Question 13 Marks
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'
- B'
- $(\text{A}\cap\text{C})'$
- $(\text{A}\cup\text{B})'$
- (A')'
- (B - C)'.
AnswerGiven: 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}
- A' = {5, 6, 7, 8, 9}
- B' = {1, 3, 5, 7, 9}
- $(\text{A}\cap\text{C})'$ = {1, 2, 5, 6, 7, 8, 9}
- $(\text{A}\cup\text{B})'$ = (5, 7, 9)
- (A')' = {1, 2, 3, 4}
- (B - C)' = {1, 3, 4, 5, 6, 7, 9}.
View full question & answer→Question 23 Marks
Let A = {x : x $\in$ N}, B = {x : x = 2n, n $\in$ N}, C = {x : x = 2n - 1, n $\in$ N} and D = {x : x is a prime natural number}. Find: $\text{B}\cap\text{C}$
AnswerWe have, A = {x : x $\in$ N} = {2, 4, 6, 8,....}, the set of even natural numbers $\text{C} = \{\text{x : x = 2n - 1, x} \in \text{ N}\}$ = {1, 3, 5,....}, the set of odd natural numbers $\therefore\text{B}\cap\text{C}=\{\text{x : x}\in\text{B and x}\in\text{C}\}$ =$\oint$ [$\therefore$ B and C are disjoint sets, i.e., have no elements in common]
View full question & answer→Question 33 Marks
In a school there are 20 teachers who teach mathematics or physics. Of these, 12 teach mathematics and 4 physics and mathematics. How many teach physics?
AnswerLet n(P) denote the number of teachers who teach Physics and n(Q) denote the number of teachers who teach Mathematics. We have, $\text{n(P or M)}= 20$ i.e., $\text{n(A}\cup\text{B)}=20$ $\text{n(M)} = 12$ and $\text{n(P}\cap\text{M)}=4$ To find: $\text{n(P)}$ We know $\text{n(P}\cup\text{M)=n(P)+n(M)}-\text{n(P}\cap\text{M})$ $\Rightarrow20 = \text{n(P)} + 12 - 4$ $\Rightarrow20 = \text{n(P)} + 8$ $\Rightarrow \text{n(P)} = 20 - 8$ $= 12$ $\therefore$ There are 12 Physics teachers.
View full question & answer→Question 43 Marks
Let A = {1, 2, 4, 5}, B = {2, 3, 5, 6}, C = {4, 5, 6, 7}. Verify the following identities: $\text{A}\cap(\text{B}-\text{C})=(\text{A}\cap\text{B})-(\text{A}\cap\text{C})$
Answer$\text{A} = \{1, 2, 4, 5\},$ $\text{B} = \{2, 3, 5, 6\},$ and $\text{C} = \{4, 5, 6, 7\}$ $\text{B}-\text{C} = \{2, 3\}$ $\text{A}\cap(\text{B}-\text{C}) = \{2\}\ .....(1)$ $(\text{A}\cap\text{B})= \{2, 5\}$ $(\text{A}\cap\text{C})= \{4, 5\}= \{2\}\ .....2)$ $(\text{A}\cap\text{B})-(\text{A}\cap\text{C})$ From $eq^n$ (1) and $eq^n$ (2), we get $\text{A}\cap(\text{B}-\text{C})=(\text{A}\cap\text{B})-(\text{A}\cap\text{C}).$
View full question & answer→Question 53 Marks
For any two sets, prove that: $\text{A}\cup(\text{A}\cap\text{B})=\text{A}.$
Answer$\text{A}\cup(\text{A}\cap\text{B})=(\text{A}\cup\text{A})\cap(\text{A}\cup\text{B})$ $[\because$ union $\cup$ is distributive over intersection $\cap]$ $=\text{A}\cap(\text{A}\cup\text{B})$ $[\because\text{A}\cup\text{A}=\text{A}]$ $=\text{A}[\because\text{A}\subset(\text{A}\cup\text{B}),$ as union of two sets is bigger then each of the individual sets$]$ Hence, $\text{A}\cup(\text{A}\cap\text{B})=\text{A}$ Proved.
