ML AGGARWAL CLASS 8 CHAPTER 6 Operation on Sets Venn Diagrams Excercise 6.1
Exercise 6.1
Question 1
Sol :
A={0,1,2,3,4,5,6,7,8} B={3,5,7,9,11}, C={0,5,10,20}
(i) $A \cup B=\{0,1,2,3,4,5,6,7,8,9,11\}$
(ii) $A \cup C=\{0,1,2,3,4,5,6,7,8,10,20\}$
(iii) $B \cup C=\{0,3,5,7,9,10,11,20\}$
(iv) $A \cap B=\{3,5,7\}$
(v) $A \cap C=\{0,5\}$
(vi) $B \cap C=\{5\}$
As B⋃C has 8 elements, $n(B \cup C)=8$
As $A \cap B$ has 3 elements, $n(A \cap B)=3$
As $A \cap C$ has 2 elements, $n(A \cap C)=2$
As $B \cap C$ has 1 element, $\eta(B \cap C)=1$
Question 2
Sol :
(i)
$A=\{0,1,4,7\}$ and $\xi=\{x \mid x \in w, x \leq 10\}$
Given $\xi_=\{0,1,2,3,4,5,6,7,8,9,10\}$
Complement of $A=A^{\prime}=\{2,3,5,6,8,9,10\}$
(ii)
$A=\{$ Consonants $\}$ and $\xi=\{$ alphabets of English $\}$
Complement of $A=A^{\prime}$, {vowels}
={a, e, i, o,u}
(iii) A = {boys in class viii of all schools in bengaluru} and
ξ = {student in class viii of all school in bengaluru}
Complement of $A=A^{\prime}$, {girls in class viii of all school in bengaluru}
(iv) $A=\{$ letter of $\mathrm{KAlKA}\}$ and $\xi=\{$ letters of $\mathrm{KOlKATA}\}$
Complement of $A=A^{\prime}=\{0, T\}$
(v) $A=\{$ odd natural numbers\} and $\xi=\{$ Whole numbers\}
Complement of $A=A^{\prime}=\{0,2,4,6,8,10,12 \ldots \cdots-\ldots .\}$
Question 3
Sol :
$A=\{x: x \in N$ and $3<x<7\}$ and $B=\{x: x \in \omega$ and $x \leq 4\}$
$A=\{4,5,6\} \quad$ and $B=\{0,1,2,3,4\}$
(i) $A \cup B=\{0,1,2,3,4,5,6\}$
(ii) $A \cap B=\{4\}$
(iii) A-B={5,6}$
$A=\{x: x \in N$ and $3<x<7\}$ and $B=\{x: x \in \omega$ and $x \leq 4\}$
$A=\{4,5,6\} \quad$ and $B=\{0,1,2,3,4\}$
(i) $A \cup B=\{0,1,2,3,4,5,6\}$
(ii) $A \cap B=\{4\}$
(iii) A-B={5,6}$
in B-A=\{0,1,2,3}
Question 4
Sol :
$p=\{0,1,2,3,4,5\} \quad Q=\{4,5,6,7,8\}$
(i) $p \cup Q=\{0,1,2,3,4,5,6,7,1\}$
(ii) $P \cap Q=\{4,5\}$
(iii) P-Q={0,1,2,3}
(iv) Q-P={6,7,8}
yes $p \cup Q$ is a proper superset of $p \cup Q$ but vice versa is not possible
since A contains elements not in B.
Question 5
Sol :
$A=\left\{\right.$ letters of ward INTEGRITY\} $B=\left\{\begin{array}{l}\text { letters of word } \\ \text { ReckonING\} }\end{array}\right.$
(i) $A \cup B=\{I, N, T, E, G, R, Y, C, 0\}$
(ii) $A \cap B=\{I, N, E, G, R\}$
(iii) A-B={T, Y}
(iv) B-A={c, k, 0}
(a) $n(A)=7 \quad \eta(B)=8 \quad n(A \cap B)=5 \quad n(A \cup B)=10$
$n(A-B)=2 \quad n(B-A)=3$
$n(A)+n(B)-n(A \cap B): 7+8-5=10=\eta(A \cup B)$
(b) $\eta(A \cup B)-n(B)=10-8=2=n(A-B)$
$\quad \eta(A)-n(A \cap B)=7-5=2=n(A-B)$
(c) $\eta(A \cup B)-n(A)=10-7=3=\eta(B-A)$
$\eta(B)-n(A \cap B)=8-5=3=\eta(B-A)$
d) $\eta(A-B)+n(B-A)+n(A \cap B)=2+3+5=10=n(A \cup B)$
Question 6
Sol :
ξ={10,11,12,13,14 .. 40}
A={5,10,15,20,2530,35,40}
B={6,12,18,24,30,36}....(3)
(i) $A \cup B=\{5,6,10,12,15,18,20,24,25,30,35,40\}$
$A \cap B=\{30\}$
(i) $\eta(A)=8, n(B)=6, n(A \cap B)=1 \quad \eta(A \cup B)=13$
$\eta(A)+\eta(B)-n(A \cap B)=8+6-1: 13=n(A \cup B)$
Question 7
Sol :
(i) $A^{\prime}=\{5,9\}$
(ii) $B^{\prime}=\{1,2,3,5,7,9\}$
(iii) $A \cup B=\{1,2,3,4,6,7,8\}$
(iv) $A \cap B=\{4,6,8\}$
(v) $A-B=A \cap B^{\prime}=\{1,2,3,7,\}$
(vi) $B-A=B \cap A^{\prime}=\{3$
(ii) $(A \cap B)^{\prime}=\{1,2,3,5,7,9\}$
(viii) $A^{\prime} \cup B^{\prime}=\{1,2,3,5,7,9\}$
(a) $(A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}=\{1,2,3,5,7,9\}$ verified
