ {

$-5,-3$ }

We see that

$A$ \

$(B$ \

$C) \neq (A$ \

$B)$ \

$C$

Therefore the set difference operator is not associative.

For A={5,10,15,20}; B={6,10,12,18,24} and C={7,10,12,14,21,28}, verify whether $A$ \ $(B$ \ $C)$ $=(A$ \ $B)$ \ $C$ . Justify your answer

The given sets are

$A=\left\{ 5,10,15,20 \right\}$ ,

$B=\left\{ 6,10,12,18,24 \right\}$ and

$C=\left\{ 7,10,12,14,21,28 \right\}$

Let us first find

$A\setminus (B\setminus C)$ as follows:

$A\setminus (B\setminus C)=\left\{ 5,10,15,20 \right\} \setminus (\left\{ 6,10,12,18,24 \right\} \setminus \left\{ 7,10,12,14,21,28 \right\} )=\left\{ 5,10,15,20 \right\} \setminus \left\{ 6,18,24 \right\}$

$=\left\{ 5,10,15,20 \right\} ........(1)$

Now we find

$(A\setminus B)\setminus C$ as follows:

$(A\setminus B)\setminus C=(\left\{ 5,10,15,20 \right\} \setminus \left\{ 6,10,12,18,24 \right\} )\setminus \left\{ 7,10,12,14,21,28 \right\} =\left\{ 5,15,20 \right\} \setminus \left\{ 7,10,12,14,21,28 \right\}$

$=\left\{ 5,15,20 \right\} ........(2)$

Since equation 1 is not equal to equation 2,

Hence,

$A\setminus (B\setminus C)\neq (A\setminus B)\setminus C$

Find
i) $A\cap B$ if (i) $A=\left \{10,11,12,13\right \}, B= \left \{12,13,14,15 \right \}$
ii) $A= \left \{5,9,11\right \} ,B=\varnothing$

(i)

$A=\left \{10,11,12,13 \right \}$ and $B=\left \{12,13,14,15\right \}$ .

12 and 13

are common in both A and B.

$\therefore A\cap B = \left \{12,13\right \}$

(ii) $A=\left \{5,9,11 \right \}$ and $B=\varnothing$ .

There is no element in common

and hence $A\cap B=\varnothing$

Find the union of the following sets.
(i) $A=\left \{1,2,3,5,6\right \}$ and $B=\left \{4,5,6,7,8\right \}$
(ii) $X =\left \{3,4,5 \right \}$ and $Y =\varnothing$

(i)

$A=\left \{1,2,3,5,6 \right \}$ and

$B=\left \{4,5,6,7,8\right \}$

1,2,3,5,6 ; 4,5,6,7,8 (repeated)

$\therefore A\cup B=\left \{1,2,3,4,5,6,7,8\right \}$

(ii) $X=\left \{3,4,5\right \}, Y=\varnothing$ There are no elements in Y $\therefore X \cup Y =\left \{3,4,5\right \}$

If $U=\left \{ x:1 \leq x\leq 10, x\in \mathbb{N} \right \}, A=\left \{ 1,3,5,7,9 \right \}$ and $B=\left \{ 2,3,5,9,10 \right \}$
Find
(i) ${A}'$
(ii) ${B}'$
(iii) ${A}'\cup {B}'$
(iv) ${A}'\cap {B}'$

(i)

${A}=\left \{ 2,4,6,8,10 \right \}$
(ii)

${B}=\left \{ 1,4,6,7,8 \right \}$
(iii)

${A}'\cup {B}'=\left \{ 1,2,4,6,7,8,10 \right \}$
(iv)

${A}'\cap {B}'=\left \{ 4,6,8 \right \}$

If $A={4, 6, 7, 8, 9}, B={2, 4, 6}$ and $C={1, 2, 3, 4, 5, 6}$ , then find
(i) $A\cup (B\cap C)$ (ii) $A\cap (B\cup C)$ (iii) $A\setminus (C\setminus B)$

$(i)B\cap C=$ {

$2,4,6$ }

$\therefore A\cup (B\cap C)=$ {

$2,4,6,7,8,9$ }

$(ii)B\cup C=$ {

$1,2,3,4,5,6$ }

$\therefore A\cap (B\cup C)=$ {

$4,6$ }

$(iii)C$ \

$B=$ {

$1,3,5$ }

$\displaystyle \therefore A$ \

$(C$ \

$B)=$ {

$4,6,7,8,9$ }

Let A = \{a, b, c, d, e, f, g, x, y, z \}, B = \{ 1, 2, c, d, e\} and C = \{ d, e, f, g, 2, y\}.
Verify $A\setminus (B \cup C)=(A \setminus B)\cap (A \setminus C)$ .

$B\cup C=$ {

$1,2,c,d,e,f,g,2,y$ }

$A$ \

$(B\cup C)=$ {

$a,b,x,z$ }

$-(i)$

$A$ \

$B=$ {

$a,b,f,g,x,y,z$ }

$A$ \

$C=$ {

$a,b,c,x,z$ }

$(A$ \

$B)\cap (A$ \

$C)=$ {

$a,b,x,z$ }

$-(ii)$

From

$(i)$ and

$(ii)$ we verify

$A$ \

$(B\cup C)=(A$ \

$B)\cap (A$ \

$C)$

For $A=\left \{ x| -3 \leq x< 4, x\in R \right \}, B=\left \{ x|x < 5, x\ on\ \mathbb{N} \right \}$ and C= { 5, 3, 1, 0, 1, 3}, show that $A \cap (B \cap C)= (A \cap B)\cup (A \cap C)$ .

First note that the set A contains all the real numbers (not just integers) that are greater than or equal to -3 and less than 4.
On the other hand the set B contains all the positive integers that are less than 5. So,

$A=\left \{ x| -3 \leq x< 4, x\in R \right \}, B=\left \{ x|x < 5, x \in \mathbb{N} \right \}$ ; that is, A consists of all real numbers from -3 upto 4 but 4 is not included.
Also,

$B=\left \{ x|x < 5, x\in \mathbb{N} \right \}$

$={1, 2, 3, 4}$ . Now, we find

$B \cup C$

$={1, 2, 3, 4}$

$\cup$

${-5, -3, -1, 0, 1, 3}$

$= { 1, 2, 3, 4, -5, -3, -1, 0}$ ; thus

$A \cup (B \cup C)$ = A

$\cap$

${1, 2, 3, 4, -5, -3, -1, 0}$

$={-3, -1, 0, 1, 2, 3}$ (1)
Next, to find

$(A \cap B)\cup (A \cap C)$ , we consider

$A \cap B =\left \{ x| -3 \leq x< 4, x\in R \right \} \cap$

$\{1, 2, 3, 4\}=\{1, 2, 3\}$

$A \cap C =\left \{ x| -3 \leq x< 4, x\in R \right \} \cap$

$\{-5, -3, -1, 0, 1, 3\}$

$=\{-3, -1, 0, 1, 2, 3\}$
Hence,

$(A \cap B)\cup (A \cap C)$

$= {1, 2, 3}$

$\cup$

$\{-3, -1, 0, 1, 3\}$

$= \{ -3, -1, 0, 1, 2, 3\}.$
Now (1) and (2) imply

$A \cap (B \cup C)=(A \cap B)\cup (A \cap C)$
1073868_622446_ans_6c5bc71aac53479abb08d059b15d1230.PNG

Let A = {10, 15, 20, 25, 30, 35, 40, 45 } 50 , B = { 1, 5,10,15, 20, 30} and C = { 1, 5,15, 20, 35, 45, }. Verify $A\setminus (B\cap C)=(A\setminus B)\cup (A\setminus C)$ .

The given sets are

$A=\left\{ 10,15,20,25,30,35,40,45,50 \right\}$ ,

$B=\left\{ 1,5,10,15,20,30 \right\}$ and

$C=\left\{ 1,5,15,20,35,45 \right\}$

Let us first find

$A\setminus (B\cap C)$ as follows:

$A\setminus (B\cap C)=\left\{ 10,15,20,25,30,35,40,45,50 \right\} \setminus (\left\{ 1,5,10,15,20,30 \right\} \cap \left\{ 1,5,15,20,35,45 \right\} )$

$=\left\{ 10,15,20,25,30,35,40,45,50 \right\} \setminus \left\{ 1,5,15,20 \right\} =\left\{ 10,25,30,35,40,45,50 \right\} ........(1)$

Now we find

$(A\setminus B)\cup (A\setminus C)$ as follows:

$(A\setminus B)\cup (A\setminus C)$

$=(\left\{ 10,15,20,25,30,35,40,45,50 \right\} \setminus \left\{ 1,5,10,15,20,30 \right\} )\cup (\left\{ 10,15,20,25,30,35,40,45,50 \right\} \setminus$

$\left\{ 1,5,15,20,35,45 \right\} )$

$=\left\{ 25,35,40,45,50 \right\} \cup \left\{ 10,25,30,40,50 \right\} =\left\{ 10,25,30,35,40,45,50 \right\} ........(2)$

Since equation 1 is equal to equation 2,

Hence,

$A\setminus (B\cap C)=(A\setminus B)\cup (A\setminus C)$

If A and B be two finite sets such that the total number of subsets of A is $960$ more than the total number of subsets of B, then $n(A)-n(B)$ (where n(x) denotes the number of elements in set x) is equal to?

let the number of elements in A be m and B be n

therefore the total number of subsets of A is

$2^{m}$ and number of subsets of B is

$2^n$

given

$2^m-2^n=960$

we know from this equation that m>n

therefore taking n common we get

$2^n(2^{m-n}-1)=960$

$2^(m-n)-1$ is odd the even part is only

$2^n$

960 can be written as

$2^6 \times 15$

therefore from above equation we can observe that

$n=5$

and $$2^(m-5)-1

$=15$

$\Rightarrow 2^(m-5)=16$ so m=9

therefore n(A)-n(B)=m-n=4

Show that the relation R defined by $\left( a,b \right) R \left( c,d \right)$ such that $a+d=b+c$ on AXA, where $A=\left\{ 1,2,....10 \right\}$ is an equivalence relation. Hence write the equivalence class $\left[ 3,4 \right] ;a,b,c,d\in A.$

R is an equivalance relation if R is reflexive,symmetric and transitive.

a)checking if it is reflexive;

Given R in

$A\times A\quad and\quad \left( a,b \right) R\left( c,d \right) \quad such\quad that\quad a+d=b+c$

For reflexive,consider

$\left( a,b \right) R\left( a,b \right) \quad \quad \left( a,b \right) \in A$

and applying given condition

$\Rightarrow a+b=b+a$ ;which is true for all A

$\therefore\quad R\quad is\quad reflexive.$

b)checking if it is symmetric;

given

$\left( a,b \right) R\left( c,d \right) \quad such\quad that\quad a+d=b+c$

consider

$\left( c,d \right) R\left( a,b \right) \quad on\quad A\times A$

applying given condition

$\Rightarrow c+b=d+a\quad which\quad satisfies\quad given\quad condition$

Hence R is symmetric.

c)checking if it is transitive;

$Let\quad \left( a,b \right) R\left( c,d \right) \quad and\quad \left( c,d \right) R\left( e,f \right)$

$and\quad \left( a,b \right) ,\left( c,d \right) ,\left( e,f \right) \in A\times A$

applying given condition:

$\Rightarrow a+d=b+c\quad \rightarrow 1\quad and\quad c+f=d+e\quad \rightarrow 2$

equation 1

$\Rightarrow a-c=b-d$

$now\quad add\quad equation\quad 1\quad and\quad 2;$

$\Rightarrow a-c+c+f=b-d+d+e$

$\Rightarrow a+f=b+e$

$\therefore\quad \left( a,b \right) R\left( e,f \right) \quad also\quad satisfies\quad the\quad condition$

Hence R is transitive.

Therfore by above inspection we can say that R is an equivalence relation.

$\Rightarrow steps\quad for\quad finding\quad equivalence\quad class\quad \left[ 3,4 \right] :$

Let

$\left( 3,4 \right) R\left( a,b \right) \quad on\quad A\times A\quad where\quad A=\left\{ 1,2,3,.......10 \right\}$

$\Rightarrow 3+b=4+a$

$let\quad a=1\Rightarrow b=2$

therfore one pair

$\left( a,b \right) =\left( 1,2 \right)$

similarly we can find the pairs

$\left( a,b \right)$

Therefore equivalence class of

$\left[ 3,4 \right] =\left\{ \left( 1,2 \right) ,\left( 2,3 \right) ,\left( 3,4 \right) ,\left( 4,5 \right) ,\left( 5,6 \right) ,\left( 6,7 \right) ,\left( 7,8 \right) ,\left( 8,9 \right) ,\left( 9,10 \right) \right\}$

$A-\left( B\cup C\cup D \right) =\left( A-B \right) \cap\ ........ \cap\ ........$

Ans.

$\left( A-C \right) \cap \left( A-D \right)$

$L.H.S.= \left( B\cup C\cup D \right) '$ w.r.t

$'A'$

$=B'\cap C'\cap D'$ w.r.t

$A$ etc.

$=\left( A-B \right) \cap \left( A-C \right) \cap \left( A-D \right)$

If $A = \{2, 3, 4, 8, 10\}, B = \{3, 4, 5, 10, 12\}$ and $C = \{4, 5, 6, 12, 14\}$
Find : $n (A \cup \,B) \, \cap \, (A \ \cup \, C)$

Given

$A=\left\{ 2,3,4,8,10 \right\}$ ,

$B=\left\{ 3,4,5,10,12 \right\}$ and

$C=\left\{ 4,5,6,12,14 \right\}$

$A\cup B=\left\{ 2,3,4,8,10 \right\}\cup \left\{ 3,4,5,10,12 \right\}$

$\therefore A\cup B=\left\{ 2,3,4,5,8,10,12 \right\}$

$A\cup C=\left\{ 2,3,4,8,10 \right\}\cup \left\{ 4,5,6,12,14 \right\}$

$\therefore A\cup C=\left\{ 2,3,4,5,6,8,10,12,14 \right\}$

$\left( A\cup B \right) \cap \left( A\cup C \right) =\left\{ 2,3,4,5,8,10,12 \right\}\cap \left\{ 2,3,4,5,6,8,10,12,14 \right\}$

$\therefore \left( A\cup B \right) \cap \left( A\cup C \right) =\left\{ 2,3,4,5,8,10,12 \right\}$

$\therefore n\left( A\cup B \right) \cap \left( A\cup C \right) =7$

$\left( A\cup B \right) -C=\left( A-C \right) \cup$ ........

Ans.

$(B-C).$

$\left( A\cup B \right) -C=\left( A\cup B \right) \cap C$

$=\left( A\cap C \right) \cup \left( B\cap C \right)$

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

Let $A=\left\{ \left( x,y \right) :x,y\ \in\ R, { x }^{ 2 }{ +y }^{ 2 }=1 \right\}$
and $B=\left\{ \left( x,0 \right) :x\ \in\ R, -1\le x\le 1 \right\}$ , then $A\cap B=$ ...........

Ans.

$\left\{(-1,0), (1,0)\right\}$

If A = {1, 2} and B = {1, 3} prove that $(A \, \times \, B) \, \cup \, (B \, \times \, A)$ = {(1, 3), (2, 3), (3, 1), (3, 2), (1, 1), (1, 2), (2, 1)}

$A = \{ 1, 2\}$

$B = \{ 1, 3\}$

$(A\times B) = \{ (1, 1), (1, 3), (2,1), (2,3)\}$

$(B\times A) = \{ (1, 1), (1, 2), (3,1), (3,2)\}$

$(A\times B)\cup (B\times A) = \{ (1, 1), (1, 3), (2,1), (2,3),$

$(1, 2), (3,1), (3,2)\}$

In a pollution study of $1500$ Indian rivers the following data were reported. $520$ were polluted by sulphur compounds, $335$ polluted by phosphate, $425$ were polluted by crude oil, $100$ were polluted by both crude oil and sulphur compounds, $180$ were polluted by both sulphur compounds and phosphates, $150$ were polluted by both phosphates and crude oil and $28$ were polluted by sulphur compounds, phosphates, and crude oil. How many of the rivers were polluted by at least one of the three impurities ? How many of the rivers were polluted by exactly one of the three impurities ?

Ans

$878,S$ only

$268$ ,

$P$ only

$33$ , C only

$203$ .
Here

$n(U)=1500$ ,

$n(S)=520$ ,

$n(P)=335$ ,

$n(C)=425$ ,

$n(C\cap S)=100$ ,

$n(S\cap P)=180$ ,

$n(P\cap C)=150$
and

$n(S\cap P\cap C)=28$ ,
Where

$S,P$ and

$C$ denote respectively the st of river polluted by sulphur compounds, phos-phates and crude oil.
Now we draw a Venn-diagram respectively the set

$U$ of

$1500$ Indian rivers by a rectangle and its three subsets

$S.P$ and

$C$ by closed curves inside

$U$ . We first write

$28$ in the region

$S\cap P\cap C$
Since

$n(C\cap S)=100$ , out of which

$28$ have

$100-28$ , That is

$72$ in the remaining par of

$C\cap S$ . Similarly we write

$152$ and

$122$ in the remaining parts of

$S\cap P$ and

$P\cap C$ respectively. Finally we put

$268$ in

$'S$ Only',

$33$ in

$P'$ Only' and

$203$ in

$C'$ Only'.
now the number of river polluted by at least one of the three impurities is given by

$n(S\cup P\cup C)=28+72+152+122+268+33+203=878$
or

${ S }_{ 1 }-{ S }_{ 2 }+{ S }_{ 3 }=(520+335+425)-(100+180+15)+28=1280-430+28=1308-430=878$
Only Sulphur

$=520-(152+72-28)=520--252=268$
Number of rivers polluted by exactly sulphur compounds

$=268$
Number of river polluted by exactly phosphates

$=33$ and the Numb, of rivers polluted by exactly crude oil

$=203$ .

Set $U={1,2,3,4,5,6,7,8,9}$
$A=\left\{ x:x\ \in N,30\le { x }^{ 2 }\le 70 \right\}$
$B=\left\{x:x\ is\ a\ prime\ number\ <10\ \right\}$
Find $A',B'\left( A\cup B \right) ,A'\cap B',\left( A-B \right) '$

Ans.

$A'=\left\{ 1,2,3,4,5,9 \right\},\ B'=\left\{ 1,4,6,8,9 \right\}$

$\left( A\cup B \right) '=A'\cap B'=\left\{ 1,4,9 \right\}$

$(A-B)=\left\{1,2,3,4,5,7,9\right\}$

Let $U=R$ (the set of all real numbers)
If $A=\left\{ x:x\in R,0<x<2 \right\},$
$B=\left\{ x:x\in R,1<x\le 3 \right\},$
Find $A',B',A\cup B,A\cap B,A-B.$

We have

$A\prime =R-A=\{ x:x\in R\ and\ x\in A\}$

$=\{ x:(x\in R\ and\ x\ge 2)$

$(x\in R\ and\ x\le 0\}$ [See the diagram also]

$=\{ x:x\in R\ and\ x\ge 2\} \cup \{ x:x\in R\ and\ x\le 0\}$

$B\prime =\{ x:x \in R\ and\ x \le 2\}$

$\cup \left\{ x:x\in R\ and\ x>3 \right\}$

Similarly

$A\cup B=\{ x:x\in R\ and\ 0<x\le 3\}$

$A\cap B=\{ x:x\in R\ and\ 1<x<2\}$

$A-B=\{ x:x\in R\ and\ 0<x\le 1\}$

Note : In such problems use of real number line is useful.

1037651_996830_ans_6c0ecc7088c14fecbca1ee2a26595e23.png

Given $A ={1,2,3}, B ={3,4}, C ={4,5,6},$
then show that $A\cup \left( B\cup C \right) =\{ 1,2,3,4,5,6\}$
and $\left( A\times B \right) \cap \left( B\times C \right) =\{ 3,4\}$

Clearly

$A\cup \left( B\cup C \right) =\{ 1,2,3,4,5,6\}$
Now

$A\times B={(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)}$
and

$B\times\ C={(3,4),(3,5),(3,6),(4,4),(4,5),(4,6)}$
Hence

$\left( A\times B \right) \cap \left( B\times C \right) =\{ (3,4)\}$ .

Show that $\left( A-B \right) \cup \left( B-A \right) =\left( A\cup B \right) \cap \left( A'\cup B' \right)$

$\left( A\cup B \right) \cap \left( A'\cup B' \right)$

$=\left\{ \left( A\cup B \right) \cap A' \right\} \cup \left\{ \left( A\cup B \right) \cap B' \right\}$ (Distributive law)

$=\left\{ \left( A\cap A' \right) \cup \left( B\cap A' \right) \right\} \cup \left\{ \left( A\cap B' \right) \cup \left( B\cap B' \right) \right\}$ (Distributive law)

$=\left\{ \phi \cup \left( B\cap A' \right) \right\} \cup \left\{ \left( A\cap B' \right) \cup \phi \right\}$

$=\left( B\cap A' \right) \cup \left( A\cap B' \right) =\left( B-A \right) \cup \left( A-B \right)$

$=\left[ \left( A-B \right) \cup \left( B-A \right) \right].$

(i) $A-\left( B-C \right) =\left( A-B \right) \cup \left( A\cap C \right)$
(ii) $A-B=\left( A\cup B \right) -B=A-\left( A\cap B \right)$
Verify these equalities if
$A=\left\{ 1,2,3,4,5,6,7 \right\} , B=\left\{ 3,5,6,7,9,11 \right\}$
and $C=\left\{ 2,5,6,9,20 \right\}$

Ans. True.
(i)

$A-\left( B-C \right) =A-\left( B\cap C' \right)$

$A\cap \left( B\cap C' \right)'$

$A\cap \left( B'\cup C \right) =A\cap B'\cup \left( A\cap C \right)$

$\left( A-B \right) \cup \left( A\cap C \right)$
(ii)

$\left( A\cup B \right) -B=\left( A\cup B \right) \cap B'$

$\left( A\cap B' \right) \cup \left( B\cap \bar { B' } \right)$

$\left( A\cap B' \right) \cup \phi$

$A\cap B'=A-B$ etc.

Let $U$ be the set of all people and $M$ = {Males}, $S$ = { College students } , $T$ = { Teenagers } , $W$ = { People having heights more than five feet }. Express each of the following in the notation of set theory.
(i) College students having heights more than five feet.
(ii) People who are not teenagers and have their heights less than five feet.
(iii) All people who are neither males nor teenagers nor college students

(i) Here we have to write down the set of all those people having both the properties:

$(1)$ college students,

$(2)$ people having their height more than five feet. Hence it is the set consisting of common elements of the sets

$S$ and

$W$ , that is, the set

$S\cap W$ .
(ii) The set of people who are not teenager is the complement of the set

$T$ , that is, it is the set

$-T$ or

$T$ . Similarly the set of people having height less than five feet is the complement of

$W$ , that is it is the set

$U-W$ or

$W$ .
Hence the required set in this case is

$\left( U-T \right) \cap \left( U-W \right)$ or

$T'\cap W'$
(iii) Arguing as in cases (i) and (ii) we see that the required set in this case is

$\left( U-M \right) \cap \left( U-T \right) \cap \left( U-S \right) =M'\cap T'\cap S'$ or using D-Morgans law, the required set is

$U-\left( M\cup T\cup S \right)$ i.e

$\left( M\cup T\cup S \right)'$

There are $240$ students in class XI of a school, $130$ play cricket, $100$ play football, $75$ play volleyball, $30$ of these play cricket and football. $25$ play volleyball and cricket, $15$ play football and volleyball. Also catch students play atleast one of the three games. How many students play all the three games?

Totl students=

$240$

$A=130$ play cricket,

$B=100$ play football,

$C=75$ play volleyball.

$30$ play cricket & football.

$25$ play volleyball & cricket.

$15$ play football & volleyball.

$\therefore A\cup B\cup C=A+B+C-A\cap B-B\cap C-A\cap C+A\cap B\cap C\\ 240=130+75+100-30-25-15+A\cap B\cap C\\ A\cap B\cap C=240-235=5$

Let $A = [x \in R:11x - 5\rangle 7x + 3]$ and $B = [x \in R:8x - 9 \ge 15x + 2]$
Find $A \cap B$ and represent it on the number line

1311877_1013429_ans_b1c033d2b59b4256800bec91ecc0ba0f.jpg

Let A and B be two sets with a finite number of elements. Assume that there is injective mapping from A to B and that there is an injective mapping from B to A. Prove that there is a bijective mapping from A to B.

A function is said to be injective if

$f(a) = f(a')$ implies that

$a = a'$ .

A function is said to be surjective if for all

$b \in B$ , there exists

$a \in A$ such

that

$f(a) = b$ .

A function is said to be bijective if it is injective and surjective.

We say that two sets A and B are equivalent,

written

$A \sim B$ if and only if there exists a function

$f : A \rightarrow B$ which is a

bijection.

Find a, b if, $\left[ 7,-\sqrt { 35 } \right]\bigcup {\left[ {5,\sqrt {35} } \right]} =[a, b]$

$\left[ 7,-\sqrt { 35 } \right] \cup \left[ 5,\sqrt { 35 } \right] =\left[ a,b \right]$

$\\ \sqrt { 35 } =5.9$

$\\ -\sqrt { 35 } =-5.9$

$\\ \left[ 7,-5.9 \right] \cup \left[ 5,5.9 \right] =\left[ 7,-5.9 \right]$

$\\ \left[ 7,-\sqrt { 35 } \right] \cup \left[ 5,\sqrt { 35 } \right] =\left[ 7,-\sqrt { 35 } \right]$

For a non-empty set $X$ , if operation $:P(X) \times P(X) \Rightarrow P(X)$ is defined as $AB=(A-B)\cup(B-A),\forall A,B \in P(X)$ then show that empty set $\phi$ is the identity for $*,$ and all elements $A$ of $P(X)$ are invertible with $A^{-1}=A$

$\pi \in P(X)$ is an identity element
To prove it let

$E \in P(X)$ be the identity element such that

$A*E=E$

$*A=A$

$\forall A \in P(X)$

$(A-E) \cup (E-A)=A \Rightarrow E = \phi$
i.e.

$(A-\phi) \cup (E- \phi)=A$

$A* \phi=\phi*A=A$

$*A=A$ , hence

$\phi$ is the identity element
Let

$Be P(X)$ be the inverse of

$A$

$\therefore A*B =B*A =\phi \forall A e P(X)$

$(A-B) \cup (B-A)=0 \Rightarrow B=A$
becasue

$A-B=\phi\ B-A=\phi \Rightarrow A=B$

$\therefore \forall A \in P(X), A*A=\phi$

$A$ ie the invertible element of

$A$

$\therefore A^{-1}=A$

If $A ={ 1, 3, 5, ..........17}$ and $B= { 2, 4, 6, ......18}$ and $N$ ( the set of natural numbers) is the universal set, then show that $A' \cup ((A\cup B )\cap B')=N$ .