View full question & answer→Question 63 Marks
Let A and B be two stes such that: $\text{n(P)} = 20,$ $\text{n(A}\cup\text{B)=42 and n(A}\cap\text{B})=4.$ Find: $\text{n(B)}.$
Answer$\text{n(P)}= 20,\text{ n(A}\cup\text{B) = 42 and n(A}\cap\text{B})=4,$ to find: $\text{n(B)}$ We know $\text{n(A}\cup\text{B) = n(A)+n(B)}-\text{n(A}\cap\text{B})$ $\Rightarrow42 = 20 + \text{n(B)} - 4$ $\Rightarrow42 = 16 + \text{n(B)}$ $\Rightarrow\text{n(B)} = 42 - 16$ $= 26$ $\therefore\text{ n(B)} = 26.$
View full question & answer→Question 73 Marks
For any two sets A and B, show that the following statements are equevalent: $\text{A}\subset\text{B}.$
AnswerIn order to show that the following four statements are equivalent, we need to show that (1) ⇒ (2), (2) ⇒ (3), (3) ⇒ (4) and (4) ⇒ (1) We first show that (1) ⇒ (2) We assume that $\text{A}\subset\text{B},$ and use this to show that $\text{A - B} =\phi$ Now $\text{A}-\text{B}=\{\text{x}\in\text{A : x}\not\in\text{B}\},\text{As A}\subset\text{B},$ $\therefore$ Each element of A is an element of B, $\therefore\text{A - B}=\phi$ Hence, we have proved that (1) ⇒ (2).
View full question & answer→Question 83 Marks
Let A = {x : x $\in$ N}, B = {x : x = 2n, n $\in$ N}, C = {x : x = 2n - 1, n $\in$ N} and D = {x : x is a prime natural number}. Find: $\text{A}\cap\text{B}$
AnswerWe have, A = {x : x $\in$ N} = {1, 2, 3,.....}, the set of natural numbers B = {x : x = 2n, n $\in$ N} = {2, 4, 6, 8,....}, the set of even natural numbers $\therefore\text{A}\cap\text{B}=\{\text{x : x}\in\text{A and x}\in\text{B}\}$ = {2, 4, 6,.....} $=\text{B }[\therefore\text{B}\subset\text{A}].$
View full question & answer→Question 93 Marks
A survey show that 76% of the Indians like oranges, where as 62% like bananas. what percentage of the Indians like both oranges and bananas?
AnswerLet n(P) denote the total percentage of Indians n(O) denotes the percentage of Indians who like oranges, and n(B) denotes the percentage of Indians who like banana. Then, $\text{n(P)} = 100,\text{ n(O)} = 76, \text{ n(B)} = 62$ To find: $\text{n(O}\cap\text{B)}$ Now, $\text{n(P) = n(O) + n(B) - n(O}\cap\text{B)}$ $\Rightarrow 100=76+62-\text{n(O}\cap\text{B)}$ $\Rightarrow 100=138-\text{n(O}\cap\text{B)}$ $\Rightarrow\text{ n(O}\cap\text{B)}=138-100$ $= 38$ $\therefore$ 38% of Indians like both orange and bananas.
View full question & answer→Question 103 Marks
Let A = {1, 2, 4, 5}, B = {2, 3, 5, 6}, C = {4, 5, 6, 7}. Verify the following identities: $\text{A}\cap(\text{B}\cup\text{C})=(\text{A}\cap\text{B})\cup(\text{A}\cap\text{C})$
Answer$\text{A} = \{1, 2, 4, 5\},$$\text{B} = \{2, 3, 5, 6\},$
and $\text{C} = \{4, 5, 6, 7\}$ $\text{B}\cup\text{C}= \{2, 3, 4, 5, 6, 7\}$ $\text{A}\cap(\text{B}\cup\text{C})= \{2, 4, 5\}\ ....(1)$ $(\text{A}\cap\text{B})= \{2, 5\}$ $(\text{A}\cap\text{C})= \{4, 5\}$ $(\text{A}\cap\text{B})\cup(\text{A}\cap\text{C})= \{2, 4, 5\} .....(2)$ From $eq^n$ (1) and $eq^n$ (2), we get $\text{A}\cap(\text{B}\cup\text{C})=(\text{A}\cap\text{B})\cup(\text{A}\cap\text{C}).$
View full question & answer→Question 113 Marks
In a group of 50 persons, 14 drink tea but not coffee and 30 drink tea. Find: How many drink coffee but not tea.
