In the previous post, Set Theory : Part-1 we have discussed about Short Answer Type questions and its solutions.
In this post, I will discuss Long Answer Type Questions of the S.N.Dey Math Solution series. Hopefully, this solution series will be helpful for my beloved students and the readers who takes interest in mathematics.
2. Given, $\,\,A=\{x: 0 \lt x \leq 2\} \,\,\text{and}\,\, B=\{x: 1 \lt x \lt 3\} ; \,\,\text{Find }\\ (i) A \cap B \,\,\,\,\,(ii)A \cup B \quad (iii)A-B \quad (iv)(A \cup B)-(A \cap B) $
Sol. 2(i) Let $\,\,A=\{x: 0 \lt x \leq 2\} \,\,\text{and}\,\, B=\{x: 1 \lt x \lt 3\} \\ \text{Then by definition, }\\ A \cap B =\{x \in A \wedge x \in B\}\\= \{x: 1 \lt x \leq 2\}.$
The Venn diagram of the intersection of sets $\,A\,$ and $\,B\,$ has been shown in the figure. In the diagram, the set $\,\, A \cap B\,\, $ has been shaded by solid yellow color.
Sol. 2(ii) Let $\,\,A=\{x: 0 \lt x \leq 2\}, \,\, B=\{x: 1 \lt x \lt 3\} \\ \text{Then by definition, }\\ A \cup B =\{x \in A \vee x \in B\}\\= \{x: 0 < x <3\}.$
Sol. 2(iii) Let $\,\,A=\{x: 0 < x \leq 2\} \,\,\text{and}\,\, B=\{x: 1 < x<3\} \\ \text{Then by definition, }\\ A -B =\{x \in A \wedge x \not \in B\}\\= \{x: 0 < x \leq 1\}.$
Sol. 2(iv) Let $\,\,A=\{x: 0 < x \leq 2\}, \,\, B=\{x: 1 < x < 3\} \\ \text{Then by definition, }\\ (A \cup B)-(A \cap B) =\{x \in (A \cup B) \wedge x \not \in (A \cap B) \}\\= \{x: 0 < x \leq 1 \wedge 2 <x<3 \}.$
3. Let $\,A=\{x:2 \leq x<5\}\,$ and $\,B = \{x:3<x<7\}\,$ be two subsets of the universal set, $\,S = \{x: 0<x \leq 10\};$ verify that, $\,\,(A \cup B)^c=A^c \cap B^c$
Sol. $\,\, A \cup B=\{x:2 \leq x<5\} \cup \{x:3<x<7\}\\=\{x: 2 \leq x < 7 \}$
So, $\,\,(A \cup B)^c =S-(A \cup B)\\=\{x: 0<x \leq 10\} - \{x: 2 \leq x < 7 \} \\ =\{x: 0<x<2, 7 \leq x \leq 10\} \cdots (1) \\ A^c=S-A=\{x: 0 <x<2, 5 \leq x \leq 10\}\\ B^c=S-B= \{x: 0 <x \leq 3, 7 \leq x \leq 10\} \\ A^c \cap B^c=\{x: 0<x<2, 7 \leq x \leq 10\} \cdots (2) $
Hence , from (1) and (2) follows the result.