$Let\quad N=\{ 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\} \\ \because A=\{ 1,3,5,7,9,........,17\} \\ B=\{ 2,4,6,8,........18\} \\ { A }^{ ' }=\{ 2,4,6,8,.......18\} \\ { B }^{ ' }=\{ 1,3,5,7,.......17\} \\ \because A={ B }^{ ' }\quad \& \quad B={ A }^{ ' }\\ (A\cup B)=\{ 1,2,3,4,5,6,............18\} \\ (A\cup B)\cap { B }^{ ' }=\{ 1,3,5,7,..........17\} \\ { A }^{ ' }\cup [(A\cup B)\cap { B }^{ ' }]=\{ 1,2,3,4,5,6,..........18\} \\ \therefore { A }^{ ' }\cup [(A\cup B)\cap { B }^{ ' }]=N$

Which of the following statements are false and why?
(a) $a\subset \{ a, b, c\}$
(b) $\varnothing \in \{c, d\}$ .

1169723_1049123_ans_0b269e1432aa4c4cb34c16996a2c93a5.jpg

Prove: $A-B= A-(A\cup B)$

$let\>A\>=\>[1,2]\>and\>B\>=\>[2,3,4]\>considering\>U\>=\>[1,2,3,4,]\\\therefore\>A\>\cap\>B\>=\>[2]\\then\>A-B=[1]\\A-(A\cap\>B)=[1]\\Hence\>A-B\>=\>A-[A\cap\>B]\\Proved$

$A=\left\{1,2,3,4\right\}$
$B=\left \{3,4,5,6\right\}$
$x=\left\{1,2,3,..10\right\}$
$(A\cup B)'=A'\cap B'$

$A = \left\{ {1,2,3,4} \right\}\,\,\,\,B = \left\{ {3,4,5,6} \right\}$

$U = \left\{ {1,2,3..........10} \right\}$

$AUB = \left\{ {1,2,3,4,5,6} \right\}$

$\left( {AUB'} \right) = \left\{ {7,8,9,10} \right\} - \left( 1 \right)$

$A' = \left\{ {5,6,7,8,9,10} \right\}$

$B' = \left\{ {1,2,7,8,9,10} \right\}$

$A' \cap B' = \left\{ {7,8,9,10} \right\} - \left( 2 \right)$

from equ 1 and 2

$\left( {AUB} \right)' = A' \cap B'$

If $A,B$ are any two sets and $C\subset\ A\bigcup B$ . Then is it true that $C\subseteq A$ or $C\subseteq B$ ?

Let

$x \in C$

Then

$x \in (A \cup B )$

$x \in A$ or

$x \in B$

$C \in A$ or

$C \in B$

Therefore,

$x$ is general element of

$C$ .

If $A \cup B = A \cup C$ and $A \cap B = A \cap C$ , then prove that $B=C$ .

Let

$x\in B......(1)$

$\Rightarrow x\in A\cup B$ [Since

$B\subset A\cup B$ , all elements of B are in

$A\cup B$ ]

$\Rightarrow x\in A\cup C$ [Given

$A\cup B=A\cup C]$

$\Rightarrow x\in A$ or

$x\in C.....(2)$

Taking

$x\in A$

$Also, x\in B$ [From (1)]

$\therefore x\in A\cap B$ [If x belongs to both A and B, it will belong to common of A and B also]

$x\in A\cap B$

So,

$x\in A\cap C$ [Given

$A\cup B=A\cup C$ ]

i.e.

$x\in A$ and

$x\in C$

i.e

$x\in C$

$\therefore$ If

$x\in B$ , then

$x\in C$

i.e. if an element belongs to set B, then it must belong to set C also

$\therefore B\subset C$

Taking

$x\in C$

Also,

$x\in B$ [From (1)]

$\therefore$ If

$x\in B$ , then

$x\in C$

i.e if an elements belongs to set B, then it must belong to set C also

$\therefore B\subset C$

Similarly, we can prove

$C\subset B$

Since

$B\subset C$ and

$C\subset B$

$\therefore B=C$

Hence proved.

For any two sets, prove that: $A\cup (A\cap B)=$ $A\cap (A\cup B)=A$

1323447_1063095_ans_56e5ad16af444d588ba413d90ca86063.jpg

Using properties of sets, show that for any two sets A and B, $(A\cup B)\cap (A\cup B')=A$

1326496_1063096_ans_84a244d7cba74f469856f01bf162bc10.jpg

In a group of 50 students, the number of student studying French, English, Sanskrit were found to be as follows:
French = 17, English = 13, Sanskrit = 15
French and English = 9, English and Sanskrit = 4
French and Sanskrit = 5, English, French and Sanskrit =Find the number of students who study
(i) French only
(ii) English only
(iii) Sanskrit only
(iv) English and Sanskrit but not French
(v) French and Sanskrit but no English
(vi) French and English but no Sanskrit
(vii) at least one of the three languages
(viii) none of the three language

Let,

$French$

$=$ set

$A$ ,

$English$

$=$ set

$B$ ,

$Sanskrit$

$=$ set

$C$

$n(A)=17,n(B)=13,n(C)=15$

$n(A\cap B)=9,n(B\cap C)=4,n(A \cap C)=5$

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

$n(u)=50$

$n(A\cap \bar{B}\cap \bar{C})=n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)$

$=17-9-5+3$

French only

$=6$

$n(\bar{A}\cap B\cap \bar{C})=n(B)-n(B\cap C)-n(A\cap B)+n(A\cap B\cap C)$

$=13-4-9+3$

English only

$=3$

$n(\bar{A}\cap \bar{B}\cap C)=n(C)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)$

$=15-4-5+3$

Sanskrit only

$=9$

$n(\bar{A}\cap B\cap C)=n(B\cap C)-n(A\cap B\cap C)$

French, English and Sanskrit

$=4-3=1$

$n(A\cap \bar{B}\cap C)=n(A\cap C)-n(A\cap B \cap C)=5-3=2$

$n(A\cap {B}\cap \bar {C})=n(A\cap B)-n(A\cap B \cap C)=9-3=6$

n(name of

$A$ or

$B$ or

$C$ )=

$5a-(A\cap B\cap C)$

$=5a-\{ n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\}$

$=50-\{ 17+13+15-9-4-5+3\}$

$=20$

$\therefore n$ (at least of

$A$ or

$B$ or

$C$ )=

$50-20=30$

If $A={x:x\in W, x<2} , \quad B={x:x\in N, 1<x<5},\quad c={3, 5}$ : then verify that: $A\times (B\cap C)=(A\times B)\cap (A\times C)$

1326501_1063160_ans_3c129f1cd6714c4f84f4746cb4760e02.jpg

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

Solution:-

Given that

$A \subset B$ , i.e., B will contain all the elements of A.

$\therefore \; A \cup B = B$

If $n(A)=15,\ n(A\cup B)=29,\ n(A\cap B)=7$ then find $n(B)$ .

$\textbf{Hint: Use the formula of the union of two sets.}$

Solution:

$\textbf{Use the formula of the union of two sets.}$

$\quad\quad\quad$ We have given

$\quad\quad\quad$ $n(A) = 15,{\rm{\;}}n(A \cup B) = 29,{\rm{\;}}n(A \cap B) = 7$

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

$\quad\quad\quad$ $\therefore \,\,29 = 15 + n\left( B \right) - 7$

$\quad\quad\quad$ $\Rightarrow n\left( B \right) = 21$ $\Rightarrow n\left( B \right) = 21$

$\textbf{Hence,}$ $\boldsymbol{n\left( B \right)}$ $\textbf{is}$ $\boldsymbol{21}$ .

A is the set of real zeroes of the polynomial $x^2+16$ . Is A an empty set? Give reasons?

1329477_1081361_ans_fc913effaa3241faa6971bc05d620488.jpg

If A = {1, 2, 3, 4} B = {3,4,5,6}, C = {4,5,6,7,8} and
Universal set n = {1,2,3,4,5,6,7,8,9,10}.
verify demorgans law.

$A=(1,2,3,4)$

$B=(3,4,5,6)$

$C=(4,5,6,7,8)$

$U=(1,2,3,4,5,6,7,8,9,10)$

According to DeMorgan's Law

$(A\vee B)'=A'\wedge B'$

$A'=(\wedge 5,6,7,8,9,10)$

$B'=(1,2,7,8,9,10)$

$(A\vee B)=3,4$

$(A\vee B)'=1,2,5,6,7,8,9,10$

$A'\wedge B'=1,2,5,6,7,8,9,10$

Hence proved

Given, $A = \left\{ {x: - 2 < x \le 5,x \in R} \right\}$ and $\left\{ {x: - 5 \le x < 3,x \in R} \right\}$

i) $A \cap B$
ii) $A' \cap B$ on different number lines.

Given that ,

$A=\{x:-2<x\le 5,x\in R\}$

$B=\{x:-5\le x<3\}$

Now ,

$U=\{x:-5\le x\le 5\}$

$A\cap B=\{-5\}$

${{A}^{'}}$ = $U-A$

$=\{x:-5\le x\le 5,x\in R\}-\{x:-2<x\le 5,x\in R\}$

$=\{x:-5\le x<3\}$

$=B$

Hence ,

${{A}^{'}}\cap B=$ $\{x:-5\le x<3\}\cap$ $\{x:-5\le x<3\}$

= $B\cap B$

Hence, this is the answer.

If $U=\left\{1,2,3,4,5,6,7,8,9\right\},A=\left\{2,4,6,8\right\}$ and $B=\left\{2,3,5,7\right\}$ verify that
$(A \cup B)'=A' \cap B'$

We have,

$U=\left\{ 1,2,3,4,5,6,7,8,9 \right\}$

$A=\left\{ 2,4,6,8 \right\}$

$B=\left\{ 2,3,5,7 \right\}$

Since,

$\left( A\cup B \right)'=U-\left( A\cup B \right)$

$\left( A\cup B \right)'=\left\{ 1,2,3,4,5,6,7,8,9 \right\}-\left( \left\{ 2,4,6,8 \right\}\cup \left\{ 2,3,5,7 \right\} \right)$

$\left( A\cup B \right)'=\left\{ 1,2,3,4,5,6,7,8,9 \right\}-\left\{ 2,3,4,5,6,7,8 \right\}$

$\left( A\cup B \right)'=\left\{ 1,9 \right\}$

Now,

$A'\cap B'=\left( U-A \right)\cap \left( U-B \right)$

$A'\cap B'=\left( \left\{ 1,2,3,4,5,6,7,8,9 \right\}-\left\{ 2,4,6,8 \right\} \right)\cap \left( \left\{ 1,2,3,4,5,6,7,8,9 \right\}-\left\{ 2,3,5,7 \right\} \right)$

$A'\cap B'=\left\{ 1,3,5,7,9 \right\}\cap \left\{ 1,4,6,8,9 \right\}$

$A'\cap B'=\left\{ 1,9 \right\}$

It is clear that,

$\left( A\cup B \right)'=A'\cap B'$

Hence, proved.

Prove that: $1.P\left( {1,1} \right) + 2.P\left( {2,2} \right) + 3.P\left( {3,3} \right) + ... + n.P\left( {n,n} \right) = P\left( {n + 1,n + 1} \right) - 1$

$LHS=1.^1P_1+2.^2P_2+3.^3P_3+......+n^nP_n$

$^nPn=n!$ , hence

$LHS=1.1!+2.2!+3.3!+....+n.n!$

$=(1.1!+1)+2.2!+3.3!+.....+n.n!-1$

$=2!+2.2!+3.3!+....+n.n!-1$

$=2!(1+2)+3.3!+....+n.n!-1$

$=3.2!+3.3!+....+n.n!-1$

$=3!+3.3!+......+n.n!-1$

$=3!(3+1)_.....+n.n!-1$

$=4.3!+.....+n.n!-1$

$=4!+.....+n.n!-1$

Similarly LHS will reduce to

$LHS=n!+n.n!-1$

$=n!(n+1)-1$

$=(n+1)!-1$

$\Rightarrow LHS=^{n+1}P_{n+1}-1=RHS$

If $U=\left\{1,2,3,4,5,6,7,8,9\right\},A=\left\{2,4,6,8\right\}$ and $B=\left\{2,3,5,7\right\}$ verfy that
$(A \cap B)'=A' \cup B'$

We have,

$U=\left\{ 1,2,3,4,5,6,7,8,9 \right\}$

$A=\left\{ 2,4,6,8 \right\}$

$B=\left\{ 2,3,5,7 \right\}$

Since,

$\left( A\cap B \right)'=U-\left( A\cap B \right)$

$\left( A\cap B \right)'=\left\{ 1,2,3,4,5,6,7,8,9 \right\}-\left( \left\{ 2,4,6,8 \right\}\cap \left\{ 2,3,5,7 \right\} \right)$

$\left( A\cap B \right)'=\left\{ 1,2,3,4,5,6,7,8,9 \right\}-\left\{ 2 \right\}$

$\left( A\cap B \right)'=\left\{ 1,3,4,5,6,7,8,9 \right\}$

Now,

$A'\cup B'=\left( U-A \right)\cup \left( U-B \right)$

$A'\cup B'=\left( \left\{ 1,3,5,7,9 \right\}-\left\{ 2,4,6,8 \right\} \right)\cup \left( \left\{ 1,4,6,8,9 \right\}-\left\{ 2,3,5,7 \right\} \right)$

$A'\cup B'=\left\{ 1,3,5,7,9 \right\}\cup \left\{ 1,4,6,8,9 \right\}$

$A'\cup B'=\left\{ 1,3,4,5,6,7,8,9 \right\}$

It is clear that,

$\left( A\cap B \right)'=A'\cup B'$

Hence, proved.

If $S$ is any set, then the family of all the subsets of $S$ is called the power set of $S$ and it is denoted by $P(S)$ . Power set of a given set is always non-empty. If $A$ has n elements, then $P(A)$ has ?

No .of subsets for a set of

$n$ elements is

$2^n$ so,

$A$ has n elements, then

$P(A)$ has

$2^{n}$ elements.

If $A'\cup B=U$ , then show that $A\subset B$

$A' \cup B = U$

we have

$A' \cup B = U'$

$(A' \cup B)' = U'$

$A \cap B' = \phi$

It means no elements is common in

$A$ &

$B$ which would only possible if

$A$ &

$B$ are disjoint sets hence for that

$A \subset B$ . It is true.

1344507_1126183_ans_b56043c48d8a44d88ddf53328fddb55c.png

$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (1 - \cos 2x)}}{{{x^4}}}$

We have,

$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (1 - \cos 2x)}}{{{x^4}}}$

We know that

$\cos 2x=1-2\sin^2x$

Therefore,

$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (1 - (1-2\sin^2 x))}}{{{x^4}}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (2\sin^2 x)}}{{{x^4}}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - (1-2\sin^2 (\sin^2 x))}}{{{x^4}}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{{2\sin^2 (\sin^2 x)}}{{{x^4}}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{2\sin^2 (\sin^2 x)}{\sin^4 x}\times \dfrac{\sin^4 x}{x^4}$

$2\mathop {\lim }\limits_{x \to 0} \left(\cfrac{\sin (\sin^2 x)}{(\sin^2 x)}\right)^2\times \dfrac{\sin^4 x}{x^4}$

We know that

$\lim_{x\to\ 0}\dfrac{\sin x}{x}=1$

Therefore,

$\Rightarrow 2\times 1\times 1$

$\Rightarrow 2$

Hence, this is the answer.

Write the following set by roaster method : The set of all natural numbers $'x'$ such that $4x+9<5$ .

$4x+9< 5$

$4x< -4$

$x< -1$

$x\in [1,2,3,4,5,....]$

Prove that $A\cup B=A\cap B$ , if $A=B$ .

$P(A)=P(B)$

inorder to prove

$A=B$ , we should prove

$A$ is a subset of

$B$ i.e.,

$A\subset B$

and

$B$ is a subset of

$A$ i.e.,

$B\subset A$

Set

$A$ is an element of power et of

$A$ as every set is a subset

i.e.,

$P\in P(A)$

$A\in P(B)$

if set

$A$ is in power set of

$B$

Set

$A$ is a subset of

$B$

$A\subset B$

Similarly we can prove

$B\subset A$

Now since

$A\subset B$ and

$B\subset A$

$\therefore$

$A=B$

Hence proved

Prove that : $A-\left( B\cup C \right) =\left( A-B \right) \cap \left( A-C \right)$ without using venn diagram.

$A-(B\cap C)=\phi$ then the result holds by properties of

$\phi$ . So assume

$A-(B\cap C)\ne \phi$ and let

$x\in A-(B\cap C)$

The

$x\in A$ and

$x\notin B\cap C$ which implies that

$x\in A$ and

$(x\notin B,x\notin C)$

Case 1:

$x\notin B$

$x\in C$ , then

$x\in A-C$ which is

$x\in (A-B)\cup (A-C)$

Case 2

$x\in B$

$x\notin C$ , then

$x\in A-C$ which means

$x\in (A-B)\cup (A-C)$

Both cases lead to the same conclusion

$x\in (A-B)\cup (A-C)$

Hence

$A-(B\cap C)\le (A-B)\cup (A-C)$

Hence proved

A and b are two sets such that $n(A)=3$ and $n(B)=6$ .
Find (i) minimum value of $n\left( A\cup B \right)$
(ii) maximum value of $n\left( A\cup B \right)$

$A\rightarrow3$

$B\rightarrow 6$

Minimum

$\rightarrow$ out of

$6$ elements of Set B,

$3$ elements are identical to the set

$A$ . In this case

$A\cup B=B$ and it will have

$6$ elements

Maximum

$\rightarrow$

$A\cup B$ elements condition,

$A$ and

$B$ are disjoint sets, that no elements will be common to them, then

$n(A\cup B)=n(A)+n(B)$

$6+3=9$

Describe the following set by set property method ${0, 3, 8, 15, 24, 35}$

$0,3,8,15,24,35$

$0\rightarrow {1}^{2}-1$

$3\rightarrow {2}^{2}-1$

$8\rightarrow {3}^{2}-1$

$15\rightarrow {4}^{2}-1$

$24\rightarrow {5}^{2}-1$

$35\rightarrow {6}^{2}-1$

$X={(6+1)}^{2}-1={7}^{2}-1=48$

$X=48$ next term

If $A=\{x:x$ is a factor of $72\}$ , $B=\{x:x$ is a multiple of $4<41\}$ .Find $(A-B)\cup(B-A)$ ?

$Given,A=\left\{ x:x\quad is\quad a\quad factor\quad of\quad 72 \right\} \\ \quad \quad \therefore \quad A=\left\{ 1,2,3,4,6,8,9,12,18,24,72 \right\} \\ \quad \quad \quad B=\quad \left\{ x:x\quad is\quad a\quad multiple\quad of\quad 4<41 \right\} \\ \therefore \quad B=\left\{ 4,8,12,16,20,24,28,32,36,40 \right\} \quad \\ \quad \quad =\left\{ 1,2,3,6,9,18,72 \right\} \quad and\quad \quad \left( B-A \right) =\quad \left\{ 16,20,28,32,36,40 \right\} \\ \quad \therefore \quad \left( A-B \right) \cup \left( B-A \right) \quad =\quad \left\{ 1,2,3,6,9,16,18,20,28,32,36,40,72 \right\}$

Prove by using venn diagram
$A-\left( B\cup C \right) =\left( A-B \right) \cap \left( A-C \right)$

$A-(B\cup B)=(A-B)\cap (A-C)$
1357843_1194277_ans_8005326c7c974d5b991c74f3d747a990.PNG

Decide which of the following are equal sets and which are not? Justify your answer.

$A=\left\{ x|3x-1=2 \right\}$

$B=\left\{ x|x\quad is\quad a\quad natural\quad number\quad but\quad x\quad is\quad neither\quad prime\quad nor\quad composite \right\}$

$C=\left\{ x|x\in N,x<2 \right\}$

1093308_1192236_ans_91e6cd7bc036475da0e811ac20fd24cb.png

Find the number of three element sets of positive integers {a,b,c} such that $a \times b \times c = 2310$

1116225_1205817_ans_b5922633b6874f4d8eb59534d9c7f047.jpg

Let $A=\left\{a, e, i, o, u\right\}$ and $B=\left\{a, b, c, d\right\}$ . Is $A$ a subset of $B$ ? No. (Why?). Is $B$ a subset of $A$ ? No. (Why?)

$A=\left\{a, e, i, o, u \right\}\quad B=\left\{ a, b, c, d\right\}$

$A$ is not Subset of

$B$ and

$B$ is not subset of

$A$ .

$A$ set is a subset of

$B$

if it contains all elements or some elements of

$B$ in the same way

$B$ is Subset of

$A$ if it contains the elements of

$A$ .

As they are not containing they are not Subsets.

$(A\cap B')'\cup (B\cap C)=?$

$(A\cap B')'=A'\cup B$

$(A\cap B')'\cup (B\cap C)=(A'\cup B)\cup (B\cap C)=(A'\cup B\cup B)\cap (A'\cup B\cup C)= (A'\cup B)\cap (A'\cup B\cup C)=(A'\cup B)$

If $A{a,b,c}$ then the number of equivalence relations on $A$ is ?

$A_{1}:\left \{ (a,a) \right \},\left \{ (b,b) \right \},\left \{ (c,c) \right \}$

because in this 3 cases they can be Reflexive, symmetric and transitive.

$A_{2}:\left \{ (a,b)(b,a)(a,a)(b,b) \right \}$ (4)

In this way we can write

for

$(b,c) \& (a,c)$

$A_{3}:\left \{ (b,c)(c,b)(b,b)(c,c) \right \}$ (5)

$A_{4}: \left \{ (a,c)(c,a)(c,c)(a,a) \right \}$ (6)

$A_{5}$ includes all i.e

$A_{5}: \{ (a,a)(b,b)(c,c)(a,b)(b,a)$

$(a,c)(c,a)(b,c)(c,b)\}$ (7)

With out including empty set we have total 7 equivalence sets by including empty set also we have

$8$ .

A and B are two sets having $3$ elements in common.If $n ( \mathrm { A } ) = 5 , n ( \mathrm { B } ) = 4$ then Find $n\left( {A \times B} \right)\,and\,n\left[ {\left( {A \times B} \right) \cap \left( {B \times A} \right)} \right]$

Consider the problem

The elements in the order pair

$\begin{array}{l} A\times B=5\times 4=20 \\ n\left( { A\times B } \right) =20 \end{array}$

Since, three elements are common in the set

$A$ and

$B$

Therefore, the number of common elements in the order pair set is

$\begin{array}{l} 3\times 3=9 \\ n\left[ { \left( { A\times B } \right) \cap \left( { B\times A } \right) } \right] =9 \end{array}$

Write the following set in the roster method: $B=\{y|y$ is a colour in the rainbow $\}$ .

1294238_1233086_ans_a9475c4d78194eebb0d702e4bf6eb6d5.jpg

Of the members of three athletic team s in a school $21$ are in cricket team, $26$ are in hockey team and $29$ are in the football team. Among them, $14$ play hockey and cricket, $15$ play hockey and football, and $12$ play football and cricket. Eight play all the three games. Find the total number of members in the three athletic team.

Let

C= Set of members of cricket team

H= Set of members of Hockey team

F= Set of members of Football team

$n(C)=21,n(H)=26,n(F)=23$

$n(H\cap C)=14,n(H\cap F)=15,n(F\cap C)=12$

$n(C\cap H\cap F)=8$

Total number of members =

$n(C\cap H\cap F)$

$=n(C)+n(H)+n(F)-n(C\cap H)-n(H\cap F)-n(C\cap F)+n(C\cap H\cap F)$

$=21+26+29-14-15-12+8=43$

1340445_1194286_ans_ce1f2f559a0a44fb80cad36d6d9bc134.png

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

${\textbf{Step-1:Prove using suitable formula of sets.}}$

$\text{It is given that, }$

${\text{A}} \cup {\text{B = A}} \cup {\text{C}}$ ${\text{…(1)}}$

$\text{and, }$

${\text{A}} \cap {\text{B = A}} \cap {\text{C}}$ ${\text{…(2)}}$

${\text{Taking}}$ $\text{'}\cap {\text{ C'}}$ ${\text{on both sides in equation (1)}}$

$\left( {{\mathrm{A\cup B}}} \right) \cap {\text{C = }}\left( {{\mathrm{A\cup C}}} \right) \cap {\text{C}}$

${\text{We know that,}}$

${\text{and}}$ $\left( {{\mathrm{A\cup C}}} \right) \cap {\text{C = C}}$

${\text{So,}}$

$\left( {{\text{A}} \cap {\text{C}}} \right)\cup \left( {{\text{B}} \cap {\text{C}}} \right) = C$

$\left( {{\text{A}} \cap \text{B}} \right)\cup \left( {{\text{B}} \cap {\text{C}}} \right) = C$ ${\text{…(3)}}$ ${\textbf{[From (2)]}}$

${\text{Again,}}$

${\text{Taking}}$ $\text{'}\cap{\text{ B'}}$ ${\text{on both side in equation (1)}}$

$\left( {{\mathrm{A\cup B}}} \right) \cap B{\text{ = }}\left( {{\mathrm{A\cup C}}} \right) \cap B$

${\text{B = }}\left( {{\text{A}} \cap {\text{B}}} \right)\cup\left( {{\text{C}} \cap {\text{B}}} \right)$

${\text{B = }}\left( {{\text{A}} \cap {\text{B}}} \right)\cup\left( {{\text{B}} \cap {\text{C}}} \right)$

${\text{B = C}}$ ${\textbf{[From (3)}]}$

${\textbf{Hence, proved.}}$

In certain town, 25% of the families own a phone and 15% own a car; 65% families own neither a phone nor a car and 2000 families own both a car and a phone. Then find the correct statement among the following;
(a) 5% families own both a car and a phone.
(b) 35% families own either a car and a phone.
(c) 40,000 families live in the town.

1204012_1249744_ans_7cb591f51d7f474e9b6650d7ec49d7f0.jpg

For two sets if $n(A)=4$ and $n(B)=3$ then find the number of elements if $A\Delta B$ .

Given two sets if

$n(A)=4$ and

$n(B)=3$ .