AnswerTo find: $\text{C} - \text{T}$ We know $\text{n(P) = n(C) + n(T) - n(T}\cap\text{C)}$ $\Rightarrow50 = \text{n(P)} + 30 - 16$ $\Rightarrow50 = \text{n(C)} + 14$ $\Rightarrow \text{n(C)} = 50 - 14$ $= 36$ New C is the disjoint union of C - T and $\text{T}\cap\text{C}$ $\therefore\text{ C = (C} - \text{T)}\cup\text{(C}\cap\text{T)}$ $\Rightarrow\text{ n(C) = n(C} - \text{T)}+\text{n(C}\cap\text{T)}$ $\Rightarrow36 = \text{n(C} - \text{T)} + 16$ $[\because\text{ n(T}\cap\text{C)}=\text{ n(C}\cap\text{T)}=16]$ $\Rightarrow\text{n(C} - \text{T)} = 36 - 16$ $= 20$ Hence, 20 persons drink coffee but not tea.
View full question & answer→Question 123 Marks
For any two sets A and B, show that the following statements are equevalent: $\text{A}-\text{B}=\phi.$
AnswerWe new show that (2) ⇒ (3) So assume that $\text{A - B}=\phi$ To show: $\text{A}\cup\text{B}=\text{B}$ $\because$ Every element of A is an element of B $[\because\text{A}-\text{B}=\phi$ only when ther is some element in A which is not in B$]$ So $\text{A}\subset\text{B}$ and therefore $\text{A}\cup\text{B}=\text{B}$ So (2) ⇒ (3) is true.
View full question & answer→Question 133 Marks
A survey of 500 television viewers produced the following information; 285 watch football, 195 watch hockey, 115 watch basketball, 45 watch football and basketball, 70 watch football and hockey, 50 watch hockey and football, 50 do not watch any of the three games. How many watch all the three games? How many watch exactly one of the three game?
AnswerLet, n(P) denote the total number of television viewers, n(F) be the number of paople who watch football, n(H) be the number of paople who watch hockey and n(B) be the number of paople who watch basketball. Then, $\text{n(P) = 500, n(F) = 285, n(H) = 195, n(B) =}\\ 115, \text{n(F}\cap\text{B)}=45, \text{n(F}\cap\text{H)}=70,\ \text{n(H}\cap\text{B)}\\=50\text{ and}\text{ n(F}\cup\text{H}\cup\text{B)}=50$ Now, $\text{n((F}\cup\text{H}\cup\text{B}'))=\text{n(P)}-\text{n(F}\cup\text{H}\cup\text{B)}$ $\Rightarrow50=500- \text{n(F) + n(H) + n(B) - n(F}\cap\text{H)} -\\ \text{n(H}\cap\text{B)}-\text{n(F}\cap\text{B) - n(F}\cap\text{H}\text{B)})$ $\Rightarrow50 =500 - (285 + 195 + 115 - 70 - 50 -45 +\\\text{n(F}\cap\text{H}\cap\text{B}))$ $\Rightarrow50=500-430-\text{n(F}\cap\text{H}\cap\text{B})$ $\Rightarrow50=70-\text{n(F}\cap\text{H}\cap\text{B})$ $\Rightarrow\text{n(F}\cap\text{H}\cap\text{B})=70-50$ $=20$ Hence, 20 people watch all the 3 games. Number of people who watch only football $= 285 - (50 + 20 + 25)$ $= 285 - 95$ $= 190$ Number of people who watch only hockey $= 195 - (50 + 20 + 30)$ $= 195 - 100$ $= 95$ And, number of people who watch only basketball $= 115 - (25 + 20 +30)$ $= 115 - 75$ $= 40$ Number of people who watch exactly one of the three game = number of people who watch either football only or hockey only or basketball only $= 190 + 95 + 40$ $= 325$ Hence, 325 people watch exactly one of the three games.
View full question & answer→Question 143 Marks
Let $A = {1, 2, 4, 5}, B = {2, 3, 5, 6}, C = {4, 5, 6, 7}$. Verify the following identities: $\text{A}-(\text{B}\cup\text{C})=(\text{A}-\text{B})\cap(\text{A}-\text{C})$
Answer$\text{A} = \{1, 2, 4, 5\},$ $\text{B} = \{2, 3, 5, 6\},$ and $\text{C} = \{4, 5, 6, 7\}$ $\text{B}\cup\text{C}= \{2, 3, 4, 5, 6, 7\}$ $\text{A}-(\text{B}\cup\text{C})= \{1\} .........(1)$
$\text{(A - B)} = \{1, 4\}$
$\text{(A - C)}= \{1, 2\}$
$(\text{A}-\text{B})\cap(\text{A}-\text{C})= \{1\} ..............(2)$ From $eq^n$ (1) and $eq^n$ (2), we get $\text{A}-(\text{B}\cup\text{C})=(\text{A}-\text{B})\cap(\text{A}-\text{C}).$
View full question & answer→Question 153 Marks
In a group of 950 person, 750 can speak Hindi and 460 can speak English. Find: how many can speak both Hindi and English.