4. If $\,\,P=\{p,q,r,s,t,u\},\,\, Q \cap R=\{q,r,v,w\},\,\,$ find
(i) $(P \cup Q) \cap (P \cap R)\,\,\,(ii) (P-Q)\cup (P-R)$
Sol. (i) $(P \cup Q) \cap (P \cap R)\\=P \cup (Q \cap R)\\=\{p,q,r,s,t,u,v,w\}\\ (ii)(P-Q)\cup (P-R)\\=P-( Q \cap R)\\= \{p,s,t,u\} $
5. If $\,\, A,B,C \,\,$ be three subsets of the universal set $\,S\,$ where $\, S=\{1,2,3,4,5,6,7\},\,\, A=\{1,3,5,6\}\,$ and $\,\,B \cap C=\{1,2,6\},$ find $\,\, (A \cup B) \cap (A \cup C),\,\,\, B^c \cup C^c.$
Sol. $\,\,(A \cup B) \cap (A \cup C)\\=A \cup (B \cap C)\\=\{1,3,5,6\} \cup \{1,2,6\}\\=\{1,2,3,5,6\} \\ B^c \cup C^c= (B \cap C)^c \\=S-(B \cap C)\\=\{1,2,3,4,5,6,7\}-\{1,2,6\}\\=\{3,4,5,7\}$
6. If $\,\,U=\{a,b,c,d,e,f\}\,\,$ be the universal set and $\,\,A,B,C\,$ are three subsets of $\,U,\,$ where $\,\,A=\{a,c,d\}$ and $\,\, B \cup C=\{a,d,c,f\},\,$
find $\, (i) (A \cap B) \cup (A \cap C) \,\, (ii) (B' \cap C')$
Sol. $\,\, (A \cap B) \cup (A \cap C)\\=A \cap (B \cup C)\\=\{a,c,d\} \cap \{a,d,c,f\}\\=\{a,c,d\} \\ \text{Now,}\,\, (ii) (B' \cap C')\\=(B \cup C)'\\=U-(B \cup C) \\=\{a,b,c,d,e,f\}- \{a,d,c,f\}\\=\{b,e\}$
7. Given, $\,X \cup Y= \{1, 2, 3, 4\}, X \cup Z =\{2, 3, 4, 5\}, X \cap Y= \{2,3\}\, \text{and}\, X \cap Z = \{2, 4\};$
Find the value of $\,X, Y ,Z$.
Sol. From the question, It is evident that $\,X,Y\,$ both
contains $\,\{2,3\}\,\cdots (1)$ and $\,X,Z\,$ both
contains $\,\{2,4\}.\,\cdots (2)$
Hence, from (1) and (2), it follows that $\,\,X=\{2,3,4\} \cdots (4)$
Now, since $ X \cup Y= \{1, 2, 3, 4\}, \cdots (5)\\ \, X \cup Z =\{2, 3, 4, 5\},\cdots (6)$
and so from (1),(4) and (5), it follows $\,\,Y=\{1,2,3\}$ .
Similarly, from (2),(4) and (6), it follows $\,\,Z=\{2,4,5\}.$
8. Verify the following relations using Venn diagrams:
(i) $(A \cup B)\cap (A \cup C)= A \cup (B \cap C)$
From figure (1), we get the L.H.S. which has been drawn by using Venn Diagram whereas
from figure (2), we get the R.H.S. which has been drawn by using Venn Diagram.
8. Verify the following relations using Venn diagrams:
(ii) $A \cap (B \cup C)= (A \cap B) \cup (A \cap C)$
From figure 1(a), we get the L.H.S. which has been drawn by using Venn Diagram whereas
from figure 1(b), we get the R.H.S. which has been drawn by using Venn Diagram.
8. Verify the following relations using Venn diagrams:
(iii) $(A \cup B)^c= A^c \cap B^c$
From figure 5(a), we get the L.H.S. which has been drawn by using Venn Diagram whereas
from figure 5(b), we get the R.H.S. which has been drawn by using Venn Diagram.
8. Verify the following relations using Venn diagrams:
(iv) $(A \cap B)^c= A^c \cup B^c$
From figure 4(a), we get the L.H.S. which has been drawn by using Venn Diagram whereas
from figure 4(b), we get the R.H.S. which has been drawn by using Venn Diagram.
From figure 2(a), we get the L.H.S. which has been drawn by using Venn Diagram whereas
from figure 2(b), we get the R.H.S. which has been drawn by using Venn Diagram.
8. Verify the following relations using Venn diagrams:
(vi) $A-(B \cup C)=(A-B) \cap (A-C)$
From figure 3(a), we get the L.H.S. which has been drawn by using Venn Diagram whereas
from figure 3(b), we get the R.H.S. which has been drawn by using Venn Diagram.