Now we have,

$A\Delta B$

$=(A-B)\cup (B-A)$ .

Since

$A,B$ are not given explicitly so it is not possible to determine

$n(A\Delta B)$ .

Let $A=\{1,2,3,4\}$ and $B=\{2,4,6,8\}$ . Then find $A\cup B$ .

Given,

$A=\{1,2,3,4\}$ and

$B=\{2,4,6,8\}$ .

Now,

$A\cup B$

$=\{1,2,3,4,6,8\}$ .

Let $A=\{1,2,3,4\}$ and $B=\{2,4,6,8\}$ . Then find $A\Delta B$ .

Given,

$A=\{1,2,3,4\}$ and

$B=\{2,4,6,8\}$ .

Now,

$A\Delta B$

$=(A-B)\cup (B-A)$

$=\{1,3\}\cup \{6,8\}$

$=\{1,3,6,8\}$ .

Let $O=$ Set of odd natural numbers $=\{1,3,5,7,9,...\}$ and $E=$ Set of even natural numbers $=\{2,4,6,8,10,........\}$ . Show that $2\in E$ .

$O = \{1,3,5,7,9,.....\}$

$E = \{ 2,4,6,8,10,..... \}$

So it is clear from the definition of

$E$ ,

$2 \in \,E\,$

Let, $A=\{1,2,3,4\}$ and $B=\{2,4,6,8\}$ . Then find $B-A$ .

Given,

$A=\{1,2,3,4\}$ and

$B=\{2,4,6,8\}$ .

Now,

$B-A$

$=\{6,8\}$ .

If $P=\left\{x\ |\ 24 < x < 30\right\}$ and $Q=\left\{x\ |\ 25 < x < 32\right\}$ , prove that $P-Q\neq Q-P$

We have

$\begin{array}{l} P=\left\{ { x\, \left| { 24<x<30 } \right| } \right\} \\ Q=\left\{ { x\, \left| { 25<x<32 } \right| } \right\} \\ P-Q\, means\, those\, element\, that\, are\, in\, set\, P\, not\, in\, set\, Q. \\ and, \\ Q-P\, means\, those\, element\, that\, are\, in\, set\, Q\, not\, in\, set\, p. \\ If\, we\, take\, 25,\, 25\, \, is\, \, in\, set\, P\, \, not\, \, in\, set\, Q \\ Hence, \\ P-Q\ne Q-P. \end{array}$

Let $O=$ Set of odd natural numbers $={1,3,5,7,9,...}$ and $E=$ Set of even natural numbers $={2,4,6,8,10,........}$ .
Fill in the blank using symbol $\in$ or $\notin$ .
$1 __ O$

$0 = \left\{ {1,3,5,7,9 - - - - } \right.$

$E = \left\{ {2,4,6,8,10 - - - - } \right.$

$1\,\underline \in \,0\,$ Answer.

Find the union and intersection of the following pair of sets
$A=\left\{ 2,\ 3,\ 5,\ 6,\ 7 \right\} ;\ B=\left\{ 4,\ 5,\ 7,\ 8 \right\}$

$\begin{array}{l} A=2,3,5,6 \\ B=4,5,7,8 \\ A\cup B=\left\{ { 2,3,4,5,6,7,8 } \right\} \\ A\cap B=\left\{ 5,7 \right\} \\ \cup \to for\, \, all,\, \, \cap \to for\, \, common\, \, elements. \end{array}$

If $f\left(x\right)=x^ {2}+1$ , find $f'\left(2\right)$ from the first principle.

Given,

$f(x)=x^2+1$

$f'(x)=\lim_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}$

$=\lim_{h\rightarrow 0}\dfrac{(x+h)^2+1-x^2-1}{h}$

$=\lim_{h\rightarrow 0}\dfrac{x^2+2hx+h^2-x^2}{h}$

$=\lim_{h\rightarrow 0}\dfrac{2hx+h^2}{h}$

$=\lim_{h\rightarrow 0}(2x+h)$

$=2x$

$\therefore f'(x)=2x$

$\Rightarrow f'(2)=2\times 2=4$

If $A \times B \subseteq C\times D$ and $A \times B\neq \phi,$ prove that $A \subseteq C$ and $B \subseteq D$ .

1861163_1354470_ans_c62098c26cdf47f6a09c38942bb730c6.jpeg

Out of $100$ students $15$ passed in English $12$ passed in Mathematics, $8$ Science, $4$ in English and Science in all the three. Find how many passed in English and Mathematics but not in Science

Given n(u)=100

$n\left( E \right) = 15,$

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

$n\left( S \right) = 8$

$n\left( {E \cap M} \right) = 6,n\left( {M \cap S} \right) = 7,$

$n\left( {E \cap S} \right)....4,\,and\,n\left( {E \cap M \cap S} \right) = 4$

$a = n\left( {E \cap M \cap S} \right) = 4$

$a + d = n\left( {M \cap S} \right) = 7$

$\therefore \,$

$\,\,d = 3$

$a + b = n\left( {M \cap E} \right) = 6$

$\therefore \,\,\,\,\,b = 2$

$a+c=n\left( { S\cap E } \right) =4$

$\therefore \,\,\,\,c = 0$

$\,\,\,\,\,\,\,a + b + d + e = n(M) = 12$

$\therefore \,\,\,4 + 2 + 3 + e = 12$

$\therefore \,e = 3$

$\,\,\,\,a + b + c + g = 15$

$\therefore \,4 + 2 + 0 + g = n(E) = 15$

$\therefore \,g = 9$

$\,\,a + c + d + f = n(S) = 8$

$\therefore \,4 + 0 + 3 + f = 8$

$\therefore \,f = 1$

The number of students passed in English and Mathematics but not in science

$b=2$

1239735_1329453_ans_4043578f65da4e5cae1797ca17deaa10.JPG

Suppose $A_{ 1 }, A_{ 2 },A_{ 30 }$ are thirty sets each with five elements and $B_{ 1 }, B_{ 2 },., B_{ n }$ are in sets each with three elements. Let $\overset { 30 }{ \underset { i=1 }{ \cup } } Ai=\overset { n }{ \underset { j=1 }{ \cup } } { B }_{ j }=S$ . Assume the each element of $S$ belongs to exactly ten of the $A_{ i }$ and exactly $9$ of $B_{ j }s$ . Find n.

Since each

$A \,A_{i}$ has

$5$ elements, we have

$\overset{30}{\underset{i = 1}{\Sigma}} n (A_i) = 5 \times 30 = 150$ ......(1)

Let S consist of m distinct elements, Since each element of S belongs to exactly

$10$ of the

$A'_is,$ we also have

$\overset{30}{\underset{i = 1}{\Sigma}} n (A_i) = 10\,m$ ....(2)

Hence from (1) and (2),

$10\,m = 150$ or

$m = 15.$

Again since each

$B_i$ has

$3$ elements and each element of S belongs to exactly

$9$ of the

$B'_iS,$ we have

$\overset{n}{\underset{j = 1}{\Sigma}} n (B_i) = 3n$ and

$\overset{n}{\underset{j = 1}{\Sigma}}n (B_j) = 9\,m$

It follows that

$3n = 9 \times 15 [\therefore m = 15]$

This gives

$n = 45$

If $A=\left\{1,2,3,4\right\},\,\,B=\left\{3,4,5,6\right\}$ and $C=\left\{1,2,4,6,7\right\}$ then find $A\cap\left(B\cap \,C\right)$

$B\cap\, C=\left\{3,4,5,6\right\}\cap\, \left\{1,2,4,6,7\right\}$

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

$A\cap\left(B\cap \,C\right)=\left\{1,2,3,4\right\}\cap\left\{4,6\right\}$

$=\left\{4\right\}$

If $A=\left\{1,2,3,4\right\},\,\,B=\left\{3,4,5,6\right\}$ and $C=\left\{1,2,4,6,7\right\}$ then find $A\cap\left(B\cup \,C\right)$

$B\cup\, C=\left\{3,4,5,6\right\}\cup\, \left\{1,2,4,6,7\right\}$

$=\left\{1,2,3,4,5,6,7\right\}$

$A\cap\left(B\cup \,C\right)=\left\{1,2,3,4\right\}\cap\left\{1,2,3,4,5,6,7\right\}$

$=\left\{1,2,3,4\right\}$

Let $A=\left\{a,c,i,o,u\right\}$ & $B=\left\{a,i,u\right\}$ show that $A \cup B=A$ .

A∪B={a,c,i,o,u}=AA∪B={a,c,i,o,u}=A

∴A∪B=A

Let $A=\{1,2,3,4,5\}$ and $B=\{2,4,6\}$ then find $A\cup B$ and $A\cap B$ .

Given,

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

$B=\{2,4,6\}$ .

Now

$A\cup B=\{1,2,3,4,5,6\}$ and

$A\cap B=\{2,4\}$ .

If $A=\left\{a,b\right\},\,\,B=\left\{x,y\right\}$ and $C=\left\{a,c,y\right\}$ , then verify that $A\times\left(B\cup \,C\right)=\left(A\times B\right)\cup\left(A\times C\right)$

$B\cup \,C=\left\{x,y\right\}\cup\left\{a,c,y\right\}=\left\{a,c,x,y\right\}$

$A\times\left(B\cup \,C\right)=\left\{a,b\right\}\times \left\{a,c,x,y\right\}$

$=\left\{\left(a,a\right),\left(a,c\right),\left(a,x\right),\left(a,y\right),\left(b,a\right),\left(b,c\right),\left(b,x\right),\left(b,y\right)\right\}$

$\therefore\,A\times\left(B\cup \,C\right)=\left\{\left(a,a\right),\left(a,c\right),\left(a,x\right),\left(a,y\right),\left(b,a\right),\left(b,c\right),\left(b,x\right),\left(b,y\right)\right\}$ ..........

$(1)$

$A\times B=\left\{a,b\right\}\times \left\{x,y\right\}=\left\{\left(a,x\right),\left(a,y\right),\left(b,x\right),\left(b,y\right)\right\}$

$A\times C=\left\{a,b\right\}\times \left\{a,c,y\right\}=\left\{\left(a,a\right),\left(a,c\right),\left(a,y\right),\left(b,a\right),\left(b,c\right),\left(b,y\right)\right\}$

$\left(A\times B\right)\cup\left(A\times C\right)=\left\{\left(a,x\right),\left(a,y\right),\left(b,x\right),\left(b,y\right)\right\}\cup\left\{\left(a,a\right),\left(a,c\right),\left(a,y\right),\left(b,a\right),\left(b,c\right),\left(b,y\right)\right\}$

$=\left\{\left(a,a\right)\left(a,x\right),\left(a,y\right),\left(a,c\right),\left(b,x\right),\left(b,y\right),\left(b,a\right),\left(b,c\right)\right\}$ .........

$(2)$

From

$(1)$ and

$(2)$ we have

$A\times\left(B\cup \,C\right)=\left(A\times B\right)\cup\left(A\times C\right)$

Hence verified.

Verify whether $A\subset B$ for the sets $A=\left\{\left\{a,b,c\right\}\right\}, B=\left\{1,\{a,b,c\},2\right\}$

The set

$A=\left\{\left\{a,b,c\right\}\right\}$ means that

$A$ is the set containing the set

$\left\{a,b,c\right\}$ as its only element.

The definition of

$A\subset B$ means that whenever

$a$ is an element of

$A$ , we have that

$a$ is also an element of

$B$ .

So by taking every element of the set

$A$ we see that it is an element of

$B$ . The only element of

$A$ is

$\left\{a,b,c\right\}$ .

$\therefore A\subset B$

Fill in the blanks
A _______ is a collection of numbers gathered to give some meaningful information.

A set is a collection of numbers gathered to give some meaningful information.

Which of the following are sets? Justify the answer. Justify your answer
The collection of all even integers.

{0,±2,±4,±6−}{0,±2,±4,±6−} is collection of well defined and so it is a set.

Which of the following are sets? Justify the answer.
The collection of ten most talented writers of India.

The term most talented is vague. So, this collection is not a set

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 nor orange juice.

Let

U=U= set of surveyed students

A=A= set of students taking apple juice.

B=B= set of students taking orange juice
Then

n(U)=400;n(A)=100;n(B)=150;n(U)=400;n(A)=100;n(B)=150;
and

n(A∩B)=75n(A∩B)=75
Now

n(A′∩B′)=n(A∪B)′n(A′∩B′)=n(A∪B)′

=n(U)−n(A∪B)=n(U)−n(A∪B)

=n(U)−n(A)−n(B)+n(A∩S)=n(U)−n(A)−n(B)+n(A∩S)

=400−100−150+75=225=400−100−150+75=225

∴255∴255 students were taking neither apple juice nor orange juice.

Write down all possible subsets of the following set.
$\{a\}$ .

The possible subsets are

$\phi, \{a\}$ .

From the diagram, relation between A and B.......

B is a subset of A

Write down all possible subsets of the following set.
$\{a, b, c\}$ .

The possible subsets are

$\phi, \{a\}, \{b\}, \{c\}, \{a, b\}, \{b, c\}, \{a, c\}, \{a, b, c\}$ .

From the diagram, relation between A and B............

B is the subset of A

Write down all possible subsets of the following set.
$\{1, \{1\}\}$ .

The possible subseis are

$\phi, \{1\}, \{\{1\}\}, \{1, \{1\}\}$ .

Decide among the following sets, which are subsets of which:
$A=\{x : x$ satisfies $x^2-8x+12=0\}$ ,
$B=\{2, 4, 6\}$ ,
$C=\{2, 4, 6, 8,.....\}$ ,
$D=\{6\}$ .

1421167_1665685_ans_dbd7549b28a24533be3af9781788b14b.jpg

Write down all possible subsets of the following set.
$\{0, 1\}$ .

The possible subsets are

$\phi, \{0\}, \{1\}, \{0, 1\}$ .

Let $A=\{3, 6, 12, 15, 18, 21\}, B=\{4, 8, 12, 16, 20\}, C=\{2, 4, 6, 8, 10, 12, 14, 16\}$ and $D=\{5, 10, 15, 20\}$ . Find $A-D$ .

$A=\{3, 6, 12, 15, 18, 21\},$

$B=\{4, 8, 12, 16, 20\},$

$C=\{2, 4, 6, 8, 10, 12, 14, 16\}$

$D=\{5, 10, 15, 20\}$ .

$A-D$ .

$\{3, 6, 12, 15, 18, 21\}-$

$\{5, 10, 15, 20\}$ .

$\{3, 6, 12, 18, 21\}$ .

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 B'.

$u=\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$

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

$B=\{2, 4, 6, 8\}$

$C=\{3, 4, 5, 6\}$

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

$\{1,2,3,4,5,6,7,8,9\}-\{2,4,6,8\}$

$\{1, 3, 5, 7, 9\}$ .

Write down all possible subsets of the following set.
$\{\phi\}$ .

We have a set -

$A = \{\phi\}$

Possible subset are

$\{\phi\} , \{\}$

Let $A=\{3, 6, 12, 15, 18, 21\}, B=\{4, 8, 12, 16, 20\}, C=\{2, 4, 6, 8, 10, 12, 14, 16\}$ and $D=\{5, 10, 15, 20\}$ . Find $C-A$ .

$A=\{3, 6, 12, 15, 18, 21\},$

$B=\{4, 8, 12, 16, 20\},$

$C=\{2, 4, 6, 8, 10, 12, 14, 16\}$

$D=\{5, 10, 15, 20\}$ .

$C-A$

$\{2,4,6,8,10,12,14,16\}-\{3,6,12,15,18,21\}$

$\{2, 4, 8, 10, 14, 16\}$ .

Let $A=\{3, 6, 12, 15, 18, 21\}, B=\{4, 8, 12, 16, 20\}, C=\{2, 4, 6, 8, 10, 12, 14, 16\}$ and $D=\{5, 10, 15, 20\}$ . Find $B-A$ .

$A=\{3, 6, 12, 15, 18, 21\},$

$B=\{4, 8, 12, 16, 20\},$

$C=\{2, 4, 6, 8, 10, 12, 14, 16\}$

$D=\{5, 10, 15, 20\}$ .

$B-A$

$\{4,8,12,16,10\}-\{3,6,12,15,18,21\}$

$\{4, 8, 16, 20\}$ .

Let $A=\{3, 6, 12, 15, 18, 21\}, B=\{4, 8, 12, 16, 20\}, C=\{2, 4, 6, 8, 10, 12, 14, 16\}$ and $D=\{5, 10, 15, 20\}$ . Find $B-D$ .

$A=\{3, 6, 12, 15, 18, 21\},$

$B=\{4, 8, 12, 16, 20\},$

$C=\{2, 4, 6, 8, 10, 12, 14, 16\}$

$D=\{5, 10, 15, 20\}$ .

$B-D$

$\{4,8,12,16,20\}-\{5,10,15,20\}$

$\{4,8,12,16\}$

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 $(A')'$ .

1421179_1665772_ans_008c5278f54c48d1972a66da4abb467c.jpg

Let $A=\{3, 6, 12, 15, 18, 21\}, B=\{4, 8, 12, 16, 20\}, C=\{2, 4, 6, 8, 10, 12, 14, 16\}$ and $D=\{5, 10, 15, 20\}$ . Find $D-A$ .

$A=\{3, 6, 12, 15, 18, 21\},$

$B=\{4, 8, 12, 16, 20\},$

$C=\{2, 4, 6, 8, 10, 12, 14, 16\}$

$D=\{5, 10, 15, 20\}$ .

$D-A$

$\{5,10,15,20\}-\{3,6,12,15,18,21\}$

$\{5, 10, 20\}$ .

Let $A=\{3, 6, 12, 15, 18, 21\}, B=\{4, 8, 12, 16, 20\}, C=\{2, 4, 6, 8, 10, 12, 14, 16\}$ and $D=\{5, 10, 15, 20\}$ . Find $A-C$ .

$A=\{3, 6, 12, 15, 18, 21\},$

$B=\{4, 8, 12, 16, 20\},$

$C=\{2, 4, 6, 8, 10, 12, 14, 16\}$

$D=\{5, 10, 15, 20\}$ .

$A-C$ .

$\{3, 6, 12, 15, 18, 21\}-$

$\{2, 4, 6, 8, 10, 12, 14, 16\}$

$\{3, 15, 18, 21\}$ .

Let $A=\{3, 6, 12, 15, 18, 21\}, B=\{4, 8, 12, 16, 20\}, C=\{2, 4, 6, 8, 10, 12, 14, 16\}$ and $D=\{5, 10, 15, 20\}$ . Find $B-C$ .

$A=\{3, 6, 12, 15, 18, 21\},$

$B=\{4, 8, 12, 16, 20\},$

$C=\{2, 4, 6, 8, 10, 12, 14, 16\}$

$D=\{5, 10, 15, 20\}$ .

$B-C$

$\{4,8,12,16,20\}-\{2,4,6,8,10,12,14,16\}$

$\{20\}$

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 $(B-C)'$ .

The given sets are:

$u=\left\{1,2,3,4,5,6,7,8,9 \right\}$

$A=\left\{1,2,3,4 \right\},\ B=\left\{2,4,6,8 \right\},\ C=\left\{3,4,5,6 \right\}$
Hence,

$B-C=\left\{2,8 \right\}$

$\therefore \ (B-C)'=u(B-C)$

$\boxed {(B-C)'=\left\{1,3,4,5,6,7,9 \right\}}$

Let $A=\{1, 2, 4, 5\}, B=\{2, 3, 5, 6\}, C=\{4, 5, 6, 7\}$ . Verify the following identity.
$A\cap (B-C)=(A\cap B)-(A\cap C)$ .

given are sets:

$A=\left\{1,2,4,5 \right\}, B=\left\{2,3,5,6 \right\}, C=\left\{4,5,6,7 \right\}$

Hence,

$B-C=\{2,3,5,6\}-\{4,5,6,7\}=\left\{2,3 \right\}$

$A\cap B=\{1,2,4,5\}\cap \{2,3,5,6\}=\left\{2,5 \right\}$

$A\cap C=\{1,2,4,5\}\cap \{4,5,6,7\}=\left\{4,5 \right\}$

$(A\cap B)-(A\cap C)=\{2,5\}-\{4,5\}=\left\{ 2\right\}$

$\dots(1)$

we have to verify,

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

$LHS=A\cap (B-C)$

$=\left\{ 1,2,4,5\right\}\cap \left\{2,3 \right\}$

$=\left\{ 2\right\}$

$RHS=(A\cap B)-(A\cap C)$

$=\left\{2 \right\}$ [from

$(1)$ ]

since

$L.H.S=R.H.S$

$\Rightarrow \ \boxed {A\cap (B-C)=(A\cap B)-(A\cap C)}$

Let $A=\{1, 2, 4, 5\}, B=\{2, 3, 5, 6\}, C=\{4, 5, 6, 7\}$ . Verify the following identity.
$A\cap(B\cup C)=(A\cap B)\cup(A\cap C)$ .

The given sets are :

$A=\left\{1, 2, 4, 5\right\}, B=\left\{2, 3, 5, 6\right\}, C=\left\{4, 5, 6, 7\right\}$

we have to prove that

$A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$

$LHS=A\cap (B\cup C)$

$=\left\{1, 2, 4, 5\right\}\cap (\left\{2, 3, 5, 6\right\}\cup \left\{4, 5, 6, 7\right\})$

$=\left\{1, 2, 4, 5\right\}\cap \left\{2, 3, 4, 5, 6, 7\right\}$

$=\left\{2, 4, 5\right\}$

$RHS=(A\cap B)\cup (A\cap C)$

$=(\left\{1, 2, 4, 5\right\}\cap \left\{2, 3, 5, 6\right\})\cup (\left\{1, 2, 4, 5\right\}\cap \left\{4, 5, 6, 7\right\})$

$=\left\{2, 5\right\}\cup \left\{4, 5\right\}$

$=\left\{2, 4, 5\right\}$

since

$L.H.S=R.H.S$

Hence,

$\boxed{A\cap (B\cup C)=(A\cap B)\cup (A\cap C)}$

Let $A=\{1, 2, 4, 5\}, B=\{2, 3, 5, 6\}, C=\{4, 5, 6, 7\}$ . Verify the following identity.
$A\cap (B\triangle C)=(A\cap B)\triangle(A\cap C)$ .

The given sets are:

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

$B = \{2,3,5,6\}$

$C= \{4,5,6,7\}$

$\because A \Delta B = (A- B) \cup (B - A)$

$\Rightarrow B \Delta C = (B - C) \cup (C - B)$

$B - C = \{2,3\}$ ,

$C- B = \{4,7\}$

$\Rightarrow B \Delta C = \{2,3\} \cup \{4,7\}$

$=\{2,3,4,7\}$

And

$A \cap B = \{ 2,5\}$ ,

$A \cap C = \{4,5\}$

Hence,

$(A\cap B) \Delta (A \cap C) = ((A\cap B)-(A\cap C)) \cup ((A\cap C)-(A\cap B))$

$\Rightarrow (A\cap B) \Delta (A\cap C) = \{2\} \cup \{4\}$

$= \{2,4\}$

We have to verify:

$A\cap (B \Delta C) = (A\cap B) \Delta (A \cap C)$

LHS

$= A\cap (B \Delta C) = \{ 1, 2, 4, 5\} \cap \{2,3,4,7\}$

$=\{2,4\}$

RHS

$= (A \cap B)\Delta (A \cap C) = \{2,4\}$

Hence,

$A\cap (B \Delta C) = (A\cap B) \Delta(A \cap C)$

Let $A=\{1, 2, 4, 5\}, B=\{2, 3, 5, 6\}, C=\{4, 5, 6, 7\}$ . Verify the following identity.
$A-(B\cup C)=(A-B)\cap (A-C)$ .

The given sets are:

$A=\left\{1,2,4,5 \right\},\ B=\left\{2,3,5,6 \right\},\ C=\left\{4,5,6,7 \right\}$

$\therefore \ B\cup C=\left\{2,3,4,5,6,7 \right\}$

$A-(B\cup C)=\left\{ 1\right\}$

$A-B=\left\{1,4 \right\}$

$A-C=\left\{1,2 \right\}$

$(A-B)\cap (A-C)=\left\{1 \right\}$

we have to verify,

$A-(B\cup C)=(A-B)\cap (A-C)$

$LHS=A-(A\cup C)$

$=\left\{1 \right\}$

$RHS=(A-B)\cap (A-C)$

$=\left\{1 \right\}$

Hence,

$\boxed {A-(B\cup C)=(A-B)\cap (A-C)}$

Let $A=\{1, 2, 4, 5\}, B=\{2, 3, 5, 6\}, C=\{4, 5, 6, 7\}$ . Verify the following identity.
$A-(B\cap C)=(A-B)\cup (A-C)$ .

The given sets are:

$A=\left\{1,2,4,5 \right\}\ B=\left\{2,3,5,6 \right\}\ C=\left\{4,5,6,7 \right\}$

$A\cap C=\left\{5,6 \right\}\ ,\ A-(B\cap C)=\left\{1,2,4 \right\}$

$A-B=\left\{1,4 \right\}\ ,\ (A-C)=\left\{1,2 \right\}$

we have to verify,

$A-(B\cap C) =(A-B)\cup (A-C)$

$LHS=(A-B)\cup (A-C)=\left\{1,4 \right\}\cup \left\{1,2 \right\}$

$=\left\{1,2,4 \right\}$

Hence,

$LHS=RHS$

$\Rightarrow \ \boxed {A-(B\cap C)=(A-B)\cup (A-C)}$

If $u=\{2, 3, 5, 7, 9\}$ is the universal set and $A=\{3, 7\}, B=\{2, 5, 7, 9\}$ , then prove that $(A\cap B)'=A'\cup B'$ .

1421385_1665784_ans_f44061a4c750429381a6b1183fb58f1d.jpg

Let $A=\{1, 2, 4, 5\}, B=\{2, 3, 5, 6\}, C=\{4, 5, 6, 7\}$ . Verify the following identity.
$A\cup(B\cap C)=(A\cup B)\cap (A\cup C)$ .