AnswerLet, n(P) denote the total number of persons, n(H) denote the total number of persons who speak Hindi and n(E) denote the total number of persons who speak English. Then, $\text{n(P)}= 950,\text{ n(H)} = 750,\text{ n(E)} = 460$ To find: $\text{n(H}\cap\text{E)}$ $\text{n(P) = n(H) + n(E)}- \text{n(H}\cap\text{E)}$ $\Rightarrow950=750+460-\text{n(H}\cap\text{E)}$ $\Rightarrow950=2110-\text{n(H}\cap\text{E)}$ $\Rightarrow\text{ n(H}\cap\text{E)}=2110-950$$= 260$
Hence, 260 persons can speak both Hindi and English.
View full question & answer→Question 163 Marks
Let 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}. Find:
-
A - B
-
A - C
-
A - D
-
B - A
-
C - A
-
D - A
-
B - C
-
B - D.
AnswerWe have, 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} If A and B are two sets, then the set A - B is defined as A - B $\{\text{x}\in \text{A : x} \not\in\text{B}\}$
-
A - B $\{\text{x}\in \text{A : x} \not\in\text{B}\}$ = {3, 6, 15, 18, 21}
-
A - C $\{\text{x}\in \text{A}\in\text{C}\}$ = {3, 15, 18, 21}
-
A - D $\{\text{x}\in \text{A : x} \not\in\text{D}\}$ = {3, 6, 12, 18, 21}
-
B - A $\{\text{x}\in \text{B : x} \not\in\text{A}\}$ = {4, 8, 16, 20}
-
C - A $\{\text{x}\in \text{C : x} \not\in\text{A}\}$ = {2, 4, 8, 10, 14, 16}
-
D - A $\{\text{x}\in \text{D : x} \not\in\text{A}\}$ = {5, 10, 20}
-
B - C $\{\text{x}\in \text{B : x} \not\in\text{C}\}$ = {20}
-
B - D $\{\text{x}\in \text{B : x} \not\in\text{D}\}$ = {4, 8, 12, 16}.
View full question & answer→Question 173 Marks
If U = {2, 3, 5, 7, 9} is the universal set and A = {3, 7}, B = {2, 5, 7, 9}, then prove that: $(\text{A}\cap\text{B})'=\text{A}'\cup\text{B}'.$
Answer$\text{U} = \{2, 3, 5, 7, 9\}$ is the universal set $\text{A} = \{3, 7\},$ $\text{B} = \{2, 5, 7, 9\}$ $\text{A}\cup\text{B}=\{\text{x : x}\in\text{A or x}\in\text{B}\}$ $= \{2, 3, 5, 7, 9\}$ $\text{LHS} =(\text{A}\cap\text{B})'$ Now, $\text{A}\cap\text{B}=\{\text{x | x}\in\text{A and x}\in\text{B}\}$ $=\{7\}$ $\therefore(\text{A}\cap\text{B})'=\{7\}'$ $=\{\text{x}\in\text{U : x}\not\in7\}$ $= \{2, 3, 5, 9\}$ $\text{RHS}=\text{A}'\cup\text{B}'$ Now, $\text{A}' = \{2, 5, 9\}\ \text{[from (i)]}$ $\text{B}' = {3}\ \text{[from (i)]}$ $\therefore\text{A}'\cup\text{B}'=\{2, 3, 5, 9\}$ LHS = RHS proved.
View full question & answer→Question 183 Marks
For any two sets A and B, show that the following statements are equevalent: $\text{A}\cap\text{B}=\text{A}.$
AnswerFinally we show that (4) ⇒ (1), which will prove the equqvalence of the 4 statement. So, assume that $\text{A}\cap\text{B}=\text{A}$ To show: $\text{A}\subset\text{B}$ $\because\text{A}\cap\text{B}=\text{A}$ therefore $\text{A}\subset\text{B},$ and so (4) ⇒ (1) is true. Hence, (1) ⇔ (2) ⇔ (3) ⇔ (4).