8. Verify the following relations using Venn diagrams:
(vii) $(A-C)\cap (B-C)=(A \cap B)-C$
From figure 4(a), we get the L.H.S. which has been drawn by using Venn Diagram whereas
from figure 4(b), we get the R.H.S. which has been drawn by using Venn Diagram.
9. (i) Draw a Venn diagram of three non-empty sets $\,\,A, B\,$ and $\,C\,$ such that $\,\, A \subset B,\,\, C \not \subset B,\,\, A \cap C =\phi.$
9. (ii) Draw a Venn diagram of three non-empty subsets $\,\,A, B\,$ of the universal set $\,\,S\,\,$ such that $\,\, A \subset B,\,\,B \cap C \neq \phi\, ,\,\, A \cap C =\phi,C \not \subset B $
10. For any three sets , $\,\,A,B,C\,\,$ prove the followings :
(i) $\,\, A \cup (B \cap C)= (A\cup B)\cap(A \cup C)$
Sol. Let $\,\, x \in A \cup (B \cap C) \\ \implies x \in A \, \,\vee x \in (B \cap C) \\ \implies x \in A \,\, \vee (x \in B \wedge x \in C) \\ \implies (x \in A \vee x \in B) \wedge (x \in A \vee x \in C) \\ \implies ( x \in A \cup B) \wedge (x \in A \cup C) \\ \implies x \in (A \cup B)\cap(A \cup C) \\ \text{Hence,}\,\, A \cup (B \cap C) \subseteq (A\cup B)\cap(A \cup C) \cdots (1)$
Now, let $\,\, y \in (A\cup B)\cap(A \cup C) \\ \implies y \in (A \cup B) \wedge y \in (A \cup C)\\ \implies (y \in A \vee y \in B) \wedge (y \in A \vee y \in C) \\ \implies y \in A \vee (y \in B \wedge y \in C) \\ \implies y \in A \cup (B \cap C) \\ \text{Hence,}\,\, (A\cup B)\cap(A \cup C) \subseteq A \cup (B \cap C) \cdots (2)$
Hence, from (1) and (2), the result follows.
10. For any three sets , $\,\,A,B,C\,\,$ prove the followings :
(ii) $\,\, A \cap (B \cup C)= (A\cap B)\cup(A \cap C)$
Sol. Let $\,\, x \in A \cap (B \cup C) \\ \implies x \in A \wedge ( x \in B \cup C)\\ \implies x \in A \wedge ( x \in B \vee x \in C) \\ \implies (x \in A \wedge x \in B)\vee (x \in A \wedge x \in C) \\ \implies x \in A \cap B \,\,\vee x \in A \cap C \\ \implies x \in (A \cap B)\cup (A \cap C)\\ \text{Hence,}\,\, A \cap (B \cup C) \subseteq (A \cap B)\cup (A \cap C) \cdots (1) $
Similarly, let $\,\,y \in (A \cap B)\cup (A \cap C) \\ \implies y \in A \cap B \vee y \in (A \cap C) \\ \implies (y \in A \wedge y \in B) \vee (y \in A \wedge y \in C) \\ \implies y \in A \wedge (y \in B \vee y \in C) \\ \implies y \in A \wedge (y \in B\cup C)\\ \implies y \in A \cap (B \cup C) \\ \text{Hence,}\,\, (A \cap B)\cup (A \cap C) \subseteq A \cap (B \cup C) \cdots (2)$
Hence, from (1) and (2), the result follows.