(b) $n(A)=7 \quad \eta\left(A^{\prime}\right)=2 \quad \eta(\xi)=9$
n(A)+n\left(A^{\prime}\right)=7+2=9=n(\xi) \text { verified }
(c)$n(A \cap B)+n\left((A \cap B)^{\prime}\right)$
$n(A \cap B)=3 ; n\left((A \cap B)^{\prime}\right)=6$
$6+3=9=\eta(\xi) .$ verified
(d) $n(A-B)=4 \quad \eta(B-A)=0 \quad n(A \cap B)=3$
$\quad \eta(A-B)+\eta(B-A)+n(A \cap B)=4+0+3=7=n(A \cup B)$
Question 8
Sol :
$\xi_{1}=\{x: x \in w, x \leq 10\}, A=\{x: x \geq 5\} \quad B=\{x: 3 \leq x<8\}$
$\xi=\{0,1,2,3,4,5,6,7,8,9,10\} \quad A=\{5,6,7,8,9,10\}$
$B=\{3,4,5,6,7\}$
(i)
$\begin{aligned} A \cup B=\{3,4,5,6,7,8,9,10\} & A^{\prime}=\{0,1,2,3,4\} \\(A \cup B)^{\prime}=\{0,1,2\} & B^{\prime}:\{0,1,2,8 ; 9,10\} \\ A^{\prime} \cap B^{\prime}=\{0,1,2\} \end{aligned}$
∴ $(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}=\{0,1,2\}$
(ii)
$A \cap B=\{5,6,7\}-1(A \cap B)^{\prime}=\{0,1,2,3,4,8,9,10\}$
$A^{\prime} \cup B^{\prime}=\{0,1,2,3,4,8,9,10\}$
∴ $(A \cap B)^{\prime}=A^{\prime} U B^{\prime}$
(iii)
$\begin{aligned} A-B &=\{8,9,10\} \\ A \cap B^{\prime} &=\{8,9,10\} \end{aligned} \quad \therefore A-B=A \cap B^{\prime}$
(iv)
$\begin{aligned} B-A &=\{3,4\} \\ B \cap A^{\prime} &=\{3,4\} \end{aligned} \quad \therefore \quad B-A=B \cap A^{\prime}$
Question 9
Sol :
$n (A)=20, \quad n (B)=16, n(A \cup B)=30 \quad n (A \cap B)=?$
we know $n (A \cup B)= n (A)+n(B)-n (A \cap B)$
$30=20+16-n (A \cap B)$
$\begin{aligned} n (A \cap B)=36-30=6 \\ n (A \cap B)=C \end{aligned}$
$n (A \cap B)=c$
Question 10
Sol :
$n(5)=20: n\left(A^{\prime}\right)=7 \quad n(A)=?$
We know $\quad n(A)+n\left(A^{\prime}\right)=n\left(\xi_{1}\right)$
n(A)=20-7=13
n(A)=13
n(A)=13
Question 11
Sol :
$n\left(\xi_{1}\right)=40 \quad n(A)=20 \quad n\left(B^{\prime}\right)=16 \quad n(A \cup B)=32$
$\begin{aligned} n(B)+n\left(B^{\prime}\right)= n (\xi) \Rightarrow & n(B)=40-16=24 . \\ & n(B)=24 \end{aligned}$
$n (A \cup B)=D(A)+\eta(B)-D(A \cap B)$
$32=20+24-n(A \cap B)$
$n(A \cap B)=44-32=12$
$n(B)=24 ; n(A \cap B)=12$
Question 12
Sol :
$n\left(\xi_{1}\right)=32, \quad n(A)=20, n(B)=16, n\left((A \cup B)^{\prime}\right)=4$
(i) $\begin{aligned} n(A \cup B) &\left.=\eta(\xi)-n(A \cup B)^{\prime}\right) \\ &=32-4=28 \end{aligned}$
$n(A \cup B)=28$
(ii) $n(A \cap B)=n(A)+n(B)-n(A \cup B)$
20+28-28
36-28=8
$n(A \cap B)=8$
(iii) $\begin{aligned} n(A-B) &=n(A)-n(A \cap B) \\ &=20-8=12 \\ & n(A-B)=12 \end{aligned}$
Question 13
Sol :
$\eta(\xi)=40 \quad: n\left(A^{\prime}\right)=15, n(B)=12 \quad n\left((A \cap B)^{\prime}\right)=32$
(i) $n(A)=n(\xi)-n(A)=40-15=25$
(ii) $n\left(B^{\prime}\right)=n\left(\Sigma_{1}\right)-n(B)=40-12=28$
(iii) $n (A \cap B)=n(\xi)-n\left((A \cap B)^{\prime}\right)=40-12=8$
(iv) $n(A \cup B)=n(A)+n(B)-n(A \cap B)=25+12-8=29$
(v) $n(A-B)=n(A)-n(A \cap B)=25-8=17$
(vi) $\eta(B-A)=n(B)-n(A \cap B)=12-8=4$
Question 14
Sol :
$\eta(A-B)=12, n(B-A)=16, \quad n(A \cap B)=5$
(i) $n(A) \quad$ ii $\quad n(B)$ iii $n(A \cup B)$
(i)$\begin{aligned} n(A-B) &=n(A)-n(A \cap B) \\ n(A) &=12+5=17 \end{aligned}$
(ii) $\begin{aligned} n(B-A) &=n(B)-n(A \cap B) \\ \eta(B) &=16+5=21 \end{aligned}$
(iii) $\begin{aligned} n(A \cup B) &=n(A)+h(B)-n(A \cap B) \\ &=17+21-5=38-5=33 \\ & n(A \cup B)=33 \end{aligned}$
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