View full question & answer→Question 193 Marks
In a survey of 100 persons it was found that 28 read magazine A, 30 read magazine B, 42 read magazine C, 8 read magazines A and B, 10 read magazines A and C, 5 read magazines B and C and 3 read all the three magazines?
- How many read none of three magazines?
- How many read magazine C only?
Answer
- Let,
n(P) denote total number of persons
n(A) denote number of people who read magazine A
n(B) denote number of people who read magazine B and
n(C) denote number of people who read magazine C.
Then,
$\text{n(P) = 100, n(A) = 28, n(B) = 30, n(C) = 42, }\text{n(A}\cap\text{B)}=8,\\\text{n(A}\cap\text{C)}=10,\text{ n(B}\cap\text{C)}=5,\text{ n(A}\cap\text{B}\cap\text{C)}=3$
Now,
$\text{n(A}\cup\text{B}\cup\text{C)}=\text{n(A) + n(B) + n(C) - n(A}\cap\text{B)}-\\\text{n(B}\cap\text{C)}-\text{n(A}\cap\text{C)}+\text{n(A}\cap\text{B}\cap\text{C)}$
$= 28 + 30 + 42 - 8 - 10 - 5 + 3$
$= 100 - 20$
$= 80$
$\therefore$ Number of people who read none of the three magazines
$=\text{n(A}\cup\text{B}\cup\text{C)}'$
$=\text{n(P)}-\text{n(A}\cup\text{B}\cup\text{C)}$
$= 100 - 80$
$= 20$
Hence, 20 people read none of the three magazines.
- n(C only) = 42 - ( 7 + 3 + 2)
$= 42 - 12$
$= 30.$ View full question & answer→Question 203 Marks
In a group of 1000 people, there are 750 who can speak Hindi and 400 who can speak Bengali. How many can speak Hindi only? How many can speak Bengali? How many can speak both Hindi and Bengali?
AnswerLet, n(P) denote total number of people, n(H) denote number of people who speak Hindi and n(B) denote number of people who speak Bengali Then, $\text{ n(P)} = 1000, \text{ n(H)} = 750, \text{ n(B)} = 400$ We have $\text{P}=(\text{H}\cup\text{B})$ $\therefore\text{ n(P)}=\text{n(H}\cup\text{B})$ $=\text{n(H)}+\text{n(B)}-\text{n(H}\cap\text{B})$ $\Rightarrow1000=750+400-\text{n(H}\cap\text{B})$ $\Rightarrow1000=1150-\text{n(H}\cap\text{B})$ $\Rightarrow\text{n(H}\cap\text{B})=1150-1000$$= 150$
Hence, 150 people can speak both Hindi and Bengali now $\text{H = (H} - \text{B)}\cup\text{(H}\cap\text{B)},$ the union being disjoint $\therefore\text{ n(H) = n(H} - \text{B)}+\text{n(H}\cap\text{B)}$ $\Rightarrow750 = \text{n(H} - \text{B}) + 150$ $\Rightarrow \text{n(H}- \text{B)}= 750 - 150$ $= 600$ Hence, 600 people can speak Hindi only. On a similar lines we have $\text{B = (B - H)}\cup\text{(H}\cap\text{B)}$ $\Rightarrow\text{n(B) = n(B - H)}+\text{n(H}\cap\text{B)}$ $\Rightarrow400 = \text{n(B}- \text{H)} + 150$ $\Rightarrow\text{n(B} - \text{H)} = 400 - 150$ $= 250$ Hence, 250 people can skeap Bengali only.
View full question & answer→Question 213 Marks
In a group of 950 person, 750 can speak Hindi and 460 can speak English. Find: how many can speak English only.
AnswerOn a similar line we have $\text{E = (E}- \text{H)}\cup\text{(H}\cap\text{E})$ i.e., E is the disjoint union of $\text{E}- \text{H}\ \&\ \text{H}\cap\text{E}$ $\therefore\text{ n(E) = n(E} - \text{H)}\cup\text{(H}\cap\text{E})$ $\Rightarrow460 = \text{n(E} - \text{H}) + 260$ $\Rightarrow\text{n(E} - \text{H)} = 460 - 260$ $= 200$ Hence, 200 persons can speak English only.
View full question & answer→Question 223 Marks
In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the two drinks. How many like both coffee and tea?