10. For any three sets , $\,\,A,B,C\,\,$ prove the followings :
(iii) $\,\, A \cup (B \cup C)= (A\cup B)\cup C$
Sol. Let $\,\, x \in A \cup (B \cup C) \\ \implies x \in A \vee ( x \in B \cup C)\\ \implies x \in A \vee ( x \in B \vee x \in C) \\ \implies (x \in A \vee x \in B)\vee x \in C \\ \implies x \in A \cup B \,\,\vee x \in C \\ \implies x \in (A \cup B)\cup C\\ \text{Hence,}\,\, A \cup (B \cup C) \subseteq (A \cup B)\cup C \cdots (1) $
Similarly, let $\,\,y \in (A \cup B)\cup C \\ \implies y \in A \cup B \vee y \in C \\ \implies (y \in A \vee y \in B) \vee y \in C \\ \implies y \in A \vee (y \in B \vee y \in C) \\ \implies y \in A \vee (y \in B\cup C)\\ \implies y \in A \cup (B \cup C) \\ \text{Hence,}\,\, (A\cup B)\cup C \subseteq A \cup (B \cup C) \cdots (2)$
Hence, from (1) and (2), the result follows.
10. For any three sets , $\,\,A,B,C\,\,$ prove the followings :
(iv) $\,\, A \cap (B \cap C)= (A\cap B)\cap C$
Sol. Let $\,\, x \in A \cap (B \cap C) \\ \implies x \in A \wedge ( x \in B \cap C)\\ \implies x \in A \wedge ( x \in B \wedge x \in C) \\ \implies (x \in A \wedge x \in B)\wedge x \in C \\ \implies x \in A \cap B \,\,\wedge x \in C \\ \implies x \in (A \cap B)\cap C\\ \text{Hence,}\,\, A \cap (B \cap C) \subseteq (A \cap B)\cap C \cdots (1) $
Similarly, let $\,\,y \in (A \cap B)\cap C \\ \implies y \in A \cap B \wedge y \in C \\ \implies (y \in A \wedge y \in B) \wedge y \in C \\ \implies y \in A \wedge (y \in B \wedge y \in C) \\ \implies y \in A \wedge (y \in B\cap C)\\ \implies y \in A \cap (B \cap C) \\ \text{Hence,}\,\, (A\cap B)\cap C \subseteq A \cap (B \cap C) \cdots (2)$
Hence, from (1) and (2), the result follows.
10. For any three sets , $\,\,A,B,C\,\,$ prove the followings :
(v) $\,\, (A \cup B)^c=A^c \cap B^c$
Sol. Let $\,\, x \in (A \cup B)^c\\ \implies x \not\in (A \cup B)\\ \implies x \not\in A \wedge x \not \in B \\ \implies x \in A^c \wedge x \in B^c \\ \implies x \in A^c \cap B^c \\ \text{Hence,}\,\, (A \cup B)^c \subseteq A^c \cap B^c \cdots (1) $
Similarly, let $\,\,y \in A^c \cap B^c \\ \implies y \in A^c \wedge y \in B^c \\ \implies y \not \in A \wedge y \not \in B \\ \implies y \not\in A \cup B \\ \implies y \in (A \cup B)^c\\ \\ \text{Hence,}\,\, A^c \cap B^c \subseteq (A \cup B)^c \cdots (2)$
Hence, from (1) and (2), the result follows.
10. For any three sets , $\,\,A,B,C\,\,$ prove the followings :
(vi) $\,\, (A \cap B)^c=A^c \cup B^c$
Sol. Let $\,\, x \in (A \cap B)^c\\ \implies x \not\in (A \cap B)\\ \implies x \not\in A \vee x \not \in B \\ \implies x \in A^c \vee x \in B^c \\ \implies x \in A^c \cup B^c \\ \text{Hence,}\,\, (A \cap B)^c \subseteq A^c \cup B^c \cdots (1) $
Similarly, let $\,\,y \in A^c \cup B^c \\ \implies y \in A^c \vee y \in B^c \\ \implies y \not \in A \vee y \not \in B \\ \implies y \not\in A \cap B \\ \implies y \in (A \cap B)^c\\ \\ \text{Hence,}\,\, A^c \cup B^c \subseteq (A \cap B)^c \cdots (2)$
Hence, from (1) and (2), the result follows.