The given sets are :

$A=\left\{1, 2, 4, 5\right\}, B=\left\{2, 3, 5, 6\right\}, C=\left\{4, 5, 6, 7\right\}$

we have to prove,

$A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$

$LHS=A\cup (B\cap C)$

$=\left\{1, 2, 4, 5\right\}\cup (\left\{2, 3, 5, 6\right\}\cap \left\{4, 5, 6, 7\right\})$

$=\left\{1, 2, 4, 5\right\}\cup \left\{ 5, 6\right\}$

$=\left\{1, 2, 4, 5, 6\right\}$

Now,

$RHS=(A\cup B)\cap (A\cup C)$

$=(\left\{1, 2, 4, 5\right\}\cup \left\{2, 3, 5, 6\right\})\cap (\left\{1, 2, 4, 5\right\}\cup \left\{4, 5, 6, 7\right\})$

$=\left\{1, 2,3, 4, 5, 6\right\}\cap \left\{1, 2, 4, 5, 6, 7\right\}$

$=\left\{1, 2, 4, 5, 6\right\}$

$\Rightarrow LHS=RHS$

$\Rightarrow \boxed{A\cup (B\cap C)=(A\cup B)\cap (A\cup C)}$

Let $u=\{1, 2, 3, 4, 5, 6, 7, 8, 9\}, A=\{2, 4, 6, 8\}$ and $B=\{2, 3, 5, 7\}$ . Verify that $(A\cup B)'=A'\cap B'$ .

The given sets are:

$u=\left\{1,2,3,4,5,6,7,8,9 \right\}$

$A=\left\{2,4,6,8 \right\},\ B=\left\{2,3,5,7 \right\}$

we have to prove that

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

Now,

$A\cup B=\left\{2,3,4,5,6,7,8 \right\}$

$L.H.S=(A\cup B)'$

$\because \ (A\cup B)'=u-(A\cup B)=\left\{1,9 \right\}$

$\Rightarrow \ (A\cup B)' =\left\{1,9 \right\}=LHS$

$R.H.S=A' \cap B'$

$\because \ A' =u-A=\left\{1,3,5,7,9 \right\}$

$B'=u-B=\left\{1,4,6,8,9 \right\}$

$\Rightarrow \ A' \cap B'=\left\{1,9 \right\}$

$\Rightarrow \ RHS=A' \cap B'=\left\{1,9 \right\}=(A\cup B),=LHS$

$\Rightarrow \ A' \cap B'=A\cup B'$

$\Rightarrow \ \boxed {A\cup B' =A' \cap B'}$

For any two sets A and B, prove that $A\cup (B-A)=A\cup B$ .

1421392_1665810_ans_5624c4b6b1b84ebbbb0b553b7bf263b7.jpg

For any two sets A and B, prove that $A-(A-B)=A\cap B$ .

1421391_1665809_ans_d7237b1e995d4cd98ee3cbb1c091a3f5.jpg

For any two sets A and B, prove that $A\cap B\subset A$ .

we have to prove

$A\cap B\subset A$

$\because \ A\cap B$ is set of all that elements that belongs to both

$A$ and

$B$ . Hence, all elements of

$A\cap B$ contained in

$A$ .

Hence,

$\boxed {A\cap B\subset A}$

For any two sets A and B, prove that $A-(A\cap B)=A-B$ .

1421390_1665808_ans_6e002e9e2a77409eaf4367e0d1dc2efe.jpg

For any two sets A and B, prove that $(A-B)\cup(A\cap B)=A$ .

1421393_1665811_ans_00449728c19c4d499fbd146fab7c0b6f.jpg

For any two sets A and B, prove that $A\subset B\Rightarrow A\cap B=A$ .

we have to that

$A\subset B \Rightarrow A\cap B=A$

$\because \ A\cap B$ is the set of all these elements that belongs to both

$A$ and

$B$

Hence

$A\cap B \subset A$ and

$A\cap B \subset B----(1)$

And we have given

$A\subset B$ i.e., every element of

$A$ is contain in both

$A$ and

$B$

Hence, every element of

$A$ is contained in

$A\cap B$

$\Rightarrow \ A\subset A\cap B---(2)$

$\therefore \$ from

$(1)$ and

$(2)$ , we have

$\boxed {A\cap B=A}$

Thus,

$\boxed {A\subset \Rightarrow A\cap B=A}$

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

1421389_1665792_ans_3df8ed4e13bf43f581248feb3336977b.jpg

For any two sets, check whether $A\cup (A\cup B)=A$ .

1443609_1665807_ans_8c2e3a34dc67429c8b382b8a0047c771.jpg

For three sets A, B and C, show that $A\subset B\Rightarrow C-B\subset C-A$ .

Let

$x\in C-B$ .

Then,

$x\in C-B$

$\Rightarrow x\in C$ and

$x\not{\in}B$

$\Rightarrow x\in C$ and

$x\not{\in}A$

$[\because A\subset B]$

$\Rightarrow x\in C-A$

$\therefore C-B\subset C-A$ .

For any two sets A and B, show that the following statements are equivalent.
(i) $A\subset B$
(ii) $A-B=\phi$
(iii) $A\cup B=B$
(iv) $A\cap B=A$ .

(i)

$\Rightarrow$ (ii)

We know that,

$A-B=\{x\in A : x\not{\in} B\}$

Since

$A\subset B$ . Therefore, there is no element in A which does not belong to B.

$\therefore A-B=\phi$

Hence, (i)

$\Rightarrow$ (ii).
(ii)

$\Rightarrow$ (ii)

We have,

$A-B=\phi\Rightarrow A\subset B\Rightarrow A\cup B=B$

Hence, (ii)

$\Rightarrow$ (iii).

(iii)

$\Rightarrow$ (iv)

We have,

$A\cup B=B\Rightarrow A\subset B\Rightarrow A\cap B=A$

Hence, (iii)

$\Rightarrow$ (iv).

(iv)

$\Rightarrow$ (ii)

We have,

$A\cap B=A\Rightarrow A\subset B$

Hence, (iv)

$\Rightarrow$ (ii)

Consequently, (ii)

$\Leftrightarrow$ (ii)

$\Leftrightarrow$ (iii)

$\Leftrightarrow$ (iv).

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

1428669_1665813_ans_b0d57c942ec6466c8a4c538a0ece77ec.jpg

Let $A=\{1, 2\}, B=\{1, 2, 3, 4\}, C=\{5, 6\}$ and $D=\{5, 6, 7, 8\}$ . Verify that $A\times C\subset B\times D$ .

1421800_1666257_ans_b4997e0528dc435d8b9ed9a1191018ed.jpg

A survey shows that $76\%$ of the Indians like oranges, whereas $62\%$ like bananas. What percentage of the Indians like both oranges and bananas?

1421405_1665818_ans_9bf51d23c5c0449498db34e952d720dc.jpg

If $A=\{1, 2\}$ , form the set $A\times A\times A$ .

1421666_1666227_ans_df2ba52bdf034007a4ed9000a73e27e5.jpg

Answer the following question in one word or one sentence or as per exact requirement of the question.
If A and B are two sets such that $A\subset B$ , then write $B'-A'$ in terms of A and B.

1443692_1665849_ans_dd7cd6c7537d46d59ba317b987c6ef6c.jpg

If $A=\{-1, 1\}$ , find $A\times A\times A$ .

1421662_1666159_ans_1d4a437829c9405eaaa651a958e7cab5.jpg

If A and B are sets, then prove that $A-B, A\cap B$ and $B-A$ are pair wise disjoint.

We have,

$A-B=\{x :x \in A$ and

$x\not{\in}B\}$ ,

$B-A=\{x\in B$ and

$x\not{\in}A\}$

$\therefore A-B$ and

$B-A$ are disjoint sets

Now,

$x\in A-B\Rightarrow x\in A$ and

$x\not{\in}B\Rightarrow x\not{\in}A\cap B$

$\therefore (A-B)$ and

$A\cap B$ are disjoint sets.

Similarly,

$B-A$ and

$A\cap B$ are disjoint sets.

If $A=\{1, 2, 3, 4, 5\}, B=\{4, 5, 6, 7, 8\}, C=\{7, 8, 9, 10, 11\}$ and $D=\{10, 11, 12, 13, 14\}$ . Find $A\cup B$ .

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

$B=\{4, 5, 6, 7, 8\}$

$C=\{7, 8, 9, 10, 11\}$

$D=\{10, 11, 12, 13, 14\}$

$A\cup B$ .

$=\{1,2,3,4,5\}\cup \{4,5,6,7,8\}$

$=\{1, 2, 3, 4, 5, 6, 7, 8\}$ .

For three sets A, B and C, show that $A\cap B=A\cap C$ need not imply $B=C$ .

Let

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

$B=\{3, 4, 5, 6\}$

$C=\{1, 3, 4, 7, 8\}$ .

Here,

$A\cap B={3,4}$ and

$B\cap C={3,4}$

Therefore,

$A\cap B=A\cap C$ , but

$B\neq C$ .

Using properties of sets, show that for any two sets A and B, $(A\cup B)\cap (A\cup B')=A$ .

Given

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

$=((A\cup B)\cap A)\cup((A\cup B)\cup B')$

$=A\cup((A\cup B)\cap B')$

$=A\cup(A\cap B')\cup(B\cap B')$

$=A\cup(A\cap B')=A$ .

For three sets A, B and C, show that $A\subset B\Rightarrow C-B\subset C-A$ .

Let

$x\in C-B$ .

Then,

$x\in C-B$

$\Rightarrow x\in C$ and

$x\not{\in}B$

$\Rightarrow x\in C$ and

$x\not{\in}A$

$[\because A\subset B]$

$\Rightarrow x\in C-A$

$\therefore C-B\subset C-A$ .

Answer the following question in one word or one sentence or as per exact requirement of the question.
Let A and B be two sets such that $n(A)=3$ and $n(B)=2$ . If $(x, 1), (y, 2), (z, 1)$ are in $A\times B$ , write A and B.

1444581_1666468_ans_4ca1772cdf4d42e8aca9e92a2e5c4f66.jpg

If $A=\{1, 2, 3, 4, 5\}, B=\{4, 5, 6, 7, 8\}, C=\{7, 8, 9, 10, 11\}$ and $D=\{10, 11, 12, 13, 14\}$ . Find $A\cup C$ .

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

$B=\{4, 5, 6, 7, 8\}$

$C=\{7, 8, 9, 10, 11\}$

$D=\{10, 11, 12, 13, 14\}$

$A\cup C$ .

$=\{1,2,3,4,5\}\cup \{7,8,9,10,11\}$

$=\{1, 2, 3, 4, 5, 7, 8,9,10,11\}$ .

For any two sets, prove that $A\cup (A\cup B)=A$ .

We have,

$A\cup (A\cup B)$

$=(A\cup A)\cap (A\cup B)$

$[\because \cup$ is distributive over

$\cap]$

$=A\cap (A\cup B)=A$

$[\because A\subset A\cup B]$ .

If $A=\{1, 2, 3, 4, 5\}, B=\{4, 5, 6, 7, 8\}, C=\{7, 8, 9, 10, 11\}$ and $D=\{10, 11, 12, 13, 14\}$ . Find $A\cup B\cup C$ .

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

$B=\{4, 5, 6, 7, 8\}$

$C=\{7, 8, 9, 10, 11\}$

$D=\{10, 11, 12, 13, 14\}$

$A\cup B\cup C$

$=\{1,2,3,4,5\}\cup\{4,5,6,7,8\}\cup\{7,8,9,10,11\}$

$=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ .

If $A=\{1, 2, 3, 4, 5\}, B=\{4, 5, 6, 7, 8\}, C=\{7, 8, 9, 10, 11\}$ and $D=\{10, 11, 12, 13, 14\}$ . Find $A\cup B\cup D$ .

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

$B=\{4, 5, 6, 7, 8\}$

$C=\{7, 8, 9, 10, 11\}$

$D=\{10, 11, 12, 13, 14\}$

$A\cup B\cup D$

$=\{1,2,3,4,5\}\cup\{4,5,6,7,8\}\cup\{10,11,12,13,14\}$

$=\{1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14\}$ .

If $A=\{2, 3\}, B=\{4, 5\}, C=\{5, 6\}$ , find $A\times (B\cup C), A\times (B\cap C), (A\times B)\cup (A\times C)$ .

$A=\{2, 3\}$

$B=\{4, 5\}$

$C=\{5, 6\}$

$A\times (B\cup C)=\{(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\}$

$A\times (B\cap C)=\{(2, 5), (3, 5)\}$ ,

$(A\times B)\cup (A\times C)=\{(2, 4), (2, 5), (3, 4), (3, 5), (2, 6), (3, 6)\}$ .

Let A and B be two sets such that: $n(A)=20, n(A\cup B)=42$ and $n(A\cap B)=4$ . Find $n(A-B)$ .

$n(A)=20$

$n(A\cup B)=42$

$n(A\cap B)=4$

$n(A-B)=n(A)-n(A\cap B)$

$n(A-B)=20-4$

$\Rightarrow n(A-B)=16$ .

Is the following represent a set? justify your answer.
The collection of good hockey players in India.

Clearly there is no specific criterion to decide whether a given hockey player of India, is good or not.
So, the given collection is not a set.

Which of the following are sets? justify your answer.
The collection of all whole numbers less than $10$

Whole number starts from

$0.$

Clearly, all whole numbers less than

$10$ are

$0,1,2,3,4,5,6,7,8,9$

If $A=\{1, 2, 3, 4, 5\}, B=\{4, 5, 6, 7, 8\}, C=\{7, 8, 9, 10, 11\}$ and $D=\{10, 11, 12, 13, 14\}$ . Find $B\cup C\cup D$ .

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

$B=\{4, 5, 6, 7, 8\}$

$C=\{7, 8, 9, 10, 11\}$

$D=\{10, 11, 12, 13, 14\}$

$B\cup C\cup D$

$=\{4,5,6,7,8\}\cup\{7,8,9,10,11\}\cup\{10,11,12,13,14\}$

$=\{4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14\}$ .

Let $X = \{n \in N : 1 \le n \le 50\}.$ If $A = \{n \in X : n$ is a multiple of $2\}$ and $B = \{n \in X : n$ is a multiple of $7\}$ , then the number of elements in the smallest subset of X containing both A and B is _____.

$X = \{x : 1 \le x \le 50 , \, x \in N\}$

$A = \{x : x$ is multiple of

$2\}$

$\therefore n(A) = 25$

$B = \{x : x$ is multiple of

$7\}$

$\therefore n(B) = 7$

$n(A \cap B) = 3$

From question

$n(A \cup B) = n(A) + n(B) - n (A \cap B)$

$= 25 + 7 - 3$

$= 29$

If $A=\{1, 2, 3, 4, 5\}, B=\{4, 5, 6, 7, 8\}, C=\{7, 8, 9, 10, 11\}$ and $D=\{10, 11, 12, 13, 14\}$ . Find $A\cap(B\cup C)$ .

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

$B=\{4, 5, 6, 7, 8\}$

$C=\{7, 8, 9, 10, 11\}$

$D=\{10, 11, 12, 13, 14\}$

$A\cap (B\cup C)$

$=\{1,2,3,4,5\}\cap\bigg(\{4,5,6,7,8\} \cup \{7,8,9,10,11\}\bigg)$

$=\{1,2,3,4,5\} \cap \{4,5,6,7,8,9,10,11\}$

$=\{4,5\}$

For any two sets of A and B, prove that $B'\subset A'\Rightarrow A\subset B$ .

Let

$x\in A$ .

Then,

$x\in A\Rightarrow x\not{\in}A'$

$\Rightarrow x\not{\in}B'$

$\Rightarrow x\in B$

$[\because B'\subset A']$

$\therefore A\subset B$ .

Let A and B be two sets such that: $n(A)=20, n(A\cup B)=42$ and $n(A\cap B)=4$ . Find $n(B)$ .

$n(A)=20$

$n(A \cup B)=42$

$n(A \cap B)=4$

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$42=20+n(B)-4$

$\implies n(B)=26$ .

Let A and B be two sets such that: $n(A)=20, n(A\cup B)=42$ and $n(A\cap B)=4$ . Find $n(B-A)$ .

$n(A)=20$

$n(A\cup B)=42$

$n(A\cap B)=4$

$n(B)=n(A\cup B)+n(A\cap B)-n(A)$

$n(B)=42+4-20$

$n(B)=26$

$n(B-A)=n(B)-n(A\cap B)$

$n(B-A)=26-4$

$\Rightarrow 22$ .

Is the following represent a set? justify your answer.
A team of $11$ best cricket players of India.

The term best is vague. So, the given collection is not a set.

Is the following represent a set? justify your answer.The collection of all question in this chapter.

The collection of all question in the chapter is a set because, if a question is given then we can easily decide whether it is a question of the chapter or not and it can be counted (so, well-defined). So, it's a set.

Which of the following are sets? justify your answer.
The collection of all persona of Kolkata whose assessed annual incomes exceed (say) $Rs\ 20\ lakh$ in the financial year $2016-17$

The given collection is clearly well defined. So it is a set

Is the following represent a set? justify your answer.
The collection of all the months of the year whose names begin with the letter $M$ .

The given collection has definite members. namely March and May. So, this collection is a set.

Which of the following are sets? justify your answer.
The collection of all interesting books.

The term interesting is vague. So, the given collection is not a set

Which of the following are sets? justify your answer.
The collection of all rich persons of Kolkata

The term rich is vague. So, the given collection is not a set

Is the following represent a set? justify your answer.
A collection of Hindi novels written by Munshi Prem Chand.

Suppose we are given a collection of Hindi novels written by Munshi Prem Chand, Now, if we take any Hindi novel then we can clearly decide whether it belongs to our collection or not.
So, the given collection is a set.

Which of the following are sets? justify your answer.
The collection of all those students of your class whose ages exceed $15$ years.

Clearly it contains definite members. So, it is a set

Is the following represent a set? justify your answer.
The collection of all difficult chapters in this book.

Clearly the term difficult is vague.So, the given collection is not well defined and therefore, it is not a set.

Which of the following are sets? justify your answer.
The collection of all short boys of your class.

The term short is vague. So the given collection is not a set

State in the case whether $A\subset B$ or $A\nsubseteq B$ .
$A=\left\{ x:x\ is\ an\ even\ natural\ number\right\}, B=\left\{ x:x\ is\ an\ integer\right\}$

$A=x\ is\ an\ even\ natural\ number$

$A=\left\{ 2, 4, 6, 8,....\right\}$

$B=$

$x\ is\ an\ integer$

$B=\left\{ ...., -1, 0, 1, 2, 3, 4, 5, 6, 7, 8,....\right\}$ .

So all the elements of A are present in B, therefore, we can write,

$A\subset B$ .

If $A=\left\{a,b,c,d,e,f\right\}, B=\left\{c,e,g,h\right\}$ and $C=\left\{a,e,m,n\right\}$ , find
$B\cup C$

The union of two sets B and C, is the set of elements which are in B or in C or in both. It is denoted by $B \cup C$ and is read "B union C".

$\therefore$ "B union C" can be written as,

$B \cup C=$

$\left\{a,c,e,g,h,m,n \right\}$

State in the case whether $A\subset$ or $A\nsubseteq B$ .
$A=\left\{ x:x\ is\ a\ real\ number\right\}, B=\left\{x:x\ is\ a\ complex\ number\right\}$

We may write a real number

$x$ as

$x+0i$ . So, every real number is a complex number.

As also can be explained from the above image that each Real number is a Complex number.

$\therefore A \subset B$
1608262_1708097_ans_70a1bb2450104cbfa363dffc04e3bd71.PNG

State in the case whether $A\subset$ or $A\nsubseteq B$ .
$A=\left\{ 0, 1, 2, 3\right\}, B=\left\{ 1, 2, 3, 4, 5\right\}$

Since, the element

$0$ of set

$A$ does not belong to set

$B$

So,

$A$ is not a subset of

$B.$

$\therefore A\nsubseteq B$

If $A=\left\{a,b,c,d,e,f\right\}, B=\left\{c,e,g,h\right\}$ and $C=\left\{a,e,m,n\right\}$ , find
$A\cup B$

The union of two sets A and B, is the set of elements which are in A or in B or in both. It is denoted by $A \cup B$ and is read "A union B".

$\therefore$ "A union B" can be written as,

$A \cup B=$

$\left\{a,b,c,d,e,f,g,h \right\}$

Which of the following are sets? justify your answer.
The collection of all interesting dramas written by Shakespeare.

The term interesting is vague. So, the given collection is not a set

State in the case whether $A\subset$ or $A\nsubseteq B$ .
$A=\left\{ x:x\ is\ an\ integer\right\}, B=\left\{ x:x\ is\ a\ rational\ number\right\}$

The symbol

$⊆$ means is a subset of. The symbol

$⊂$ means is a proper subset of.

Since, every integer is a rational number.

So, in

$A=\{x:x$ is an integer

$\},$

$B=\{x:x$ is a rational number

$\}$

So,

$A \subset B$

1608218_1708088_ans_c335a5f1dcc04613946c0741a270a72d.PNG

State in the case whether $A\subset$ or $A\nsubseteq B$ .
$A=\phi, B=\{0\}$

The empty set

$\{ \},$ denoted by

$\phi$ , is also a subset of any given set

$X.$ It is also always a proper subset of any set except itself.

So in,

$A=\phi, B=\{0\}$

$A\subset B$

State in the case whether $A\subset$ or $A\nsubseteq B$ .
$A=\{1, 2, 3\}, B=\{1, 2, 4\}$

The symbol

$⊆$ means "is a subset of". The symbol

$⊂$ means "is a proper subset of".

Since,

$3 \in A$ does not belong to

$B$

So, in

$A=\{1, 2, 3\}, B=\{1, 2, 4\}$

$A \nsubseteq B$

State in the case whether $A\subset$ or $A\nsubseteq B$ .
$A=\left\{ x:x\ is\ an\ isosceles\ triangle\ in\ a\ plane\right\}$ ,
$B=\left\{ x:x\ is\ an\ equilateral\ triangle\ in\ the\ same\ plane\right\}$

The symbol

$⊆$ means is a subset of. The symbol

$⊂$ means is a proper subset of.

$A=\left\{ x:x\ is\ an\ isosceles\ triangle\ in\ a\ plane\right\}$ ,

$B=\left\{ x:x\ is\ an\ equilateral\ triangle\ in\ the\ same\ plane\right\}$

Since, no isosceles triangle is a equilateral triangle.

$A\nsubseteq B$

State in the case whether $A\subset$ or $A\nsubseteq B$ .
$A=\left\{ x:x\ is\ a\ square\ in\ a\ plane\right\}, B=\left\{ x:x\ is\ a\ rectangle\ in\ the\ same\ plane\right\}$

The symbol

$⊆$ means is a subset of. The symbol

$⊂$ means is a proper subset of.

Since, every square is a rectangle.

So, in

$A=\left\{ x:x\ is\ a\ square\ in\ a\ plane\right\}, B=\left\{ x:x\ is\ a\ rectangle\ in\ the\ same\ plane\right\}$

So,

$A\subset B$

If $A=\left\{1,2,3,4,5 \right\}, B=\left\{4,5,6,7,8 \right\}, C=\left\{7,8,9,10,11 \right\}$ and $D=\left\{10,11,12,13,14 \right\}$ , find :
$A\cup C$

Union of two sets. The union of two sets A and B is the set of elements which are in A, in B, or in both A and B.

So in,

$A=\left\{1,2,3,4,5 \right\}, B=\left\{4,5,6,7,8 \right\}, C=\left\{7,8,9,10,11 \right\}$ and

$D=\left\{10,11,12,13,14 \right\}$

$A\cup C=$

$\left\{1,2,3,4,5,7,8,9,10,11 \right\}$

If $A=\left\{1,2,3,4,5 \right\}, B=\left\{4,5,6,7,8 \right\}, C=\left\{7,8,9,10,11 \right\}$ and $D=\left\{10,11,12,13,14 \right\}$ , find :
$B\cup D$

Union of two sets. The union of two sets A and B is the set of elements which are in A, in B, or in both A and B.

So, in

$A=\left\{1,2,3,4,5 \right\}, B=\left\{4,5,6,7,8 \right\}, C=\left\{7,8,9,10,11 \right\}$ and

$D=\left\{10,11,12,13,14 \right\}$

$B\cup D=$

$\left\{4,5,6,7,8,10,11,12,13,14 \right\}$

Examine whether the following statements is true or false:
$\left\{ a, b\right\} \nsubseteq \left\{ b, c,a\right\}$

1972980_1708129_ans_fecf31da897844de8eda0aeb8053aaf0.jpeg

State in the case whether $A\subset$ or $A\nsubseteq B$ .
$A=\left\{x:x\ is\ a\ square\ in\ a \ plane\right\}, B=\left\{x:x\ is\ a\ rectangle\ in\ the\ same\ plane\right\}$

The symbol

$⊆$ means is a subset of. The symbol

$⊂$ means is a proper subset of.

Since, every square is a rectangle.

$So,\,in$

$A=\{x:x\text{ is a square in plane }\!\!\}\!\!\text{ }$

$\text{B= }\!\!\{\!\!\text{ x: x is a rectangle in the same plane }\!\!\}\!\!\text{ }$

$\text{So,}\,\text{A}\subset \text{B}$

Examine whether the following statements are true or false:
$\left\{ a\right\} \in \left\{ a, b,c\right\}$

$\{a\}\in \{a,b,c\}$

Given set

$\{a\}$ belongs to

$\{a,b,c\}$

Here, set

$\{a\}$ is not an elements of

$\{a,b,c\}$

As elements of

$\{a,b,c\}$ are

$a,b,c$

$\therefore$ The statement is false.

Examine whether the following statements are true or false:
$\left\{ \phi \right\} \subset \left\{ a, b,c\right\}$

The empty set is a subset of every set.

The empty set is a proper subset of every set except for the empty set.

$\phi \subset \left\{ a, b, c\right\}$ is true and

$\{\ \phi \}\ \subset \left\{ a, b, c\right\}$ is false.

If $A=\left\{1,2,3,4,5 \right\}, B=\left\{4,5,6,7,8 \right\}, C=\left\{7,8,9,10,11 \right\}$ and $D=\left\{10,11,12,13,14 \right\}$ , find :
$(A\cup B) \cup C$

Union of two sets. The union of two sets A and B is the set of elements which are in A, in B, or in both A and B.

$A=\left\{1,2,3,4,5 \right\}, B=\left\{4,5,6,7,8 \right\}, C=\left\{7,8,9,10,11 \right\}$ and

$D=\left\{10,11,12,13,14 \right\}$

$A \cup B=\{1,2,3,4,5,6,7,8 \}$

$(A\cup B) \cup C=$

$\left\{1,2,3,4,5,6,7,8,9,10,11 \right\}$

State in the case whether $A\subset$ or $A\nsubseteq B$ .
$A=\left\{x:x\ is\ an\ even\ natural\ number\ less\ than\ 8\right\}, B=\left\{ x:x\ is\ a \ natural\ number\ which\ divides\ 32\right\}$

$A=\left\{x:x\ is\ an\ even\ natural\ number\ less\ than\ 8\right\}$

$A=\{2,4,6\}$

$B=\left\{ x:x\ is\ a \ natural\ number\ which\ divides\ 32\right\}$

$B=1, 2, 4, 8, 16, 32$

$A=\left\{ 2, 4, 6\right\}$ and

$B=\left\{ 1, 2, 4, 8, 16, 32\right\}$ . So

$A\nsubseteq B$ .