AnswerLet, n(P) denote the total number of people n(C) denote the total number of people who like coffee and n(T) denote the total number of people who like tea. then, $\text{n(P)} = 70$ $\text{n(C)} = 37$$\text{n(T)} = 52$
We are given that each person likes at leats one of the two drinks, i.e., $\text{P=C}\cup\text{T}$ To find: $\text{n(C}\cap\text{T})$ We know $\text{n(P)=n(C)+n(T)-n(C}\cap\text{T})$ $\Rightarrow70=37+52-\text{n(C}\cap\text{T})$ $\Rightarrow70=89-\text{n(C}\cap\text{T})$ $\Rightarrow\text{n(C}\cap\text{T})=89-70$ $= 19$ Hence, 19 people like both coffee and tea.
View full question & answer→Question 233 Marks
In a survey of 100 students, the number of students studying the various languages were found to be: English only 18, English but not Hindi 23, English and Sanskrit 8, English 26, Sanskrit 48, Sanskrit and Hindi 8, no language 24. Find:
- How many students were studying Hindi?
- How many students were studying English and Hindi?
Answer
- Let,
n(P) denote total number of students
n(E) denote total number of students studying English language
n(H) denote total number of students studying Hindi language and
n(S) denote total number of students studying Sanskrit language.
Then,
$\text{n(P) = 100, n(E - H) = 23, n(E}\cap\text{S) = 8, n(E) = 26,}\\\text{n(S) = 48, n(S}\cap\text{H) = 8, n((E}\cup\text{H}\cup\text{S})')=24$
Number of students studying English only = 18
We have,
$\text{n((E}\cup\text{H}\cup\text{S})')=24$
$\Rightarrow\text{n(P)}-\text{n(E}\cup\text{H}\cup\text{S})=24$
$\Rightarrow100-24=\text{n(E}\cup\text{H}\cup\text{S})$
$\Rightarrow\text{n(E}\cup\text{H}\cup\text{S})=76$
We have $\text{n(E}\cup\text{H}\cup\text{S})=\text{n(E) + n(H) + n(S) - n(E}\cap\text{H)}-\\\text{n(H}\cap\text{S)}-\text{n(E}\cap\text{S)}+\text{n(E}\cap\text{H}\cap\text{S})$
$\Rightarrow76 = 26 + \text{n(H)} + 48 - 3 - 8 - 8 + 3$
$\Rightarrow76 = 26 + \text{n(H)} + 48 - 16$
$\Rightarrow76 = 26 + 32 +\text{n(H)}$
$\Rightarrow\text{n(H)} = 76 - 58$
$= 18$
$\therefore$ 18 students were studying Hindi.
- From (i) we have $\text{n(E}\cap\text{H})=3$
$\therefore$ 3 students were studying both English and Hindi. View full question & answer→Question 243 Marks
If A and are sets, then prove thet $\text{A - B},\text{ A}\cap\text{B}$ and $\text{A - B}$ are pair wise disjoint.
AnswerWe need to show that $(\text{A - B})\cap(\text{A}\cap\text{B})=\phi,\ (\text{A}\cap\text{B})\cap(\text{A - B})=\phi$ and $(\text{A - B})\cap\text{B - A}=\phi$ The 3 sets $\text{A} - \text{B}, \text{ A}\cap\text{B}$ and B - A may be represented by venn diagram as follows It is clear from the diagram thet the 3 sets are pairwise disjoint, but we shell give a proff of it. We first show that $(\text{A - B})\cap(\text{A}\cap\text{B})=\phi$ Let $\text{x}\in(\text{A - B})$ $\Rightarrow\text{x}\in\text{A and x}\not\in\text{B}$ [by definition of A - B] $\Rightarrow\text{x}\not\in\text{A}\cap\text{B.}$ This is true for all $\text{x}\in(\text{A - B)}$ Hence $(\text{A - B})\cap(\text{A}\cap\text{B})=\phi$ On a similar lines, it can be seen that $(\text{A}\cap\text{B})\cap(\text{A}-\text{B})=\phi$ Finally, we show that $\text{(A - B)}\cap\text{(A - B)}=\phi$ We have, $\text{A - B}=\{\text{x}\in\text{A : x}\not\in\text{B}\}$ and $\text{B - A}=\{\text{x}\in\text{B : x}\not\in\text{A}\}$ Hence, $(\text{A - B})\cap(\text{B - A})=\phi.$
View full question & answer→Question 253 Marks
For any two sets, prove that: $\text{A}\cap(\text{A}\cup\text{B})=\text{A}.$