10. For any three sets , $\,\,A,B,C\,\,$ prove the followings :
(vii) $\,\, A-(B \cup C)=(A-B) \cap (A-C) $
Sol. $\,\, (A-B) \cap (A-C) \\=(A \cap B^c) \cap (A \cap C^c)\\ =(B^c \cap A) \cap (A \cap C^c) \\= B^c \cap (A \cap A) \cap C^c \\= B^c \cap A \cap C^c \quad [\therefore A \cap A=A]\\= A \cap (B^c \cap C^c)\\=A \cap (B \cup C)^c \quad [\text{By De Morgan's Law}] \\=A-(B \cup C)\quad [\text{Since,}\,\,A-B= A \cap B^c]$
10. For any three sets , $\,\,A,B,C\,\,$ prove the followings :
(viii) $\,\, A-(B \cap C)=(A-B) \cup (A-C) $
Sol. $\,\, A-(B \cap C) \\= A \cap (B \cap C)^c \quad [\therefore A-B=A \cap B^c]\\= A \cap (B^c \cup C^c) \quad [\text{By De Morgan's Law}]\\=(A \cap B^c) \cup (A \cap C^c)\\=(A-B) \cup (A-C)$
10. For any three sets , $\,\,A,B,C\,\,$ prove the followings :
(ix) $\,\, (A \cap B)-C=(A-C) \cap (B-C) $
Sol. $\,\, (A-C) \cap (B-C) \\=(A \cap C^c) \cap (B \cap C^c)\\=(A \cap C^c) \cap (C^c \cap B)\\=A \cap (C^c \cap C^c)\cap B\\=A \cap C^c \cap B\\=A \cap B \cap C^c \\\quad [\text{Since,}\,\,A \cap B =B \cap A]\\=(A \cap B)\cap C^c\\=(A \cap B)-C \\ [\text{Since,}\, P \cap Q^c=P-Q] $
10. For any three sets , $\,\,A,B,C\,\,$ prove the followings :
(x) $\,\, (A \cup B)-C=(A-C) \cup (B-C) $
Sol. $\,\,(A-C) \cup (B-C) \\=(A \cap C^c)\cup (B \cap C^c) \quad [\therefore P-Q=P \cap Q^c]\\=(C^c \cap A) \cup (C^c \cap B)\\=C^c \cap (A \cup B)\\=(A \cup B) \cap C^c\\=(A \cup B)-C \quad (\text{proved})$
11. Applying Set Algebra, prove the followings:
$11(i)\, A \cap (B-A)=\phi, \,\, 11(ii) A \cup (B-A), \\ 11(iii)(A \cap B)-C=(A-C)\cap (B-C),\,\,11(iv)\,(A \cup B)-C=(A-C)\cup (B-C) $
Sol. $\,\,11(i)\,A \cap (B-A)\\=A \cap (B \cap A^c)\\=(A \cap A^c)\cap B\\=\phi \cap B\\=\phi \\ 11(ii)\,\, A \cup (B-A)\\= A \cup (B \cap A^c)\\=(A \cup B) \cap (A \cup A^c)\\=(A \cup B)\cap S\\=A \cup B \\ 11(iii)\,\, (A \cap B)-C \\=(A \cap B)\cap C^c\\=(A \cap B)\cap (C^c \cap C^c)\\=[(A \cap B) \cap C^c] \cap C^c\\=[A \cap (B \cap C^c)] \cap C^c \\=(B \cap C^c)\cap (A \cap C^c)\\=(B-C) \cap (A-C)\\=(A-C) \cap (B-C)\\ 11(iv)\,\,(A \cup B)-C\\=(A \cup B)\cap C^c\\=(A \cap C^c)\cup (B \cap C^c)\\=(A-C) \cup (B-C)$
12. In an engineering college, $\,80\,$ students get chance for Computer Science, $\,75\,$ for Information Technology, $\,72\,$ for Electronics. If $\,60\,$ students get chance in 1st and 2nd; $\,50\,$ in 2nd and 3rd, $\,40\,$ in 1st and 3rd and $\,30\,$ get chance in all three branches, how many seats are there in the engineering college? [The college has only three disciplines.]