If $A=\left\{1,2,3,4,5 \right\}, B=\left\{4,5,6,7,8 \right\}, C=\left\{7,8,9,10,11 \right\}$ and $D=\left\{10,11,12,13,14 \right\}$ , find :
$B\cup C$

Union of two sets. The union of two sets A and B is the set of elements which are in A, in B, or in both A and B.

So in,

$A=\left\{1,2,3,4,5 \right\}, B=\left\{4,5,6,7,8 \right\}, C=\left\{7,8,9,10,11 \right\}$ and

$D=\left\{10,11,12,13,14 \right\}$

So,

$B\cup C=$

$\left\{4,5,6,7,8,9,10,11 \right\}$

If $A$ and $B$ are two sets such that $A\subseteq B$ then find :
$A\cup B$

1972436_1708198_ans_eb3d61f2f5104a9cb76ebf37d53da91e.jpeg

Examine whether the following statements are true or false:
$A=$ set of all circles of unit radius in a plane and $B=$ set of all circles in the same plane then $A\subset B$ .

1972417_1708164_ans_29b1b78e012c4d2a91ffcd47961bd0b2.jpeg

If $A=\left\{2x : x \in N\ and\ 1 \le x < 4 \right\}, B=\left\{(x+2): x \in N\ and\ 2\le x < 5 \right\}$ and $C=\left\{x : x \in N\ and\ 4 < x < 8 \right\}$ , find
$A\cup B$

Union of two sets. The union of two sets A and B is the set of elements which are in A, in B, or in both A and B.

In mathematics, the intersection of two sets

$A$ and

$B,$ denoted by

$A ∩ B,$ is the set containing all elements of

$A$ that also belong to

$B ($ or equivalently, all elements of

$B$ that also belong to

$A).$

$A=\left\{2x : x \in N\ and\ 1 \le x < 4 \right\}=\{2,4,6\}$

$B=\left\{(x+2): x \in N\ and\ 2\le x < 5 \right\}=\{4,5,6\}$

$C=\left\{x : x \in N\ and\ 4 < x < 8 \right\}=\{5,6,7\}$

$A\cup B=\{2,4,5, 6\}$

Examine whether the following statements are true or false:
$\left\{ a, e\right\} \subset \left\{ x:x\ is\ a\ vowel\ in\ the\ English\ alphabet \right\}$

we know that

english vowels

$=a, e, i, o, u$

$\left\{ a, e\right\} \subset \left\{ a, e, i, o, u \right\}$ is true.

If $A=\left\{3, \left\{ 4, 5\right\}, 6\right\}$ .
State whether the following statement is true or not.
$\left\{ \left\{ 4, 5\right\} \right\} \subseteq A$

The given statement is true.

If $A=\left\{3, \left\{ 4, 5\right\}, 6\right\}$ ,
$\left\{4, 5\right\}\subseteq A$

The given statement is true.

If $A=\left\{2,4,6,8,10,12 \right\}$ and $B=\left\{3,4,5,6,7,8,10 \right\}$ , find
$(A-B)\cup (B-A)$

Given:

A = { 2,4,6,8,10,12} and B = {3,4,5,6,6,8,10}

A - B = Elements in set A but not in set B

Hence A - B = { 2, 12 }

B - A = Elements in set B but not in set A

Hence B - A = { 3, 5, 7 }

Now, ( A - B ) U ( B - A ) is the set of those elements which belongs to both the sets.

Hence ( A - B ) U ( B - A ) = { 2, 3, 5, 7, 12 }

State whether the given statement is true or false:
If $A\subset B$ and $x\notin B$ then $x\notin A$ .

1972445_1708264_ans_286e421732b647e5a445510575c38db6.jpeg

If $A=\left\{a,b,c,d,e\right\},B=\left\{a,c,e,g\right\}$ and $C=\left\{b,c,f,g\right\}$ , verify that:
$B\cup C=C\cup B$

Given

$B=\left\{ a,c,e,g \right\}$ and

$C\left\{ b,c,f,g \right\}$

$B\cup C=\left\{ a,c,e,g,b,f \right\}$ Equation (1)

$C\cup B=\left\{ b,c,f,g,a,e \right\}$ Equation (2)

As sequence of elements of sets is immaterial, from equation (1) and (2), we can conclude that,

$B\cup C=C\cup B$

Prove that $A\subseteq B, B\subseteq C$ and $C\subseteq A\Rightarrow A=C$ .

${\textbf{Step -1: Observe the cases given .}}$

${\text{A is subset of B}}$

$A \subseteq B..............(1)$

${\text{B is subset of C}}$

$B \subseteq C..............(2)$

${\text{C is subset of A}}$

$C \subseteq A...............(3)$

${\textbf{Step -2: From equation 1 and 2 we get}}$

$A\subseteq B$ ${\text{and}}$ $B\subseteq C$

${\text{so,}}$ $A\subseteq C$

${\textbf{Step-3: Now from equation 3 and 4 we get}}$

$C\subseteq A$ ${\text{and}}$ $A\subseteq C$

${\textbf{Hence, A=C}}$

If $U$ is the universal set and $A\subset U$ then fill in the blanks :
$\phi ' \cup A =......$

$\phi ' \cup A =$

$A$

If $A=\left\{3, \left\{ 4, 5\right\}, 6\right\}$ ,
$\left\{ 3, 4, 5\right\} \subseteq A$

The given statement is False.

If $A=\left\{3, \left\{ 4, 5\right\}, 6\right\}$ ,
$\left\{ 3, 6\right\} \subseteq A$

The given statement is True.

If $A=\left\{3, \left\{ 4, 5\right\}, 6\right\}$ ,
$\phi \subseteq A$

The given statement is True.

If $U$ is the universal set and $A\subset U$ then fill in the blanks :
$A\cup A' =......$

$A'=$ complement of

$A=U-A$

$\therefore A\cup A'=U$

If $A=\left\{3, \left\{ 4, 5\right\}, 6\right\}$ , State whether the following statement is true or not.
$\left\{ \phi \right\} \subseteq A$

The given statement is False.

For any sets $A$ and $B$ . prove that:
$A\cap B'=\phi\Rightarrow A\subset B$

1972443_1708325_ans_712125b417814a6ea05a2210f9832ee4.jpeg

State whether the given statement is true or false:
If $A, B$ and $C$ are three sets such that $A\in B$ and $B\subset C$ then $A\subset C$ .

1972440_1708280_ans_0c53b9a21fd2445c9ac5ad14f8a95f6a.jpeg

State whether the following pairs of sets are equal or not:
$A =\{2, 4, 6, 8,10\},\quad B = \{\text{even natural numbers}\}$

Here we have,

$A = \{2, 4, 6, 8, 10\}$

$B = \{0,2,4,6,8,10,12.........\}$

$\therefore A\neq B$

For any sets $A$ and $B$ . prove that:
$A'\cup B=U\Rightarrow A\subset B$

$A'\cup B=U\Rightarrow A\cap (A'\cup B)=A\cap U\Rightarrow (A\cap A')\cup (A\cap B)=A$

$\Rightarrow \phi\cup (A\cap B)=A\Rightarrow A\cap B=A\Rightarrow A\subset B$ .

State whether the following pairs of sets are equal or not:
$A = \{x|x\text{ is a letter in the word SOPHIA}\}$
$B = \{x|x\text{ is a letter in the word MUMTAZ}\}$

$A = \{x|x\text{ is a letter in the word SOPHIA}\}$

$\Rightarrow A= \{S, 0, P, H, I, A\}$

$B = \{x|x\text{ is a letter in the word MUMTAZ}\}$

$\Rightarrow B = \{M, U, T, A, Z\}$

$\therefore$

$A\neq B$

State whether the given statement is true or false:
If $A\subseteq \phi$ then $A=\phi$ .

The given statement is False.

If $A\subset B$ , show that $(B' -A')=\phi$

1972449_1708379_ans_4668802a17614e4281e9f84bb0ce285b.jpeg

State whether the given statement is true or false:If $A, B$ and $C$ are three sets such that $A\nsubseteq B$ and $B\nsubseteq C$ then $A\nsubseteq C$ .

1972984_1708291_ans_29b6237514fd400ab5f09859786ea045.jpeg

State whether the given statement is true or false:
If $A, B$ and $C$ are three sets such that $A\subset B$ and $B\in C$ then $A\in C$ .

Let

$A=\{\ a\}\, B=\left\{ \left\{ a\right\}, b\right\}$ and

$C=\left\{\left\{ a \right\}, b, c\right\}$ .

Then,

$A\subset B$ and

$B\in C$ . But,

$A\notin C$ .

Hence given statement is false.

If $A\subset B$ , prove that $B' \subset A'$ .

$Let\,$

$x\in A$

$Then$

$x\in A\Rightarrow x\notin A'$

$\Rightarrow x\notin B'$

$\Rightarrow x\in A'\,\,\,\,\left[ B'\subset A' \right]$

$\therefore A\subset B$

Let $A= (1, 2, (3, 4), 5)$ . Which of the following statements are incorrect and why?
(i) $(3, 4) A$
(ii) $(3, 4)\epsilon A$
(iii) $((3, 4)) A$
(iv) $1 A$
(v) $1 A$
(vi) $(1, 2, 5) A$
(vii) $(1, 2, 5) A$
(viii) $(1, 2, 3) A$
(ix) $\not{\bigcirc} A$
(x) $\not{\bigcirc} A$
(xi) $(\not{\bigcirc}) A$

It is given that

$A= (1, 2, (3, 4), 5)$
(i)The statement

$(3, 4) ⊂ A$ is incorrect because

$3 ∈ (3, 4)$ ; however,

$3∉A$ .
(ii) The statement

$(3, 4) ∈ A$ is correct because

$(3, 4)$ is an element of

$A$ .
(iii) The satement

$((3, 4)) ⊂ A$ is correct because

$(3, 4) ∈ ((3, 4))$ and

$(3, 4) ∈ A$ .
(iv) The statement

$1∈A$ is correct because

$1$ is an element of

$A$ .
(v) The statement

$1⊂ A$ is incorrect because an element of a set can never be a subset of itself.
(vi) The statement

$(1, 2, 5) ⊂ A$ is correct because each element of

$(1, 2, 5)$ is also an element of

$A$ .
(vii) The statement

$(1, 2, 5) ∈ A$ is incorrect because

$(1, 2, 5)$ is not an element of

$A$ .
(viii) The statement

$(1, 2, 3) ⊂ A$ is incorrect because

$3 ∈ (1, 2, 3)$ ; however,

$3 ∉ A$ .
(ix) The statement

$\not{\bigcirc} ∈ A$ is incorrect because

$\not{\bigcirc}$ is not an element of

$A$ .
(x) The statement

$\not{\bigcirc} ⊂ A$ is correct because

$\not{\bigcirc}$ is a subset of every set.
(xi) The statement

$(\not{\bigcirc}) ⊂ A$ is incorrect because

$\not{\bigcirc}$ is a subset of

$A$ and it is not an element of

$A$ .

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

2172833_1797543_ans_737b72ed74964dc5b0c4aae29661e859.jpg

The collection of all months of a year beginning with the letter $J$

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.

If $A$ = { $x: x$ is a natural number}, $B$ ={ $x: x$ is an even natural number}, $C$ = { $x: x$ is an odd natural number} and $D$ = { $x: x$ is a prime number}, find
$A B$

It can be written as

$A$ = {

$x: x$ is a natural number} =

$\left \{ 1, 2, 3, 4, 5 … \right \}$

$B$ = {

$x: x$ is an even natural number} =

$\left \{ 2, 4, 6, 8 … \right \}$

$C$ = {

$x: x$ is an odd natural number} =

$\left \{ 1, 3, 5, 7, 9 … \right \}$

$D$ = {

$x: x$ is a prime number} =

$\left \{ 2, 3, 5, 7 … \right \}$

$A ∩B$ = {

$x: x$ is a even natural number} =

$B$

State whether the following pairs of sets are equal or not:
$A = \{\text{kids 5 metres tall}\}, B = \{x : x \in N \text{ and }\,2x = 3\}$

We have,

$A = \{\text{kids 5 metres tall}\}$

$\Rightarrow A = \{\}\qquad$ [ As kids are never that tall.]

$\Rightarrow$ A is empty set

$B= \{X : X \in N \,and \,2x = 3\}$

$\Rightarrow B = \{\} \Rightarrow$ B is empty set

$\therefore A = B$

Which of the following are sets? Justify our answer.
The collection of ten most talented writers of India.

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.

Let $U = \left \{ 1, 2, 3; 4, 5, 6, 7, 8, 9 \right \}, A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 2, 4, 6, 8 \right \}$ and $C = \left \{ 3, 4, 5, 6 \right \}$ . Find
$(B C)$

$U = \left \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \right \}$

$A = \left \{ 1, 2, 3, 4 \right \}$

$B = \left \{ 2, 4, 6, 8 \right \}$

$C = \left \{ 3, 4, 5, 6 \right \}$

$B – C = \left \{ 2, 8 \right \}$

$\therefore (B – C)’ = \left \{ 1, 3, 4, 5, 6, 7, 9 \right \}$

If $A$ = { $x: x$ is a natural number}, $B$ ={ $x: x$ is an even natural number}, $C$ = { $x: x$ is an odd natural number} and $D$ = { $x: x$ is a prime number}, find
$A D$

It can be written as

$A$ = {

$x: x$ is a natural number} =

$\left \{ 1, 2, 3, 4, 5 … \right \}$

$B$ = {

$x: x$ is an even natural number} =

$\left \{ 2, 4, 6, 8 … \right \}$

$C$ = {

$x: x$ is an odd natural number} =

$\left \{ 1, 3, 5, 7, 9 … \right \}$

$D$ = {

$x: x$ is a prime number} =

$\left \{ 2, 3, 5, 7 … \right \}$

$A ∩ D$ = {

$x: x$ is a prime number} =

$D$

If $A$ = { $x: x$ is a natural number}, $B$ ={ $x: x$ is an even natural number}, $C$ = { $x: x$ is an odd natural number} and $D$ = { $x: x$ is a prime number}, find
$B C$

It can be written as

$A$ = {

$x: x$ is a natural number} =

$\left \{ 1, 2, 3, 4, 5 … \right \}$

$B$ = {

$x: x$ is an even natural number} =

$\left \{ 2, 4, 6, 8 … \right \}$

$C$ = {

$x: x$ is an odd natural number} =

$\left \{ 1, 3, 5, 7, 9 … \right \}$

$D$ = {

$x: x$ is a prime number} =

$\left \{ 2, 3, 5, 7 … \right \}$

$B ∩ C = \phi$

If $A$ = { $x: x$ is a natural number}, $B$ ={ $x: x$ is an even natural number}, $C$ = { $x: x$ is an odd natural number} and $D$ = { $x: x$ is a prime number}, find
$C D$

It can be written as

$A$ = {

$x: x$ is a natural number} =

$\left \{ 1, 2, 3, 4, 5 … \right \}$

$B$ = {

$x: x$ is an even natural number} =

$\left \{ 2, 4, 6, 8 … \right \}$

$C$ = {

$x: x$ is an odd natural number} =

$\left \{ 1, 3, 5, 7, 9 … \right \}$

$D$ = {

$x: x$ is a prime number} =

$\left \{ 2, 3, 5, 7 … \right \}$

$C ∩ D$ = {

$x: x$ is odd prime number}

If $A$ = { $x: x$ is a natural number}, $B$ ={ $x: x$ is an even natural number}, $C$ = { $x: x$ is an odd natural number} and $D$ = { $x: x$ is a prime number}, find
$A C$

It can be written as

$A$ = {

$x: x$ is a natural number} =

$\left \{ 1, 2, 3, 4, 5 … \right \}$

$B$ = {

$x: x$ is an even natural number} =

$\left \{ 2, 4, 6, 8 … \right \}$

$C$ = {

$x: x$ is an odd natural number} =

$\left \{ 1, 3, 5, 7, 9 … \right \}$

$D$ = {

$x: x$ is a prime number} =

$\left \{ 2, 3, 5, 7 … \right \}$

$A ∩ C$ = {

$x: x$ is an odd natural number} =

$C$ .

Let $U = \left \{ 1, 2, 3; 4, 5, 6, 7, 8, 9 \right \}, A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 2, 4, 6, 8 \right \}$ and $C = \left \{ 3, 4, 5, 6 \right \}$ . Find
$B$

$U = \left \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \right \}$

$A = \left \{ 1, 2, 3, 4 \right \}$

$B = \left \{ 2, 4, 6, 8 \right \}$

$C = \left \{ 3, 4, 5, 6 \right \}$

$B’ = \left \{ 1, 3, 5, 7, 9 \right \}$

If $U = \left \{ 1, 2, 3, 4, 5,6,7,8, 9 \right \}, A = \left \{ 2, 4, 6, 8 \right \}$ and $B = \left \{ 2, 3, 5, 7 \right \}$ . Verify that
$(A \cup B)’ = A’ ∩ B’$

It is given that

$U = \left \{ 1, 2, 3, 4, 5,6,7,8, 9 \right \}$

$A = \left \{ 2, 4, 6, 8 \right \}$

$B = \left \{ 2, 3, 5, 7 \right \}$

$(A \cup B)’ = \left \{ 2, 3, 4, 5, 6, 7, 8 \right \}' = \left \{ 1, 9 \right \}$

$A’ ∩ B’ = \left \{ 1, 3, 5, 7, 9 \right \} ∩ \left \{ 1, 4, 6, 8, 9 \right \} = \left \{ 1, 9 \right \}$
Therefore,

$(A \cup B)’ = A’ ∩ B’$ .

Let $U = \left \{ 1, 2, 3; 4, 5, 6, 7, 8, 9 \right \}, A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 2, 4, 6, 8 \right \}$ and $C = \left \{ 3, 4, 5, 6 \right \}$ . Find
$(A \cup B)$

$U = \left \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \right \}$

$A = \left \{ 1, 2, 3, 4 \right \}$

$B = \left \{ 2, 4, 6, 8 \right \}$

$C = \left \{ 3, 4, 5, 6 \right \}$

$A \cup B = \left \{ 1, 2, 3, 4, 6, 8 \right \}$

$\therefore (A \cup B)’ = \left \{ 5, 7, 9 \right \}$

Let $U = \left \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \right \}, A = \left \{ 1, 2, 3, 4 \right \}, B = \left \{ 2, 4, 6, 8 \right \}$ and $C = \left \{ 3, 4, 5, 6 \right \}$ . Find
$(A’)’$

$U = \left \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \right \}$

$A = \left \{ 1, 2, 3, 4 \right \}$

$B = \left \{ 2, 4, 6, 8 \right \}$

$C = \left \{ 3, 4, 5, 6 \right \}$

$(A’)’ = A = \left \{ 1, 2, 3, 4 \right \}$

If $A$ = { $x: x$ is a natural number}, $B$ ={ $x: x$ is an even natural number}, $C$ = { $x: x$ is an odd natural number} and $D$ = { $x: x$ is a prime number}, find
$B D$

It can be written as

$A$ = {

$x: x$ is a natural number} =

$\left \{ 1, 2, 3, 4, 5 … \right \}$

$B$ = {

$x: x$ is an even natural number} =

$\left \{ 2, 4, 6, 8 … \right \}$

$C$ = {

$x: x$ is an odd natural number} =

$\left \{ 1, 3, 5, 7, 9 … \right \}$

$D$ = {

$x: x$ is a prime number} =

$\left \{ 2, 3, 5, 7 … \right \}$

$B ∩ D = \left \{ 2 \right \}$ .

Show that if $A B$ , then $C B C A$ .

Let

$A\subset B$
To show:

$C – B ⊂ C – A$
Let

$x ∈ C – B$

$\Rightarrow x ∈ C$ and

$x ∉ B$

$\Rightarrow x ∈ C$ and

$x ∉ A[A ⊂ B]$

$\Rightarrow x ∈ C – A$

$\therefore C – B ⊂ C – A$ .

State which of the following collections are set:
All the beautiful flowers.

It is not a set because the collection is not defined.

State which of the following collections are set:
Four colours of a rainbow.

It is not a set because the collection is not defined.

State which of the following collections are set:
All clever people of Lucknow.

It is not a set because the collection is not defined.

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 $X Y$ have?

It is given that,

$n (X \cup Y) = 18, n (X) = 8,n (Y) = 15$

$n(X\cap Y)=$ ?
We know that,

$n (X \cup Y) = n (X) + n (Y) – n (X \cap Y)$

$\therefore 18 = 8 + 15 - n (X ∩ Y)$

$\Rightarrow n (X ∩ Y) = 23 – 18 = 5$

$\therefore n (X ∩ Y) = 5$

State which of the following collections are set:
All tall students of your school.

It is not a set because the collection is not defined.

State which of the following collections are set:
Four cities of India having more than one lakh population.

It is not a set because the collection is not defined

Fill in the blanks to make each of the following a true statement:
$A \cup A = .$ .

$A \cup A’ = U$

If $\xi=\{1,2,3, \ldots .9\}, A=\{1,2,3,4,6,7,8\}$ and $B=\{4,6,8\},$ then find.
(i) $A^{\prime}$
(ii) $B^{\prime}$
(iii) $A \cup B$
(iv) $\mathrm{A} \cap \mathrm{B}$
(v) $A-B$
(vi) $B-A$
(vii) $(A \cap B)^{\prime}$
(viii) $A^{\prime} \cup B^{\prime}$
Also verify that:
(a) $(A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}$
(b) $n(A)+n\left(A^{\prime}\right)=n(\xi)$
(c) $n(A \cap B)+n\left((A \cap B)^{\prime}\right)=n(\xi)$
(d) $n(A-B)+n(B-A)+n(A \cap B)=n(A \cup B)$

From the question it is given that,

$\xi=\{1,2,3,4,5,6,7,8,9\}$

$A=\{1,2,3,4,6,7,8\}$

$B=\{4,6,8\}$
(i)

$\mathrm{A}^{\prime}=\xi-\mathrm{A}=\{1,2,3,4,5,6,7,8,9\}-\{1,2,3,4,6,7,8\}$

$\mathrm{A}^{\prime}=\{5,9\}$
(ii)

$\mathrm{B}^{\prime}=\xi-\mathrm{B}=\{1,2,3,4,5,6,7,8,9\}-\{4,6,8\}$

$B^{\prime}=\{1,2,3,5,7,9\}$
(iii)

$A \cup B=\{1,2,3,4,6,7,8\} \cup\{4,6,8\}$

$A \cup B=\{1,2,3,4,6,7,8\}$
(iv)

$\mathrm{A} \cap \mathrm{B}=\{1,2,3,4,6,7,8\} \cap\{4,6,8\}$

$A \cap B=\{4,6,8\}$
(v)

$\mathrm{A}-\mathrm{B}=\{1,2,3,4,6,7,8\}-\{4,6,8\}$

$A-B=\{1,2,3,7\}$
(vi)

$\mathrm{B}-\mathrm{A}=\{4,6,8\}-\{1,2,3,4,6,7,8\}$

$B-A=\{\}$
(vii)

$(\mathrm{A} \cap \mathrm{B})^{\prime}=\xi-(\mathrm{A} \cap \mathrm{B})$

$(\mathrm{A} \cap \mathrm{B})^{\prime}=\{1,2,3,4,5,6,7,8,9\}-\{4,6,8\}$

$(\mathrm{A} \cap \mathrm{B})^{\prime}=\{1,2,3,5,7,9\}$
(viii)

$\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}=\{5,9\} \cup\{1,2,3,5,7,9\}$

$\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}=\{1,2,3,5,7,9\}$
Then,

$n(\xi)=9$

$n(A)=7$

$n\left(A^{\prime}\right)=2$

$n\left(B^{\prime}\right)=6$

$n(A \cap B)=3$

$n\left((A \cap B)^{\prime}\right)=6$

$n\left(A^{\prime} \cup B^{\prime}\right)=6$

$n(A-B)=4$

$n(B-A)=0$

$n(A \cup B)=7$
(a)

$(A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}$

$(\mathrm{A} \cap \mathrm{B})^{\prime}=\{1,2,3,5,7,9\}$

$\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}=\{1,2,3,5,7,9\}$
By comparing the results,

$(A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}$
(b)

$n(A)+n\left(A^{\prime}\right)=n(\xi)$

$7+2=9$

$9=9$
Therefore, by comparing the results,

$n(A)+n\left(A^{\prime}\right)=n(\xi)$
(c)

$n(A \cap B)+n\left((A \cap B)^{\prime}\right)=n(\xi)$

$3+6=9$

$9=9$
Therefore, by comparing the results,

$n(A \cap B)+n\left((A \cap B)^{\prime}\right)=n(\xi)$
(d)

$n(A-B)+n(B-A)+n(A \cap B)=n(A \cup B)$

$4+0+3=7$

$7=7$
Therefore, by comparing the results,

$n(A-B)+n(B-A)+n(A \cap B)=n(A \cup B)$

If $\mathrm{A}\{\mathrm{x}: \mathrm{x} \in \mathrm{N} \text { and } 3<\mathrm{x}<7\}$ and $\mathrm{B}=\{\mathrm{x}: \mathrm{x} \in \mathrm{W} \text { and } \mathrm{x} \leq 4\},$ find
$\mathrm{B}-\mathrm{A}$

From the question it is given that,

$A=\{x: x \in N \text { and } 3<x<7\}$

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

$B=\{x: x \in W \text { and } x \leq 4\}$

$B=\{0,1,2,3,4\}$ Then,

$\mathrm{B}-\mathrm{A}$

$B-A=\{0,1,2,3\}$

State which of the following collections are set:
All months of a year having at least 30 days.

It is a set.