Answer$\text{A}\cap(\text{A}\cup\text{B})=(\text{A}\cap\text{A})\cup(\text{A}\cap\text{B})$ $[\because\cap$ is distributive over $\cup]$ $=\text{A}\cup(\text{A}\cap\text{B})$ $[\because\text{A}\cap\text{A}=\text{A}]$ $=\text{A [using (i)]}.$
View full question & answer→Question 263 Marks
Let A = {x : x $\in$ N}, B = {x : x = 2n, n $\in$ N}, C = {x : x = 2n - 1, n $\in$ N} and D = {x : x is a prime natural number}. Find: $\text{C}\cap\text{D}$
AnswerHere, c = {x : x = 2n - 1, x $\in$ N} = {1, 3, 5,....}, the set of odd natural numbers D = {x : x is a prime natural number} = {2, 3, 5, 7....} $\therefore\text{C}\cap\text{D}=\{\text{x : x}\in\text{C and x}\in\text{D}\}$ We observe that except, the element 2, every other element in 0 is an odd natural number. Hence, $\text{C}\cap\text{D}=\text{D}-\{2\}$ $=\{\text{x}\in\text{D}:\text{x}\not=2\}.$
View full question & answer→Question 273 Marks
Show that for any sets A and B,$\text{A}=(\text{A}\cap\text{B})\cap(\text{A - B})$
AnswerWe know that $(\text{A}\cap\text{B})\subset\text{A and (A - B)}\subset\text{A}$ $\Rightarrow(\text{A}\cap\text{B})\cap(\text{A - B})\subset\text{A}......(\text{i})$ Let and $\text{x}\in(\text{A}\cap\text{B})\cap(\text{A - B})$ $\Rightarrow\text{x}\in(\text{A}\cap\text{B})\text{ and x}\in(\text{A}-\text{B})$ $\Rightarrow\text{x}\in\text{A and}\text{ x}\in\text{B and }\text{x}\in\text{A and }\text{x}\not\in\text{B}$ $\Rightarrow\text{x}\in\text{A and }\text{x}\in\text{A}$ [$\because\text{ x}\in\text{A}\text{ x}\not\in\text{A}$ are not possible simutaneusly] $\Rightarrow\text{x}\in\text{A}$ $\therefore(\text{A}\cap\text{B})\cap\text{(A - B)}\subset\text{A}..........\text{(ii)}$ From (i) and (ii), we get $\text{A}=(\text{A}\cap\text{B})\cap(\text{A - B}).$
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Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that: $(\text{A}\cap\text{B})'=\text{A'}\cup\text{B'}$
Answer$\text{A}\cap\text{B}=\{\text{x : x} \in \text{A and x }\in\text{B}\}$ $= \{2\}$ $\therefore(\text{A}\cap\text{B})'=\{\text{x}\in\text{U : x}\not\in\text{A}\cap\text{B}\}$ $=\{1, 3, 4, 5, 6, 7, 8, 9\}$ Also, $\text{A}' ∪ \text{B}' = \{\text{x : x} \in \text{A' or x} \in\text{ B}'\}$ $=\{1, 3, 4, 5, 6, 7, 8, 9\}$ Hence, $(\text{A} \cap \text{B})' = \text{A}' \cup\text{ B}'=\{1, 3, 4, 5, 6, 7, 8, 9\}.$
View full question & answer→Question 293 Marks
For any two sets A and B, prove that. $\text{A}\cap\text{B}\subset\text{A}.$
AnswerLet $\text{x}\in\text{A}\cap\text{B}.$ Then $\Rightarrow\text{x}\in\text{A}\text{ and x}\in\text{B}$ $\Rightarrow\text{x}\in\text{B}$ $\therefore(\text{A}\cap\text{B})\subset\text{B}.$
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If U = {2, 3, 5, 7, 9} is the universal set and A = {3, 7}, B = {2, 5, 7, 9}, then prove that: $(\text{A}\cup\text{B})'=\text{A}'\cap\text{B}'.$
Answer$\text{U} = \{2, 3, 5, 7, 9\}$ is the universal set $\text{A} = \{3, 7\},$ $\text{B} = \{2, 5, 7, 9\}$ $\text{A}\cup\text{B}=\{\text{x : x}\in\text{A or x}\in\text{B}\}$ $= \{2, 3, 5, 7, 9\}$ $\text{LHS}=(\text{A}\cup\text{B})'$ $=\{2, 3, 5, 7, 9\}'$ $=\text{U}-\text{A}\cup\text{B}$ $=\phi$ $\text{RHS} =\text{A}'\cap\text{B}'$ $\text{A}'=\{\text{x}\in\text{U : x}\not\in\text{A}\}$ $= \{2, 5, 9\}$ $\text{B}'=\{\text{x}\in\text{U : x}\not \in \text{B}\}$ $=\{3\}$ $\therefore\text{A}'\cap\text{B}'=\{2, 5, 9\}\cap\{3\}$ $=\phi$ $[\because$ the two sets are disjoint$]$ $\therefore$ LHS = RHS proved.