Sol. Let C $=80 \,\,$is the no. of students who took Computer Science,
I $=75 \,\,$is the no. of students who took Information Technology,
E $=72 \,\,$is the no. of students who took Electronics.
Then, We know, $\,\,n(C \cup I \cup E) \\=n(C)+n(I)+n(E)-n(C \cap I)- \\n(I \cap E)-n(E \cap C)+n(C \cap I \cap E)\\= 80+75+72-60-50-40+30 \\=107.$
So, there are 107 seats are there in the aforementioned engineering college.
13. In a survey of Jalpaiguri government engineering college students, it was found that $\,40 \%\,$ use their own books, $\,50 \%\,$ use library books, $\,30 \%\,$ use borrowed books, $\,20 \%\,$ use both their own books and library books, $\,15 \%\,$ use their own books and borrowed books, $\,10 \%\,$use library books and borrowed books, and $\,4 \%\,$ use their own books, library books and borrowed books. Calculate the percentage of students who do not use a book at all.
Sol. Let A is the no. of students who use their own books,
B is the percentage of students who use library books,
C is the percentage of students who use borrowed books,
x is the percentage of students who do not use a book at all.
Then, We know, $\,\,n(A \cup B \cup C) \\=n(A)+n(B)+n(C)-n(A \cap B)- \\n(B \cap C)-n(C \cap A)+n(A \cap B \cap C)\\ \implies 100-x= 40+50+30-20-10-15+4 \\ \implies 100-x=79 \\ \implies x=100-79=21.$
Hence follows the result.
14. A company studies the product preferences of $\,300\,$consumers. It was found that $\,226\,$ liked product A, $\,51\,$ liked product B, $\,54\,$ liked product C; $\,21\,$ liked products A and B, $\,54\,$ liked products A and C, $\,39\,$ liked products B and C and $\,9\,$ liked all the three products. Prove that, the study results are not correct. [Assume that each consumer likes at least one of the three products.]
Sol. $\,\,n(A \cup B\cup C)=n(A)+n(B)+n(C) \\-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A \cap B \cap C)\\=226+51+54-21-39-54+9\\=226$
But according to the problem, total number of consumers is: $\,\,300.$
Hence, we can conclude that the study results are not correct.
15. The production manager of Sen, Sarkar & Lahiri Company examined $\,100\,$ items produced by the workers and furnished the following report to his boss:
Defect in measurement $\,50\,$, defect in colouring $\,30\,$, defect in quality $\,23\,$, defect in quality and colouring $\,10\,$, defect in measurement and colouring $\,8\,$ defect in measurement and quality $\,20\,$ and $\,5\,$ are defective in all respects. The manager was penalised for the report. Using appropriate result of set theory explain the reason for the penal measure.
Sol. Let defect in measurement is denoted by the set $\,A,\,$ defect in colouring is denoted by the set $\,B\,$ and defect in quality is denoted by the set $\,C.\,$
So, from the given condition we compute
$\,n(C \cap A^c\cap B^c) \\=n(C)-n(C\cap A)-n(C \cap B)+n(A \cap B\cap C) \\=23-20-10+5\\=-2,\,\, \text{which is impossible.}$
16. In a city three daily newspapers $\,X, Y, Z\,$ are published; $\,65\%\,$ of the citizens read X, $\,54\%\,$ read Y, $\,45\%\,$read Z; $\,38\%\,$ read X and Y; $\,32\%\,$ read $\,Y\,$ and $\,Z\,$ ; $\, \,28\%\,\,$ read $\,X\,$ and $\,Z\,$; $\,12\%\,$ do not read any one of these three papers. If the total number of people in the city be $\,1000000\,$ find the number of citizens who read all the three newspapers. [You may use a Venn diagram or a standard formula for the enumeration of elements of sets.]