If $\xi=\{x: x \in W, x \leq 10\}, A .=\{x: x \geq 5\}$ and $B=\{x: 3 \leq x<8\},$ then verify that:
$B-A=B \cap A^{\prime}$

$\xi=\{0,1,2,3,4,5,6,7,8,9,10\}$

$A=\{5,6,7,8,9,10\}$

$B=\{3,4,5,6,7\}$

$\mathrm{B}-\mathrm{A}=\mathrm{B} \cap \mathrm{A}^{\prime}$
Consider the

$\mathrm{LHS}=\mathrm{B}-\mathrm{A}$

$\mathrm{B}-\mathrm{A}=\{3,4,5,6,7\}-\{5,6,7,8,9,10\}$

$B-A=\{3,4\}$
Therefore,

$L H S=B-A=\{3,4\}$
Now, consider

$\mathrm{RHS}=\mathrm{B} \cap \mathrm{A}^{\prime}$

$\mathrm{B} \cap \mathrm{A}^{\prime}=\{3,4,5,6,7\} \cap\{5,6,7,8,9,10\}$

$B \cap A^{\prime}=\{3,4\}$
Therefore,

$\mathrm{RHS}=\mathrm{B} \cap \mathrm{A}^{\prime}=\{3,4\}$
By comparing the

$L H S$ and

$R H S, B-A=B \cap A^{\prime}$

If $\mathrm{A}\{\mathrm{x}: \mathrm{x} \in \mathrm{N} \text { and } 3<\mathrm{x}<7\}$ and $\mathrm{B}=\{\mathrm{x}: \mathrm{x} \in \mathrm{W} \text { and } \mathrm{x} \leq 4\},$ find
$\mathrm{A} \cup \mathrm{B}$

From the question it is given that,

$A=\{x: x \in N \text { and } 3<x<7\}$

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

$B=\{x: x \in W \text { and } x \leq 4\}$

$B=\{0,1,2,3,4\}$ Then,

$\mathrm{A} \cup \mathrm{B}$

$A \cup B=\{0,1,2,3,4,5,6\}$

State which of the following collections are set:
Last three days of a week.

$A=\{Friday,Saturday, Friday\}$
It is a set

If $\xi=\{x: x \in W, x \leq 10\}, A .=\{x: x \geq 5\}$ and $B=\{x: 3 \leq x<8\},$ then verify that:
$A-B=A \cap B^{\prime}$

$\xi=\{0,1,2,3,4,5,6,7,8,9,10\}$

$A=\{5,6,7,8,9,10\}$

$B=\{3,4,5,6,7\}$

$A-B=A \cap B^{\prime}$
Consider the

$\mathrm{LHS}=\mathrm{A}-\mathrm{B}$

$A-B=\{5,6,7,8,9,10\}-\{3,4,5,6,7\}$

$A-B=\{8,9,10\}$
Therefore,

$\mathrm{LHS}=\mathrm{A}-\mathrm{B}=\{8,9,10\}$
Now, consider RHS

$=\mathrm{A} \cap \mathrm{B}^{\prime}$

$\left(A \cap B^{\prime}\right)=\{5,6,7,8,9,10\} \cap\{0,1,2,8,9,10\}$

$\left(A \cap B^{\prime}\right)=\{8,9,10\}$
Therefore,

$\mathrm{RHS}=\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)=\{8,9,10\}$
By comparing LHS and RHS,

$A-B=A \cap B^{\prime}$

If $\mathrm{A}\{\mathrm{x}: \mathrm{x} \in \mathrm{N} \text { and } 3<\mathrm{x}<7\}$ and $\mathrm{B}=\{\mathrm{x}: \mathrm{x} \in \mathrm{W} \text { and } \mathrm{x} \leq 4\},$ find
$A-B$

From the question it is given that,

$A=\{x: x \in N \text { and } 3<x<7\}$

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

$B=\{x: x \in W \text { and } x \leq 4\}$

$B=\{0,1,2,3,4\}$ Then,

$A-B$

$A-B=\{5,6\}$

If $n(\xi)=32, n(A)=20, n(B)=16$ and $n\left((A \cup B)^{\prime}\right)=4,$ find :
$n(A-B)$

From the question it is given that,

$\begin{array}{l}n(\xi)=32 \\n(A)=20 \\n(B)=16 \\n\left((A \cup B)^{\prime}\right)=4\end{array}$
Then,

$n(A-B)=n(A)-n(A \cap B)$

$\begin{array}{l}=20-8 \\=12\end{array}$
Therefore,

$n(A-B)=12$

Let set A = $\text{{6, 8, 10, 12}}$ and set B = $\text{{3, 9, 15, 18}}.$
Insert the symbol $\in$ or $\notin$ to make each of the following true :
$(i) 6 . A$
$(ii) 10 . B$
$(iii) 18 . B$
$(iv) (6 + 3) . B$
$(v) (15 9) . B$
$(vi) 12 . A$
$(vii) (6 + 8) . A$
$(viii) 6$ and $8 . A$

$(i) 6 \in A$

$(ii)10 \notin B$

$(iii)18 \in B$

$(iv) (6+3)$ or

$9 \in B$

$(v)(15-9)$ or

$6 \notin B$

$(vi)12 \in A$

$(vii)(6+8)$ or

$14 \notin A$

$(viii) 6$ and

$8 \in A$

State whether true or false:
(i) Set $\text{{4, 5, 8}}$ is same as the set $\text{{5, 4, 8}}$ and the set $\text{{8, 4, 5}}$
(ii) Sets $\text{{a, b, m, n}}$ and $\text{{a, a, m, b, n, n}}$ are same.
(iii) Set of letters in the word $suchismita$ is $\text{{s, u, c, h, i, m, t, a}}$
(iv) Set of letters in the word $MAHMOOD$ is $\text{{M, A, H, O, D}}$ .

(i) It is true.

(ii) It is true.

(iii) It is true as

$\text{{s, u, c, h, i, s, m, i, t, a}} = \text{{s, u, c, h, i, m, t, a}}$

(iv) It is true as it has the same elements.

$M = \{x : x \,is \,a \,natural \,number \,between \,0 \,and \,8)$ and $N = \{x : x \,is \,a \,natural \,number \,from \,5 \,to \,10\}$ . Find:
(i) $M-N$ and $n(M-N)$
(ii) $N-M$ and $n(N-M)$

We know that

Natural numbers between

$0$ and

$8 \,M=\{0, 1, 2, 3, 4, 5, 6, 7\}$

Natural numbers between

$5$ and

$10 \,N = \{6, 7, 8, 9, 10\}$

(i)

$M-N =\{1, 2, 3, 4\}$

$n(M-N) = 4$

(ii)

$N-M =\{8, 9, 10\}$

$n(N-M) = 3$

If $n(A-B)=12, n(B-A)=16$ and $n(A \cap B)=5,$ find:
(i) $n(A)$
(ii) $n(B)$
(iii) $n(A \cup B)$

From the question it is given that,

$\begin{array}{l}n(A-B)=12 \\n(B-A)=16 \\n(A \cap B)=5\end{array}$
(i)

$n(A)$
We know that,

$n(A)=n(A-B)+n(A \cap B)$

$\begin{array}{l}=12+5 \\=17\end{array}$
(ii)

$n(B)$
We know that,

$n(B)=n(B-A)+n(A \cap B)$

$\begin{aligned}&=16+5 \\n(B) &=21\end{aligned}$
(iii)

$n(A \cup B)$
We know that,

$n(A \cup B)=n(A)+n(B)-n(A \cap B)$

$\begin{array}{c}=17+21-5 \\=38-5 \\n(A \cup B)=33\end{array}$

Find, whether or not, each of the following collections represent a set:
(i) The collection of good students in your school.
(ii) The collection of the numbers between 30 and 45.
(iii) The collection of fat-people in your colony.
(iv) The collection of interesting books in your school library.
(v) The collection of books in the library and are of your interest.

(i) The collection of good students in your school is not a set as it is not well defined.
(ii) The collection of the numbers between 30 and 45 is a set.
(iii) The collection of fat-people in your colony is not a set as it is not well defined.
(iv) The collection of interesting books in your school library is not a set as it is not well defined.
(v) The collection of books in the library and are of your interest is a set.

If $n(\xi)=32, n(A)=20, n(B)=16$ and $n\left((A \cup B)^{\prime}\right)=4,$ find :
$n(A \cup B)$

From the question it is given that,

$\begin{array}{l}n(\xi)=32 \\n(A)=20 \\n(B)=16 \\n\left((A \cup B)^{\prime}\right)=4\end{array}$
Then,

$n(A \cup B)=n(\xi)-n\left((A \cup B)^{\prime}\right)$

$=32-4$

$n(A \cup B)=28$

Given
A = {Natural numbers less than $10$ }
B = {Letters of the word PUPPET}
C = {Squares of first four whole numbers}
D = {Odd numbers divisible by $2$ }
Find: $A \cap B$ and $(A \cup B)$

Given
A = {Natural numbers less than

1010}
B = {Letters of the word PUPPET}
C = {Squares of first four whole numbers}
D = {Odd numbers divisible by

22}

If A = { a, b, c, d, e}, B = { c, d, e, f }, C = {b, d}, D = { a, e}
then the following statement is true or false?
$B \ \subseteq \ A$

Since,

$f \ \epsilon \ B$ but f \ \not{\epsilon \ A$$
So,

$B \ \subseteq \ A$ is false.

If A = { a, b, c, d, e}, B = { c, d, e, f }, C = {b, d}, D = { a, e}
then the following statement is true or false?
$D \ \subseteq \ A$

Since, all elements of set

$D$ is in set

$A$ .
So,

$D \ \subseteq \ A$ is true.

If A = { a, b, c, d, e}, B = { c, d, e, f }, C = {b, d}, D = { a, e}
then the following statement is true or false?
$D \ \subseteq \ B$

Since,

$a \ \epsilon \ D$ but

$a \not{\epsilon} \ B$
So,

$D \ \subseteq \ B$ is false.

Given $A = {a, c}, B = {p, q, r}$ and $C =$ Set of digits used to form number $1351$ .
Write all the subsets of sets $A, B$ and $C.$

A =

$\text{{a, c}}$
Hence, the subsets are

${ }$ or

$\phi, {a}, {c}$ and

${a, c}$ .
B =

$\text{{p, q, r}}$

Hence, the subsets are

$\text{{ } or Φ, {p}, {q}, {r}, {p, q}, {p, r}, {q, r}}$ and

$\text{{p, q, r}}$ .

C = Set of digits used to form number

$1351$

Hence, the subsets are

$\text{{ } or Φ, {1}, {3}, {5}, {1, 3}, {3, 5}, {1, 5}}$ and

$\text{{1, 3, 5}}$ .

If A = { a, b, c, d, e}, B = { c, d, e, f}, C = {b, d}, D = { a, e} then which of the following statement is true or false?
$C \ \subseteq \ B$

Since,

$b \ \epsilon \ C$ but

$b \ \not\epsilon \ B$
So,

$C \ \subseteq \ B$ is false.

If A = { a, b, c, d, e}, B = { c, d, e, f }, C = {b, d}, D = { a, e}
then the following statement is true or false?
$A \ \subseteq \ D$

Since,

$b, \ \epsilon \ A$ but

$b \not{\epsilon} \ D$
So,

$A \ \subseteq \ D$ is false.

If $A =\text{{even numbers less than 12}}$ ,
$B = \text{{2, 4}}$ ,
$C = \text{{1, 2, 3}}$ ,
$D = \text{{2, 6}}$ and $E = \text{{4}}$
State which of the following statements are true :
(i) $B\subset A$
(ii) $C\subseteq A$
(iii) $D\subset C$
(iv) $D A$
(v) $E\supseteq B$
(vi) $A \supseteq B\supseteq E$

$A =\text{{even numbers less than 12}} = {{2, 4, 6, 8, 10}}$

$B = \text{{2, 4}}$

$C = \text{{1, 2, 3}}$

$D =\text{{2, 6}}$ and

$E = \text{{4}}$
(i) B

$\subset$ A is true.
(ii) C

$\subseteq$ A is false.
(iii) D

$\subset$ C is false.
(iv) D

$⊄$ A is false.
(v) E

$\supseteq$ B is false
(vi) A

$\supseteq$ B

$\supseteq$ E is true.

For the universal set $\text{{4, 5, 6, 7, 8, 9, 10, 11,12,13}}$ ; find its subsets $A, B, C$ and $D$ such that

(i) $A = \text{{even numbers}}$

(ii) $B =\text{{odd numbers greater than 8}}$

(iii) $C = \text{{prime numbers}}$

(iv) $D = \text{{even numbers less than 10}}$

(i)

$A = \text{{even numbers} = {4, 6, 8, 10, 12}}$

(ii)

$B =\text{{odd numbers greater than 8} = {9, 11, 13}}$

(iii)

$C =\text{{prime numbers} = {5, 7, 11, 13}}$

(iv)

$D = \text{{even numbers less than 10} = {4, 6, 8}}$

If A = { a, b, c, d, e}, B = { c, d, e, f }, C = {b, d}, D = { a, e}
then the following statement is true or false?
$C \ \subseteq \ A$ .

Since all elements of set

$C$ is in set

$A$ .
So,

$C \ \subseteq \ A$

Write the subset relations between the following sets.

X = set of all quadrilaterals.
Y = set of all rhombuses.
S = set of all squares.
T = set of all parallelograms.
V = set of all rectangles.

The subset relations are:
S < V < T < X and S < Y < T < X

$A= \{ 1,3,2,7 \}$ then write any three subsets of $A$ .

We have,

$A= \{1,3,2,7 \}$
The three subsets of

$A$ are:

$\{1\}, \{2\}$ and

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

Which of the following are sets? Justify your answer.
A team of eleven best-cricket batsmen of the world.

The terms best in vague. So, this collection is not a set.

If $n (A) = 15, n ( A \cup B)= 29, n (A \cap B) = 7$ then $n (B)=$ ?

Since,

$n(B) = n( A \cup B) + n( A \cap B) - n(A) = 29 + 7 -15$
So,

$n(B) = 21$

Which of the following are sets? Justify your answer.A collection of most dangerous animals o fthe world.

The terms most dangerous is vague. So, this collection is not a set.

If n(A) = 7, n(B) = 13, n(A $\cap$ B) = 4, then n(A $\cup$ B) = ?

Since, n(A

$\cup$ B) = n(A) + n(B) - n(A

$\cap$ B) = 7 + 13 - 4

$\therefore$ n(A

$\cup$ B) = 16

Write the subset relation between the sets.
P is the set of all residents in Pune.
M is the set of all residents in Madhya Pradesh.
I is the set of all residents in Indore.
B is the set of all residents in India.
H is the set of all residents in Maharashtra.

Since Pune is a city in Maharashtra, and Maharashtra is a state in India.
So,

$P \ \subset \ H \ \subset \ B$
Similarly,
Indore is a city in Madhya Pradesh, and Madhya Pradesh is a state in India.
So,

$I \ \subset \ M \ \subset \ B$

Let $A=(1, \left.2\right\}, B=[1, 2, 3, \left.4\right\}, C=\left\{5, 6\right\}$ and $D=( 5, 6, 7, 8 \left. \right\}$ . Verify that
$A\times C$ is a subset

$A \times C=\left\{ (1, 5), (1,6), (2, 5), (2, 6) \right\}$

$B \times D =\left\{ (1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8) \right\}$ .
Clearly every elements of

$A\times C$ is an element of

$A \times C$ is an element of

$B \times D$ .

$A \times C \subset B \times D$ .

Let $A=\left\{1,2, \left\{3,4 \right\}, 5\right\}.$ Which of the following statements are incorrect and why?
$\left\{\varphi\right\} \subset A$

correct

∵φ∵φ is a subset o every set

Let $A=\left\{1,2, \left\{3,4 \right\}, 5\right\}.$ Which of the following statements are incorrect and why?
$\left\{3,4 \right\} \subset A$

Incorrect

∵{3,4}∵{3,4}

∈A

Let $A=\left\{1,2, \left\{3,4 \right\}, 5\right\}.$ Which of the following statements are incorrect and why?
$\left\{\left\{3,4 \right\} \right\}\subset A$

correct

∵{3,4}∵{3,4}

∈A

Let $A=\left\{1, 2, 3\right\}, B =\left\{ 3, 4\right\}$ and $C=\left\{ 4, 5, 6\right\}$ . Find
$(A \times B) \cup (A \times C)$

$A \times B =\left\{ (1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)\right\}$

$A \times C=\left\{ (1, 4), (1, 5),(1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\right\}$

$(A \times B) \cup (A \times C)=\left\{ (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6),(3, 3), (3, 4), (3, 5), (3, 6)\right\}$ .

Let $A=\left\{1,2, \left\{3,4 \right\}, 5\right\}.$ Which of the following statements are incorrect and why?
$1 \subset A$

Incorrect

∵1∵1 is

AA bot not

1⊂A

Assume that $P(A) P(B)$ . Show that $A-B$

Let

x∈A∴(X)∈P(A)x∈A∴(X)∈P(A)

={x}∈P(B)∵P(A)=P(B)={x}∈P(B)∵P(A)=P(B)

∴x∈B∴x∈B

∴A⊂B∴A⊂B
Let

y∈B∴(y)∈P(B)y∈B∴(y)∈P(B)

(y)∈P(A)∵P(A)=P(B)(y)∈P(A)∵P(A)=P(B)

y∈Ay∈A

B⊂AB⊂A
From

(1)(1) and

(2)(2), we get

A=B

If $A=\left\{3,6,9,12,15,18,21 \right\}, B=\left\{4,8,12,16,20 \right\}, C=\left\{2,4,6,8,10,12,14,16 \right\} , D= \left\{5,10,15,20 \right\}$ find
$A-B$

A−B={3.6.9.15.18.21}

Using properties of sets, show that
$A \cap (A \cup B)=A$

A∩(A∪B)=(A∩A)∪(A∩B)∵A∩(A∪B)=(A∩A)∪(A∩B)∵ distributive law

=A∪(A∩B)=A

Using properties of sets, show that
$A \cup (A \cap B)=A$

A∪(A∩B)=(A∪A)∩(A∪B)∵A∪(A∩B)=(A∪A)∩(A∪B)∵ distributive law

=A∩(A∪B)=A

If $A=\left\{1,2,3,4 \right\}, B=\left\{3,4,5,6 \right\}, C=\left\{5,6,7,8\right\}$ and $D=\left\{7,8,9,10 \right\}$ ; find
$A \cup B \cup D$

A∪B∪D={1,2,3,4,5,6,7,8,9,10}

Find the union and the intersection o for each of the following pairs of sets:
$X= \left\{x : x\ is\ a\ natural\ number\ of\ and\ 1 < x \le 6 \right\}, Y=\left\{x:x\ is\ a\ natural \ numbers\ and\ 6 < x < 10\right\}$

A={2,3,4,5,6}A={2,3,4,5,6}

B={7,8,9}B={7,8,9}

A∪B={2,3,4,5,6,7,8,9}A∪B={2,3,4,5,6,7,8,9}

A∩B={2,3,4,5,6,7,8,9}

If $A=\left\{3,5,7,9,11 \right\}, B=\left\{7,9,11,13 \right\}, C=\left\{11,13,15\right\}$ and $D=\left\{15,17 \right\}$ ; find
$(A \cap B) \cap (B \cup C)$

(A∩B)∩(B∪C)={7,9,11}∩{7,9,11,13,15,}={7,9,11}(A∩B)∩(B∪C)={7,9,11}∩{7,9,11,13,15,}={7,9,11}

If $A=\left\{1,2,3,4 \right\}, B=\left\{3,4,5,6 \right\}, C=\left\{5,6,7,8\right\}$ and $D=\left\{7,8,9,10 \right\}$ ; find
$A \cup B \cup C$

A∪B∪C={1,2,3,4,5,6,7,8}

Let $A=\left\{1,2, \left\{3,4 \right\}, 5\right\}.$ Which of the following statements are incorrect and why?
$\varphi \subset A$

correct

∵φ∵φ is a subset o every set

If $A=\left\{3,6,9,12,15,18,21 \right\}, B=\left\{4,8,12,16,20 \right\}, C=\left\{2,4,6,8,10,12,14,16 \right\} , D= \left\{5,10,15,20 \right\}$ find
$B-A$

B−A={4,8,15,20}

If $A=\left\{3,6,9,12,15,18,21 \right\}, B=\left\{4,8,12,16,20 \right\}, C=\left\{2,4,6,8,10,12,14,16 \right\} , D= \left\{5,10,15,20 \right\}$ find
$D-B$

D−B={5,10,15}

If $A=\left\{3,6,9,12,15,18,21 \right\}, B=\left\{4,8,12,16,20 \right\}, C=\left\{2,4,6,8,10,12,14,16 \right\} , D= \left\{5,10,15,20 \right\}$ find
$C-B$

C−B={2,6,10,14}

If $A=\left\{3,6,9,12,15,18,21 \right\}, B=\left\{4,8,12,16,20 \right\}, C=\left\{2,4,6,8,10,12,14,16 \right\} , D= \left\{5,10,15,20 \right\}$ find
$C-D$

C−D={2,4,6,8,12,14,16}

If $A=\left\{3,6,9,12,15,18,21 \right\}, B=\left\{4,8,12,16,20 \right\}, C=\left\{2,4,6,8,10,12,14,16 \right\} , D= \left\{5,10,15,20 \right\}$ find
$D-C$

D−C={5,15,20}

$IV X=\left\{a,b,c,d \right\}$ and $Y=\left\{f,b,d,g \right\}$ find
$X-Y$

X−Y={a,c}

If $A=\left\{3,6,9,12,15,18,21 \right\}, B=\left\{4,8,12,16,20 \right\}, C=\left\{2,4,6,8,10,12,14,16 \right\} , D= \left\{5,10,15,20 \right\}$ find
$C-A$

$C-A = \left\{2,4,8,10,14,16 \right\}$

If $A=\left\{3,6,9,12,15,18,21 \right\}, B=\left\{4,8,12,16,20 \right\}, C=\left\{2,4,6,8,10,12,14,16 \right\} , D= \left\{5,10,15,20 \right\}$ find
$B-C$

B−C={20}

Fill in the blanks to make each of the following a true statement:
$A \cup A^{'}=$ ___

let

$U$ = universal set

Let

$A=\left\{ x:x\in U \right\}$

$\therefore { A }^{ \prime }=\left\{ x:x\notin A,x\in U \right\}$

Thus,

$A\cup { A }^{ \prime }=\left\{ x:x\in A\quad or\quad x\in { A }^{ \prime } \right\}$

$\therefore A\cup { A }^{ \prime }=U$

Which of the following are sets ? Justify your answer
The collection of Greatmen contributed to Indian culture.

The collection of Greatmen contributed in Indian culture is not a set because we cannot give proper credit for their contribution .

Which of the following are sets ? Justify your answer
The collection of all even integers.

The collection of all even integer is a set.

because all even integers are

$2, 4, 6, 8, 10 , ..................$

Which of the following are sets ? Justify your answer
The collection of various geometrical figures.

The collection of various geometrical figures is not set because these figures are not in particular shape

Which of the following are sets ? Justify your answer
The collection of all integers which divides 46

The number which device the number

$46$ are

$1 , 2$ and

$23 .$

Hence

$\{ 1 , 2, 23\}$ is a set .

Which of the following are sets ? Justify your answer
The collection of the best 20 cricket batsman of the world .

The collection of the best

$20$ cricket batsman of the world is not a set.

because there is no scale to measure their qualities

If $A = \left \{ x : x \epsilon N , 2 \leq x \leq 9 \right \}$ and B = { x : x is 2 digits natural numbers , sum of which digits is 8 } , then find the following sets:
$A-B$

$A = \left \{ x : x \epsilon N , 2 \leq x \leq 9 \right \} = (2, 3, 4, 5, 6, 7, 8, 9$ }
B = { x : x two digit natural number, sum of whose digits is 8 } =

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

$A - B = \left \{ 2, 3, 4, 5, 6, 7, 8, 9 \right \} \cap \left \{ 17, 26, 35, 44, 53, 62, 71 \right \}$
= All elements of A which arc not present in B

$=\left \{ 2, 3, 4, 5, 6, 7, 8,9 \right \}$

$= A$

Whether following subsets of N are total ordered set for relaton " x divides y or not ".
{ 2 , 4 , 6 , 8......}

Let subset

$A = { 2 , 4 , 6 , 8......} \in N$
A relation given in N

$xRy \Rightarrow x$ divides y

$x , y \in A$

If $A = \left \{ x : x \epsilon N , 2 \leq x \leq 9 \right \}$ and B = { x : x is digits natural numbers , sum of which digits is 8 } , then find the following sets:
$A \cup B$

$A = \left \{ x : x \epsilon N , 2 \leq x \leq 9 \right \} = (2, 3, 4, 5, 6, 7, 8, 9$ }

$B = \{ x : x$ two digit natural number, sum of whose digits is

$8 \}$ =

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

$A \cup B = \left \{ 2, 3, 4, 5, 6, 7, 8, 9, \right \} \cup \left \{ 17, 26, 35, 44, 53, 62, 71 \right \}$

$A \cup B = \left \{ 2, 3, 4, 5, 6, 7, 8, 9, 17, 26, 35 , 44, 53, 62, 71 \right \}$

Which of the following are sets ? Justify your answer
The collection of literature written by the poet Kalidas.

The collection of literature written by the poet Kalidas is a set because it is renowned literature.