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If P and Q are two sets such that P has 40 elements, $\text{P}\cup\text{Q}$ has 60 elements and $\text{P}\cap\text{Q}$ has 10 elements, how many elements does Q have?
AnswerWe have, $\text{n(P)} = 40,\text{n(P}\cup\text{Q)}=60,\text{ n(P}\cap\text{Q})=10,$ to find $\text{n(Q)}.$ $\Rightarrow60 = 40 + \text{n(Q)} - 10$ $\Rightarrow60 = 30 +\text{n(Q)}$ $\Rightarrow\text{n(Q)} = 60 - 30$ $= 30$ Hence, Q has 30 elements.
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If A and B are two sets such that $\text{n}(\text{A}\cup\text{B})=50,\text{n(A)} = 28$ and $\text{n(B)} = 32,$ find $\text{n(A}\cap\text{B}).$
Answer$\text{n}(\text{A}\cup\text{B})=50,\text{n(A)= 28 and n(B)} = 32,$ where n(X) does not the cardinal number of the set x. We know that $\text{n}(\text{A}\cup\text{B})=\text{n(A)+n(B)-n(A}\cap\text{B)}$ $\Rightarrow50=28+32-\text{n(A}\cap\text{B})$ $\Rightarrow50=60-\text{n(A}\cap\text{B})$ $\Rightarrow\text{n(A}\cap\text{B})=60-50$ $= 10$ $\therefore\text{n(A}\cap\text{B})=10.$
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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: $(\text{A}\cup\text{D})\cap(\text{B}\cup\text{C})$
Answer$(\text{A}\cup\text{D})\cap(\text{B}\cup\text{C})=\{\text{x | x}\in(\text{A}\cup\text{D) or x}\in(\text{B}\cup\text{C})\}$ Now, $\text{A}\cup\text{D}$ = {1, 2, 3, 4, 5, 10, 11, 12, 13, 14} and $\text{B}\cup\text{C}$ = {4, 5, 6, 7, 8, 9, 10, 11} $(\text{A}\cup\text{D})\cap(\text{B}\cup\text{C})$ = {4, 5, 10, 11}.
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Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that: $(\text{A}\cup\text{B})'=\text{A'}\cap\text{B'}$
AnswerU = {1, 2, 3, 4, 5, 6, 7, 8, 9} B = {2, 3, 5, 8} B = {2, 3, 5, 7} We have, $\text{A}\cup\text{B}=\{\text{x : x}\in\text{A or x}\in \text{B}\}$ $= \{2, 3, 4, 5, 6, 7, 8\}$ $\therefore(\text{A}\cup\text{B})'=\{\text{x}\in\text{U}:\text{x}\not\in\text{A}\cup\text{B}\}$ $= \{1, 9\}$ $\text{A'}=\{\text{x}\in\text{U}:\text{x}\not \in,\text{A}\}$ $= \{1, 3, 5, 7, 9\}$ $\text{B'}=\{\text{x}\in\text{U}:\text{x}\not \in,\text{B}\}$ $=\{1, 4, 6, 8, 9\}$ Hence, $\text{A}'\cup\text{B}' =\{1, 9\}$ Hence, $(\text{A}\cup\text{B})'=\text{A}'\cap\text{B}'= \{1, 9\}.$
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Let A = {x : x $\in$ N}, B = {x : x = 2n, n $\in$ N}, C = {x : x = 2n - 1, n $\in$ N} and D = {x : x is a prime natural number}. Find: $\text{B}\cap\text{D}$
AnswerHere, B = {x : x = 2n, x $\in$ N} = {2, 4, 6, 8,....}, the set of even natural numbers D = {x : x is a prime natural number} = {2, 3, 5, 7....} $\therefore\text{B}\cap\text{D}=\{\text{x : x}\in\text{B and x}\in\text{D}\}$ = {2}.
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