Sol. From the given condition, we notice
$n(X)=65\%=0.65 ,\,\,n(Y)=54 \%=0.54, \\n(Z)=45\%=0.45,\\ n(X \cap Y)=38\%=0.38,\,n(Y \cap Z)=32\%=0.32, \\ n(X \cap Z)=28\%=0.28.$
Now, since $\,12\%\,$ do not read any one of these three papers,
rest of $\,\,(100-12)\%=88\%\,\,$ read at least one of the three newspapers and
those readers can be denoted by $\,\,n(X \cup Y \cup Z)=88\%=0.88.$
We know that $n(X \cup Y\cup Z)=n(X)+n(Y)+n(Z) \\-n(X\cap Y)-n(Y\cap Z)-n(Z\cap X)+n(X \cap Y \cap Z) \\ \implies 0.88=0.65+0.54+0.45-0.38-0.32-0.28+n(X \cap Y \cap Z) \\ \implies 0.88=0.66+n(X \cap Y \cap Z) \\ \implies n(X \cap Y \cap Z)=0.88-0.66=0.22$
So, the number of citizens who read all the three newspapers $\\=1000000 \times 0.22\\=220000$
17. In Kalyani Govt. Engineering College out of $\,1000\,$ students , $\,540\,$ played football, $\,465\,$ played cricket and $\,370\,$ played volleyball; of the total $\,325\,$ played both football and cricket, $\,260\,$ played football and volleyball, $\,235\,$ played cricket and volleyball, $\,125\,$ played all the three games. How many students (i) did not play any game (ii) played only one game and (iii) played just two games?
Sol. Let A is the no. of students who played football,
B is the no. of students who played cricket ,
C is the no. of students who played volleyball,
Then, We know, $\,\,n(A \cup B \cup C) \\=n(A)+n(B)+n(C)-n(A \cap B)- \\n(B \cap C)-n(C \cap A)+n(A \cap B \cap C)\\ =540+465+370-325-235-260+125 \\ =680$
So, $\,680\,$ students played at least one game.
Therefore, the no. of students who did not play any game is: $\\=1000-680=320.\cdot (i)$
If we are to consider the students who played only one game, then we have to consider the following sets $\,\, A \cap B^c \cap C^c,\,\, A ^c \cap B \cap C^c,\,\,A^c \cap B^c \cap C. \\ n(A \cap B^c \cap C^c) \\=n(A)-n(A \cap B)-n(A \cap C)+n(A \cap B \cap C)\\=540-325-235+125\\=105\\ n(A^c \cap B \cap C^c) \\= n(B \cap A^c \cap C^c)\\=n(B)-n(B \cap A)-n(B \cap C)+n(A \cap B \cap C)\\=465-325-260+125\\=5 \\ n(A^c \cap B^c \cap C) \\=n(C \cap A^c \cap B^c)\\=n(C)-n(C \cap A)-n(C \cap B)+n(A \cap B \cap C)\\=370-235-260+125\\=0$
So, the no. of students who played only one game is : $\\=105+5+0=110 \cdots (ii)$
If we are to consider the students who played only two games, then we have to consider the following sets $\,\, A \cap B \cap C^c,\,\, A ^c \cap B \cap C,\,\,A \cap B^c \cap C. \\ n(A \cap B \cap C^c)=n(A \cap B)-n(A \cap B \cap C)\\=325-125\\=200\\ n(A^c \cap B \cap C)= n(B \cap C)-n(A \cap B \cap C)\\=235-125\\=110 \\ n(A \cap B^c \cap C)=n(A \cap C)-n(A \cap B \cap C)\\=260-125\\=135$
So, the no. of students who played two game is : $\\=200+110+135=445 \cdots (iii)$
18. A group consists of a number of students and each student of the group can speak at least one of the languages Bengali, Hindi and English. $\,65\,$ can speak Bengali, $\,54\,$ Hindi and $\,37\,$ English; $\,31\,$ can speak both Bengali and Hindi; $\,17\,$ both Hindi and English, and $\,18\,$ both Bengali and English. Determine the greatest and least number of students in the group.