If $A = \left \{ x : x \epsilon N , 2 \leq x \leq 9 \right \}$ and B = { x : x is 2 digits natural numbers , sum of which digits is 8 } , then find the following sets:
$(A - B ) \cup (B - A )$

$A = \left \{ x : x \epsilon N , 2 \leq x \leq 9 \right \} = (2, 3, 4, 5, 6, 7, 8, 9$ }
B = { x : x two digit natural number, sum of whose digits is 8 } =

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

$( A - B ) \cup ( B - A)$

$A - B = \left \{ 2, 3, 4, 5, 6, 7, 8, 9 \right \}$

$B - A = \left \{ 17, 26, 35, 44, 53, 62, 71 \right \}$

$(A-B) \cup (B-A) = \left \{ 2, 3, 4, 5, 6, 7, 8, 9 , 17, 26, 35, 44, 53, 62, 71 \right \} = A \cup B$

If $u = \left \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \right \} , A = \left \{ 1, 2, 3, 4, 5, 6 \right \} B = \left \{ 2, 3, 4 \right \}$ and $C= \left \{ 4, 6, 8, 10 \right \}$ then find the value of following sets:
$( A \cap B ) \cup B$

$( A \cap B ) \cup B$

$A \cap B = \left \{ 1, 2, 3, 4, 5, 6 \right \} \cap \left \{ 2, 3, 4 \right \} = \left \{ 2, 3, 4 \right \}$

$(A\cap B ) \cup C = \left \{ 2, 3, 4 \right \} \cup \left \{ 4, 6, 8, 10 \right \} = \left \{ 2, 3, 4, 6, 8, 10 \right \}$

If $u = \left \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \right \} , A = \left \{ 1, 2, 3, 4, 5, 6 \right \} B = \left \{ 2, 3, 4 \right \}$ and $C= \left \{ 4, 6, 8, 10 \right \}$ then find the value of following sets:
$A' \cup B'$

$A' \cup B'$

$A = \left \{ 1, 2, 3, 4, 5, 6 \right \}$

$U = \left \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \right \}$

$A'= U - A = \left \{ 1, 2,......10 \right \} - \left \{ 1, 2 ....6 \right \} = \left \{ 7, 8, 9, 10 \right \}$
Similarly ,

$B' = \left \{ 1, 5, 6, 7, 8, 9, 10 \right \}$

$A' \cup B' = \left \{ 7, 8, 9, 10 \right \} \cup \left \{ 1, 5, 6, 7, 8, 9, 10 \right \} = \left \{ 1, 5, 6, 7, 8, 9, 10 \right \}$

If $u = \left \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \right \} , A = \left \{ 1, 2, 3, 4, 5, 6 \right \} B = \left \{ 2, 3, 4 \right \}$ and $C= \left \{ 4, 6, 8, 10 \right \}$ then find the value of following sets:
$( A \cup B ) \cap B$

$( A \cup B ) \cap B$

$A \cup B = \left \{ 1, 2, 3, 4, 5, 6 \right \} \cup \left \{ 2, 3, 4 \right \} = \left \{ 1, 2, 3, 4, 5, 6 \right \}$

$(A\cup B) \cap B = \left \{ 1, 2, 3, 4, 5, 6 \right \} \cap \left \{ 2, 3, 4 \right \} = \left \{ 2, 3, 4 \right \} = B$

If $U = \left \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \right \} , A = \left \{ 1, 2, 3, 4, 5, 6 \right \} B = \left \{ 2, 3, 4 \right \}$ and $C= \left \{ 4, 6, 8, 10 \right \}$ then find the value of following sets:
$(A \cup B')$

$(A \cup B')$

$A \cup B = \left \{ 1, 2, 3, 4, 5, 6 \right \} \ \cup \left \{ 2, 3, 4 \right \} = \left \{ 1, 2, 3, 4, 5, 6 \right \}$

$(A \cup B') = U - (A \cup B ) = \left \{ 1, 2,,....10 \right \} - \left \{ 1, 2, 3, 4, 5, 6 \right \} = \left \{ 7, 8, 9, 10 \right \}$

P $\cup$ Q and P $\cap$ Q.

$= \{4,5,6,7,8 \}$ as

$x$ is greater than equal to

$4$ but less than or equal to

$8$

And

$Q=\{ 1,2,3,4,5 \}$ as

$x$ is a natural number less than

$6$

Union of two sets has all the elements of both the sets

So,

$P \cup Q =\{ 1,2,3,4,5,6,7,8 \}$

And Intersection of two sets has the elements common in both the sets

So,

$P \cap Q = \{4,5 \}$ .

In a pollution study of 1500 Indian rivers the following data were reported.520 were polluted by sulphur compounds, 335 were polluted by phosphates, 425 were polluted by crude oil, 100 were polluted by both crude oil and sulphur compounds, 180 were polluted by both sulphur compound and phosphates, 150 were polluted by both phosphates and crude oil and 28 were polluted by sulphur compounds, phosphates and crude oil. Let "n" be the number of the rivers were polluted by at least one of the three impurities.Find sum of digits of "n" ?

Given,

$n(S)=520, n(P)=335,n(C)=425,n(S\cap P)=180,n(C\cap S)=100,n(P\cap C)=150,n(C\cap P \cap S)=28$
Now, number of rivers polluted by at least one of the impurities is

$n(P\cup S\cup C)=n(P)+n(S)+n(C)-n(P\cap S)-n(S\cap C)-n(P\cap C)+n(P\cap S\cap C)$

$=520+335+425-100-180-150+28=878$

Now, number of rivers polluted by exactly sulphur compounds

$n(S\cap P'\cap C')=n(S\cap (P\cup C)')$

$=n(S)-n(S\cap (P\cup C))$

$=n(S)-[n(S\cap P)\cup n(S\cap C)]$

$=n(S)-[n(S\cap P)+n(S\cap C)-n(S\cap P\cap C)$

$=520-180-100+28=268$

Now, number of rivers polluted by exactly phosphorous compounds

$n(P\cap S'\cap C')=n(P\cap (S\cup C)')$

$=n(P)-[n(P\cap (S\cup C))]$

$=n(P)-[n(P\cap S)\cup n(P\cap C)]$

$=n(P)-n(P\cap S)-n(P\cap C)+n(P\cap C\cap S)$

$=335-180-150+28=33$

Now, number of rivers polluted by exactly crude oil

$n(C\cap S'\cap P')=n(C\cap (S\cup P)')$

$=n(C)-[n(C\cap (S\cup P))]$

$=n(C)-[n(C\cap S)\cup n(C\cap P)]$

$=n(C)-n(C\cap S)-n(C\cap P)+n(P\cap C\cap S)$

$=425-150-100+28=203$

So, the number of rivers polluted by exactly one of the three impurities

$=268+33+203=504$

Enter $1$ if the below relation holds true and $0$ if the above relation doesnot hold true:
A $\cup$ (B $^{'}$ $\cup$ A $^{'}$ ) = A $\cup$ (B $\cap$ A) $^{'}$

$\xi =$ {

$11,12,13,14,15, 16, 17, 18, 19, 20, 21,22,23,24,25,26,27,28,29,30$ } as

$x$ is a natural number more than

$10$ and less than equal to

$30$

$A =$ {

$11,12,13,14,15, 16, 17, 18, 19$ } as

$x$ is a number greater than or equal to

$20$ with the limit of the universal set already defined as

$\xi$

And,

$B =$ {

$16, 17, 18, 19, 20, 21,22,23,24$ } as

$x$ is a natural number less than

$25$ and more than

$15$

For LHS:

${ (A )}^{'} = \xi - A =$ {

$20, 21,22,23,24,25,26,27,28,29,30$ } (Since,

$\xi - A$ will have the elements of

$\xi$ which are not in

$A$ )

Similarly,

${ ( B )}^{'} = \xi - B =$ {

$11,12,13,14,15, 25,26,27,28,29,30$ }
Union of two sets has the elements of both the sets.
So,

${ B }^{'} \cup { A}^{'} =$ {

$11,12,13,14,15, 20, 21,22,23,24,25,26,27,28,29,30$ }

And

$A \cup ({ B }^{'} \cup { A}^{'}) =$ {

$11,12,13,14,15, 16, 17, 18, 19 20, 21,22,23,24,25,26,27,28,29,30$ } - (1)

For RHS:

Intersection of two sets has the elements common in both the sets.

$=> B \cap A =$ {

$16, 17, 18, 19$ }

$=> { (B \cap A )}^{'} = \xi - (B \cap A ) =$ {

$11,12,13,14,15, 20, 21,22,23,24,25,26,27,28,29,30$ } (Since,

$\xi - (B \cap A )$ will have the elements of

$\xi$ which are not in

$(B \cap A )$ )

So,

$A \cup { (B \cap A )}^{'} =$ {

$11,12,13,14,15,16, 17, 18, 19 , 20, 21,22,23,24,25,26,27,28,29,30$ } -- (2)

From 1, 2

$A \cup ({ B }^{'} \cup { A}^{'}) =A \cup { (B \cap A )}^{'}$

(A $\cup$ B) $^{'}$ = A $^{'}$ $\cap$ B $^{'}$

$\xi =$ {

$11,12,13,14,15, 16, 17, 18, 19, 20, 21,22,23,24,25,26,27,28,29,30$ } as

$x$ is a natural number more than

$10$ and less than equal to

$30$

$A =$ {

$11,12,13,14,15, 16, 17, 18, 19$ } as

$x$ is a number greater than or equal to

$20$ with the limit of the universal set already defined as

$\xi$

And,

$B =$ {

$16, 17, 18, 19, 20, 21,22,23,24$ } as

$x$ is a natural number less than

$25$ and more than

$15$

For LHS:
Union of two sets has the elements of both the sets.
So,

$A \cup B =$ {

$11,12,13,14,15, 16, 17, 18, 19, 20, 21,22,23,24$ }

And

${ (A \cup B)}^{'} = \xi - (A \cup B) =$ {

$25,26,27,28,29,30$ } (Since,

$\xi - (A \cup B)$ will have the elements of

$\xi$ which are not in

$(A \cup B)$ ) - (1)

For RHS:

${ (A )}^{'} = \xi - A =$ {

$20, 21,22,23,24,25,26,27,28,29,30$ } (Since,

$\xi - A$ will have the elements of

$\xi$ which are not in

$A$ )

Similarly,

${ ( B )}^{'} = \xi - B =$ {

$11,12,13,14,15, 25,26,27,28,29,30$ }

Intersection of two sets has the elements common in both the sets.
So,

${ (A )}^{'} \cap { (B )}^{'} =$ {

$25,26,27,28,29,30$ } -- (2)

From 1, 2

${ (A \cup B)}^{'} = { (A )}^{'} \cap { (B )}^{'}$

A $\cup$ B.

$A=\left\{ x:x\in N,3<x\le 6 \right\}$

$\therefore A=\left\{ 4,5,6 \right\}$

$B=\left\{ x:x\in W,x<4 \right\}$

$\therefore B=\left\{ 0,1,2,3 \right\}$ [

$x\in W$ i.e., Whole Number]

$A\cup B=$ contains all the elements of

$A$ and

$B$ But elements are not repeated.

$\therefore A\cup B=\left\{ 0,1,2,3,4,5,6 \right\}$

If $A = \{x : x \,\,\epsilon\,\, W, 3 \leq x < 6\}, B = \{3, 5, 7\}$ and $C = \{2, 4\}$ ;
find : $A \times (B - C)$
Find the number of elements in such a set.

$A=\left\{ x:x\in W,3\le x<6 \right\}$

$\Longrightarrow A=\left\{ 3,4,5 \right\}$

$B=\left\{ 3,5,7 \right\}$ and

$C=\left\{ 2,4 \right\}$

Now,

$B-C=$ contains all elements present only in

$B$ NOT in

$C$

$B-C=\left\{ 3,5,7 \right\} -\left\{ 2,4 \right\}$

$B-C=\left\{ 3,5,7 \right\}$

$A\times \left( B-C \right) =\left\{ 3,4,5 \right\} \times \left\{ 3,5,7 \right\}$

$=\left\{ \left( 3,3 \right) ,\left( 3,5 \right) ,\left( 3,7 \right) ,\left( 4,3 \right) ,\left( 4,5 \right) ,\left( 4,7 \right) ,\left( 5,3 \right) ,\left( 5,5 \right) ,\left( 5,7 \right) \right\}$

Thus there are a total of

$9$ elements in the above set.

Given A = { $x$ : $x \,\,\in\,\, N$ and $3 < x \leq 6$ } and B = { $x$ : $x \,\,\in\,\, W$ and $x < 4$ }, then find : A - B.

$A=\left\{ x:x\in N,3<x\le 6 \right\}$

$\therefore A=\left\{ 4,5,6 \right\}$

$B=\left\{ x:x\in W,x<4 \right\}$

$\therefore B=\left\{ 0,1,2,3 \right\}$ [

$\because x\in W$ i.e., Whole Number]

$A-B=$ Contains elements present only in

$A$ and NOT in

$B$ .

$\therefore A-B=\left\{ 4,5,6 \right\}$

Verify $n(A\cup B \cup C)=n(A)+n(B)+n(c)-n(A\cap B)-n(B\cap C)-n(A \cap C)+n(A \cap B\cap C)$ for the sets given below:
(i) A={4,5,6}, B={5,6,7,8} and C={6,7,8,9}
(ii) A={a, b, c, d, e}, B={x, y, z} and C={a, e, x}

(i) The given sets are

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

$B=\left\{ 5,6,7,8 \right\}$ and

$C=\left\{ 6,7,8,9 \right\}$

Now we find the following:

$A=\left\{ 4,5,6 \right\} \Rightarrow n(A)=3\\ B=\left\{ 5,6,7,8 \right\} \Rightarrow n(B)=4\\ C=\left\{ 6,7,8,9 \right\} \Rightarrow n(C)=4$

$A\cap B=\left\{ 4,5,6 \right\} \cap \left\{ 5,6,7,8 \right\} =\left\{ 5,6 \right\} \Rightarrow n(A\cap B)=2\\ B\cap C=\left\{ 5,6,7,8 \right\} \cap \left\{ 6,7,8,9 \right\} =\left\{ 6,7,8 \right\} \Rightarrow n(B\cap C)=3\\ A\cap C=\left\{ 4,5,6 \right\} \cap \left\{ 6,7,8,9 \right\} =\left\{ 6 \right\} \Rightarrow n(A\cap C)=1$

$A\cap B\cap C=\left\{ 4,5,6 \right\} \cap \left\{ 5,6,7,8 \right\} \cap \left\{ 6,7,8,9 \right\} =\left\{ 6 \right\} \Rightarrow n(A\cap B\cap C)=1\\ A\cup B\cup C=\left\{ 4,5,6 \right\} \cup \left\{ 5,6,7,8 \right\} \cup \left\{ 6,7,8,9 \right\} =\left\{ 4,5,6,7,8,9 \right\} \Rightarrow n(A\cup B\cup C)=6$

Now consider

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\ \Rightarrow 6=3+4+4-2-3-1+1\\ \Rightarrow 6=3+4+4+1-(2+3+1)\\ \Rightarrow 6=12-6\\ \Rightarrow 6=6$

which is true.

Hence,

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$

(ii) The given sets are

$A=\left\{ a,b,c,d,e \right\}$ ,

$B=\left\{ x,y,z \right\}$ and

$C=\left\{ a,e,x \right\}$

Now we find the following:

$A=\left\{ a,b,c,d,e \right\} \Rightarrow n(A)=5\\ B=\left\{ x,y,z \right\} \Rightarrow n(B)=3\\ C=\left\{ a,e,x \right\} \Rightarrow n(C)=3$

$A\cap B=\left\{ a,b,c,d,e \right\} \cap \left\{ x,y,z \right\} =\left\{ \phi \right\} \Rightarrow n(A\cap B)=0\\ B\cap C=\left\{ x,y,z \right\} \cap \left\{ a,e,x \right\} =\left\{ x \right\} \Rightarrow n(B\cap C)=1\\ A\cap C=\left\{ a,b,c,d,e \right\} \cap \left\{ a,e,x \right\} =\left\{ a,e \right\} \Rightarrow n(A\cap C)=2$

$A\cap B\cap C=\left\{ a,b,c,d,e \right\} \cap \left\{ x,y,z \right\} \cap \left\{ a,e,x \right\} =\left\{ \phi \right\} \Rightarrow n(A\cap B\cap C)=0\\ A\cup B\cup C=\left\{ a,b,c,d,e \right\} \cup \left\{ x,y,z \right\} \cup \left\{ a,e,x \right\} =\left\{ a,b,c,d,e,x,y,z \right\} \Rightarrow n(A\cup B\cup C)=8$

Now consider

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\ \Rightarrow 8=5+3+3-0-1-2+0\\ \Rightarrow 8=5+3+3+0-(0+1+2)\\ \Rightarrow 8=11-3\\ \Rightarrow 8=8$

which is true.

Hence,

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$

Given a non empty set X, consider $P(X)$ which is set of all subsets of $X$ . Define the relation $R$ is $P(X)$ as follows:
For subsets $A, B$ in $P(X), ARB$ if and only if $A\subset B$ . Is R an equivalence relation on $P(X)$ ? Justify your answer

$Let\quad X=\{ 1,2,3\}$

$P(X)$ =Power set of

$X$ =Set of all subsets of

$X$ .

$\left\{ \phi ,\{ 1\} \right\} ,\{ 2\} ,\{ 3\} ,\{ 1,2\} ,\{ 2,3\} ,\{ 1,3\} ,\{ 1,2,3\} \} \\ Since\quad \{ 1\} \subset \{ 1,2\} \\ \{ 1\} R\{ 1,2\} \\$

${ \because of\quad ABCD,all\quad elements\quad of \quad A\quad are\quad in\quad B}$

$ARB$ means

$A\subset B$

here,relation is

$R=\{ (A,B):A\quad and\quad B\quad are\quad sets,ACB\}$

Since every set is a subset of itself.

$ACA\quad \therefore (A,A)\epsilon R,\quad$ R is reflexive.

To check whether symmetric or not,

$(A,B)\epsilon \quad ,then\quad (B,A) \epsilon R.$

$If(A,B)\epsilon R,\quad A\subset B\\ but\quad B\subset A\quad is\quad not\quad true.\\ eg:-Let\quad A=\{ 1\} \quad and\quad B=\{ 1,2\}$

As all elements of

$A$ are in

$B \quad A\subset B$ .

But all elements of

$B$ are not in

$A$

$\therefore B\subset A$ is not true.

$\therefore R$ is not symmetric.

Since

$(A,B) \epsilon R \quad and (B,C) \epsilon R of A\subset B\quad and\quad B\subset C$

then

$A\subset B$

$\Rightarrow (A,C)\epsilon R\\ So,\quad If\quad (A,B)\epsilon R\quad and\quad (B,C)\epsilon R,\\ then\quad (A,C)\epsilon R$

$\therefore R$ is transitive.

hence

$R$ is reflexive and transitive but not symmetric.

hence,

$R$ is not an equivalence relation since it is not symmetric.

If $A=\{x:x^2-5x+6=0\}, B=\{2,3\}$ and $C=\{2,3,3\}$ then prove that $A=B=C$ .

$A=\{x:\,x^2-5x+6=0\}$

$=\{x\,:\,x^2-3x-2x+6=0\}$

$=\{x\,:\,(x-2)(x-3)=0\}$

$=\{x\,:x=2, 3\}=\{2,3\}$

Therefore,

$A=B=C$ as

$B=\{2, 3\}$ and

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

Let $X$ be a non- empty set and $$ be a binary operation on $P(x)$ defined by $AB = A \cap B$ for all $A,B \in P\left( x \right)$ . Find the identity and invertible elements

Let

$X=\left\{1,2,3\right\}$

$P\left(X\right)=\left\{\phi,\left\{1\right\},\left\{2\right\},\left\{3\right\},\left\{1,2\right\},\left\{2,3\right\},\left\{1,3\right\},\left\{1,2,3\right\}\right\}$

$A\ * x=A\cap X=A$

$X\ * A=X\cap A=A$

$(i)e$ is the identity of

$\ *$

$a\ * X=A\cap X=A$ for

$a\in A$

$X\ * A=X\cap A=A$

$\therefore\,A\ * X=X\ * A=A$ for

$A\in P\left(x\right)$

Hence

$X$ is an identity element.

$(ii)a\ * b=e=b\ * a$

We have

$e=X$

$\Rightarrow\,A\ * B=e=B\ * A$

$\Rightarrow\,A\cap B=X$

This is possible if

$A=B=X$

$A\ * X=A=X\ * A$ for all

$A\in\,P\left(X\right)$

Hence

$X$ is the only invertible element in

$P\left(X\right)$

If $^{n-1}C_3 + ^{n-1}C_4 > ^nC_3$ , then set of complete values of n is:

$\displaystyle ^{n-1}C_{3}+^{n-1}C_{4}> ^{n}C_{3}$

we,

$\displaystyle ^{n}C_{r}+^{n}C_{r-1} = ^{n+1}C_{r}$

$\displaystyle ^{n+1}C_{4}+^{n+1}C_{4-1}> ^{n}C_{3}$

$\displaystyle ^{n-1+1}C_{4}> ^{n}C_{3}$

$\displaystyle ^{n}C_{4}> ^{n}C_{3}$

$\displaystyle \frac{n!}{(n-4)!4!}> \frac{n!}{(n-3)!3!}$

$\displaystyle \frac{1}{4}> \frac{1}{n-3}$

$\displaystyle \frac{1}{n-1}-\frac{1}{4} < 0$

$\displaystyle \frac{4-n+1}{4(n-1)}< 0$

$\displaystyle \frac{n-3}{(n-1)}> 0 \frac{+-+}{13}$

$\displaystyle x\epsilon (-\infty ,1)\cup (3,\infty )$

1097482_1138149_ans_d9cb5dbef36341b49f2f4b4954e68cc2.jpg

If $X=\left\{ 1,2,3,4,5,6,7,8,9,10 \right\}$ is the universal set & $A=\left\{ 1,2,3,4\right\},\ B=\left\{ 2,4,6,8\right\},\ C=\left\{ 3,4,5,6\right\}$ verify the following.
a) $A\cup \left( B\cup C \right) =\left( A\cup B \right) \cup C$
b) $A\cap \left( B\cup C \right) =\left( A\cap B \right) \cup \left( A\cap C \right)$
c) $\left( A' \right) '=A$

$X=\left\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \right\}$

$A=\left\{1, 2, 3, 4 \right\}$

$a$

$(B\cup C)=\left\{2, 3, 4, 5, 6, 8 \right\}$

$A\cup (B\cup C) =\left\{1, 2, 3, 4, 5, 6, 8 \right\}----(1)$

$(A\cup B)=\left\{1. 2. 3. 4. 6. 8 \right\}$

$(A\cup B)\cup C=\left\{ 1, 2, 3, 4, 5, 6, 8\right\}----(2)$

$\therefore \$ from

$(1)$ and

$(2)$

$A\cup (B\cup C)=(A\cup B)\cup C$

$b$

$(B \cup C)=\left\{2, 3, 4, 5, 6, 8 \right\}$

$A\cap (B\cup C)=\left\{2, 3, 4 \right\}---(3)$

$(A\cap B)=\left\{2, 4 \right\}$

$(A\cap C)=\left\{3, 4 \right\}$

$\therefore \ (A\cap B)\cup (A\cap C)=\left\{ 2, 3, 4\right\}---(4)$

$\therefore \$ from

$(3)$ and

$(4)$

$A\cap (B \cup C)=(A\cap B)\cup (A\cap C)----proved$

$c$

$A=\left\{5, 6, 7, 8, 9, 10 \right\}$

$(A')' =\left\{1, 2, 3, 4, \right\}=A---Proved$

If sum of all $x$ in $[0,2\pi]$ such that $3\cot^{2}{x}+8\cot{x}+3=0$ is $K\pi$ then find $K$ .

$3\cot^{2}x+8\cot x+3=0$

Let

$\cot x=y$

$3y^{2}+8y+3=0$

Let the roots be

$y_{2}$ and

$y_{2}$

Sum of roots :

$y_{1}+y_{2}=\cfrac{-8}{3}$

Product of roots:

$4y_{1}y_{2}=\cfrac{3}{3}=1$ $

$y_{1}=\cfrac{1}{y_{2}}$

Since,

$y=\cot x$

$\therefore \cot x_{1}=\cfrac{1}{\cot x_{2}}$

$\cot x_{1}=\tan x_{2}$

$\tan x_{2}=\tan(\cfrac{3z}{2}-x_{1})$

So,

$x_{2}=\cfrac{3z}{2}-x_{1}$

$x_{2}+x_{1}=\cfrac{3z}{2}$

Value of

$K=\cfrac{3}{2}$

If $A=\left\{ x:x=4n+1,n\le 5,n\epsilon N \right\}$ and $B=\{ 3n:n\le 8,n\epsilon N\}$ , then find $A-(A-B)$ .

Hint : Write the set in roster form from given in set builder form

A={5,9,13,17,21}

B={3,6,9,12,15,18,21,24}

A-B={ 5,13,17,21}

A−(A−B) = {9}

If $A=\left\{1,2,3\right\},\,B=\left\{1,2,3,4\right\}$ , then find the number of possible one-one functions from $A$ to $B$ .

One-one functions are possible from

$A$ to

$B$ when

$n\left(A\right)\le\,n\left(B\right)$

Here,

$n\left(A\right)=3,\,n\left(B\right)=4$

Number of possible one-one functions from

$A$ to

$B$ .

$^{n\left(B\right)}{{P}_{n\left(A\right)}}=\dfrac{n\left(B\right)!}{\left(n\left(B\right)-n\left(A\right)\right)!}$ if

$n\left(A\right)\le \,n\left(B\right)$

$\therefore$

$Number \,of\, possible\, one-one\, functions\, from\, A\, to\, B$ .

$^{4}{{P}_{3}}=\dfrac{4!}{\left(4-3\right)!}$ since

$n\left(A\right)\le \,n\left(B\right)$

$=\dfrac{4\times 3\times 2\times 1}{1}=24$

$\therefore\,$ Number of possible one-one functions from

$A$ to

$B$ is

$24$

Let $u=\{1, 2, 3, 4, 5, 6, 7, 8, 9\}, A=\{2, , 6, 8\}$ and $B=\{2, 3, 5, 7\}$ . Verify that $(A\cap B)'=A'\cup B'$ .

The given sets are:

$u=\left\{1,2,3,4,5,6,7,8,9 \right\}$

$A=\left\{2,6,8 \right\},\ B=\left\{2,3,5,7 \right\}$

we have to prove

$(A\cap B)'=A'\cup B'$

$LHS=(A\cap B)'$

$\because \ A\cap B=\left\{2 \right\}\ \Rightarrow (A\cap B)'=\left\{1,3,4,5,6,7,8,9 \right\}$

$RHS=A' \cup B'$

$\because A'=u-A=\left\{1,3,4,5,7,9 \right\}$

and

$B'=u-B=\left\{1,4,6,8,9 \right\}$

$\Rightarrow \ A'\cup B'=\left\{1,3,4,5,6,7,8,9 \right\}$

$\Rightarrow \ \boxed {(A\cap B)'=A' \cup B'}$

If $A=\{1, 2, 3\}, B=\{4\}, C=\{5\}$ , then verify that $A\times (B-C)=(A\times B)-(A\times C)$ .

1421799_1666250_ans_4fb1061461b44155aaf52268d5d59ace.jpg

Let A and B be two sets such that $n(A)=3$ and $n(B)=2$ .
If $(x, 1), (y, 2), (z, 1)$ are in $A\times B$ , find A and B, where x, y, z are distinct elements.

1421615_1666151_ans_8ed4ed9f12d74b4fb41e27b735cd9b69.jpg

Sets - Class 11 Engineering Maths - Extra Questions

Let A={1,2,3,4}A=\left \{ 1, 2, 3, 4 \right \}S={(a,b);a,b∈A,a divides b}S=\left \{ (a, b); a,b\in A, a\ \mbox{divides}\ b \right \}, write down SS explicitly

If SS and TT are two sets such that SS has 2121 elements. T has 3232 elements and SS ∩\displaystyle \cap TT has 1111 elements, then find the number of elements in SS ∪\displaystyle \cup TT.