Sol. Let $\,\, A,B,C \,\,$ denotes the no. of students who can speak Bengali, Hindi and English.
Then, $\,\,n(A\cup B\cup C)\\=n(A)+n(B)+n(C)-n(A \cap B)\\ -n(B \cap C) -n(C \cap A) + n(A \cap B\cap C)\\=65+54+37-31-17-18 + n(A \cap B\cap C)\\=90 + n(A \cap B\cap C)$
Now, the value of $\,n(A \cap B\cap C)\,\,$ is minimum, if $\,n(A \cap B\cap C)=0\,\,$ and vice versa.
Now, the maximum value of $\,\,n(A\cup B\cup C) \\= \text{min} \{n(A \cap B),n(B \cap C),n(C \cap A)\}=17.$
So, the greatest number of students in the group= $90+17=107$
and the least number of students in the group= $90+0=90$
19. Using set operations show that the numbers $\,231\,$ and $\,260\,$ are prime to each other.
Sol. Let $\,A\,$ and $\,B\,$ be the set of factors of $\,231\,$ and $\,260\,$ respectively.
So, $\,\,A=\{1,3,7,11,21,33,77,231\} \cdots (1) \quad$ and
$\,\,B=\{1,2,4,5,10,13,20,26,52,65,130,260\} \cdots (2)$
So, from (1) and (2), it follows that $\,\, A \cap B=\{1\}\cdots (3)$
Hence , from (3) , the desired result follows.
20. Suppose $\,\, A_1, A_2, A_3,.......,A_{30}\,\,$ are thirty sets each with five elements and $\,\,B_1,B_2,.....,B_n\,\,$ are $\,n\,$ sets each with $\,3\,$ elements. Let $\,A_1 \cup A_2 \cup A_3 \cup ......\cup A_{30}=B_1\cup B_2\cup ....\cup B_n=S. \,$ Assume that each element of $\,S\,$ belongs to exactly ten of the A's and to exactly nine of the B's. Find $\,n.$
Sol. Total no. of elements of the set $\,\,A= \frac{30 \times 5}{10} $ and
Total no. of elements of the set $\,\,A= \frac{n \times 3}{9} $
According to the condition, we get
$ \frac{30 \times 5}{10} = \frac{n \times 3}{9} \\ \implies n=45.$
21. 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 there in the conference?
Sol. Let $\,\,F\,$ denotes the number of Indian women, $\,\,M\,$ denotes the number of Indian men,
$\,\,D\,$ denotes the number of Indian doctors.
Total no. of Indians $=n(F)+n(M)=29+23=52$
So, total no. of foreigners$\\ =100-52\\=48 .$
Now, $\,\,n(D)=4, n(M \cup D)=24 \\ n(M \cap D)=n(M)+n(D)-n(M \cup D)\\ \implies n(M \cap D)=23+4-24=3 $
So, total no. of women doctors are : $=4-3=1$
22. Suppose that there are two sets $\,A\,$ and $\,B\,$ having $\,99\,$ elements in common, then find the no. of elements which are common to each of the sets $\,A \times B\,$and $\,B \times A.\,$
Sol. Since there are two sets $\,A\,$ and $\,B\,$ having $\,99\,$ elements in common, we have
$\,\,n(A \cap B)=99. \\ \text{Now , we know}\,\, (A \times B) \cap (C \times D)\\=(A \cap C) \times (B \cap D)\\ \text{And so,}\,\,(A \times B) \cap (B \times A)\\=(A \cap B) \times (B \cap A)\,\,\,\,\, \text{and so,}\\ n[(A \times B) \cap (B \times A)] = 99 \times 99=99^2$
>>>>>>>>>>>>>>>>>>>>To download full PDF Chhaya Math Solution on Set Theory , click here.
To continue with SET THEORY (Part-6), click here .
Please do not enter any spam link in the comment box