{a,b}\displaystyle \left \{ a,b \right \} is a subset of {b,c,a}\left \{ b,c,a \right \}.If true enter 1 else 0.

a,e\displaystyle {a,e} is a subset of {x:x is a vowel in the English alphabet} \left \{ x:x \text{ is a vowel in the English alphabet}\right \}.( Enter 1 if true or 0 otherwise)

Let mm be a number in a set A={1,2,3} such that A⊈{1,3,5}A=\displaystyle \left \{ 1,2,3 \right \} \ such \ that \ A\nsubseteq \left \{ 1,3,5 \right \} and m∉{1,3,5}m\notin \{1,3,5\}. Then mm is:

Let A={1,2,3,4,5,6}A = \{1, 2, 3, 4, 5, 6\}, B={2,4,6,8}B = \{2, 4, 6, 8\}. Find A−BA - B and B−AB - A.

Find n(B−C)cn(B-C)^c.

If A={2,3,4,5,6,7}A=\left \{2, 3,4, 5, 6, 7\right \} and B={3,5,7,9,11,13}B =\left \{3, 5, 7, 9,11, 13\right \}, then find A−BA-B and B−AB -A

In a group of 5050 persons, 1414 drink tea but not coffee and 3030 drink tea. Find how many drink coffee but not tea?

Let A={2,4,6,8}A = \{2, 4, 6, 8\} and B={6,8,10,12}B = \{6, 8, 10, 12\} then find A∪B\displaystyle A\cup B.

Find n(A∪B)c(A\cup B)^c

aϵ{a,b,c}\displaystyle a \epsilon \left \{ a,b,c \right \}, then {a}\{a\} is a subset of {a,b,c}\{a,b,c\}.( Enter 1 if true or 0 otherwise)

If A{1,2,3}A\left \{1, 2,3\right \} and B={1,3,5,7}B =\left \{1, 3, 5, 7\right \}, then find A∪BA\cup B

If RR is the set of real numbers and QQ is the set of rational numbers, then what is R−QR-Q?

Let A={1,2,3,4,5,6}A = \left \{1, 2, 3, 4, 5, 6\right \}. Insert the appropriate symbol ∈\in or ∉\not\in in the blank spaces:(i). .A (ii) 8 . . . A (iii). .A (iv). . A (v). .A (vi). .A

(P∪Q)∩R(P\cup Q)\cap R

Decide, among the following sets, which sets are subsets of one and another:A={x:xϵ R and x satisfy x2−8x+12=0},A =\{ x : x \epsilon\ \text{R and x satisfy} \ x^2- 8x + 12 = 0\},B={2,4,6}, B = \{ 2, 4, 6\}, C={2,4,6,8,...},D={6}C = \left \{ 2, 4, 6, 8, . . . \right \}, D = \left \{ 6 \right \}

Represent the following set by using their standard notations: Set of integers

Find out the set, which is defined as {x:x−3<0}\{x : x - 3 < 0\}; x is any integer.

If A={3k|k ϵZ}A = \left \{3k | k \ \epsilon \mathbb {Z}\right \} and B=NB = \mathbb {N}, find A△BA\triangle B

Find the set if it is defined as {x:|x2−9|=0}\{x : |x^2 - 9| = 0\}.

Let A={0,1,2,3,4,5}A=\left \{ 0,1,2,3,4,5 \right \}. Insert the appropriate symbol ∈\in or ∉\notin in the blank spaces(i) 0−−−−A0 ---- A(ii) 6−−−−A6 ---- A(iii) 3−−−−A3 ---- A(iv) 4−−−−A4 ---- A(v) 7−−−−A7 ---- A

Represent the following set by using their standard notations:Set of natural numbers

If n(A)=26,n(B)=10,n(A∪B)=30,n(A′)=17n\left ( A \right )=26, n\left ( B \right )=10, n\left ( A\cup B \right )=30,n\left ( {A}' \right )=17, find n(A∩B)n\left ( A\cap B \right ) and n(∪)n\left ( \cup \right ).

Examine whether A=A= {\{ x:xx:x is a positive integer divisible by 3}3 \} is a subset of B=B= {\{ x:xx:x is a multiple of 55, x∈Nx\in N }\}.

Is ϕ={ϕ}\phi =\left \{ \phi \right \}? Why?

If AA and BB are two sets such that AA has 5050 elements, BB has 6565 elements and A∪BA\cup B has 100100 elements, how many elements does A∩BA\cap B have?

If AA and BB are two sets containing 1313 and 1616 elements respectively, then find the minimum and maximum number of elements in A∪BA\cup B?

Let X={−3,−2,−1,0,1,2}X=\left \{ -3,-2,-1,0,1,2 \right \} and Y=Y= {\{ x:xx:x is an integer and −3≤x<2 -3\leq x< 2 }\}(i) Is XX a subset of YY?(ii) Is YY a subset of XX?

If n(A∩B)=5,n(A∪B)=35,n(A)=13,n\left ( A\cap B \right )=5, n\left ( A\cup B \right )=35, n\left ( A \right )=13, find n(B)n\left ( B \right ).

Which of the sets are equal sets? State the reason.ϕ,{0},{ϕ}\phi ,\left \{ 0 \right \},\left \{ \phi \right \}

If A={−2,−1,0,3,4},B={−1,3,5}A=\left \{-2,-1,0,3,4 \right \}, B=\left \{-1,3,5\right \}, find (i) A−BA-B (ii) B−AB-A

Let AA and BB be two finite sets such that n(A−B)=30,n(A∪B)=180n\left ( A- B \right )=30, n\left ( A\cup B \right )=180. Find n(B)n\left ( B \right ).

If n(∪)=38,n(A)=16,n(A∩B)=12,n(B′)=20n\left ( \cup \right )=38, n\left ( A \right )=16, n\left ( A\cap B \right )=12,n\left ( {B}' \right )=20, find n(A∪B)n\left ( A\cup B \right ).

Given n(A)=285, n(B)=195, n(U)=500, n(A∪B)=410n(A \cup B)=410, find n(A′∪B′)n(A' \cup B').

Giventhe sets A={4,5,6,7}A=\left \{4,5,6,7\right \} and B={1,3,8,9}B=\left \{1,3,8,9\right \}. Find A∩B A\cap B

Let A = {4, 6, 8, 10 } and B = { 3, 4, 5, 6, 7 }. If f:A→Bf: A \rightarrow B is defined by f(x)=12x+1f(x)=\frac{1}{2}x+1 then represent f(x) by a set of ordered pairs

Given A ={1,2,3,4}= \{1, 2, 3, 4\}, B ={3,4,5,6}= \{3, 4, 5, 6\} and C ={6,7}= \{6, 7\}. Verify that (A ∩\cap B) ∩\cap C = A ∩\cap (B ∩\cap C).

If A = {-1, 1}, find A ×\times A ×\times A.

For the given sets A= { 10, 0, 1, 9, 2, 4, 5} and B= { -1, -2, 5, 6, 2, 3, 4}, verify that set union is commutative. Also verify it by using Venn diagram.

Let A = {1, 2, 3}, B = {2, 4, 6, 8}, C = {3, 4, 5, 6} then prove the following :-A∪BA \, \cup \, B = {1, 2, 3, 4, 6, 8}

Using properties of sets, prove that A∪(A∩B)=AA \cup(A \cap B)=A

A−(A−B)=A-( A-B )= ...........

Given : A = {2, 3}, B = {4, 5}, C = {5, 6}, then prove the following :A×(B∩C)A \, \times \, (B \, \cap \, C) = {(2, 5), (3, 5)}

Given : A = {2, 3}, B = {4, 5}, C = {5, 6}, then prove the following :A×(B∪C)A \, \times \, (B \, \cup \, C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

Verify the following identities where A = {1, 2, 3, 4, 5}, B = {2, 3, 5, 6}, C = {4, 5, 6, 7}A∩(B∪C)=(A∩B)∪(A∩C)A \, \cap \, (B \, \cup \, C) \, = \, (A \, \cap \, B) \, \cup \, (A \, \cap \, C)

Find the smallest set a such that a union {1}={1,2,3,5,9}

Let AA={1,2},=\{1,2\}, and BB={1,3},=\{1,3\}, then prove that (A×B)∪(B×A)={(1,3),(2,3),(3,1),(3,2),(1,1),(1,2),(2,1)}\left( A\times B \right) \cup \left( B\times A \right)=\left\{(1,3),(2,3),(3,1),(3,2),(1,1),(1,2),(2,1)\right\}

Given A=2,3,B=4,5,C=5,6,A ={2,3}, B = {4,5}, C ={5,6}, find(i) A×(B∩C)= .......A\times \left( B\cap C \right) =\ .......(ii) (A×B)∪(B×C)= .......\left( A\times B \right) \cup \left( B\times C \right) =\ .......

Comment on the following statementsA−B=A∩B′=B′−A′ A-B=A\cap B'=B'-A'

Comment on the following statementsA−(A−B)=A∩B A-\left( A-B \right) =A\cap B

Let A={1,2,3},A=\left\{1,2,3 \right\}, B={2,4,6,8},B=\left\{2,4,6,8 \right\}, C={2,3,5,6}C=\left\{2,3,5,6 \right\}. Then A∩(B∪C). ........... A\cap \left( B\cup C \right).\ ...........

Set A={x:x is a digit in the number 3591}A=\left\{ x:x\ is\ a\ digit\ in\ the\ number\ 3591 \right\}B={x:x∈N,x<10} B=\left\{ x:x\in N,x<10 \right\} . Find A∪B,A∩B,A−B A\cup B,A\cap B,A-B and B−AB-A.

Let A and B be two sets such that A×BA\times B consists of 6 elements. If three elements of A×BA\times B are (1,4), (2,6), (3, 6), find A×BA\times B and B×AB\times A.

Write the following set in roaster form: A={x:xA=\{x\,:\,x is an integer and −3≤x<7}-3\leq x\,<\,7\}

List all the elements of the following sets :B={x:x=12n−1,1≤n≤5}B = \left\{ {x:x = {\dfrac{1} {{2n - 1}}},1 \le n \le 5} \right\}

Describe the following sets in Roster form:The sets of all letters in the word 'BETTER'.

List all the elements of the following sets :F = { xx : xx is a letter in the word "MISSISSIPPI"}

List all the elements of the following sets :A={x:x2≤10,x∈Z}A = \left\{ {x:{x^2} \le 10,x \in Z} \right\}

List all the elements of the following sets :D = { xx : xx is a vowel in the word "EQUATION"}

For any two sets of A and B, prove thatB′⊂A′B' \subset A' if A⊂BA \subset B

List all the elements of the following sets :{C=x:x{C = x:x} is an integar , −12<x<92{ - {1 \over 2} < x < {9 \over 2}}}

In a group of 7070 people, 3737 like coffee, 5252 like tea and each person like at least one of the two drinks. How many people like both coffee and tea?

Let AA and BB be two sets. If A∪X=B∩X=ϕA\cup X=B\cap X=\phi and A∪X=B∪XA\cup X=B\cup X for sum set XX that A=BA=B.

If A={6,9,11}A=\left\{6,9,11\right\} then find A∪ϕA\cup \phi

Show that for any sets A and B,AA=(A∩B)∪(A−B)A=(A\cap B)\cup (A-B) and A∪(B−A)=A∪BA\cup (B-A)=A\cup B

If A=(1,2,3,4,5,6,7,8,9)A=(1,2,3,4,5,6,7,8,9)B=(1,3,5,,7,9,11,13)B=(1,3,5,,7,9,11,13)C=(2,4,6,8,10,12,14)C=(2,4,6,8,10,12,14)D=(2,3,5,7,11,13,17)D=(2,3,5,7,11,13,17) then find B∪CB \cup C

If AA and BB are two disjoint sets n(A)=15n(A)=15 and n(B)=10n(B)=10 find n(A∪B)n\left( {A \cup B} \right) and n(A∩B)n\left( {A \cap B} \right).

Write all the elements of the set S={1,{3},{5,7},9,10}S = \{ 1,\{ 3\} ,\{ 5,7\} ,9,\,10\} .

Let AA and BB be events such that p(A)=511,P(B)=611p\left( A \right) = \dfrac{5}{{11}},\,P\left( B \right) = \dfrac{6}{{11}}, and P(A∪B)=711P\left( {A \cup B} \right) = \dfrac{7}{{11}} and P(B/A).P(B/A).

Let A={a,e,i,o,u}=\{a, e, i, o, u\} and B={a,b,c,d}=\{a, b, c, d\}. Is A a subset of B? No(why?). Is B a subset of A? No. (Why?)

If f={(4,5),(5,6),(6,−4)}f=\left\{(4,5),(5,6),(6,-4)\right\} and g={(4,−4),(6,5),(8,5)}g=\left\{(4,-4),(6,5),(8,5)\right\}.Then |f−g|=?|f-g|=?

find the range of the following data 18,10,8,16,9,24,12,25,13.19.

If A=(1,2,3,4,5,6,7,8,9)A=(1,2,3,4,5,6,7,8,9)B=(1,3,5,,7,9,11,13)B=(1,3,5,,7,9,11,13)C=(2,4,6,8,10,12,14)C=(2,4,6,8,10,12,14)D=(2,3,5,7,11,13,17)D=(2,3,5,7,11,13,17) then find (A−D)∪(B−A)(A-D) \cup \left( {B - A} \right)

If A=(1,2,3,4,5,6,7,8,9)A=(1,2,3,4,5,6,7,8,9)B=(1,3,5,,7,9,11,13)B=(1,3,5,,7,9,11,13)C=(2,4,6,8,10,12,14)C=(2,4,6,8,10,12,14)D=(2,3,5,7,11,13,17)D=(2,3,5,7,11,13,17) then find A-DA-D

18001800 boys and 900900 girls appeared for an examination. If 42\%42\% of the boys and 30\%30\% of the girls passed, find(i) number of boys passed(ii) number of girls passed(iii) total number of students passed(iv) number of students failed(v) percentage of student failed.

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

If A=\{ 3,4,6\} , B=\{ 1,3\} A=\{ 3,4,6\} , B=\{ 1,3\} and C=\{ 1,2,6\} C=\{ 1,2,6\} then find \left( A-B \right) \times \left( A-C \right) \left( A-B \right) \times \left( A-C \right)

If A=\left\{4,5,6\right\};B=\left\{7,8\right\}A=\left\{4,5,6\right\};B=\left\{7,8\right\} then show that A\bigcup B=B\bigcup AA\bigcup B=B\bigcup A

Let $A=\left \{ 1, 2, 3, 4 \right \}$
$S=\left \{ (a, b); a,b\in A, a\ \mbox{divides}\ b \right \}$ , write down $S$ explicitly

If $S$ and $T$ are two sets such that $S$ has $21$ elements. T has $32$ elements and $S$ $\displaystyle \cap$ $T$ has $11$ elements, then find the number of elements in $S$ $\displaystyle \cup$ $T$ .

$\displaystyle \left \{ a,b \right \}$ is a subset of $\left \{ b,c,a \right \}$ .If true enter 1 else 0.

$\displaystyle {a,e}$ is a subset of $\left \{ x:x \text{ is a vowel in the English alphabet}\right \}$ .( Enter 1 if true or 0 otherwise)

Let $m$ be a number in a set $A=\displaystyle \left \{ 1,2,3 \right \} \ such \ that \ A\nsubseteq \left \{ 1,3,5 \right \}$ and $m\notin \{1,3,5\}$ . Then $m$ is:

Let $A = \{1, 2, 3, 4, 5, 6\}$ , $B = \{2, 4, 6, 8\}$ . Find $A - B$ and $B - A$ .

Find $n(B-C)^c$ .

If $A=\left \{2, 3,4, 5, 6, 7\right \}$ and $B =\left \{3, 5, 7, 9,11, 13\right \}$ , then find $A-B$ and $B -A$

In a group of $50$ persons, $14$ drink tea but not coffee and $30$ drink tea. Find how many drink coffee but not tea?

Let $A = \{2, 4, 6, 8\}$ and $B = \{6, 8, 10, 12\}$ then find $\displaystyle A\cup B$ .

Find n $(A\cup B)^c$

$\displaystyle a \epsilon \left \{ a,b,c \right \}$ , then $\{a\}$ is a subset of $\{a,b,c\}$ .( Enter 1 if true or 0 otherwise)

If $A\left \{1, 2,3\right \}$ and $B =\left \{1, 3, 5, 7\right \}$ , then find $A\cup B$

If $R$ is the set of real numbers and $Q$ is the set of rational numbers, then what is $R-Q$ ?

Let $A = \left \{1, 2, 3, 4, 5, 6\right \}$ . Insert the appropriate symbol $\in$ or $\not\in$ in the blank spaces:
(i). .A (ii) 8 . . . A (iii). .A (iv). . A (v). .A (vi). .A

$(P\cup Q)\cap R$

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

Find out the set, which is defined as $\{x : x - 3 < 0\}$ ; x is any integer.

If $A = \left \{3k | k \ \epsilon \mathbb {Z}\right \}$ and $B = \mathbb {N}$ , find $A\triangle B$

Find the set if it is defined as $\{x : |x^2 - 9| = 0\}$ .

Let $A=\left \{ 0,1,2,3,4,5 \right \}$ . Insert the appropriate symbol $\in$ or $\notin$ in the blank spaces
(i) $0 ---- A$
(ii) $6 ---- A$
(iii) $3 ---- A$
(iv) $4 ---- A$
(v) $7 ---- A$

If $n\left ( A \right )=26, n\left ( B \right )=10, n\left ( A\cup B \right )=30,n\left ( {A}' \right )=17$ , find $n\left ( A\cap B \right )$ and $n\left ( \cup \right )$ .

Examine whether $A=$ $\{$ $x:x$ is a positive integer divisible by $3 \}$ is a subset of $B=$ $\{$ $x:x$ is a multiple of $5$ , $x\in N$ $\}$ .

Is $\phi =\left \{ \phi \right \}$ ? Why?

If $A$ and $B$ are two sets such that $A$ has $50$ elements, $B$ has $65$ elements and $A\cup B$ has $100$ elements, how many elements does $A\cap B$ have?

If $A$ and $B$ are two sets containing $13$ and $16$ elements respectively, then find the minimum and maximum number of elements in $A\cup B$ ?

Let $X=\left \{ -3,-2,-1,0,1,2 \right \}$ and $Y=$ $\{$ $x:x$ is an integer and $-3\leq x< 2$ $\}$
(i) Is $X$ a subset of $Y$ ?
(ii) Is $Y$ a subset of $X$ ?

If $n\left ( A\cap B \right )=5, n\left ( A\cup B \right )=35, n\left ( A \right )=13,$ find $n\left ( B \right )$ .

Which of the sets are equal sets? State the reason.
$\phi ,\left \{ 0 \right \},\left \{ \phi \right \}$

If $A=\left \{-2,-1,0,3,4 \right \}, B=\left \{-1,3,5\right \}$ , find (i) $A-B$
(ii) $B-A$

Let $A$ and $B$ be two finite sets such that $n\left ( A- B \right )=30, n\left ( A\cup B \right )=180$ . Find $n\left ( B \right )$ .

If $n\left ( \cup \right )=38, n\left ( A \right )=16, n\left ( A\cap B \right )=12,n\left ( {B}' \right )=20$ , find $n\left ( A\cup B \right )$ .

Given n(A)=285, n(B)=195, n(U)=500, $n(A \cup B)=410$ , find $n(A' \cup B')$ .

Given

the sets $A=\left \{4,5,6,7\right \}$ and $B=\left \{1,3,8,9\right \}$ . Find $A\cap B$

Let A = {4, 6, 8, 10 } and B = { 3, 4, 5, 6, 7 }. If $f: A \rightarrow B$ is defined by $f(x)=\frac{1}{2}x+1$ then represent f(x) by a set of ordered pairs

Given A $= \{1, 2, 3, 4\}$ , B $= \{3, 4, 5, 6\}$ and C $= \{6, 7\}$ . Verify that (A $\cap$ B) $\cap$ C = A $\cap$ (B $\cap$ C).

If A = {-1, 1}, find A $\times$ A $\times$ A.

Let A = {1, 2, 3}, B = {2, 4, 6, 8}, C = {3, 4, 5, 6} then prove the following :-
$A \, \cup \, B$ = {1, 2, 3, 4, 6, 8}

Using properties of sets, prove that $A \cup(A \cap B)=A$

$A-( A-B )=$ ...........

Given : A = {2, 3}, B = {4, 5}, C = {5, 6}, then prove the following :
$A \, \times \, (B \, \cap \, C)$ = {(2, 5), (3, 5)}

Given : A = {2, 3}, B = {4, 5}, C = {5, 6}, then prove the following :
$A \, \times \, (B \, \cup \, C)$ = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

Verify the following identities where A = {1, 2, 3, 4, 5}, B = {2, 3, 5, 6}, C = {4, 5, 6, 7}
$A \, \cap \, (B \, \cup \, C) \, = \, (A \, \cap \, B) \, \cup \, (A \, \cap \, C)$

Let $A$ $=\{1,2\},$ and $B$ $=\{1,3\},$ then prove that $\left( A\times B \right) \cup \left( B\times A \right)=\left\{(1,3),(2,3),(3,1),(3,2),(1,1),(1,2),(2,1)\right\}$

Given $A ={2,3}, B = {4,5}, C ={5,6},$ find
(i) $A\times \left( B\cap C \right) =\ .......$
(ii) $\left( A\times B \right) \cup \left( B\times C \right) =\ .......$

Comment on the following statements
$A-B=A\cap B'=B'-A'$

Comment on the following statements
$A-\left( A-B \right) =A\cap B$

Let $A=\left\{1,2,3 \right\},$ $B=\left\{2,4,6,8 \right\},$ $C=\left\{2,3,5,6 \right\}$ . Then $A\cap \left( B\cup C \right).\ ...........$

Set $A=\left\{ x:x\ is\ a\ digit\ in\ the\ number\ 3591 \right\}$
$B=\left\{ x:x\in N,x<10 \right\}$ . Find $A\cup B,A\cap B,A-B$ and $B-A$ .

Let A and B be two sets such that $A\times B$ consists of 6 elements. If three elements of $A\times B$ are (1,4), (2,6), (3, 6), find $A\times B$ and $B\times A$ .

Write the following set in roaster form:
$A=\{x\,:\,x$ is an integer and $-3\leq x\,<\,7\}$

List all the elements of the following sets :
$B = \left\{ {x:x = {\dfrac{1} {{2n - 1}}},1 \le n \le 5} \right\}$

Describe the following sets in Roster form:
The sets of all letters in the word 'BETTER'.

List all the elements of the following sets :
F = { $x$ : $x$ is a letter in the word "MISSISSIPPI"}

List all the elements of the following sets :
$A = \left\{ {x:{x^2} \le 10,x \in Z} \right\}$

List all the elements of the following sets :
D = { $x$ : $x$ is a vowel in the word "EQUATION"}

For any two sets of A and B, prove that
$B' \subset A'$ if $A \subset B$

List all the elements of the following sets :
{ ${C = x:x}$ is an integar , ${ - {1 \over 2} < x < {9 \over 2}}$ }

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

Let $A$ and $B$ be two sets. If $A\cup X=B\cap X=\phi$ and $A\cup X=B\cup X$ for sum set $X$ that $A=B$ .

If $A=\left\{6,9,11\right\}$ then find $A\cup \phi$

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

If $A=(1,2,3,4,5,6,7,8,9)$
$B=(1,3,5,,7,9,11,13)$
$C=(2,4,6,8,10,12,14)$
$D=(2,3,5,7,11,13,17)$
then find $B \cup C$

If $A$ and $B$ are two disjoint sets $n(A)=15$ and $n(B)=10$ find $n\left( {A \cup B} \right)$ and $n\left( {A \cap B} \right)$ .

Write all the elements of the set $S = \{ 1,\{ 3\} ,\{ 5,7\} ,9,\,10\}$ .

Let $A$ and $B$ be events such that $p\left( A \right) = \dfrac{5}{{11}},\,P\left( B \right) = \dfrac{6}{{11}}$ , and $P\left( {A \cup B} \right) = \dfrac{7}{{11}}$ and $P(B/A).$

Let A $=\{a, e, i, o, u\}$ and B $=\{a, b, c, d\}$ . Is A a subset of B? No(why?). Is B a subset of A? No. (Why?)

If $f=\left\{(4,5),(5,6),(6,-4)\right\}$ and $g=\left\{(4,-4),(6,5),(8,5)\right\}$ . $|f-g|=?$

If $A=(1,2,3,4,5,6,7,8,9)$
$B=(1,3,5,,7,9,11,13)$
$C=(2,4,6,8,10,12,14)$
$D=(2,3,5,7,11,13,17)$
then find $(A-D) \cup \left( {B - A} \right)$

If $A=(1,2,3,4,5,6,7,8,9)$
$B=(1,3,5,,7,9,11,13)$
$C=(2,4,6,8,10,12,14)$
$D=(2,3,5,7,11,13,17)$
then find $A-D$

$1800$ boys and $900$ girls appeared for an examination. If $42\%$ of the boys and $30\%$ of the girls passed, find
(i) number of boys passed
(ii) number of girls passed
(iii) total number of students passed
(iv) number of students failed
(v) percentage of student failed.

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

If $A=\{ 3,4,6\} , B=\{ 1,3\}$ and $C=\{ 1,2,6\}$ then find $\left( A-B \right) \times \left( A-C \right)$

If $A=\left\{4,5,6\right\};B=\left\{7,8\right\}$ then show that $A\bigcup B=B\bigcup A$

If A={1,2,3,4,5} , B={3,4,7,8} then find $A \cup B$ and $A \cap B$ .

If $A=\{6,9,11\}$ and $B=\phi$ ,find $A\cup B$ .

Let $O=$ Set of odd natural numbers $=\{1,3,5,7,9,...\}$ and $E=$ Set of even natural numbers $=\{2,4,6,8,10,........\}$ . Then show that $-1 \notin E$ .

Let $O=$ Set of odd natural numbers $=\{1,3,5,7,9,...\}$ and $E=$ Set of even natural numbers $=\{2,4,6,8,10,........\}$ . Then show that $3\in E$ .

Let $O=$ Set of odd natural numbers $=\{1,3,5,7,9,...\}$ and $E=$ Set of even natural numbers $=\{2,4,6,8,10,........\}$ . Then show that $0.1\notin E$ .

Let $A=\{1,2,3,4\}$ and $B=\{2,4,6,8\}$ . Then find $A-B$ .