(Each question 3 marks)

Question 13 Marks

Show that the following four conditions are equivalent :

A $\subset$ B
A – B = $\phi$
A $\cup$B = B
A $\cap$B = A

Answer

Here, first we will prove (i) $\Leftrightarrow$(ii)
Where, (i) = A $\subset $Band (ii) = A - B $\neq$ $\phi$
Suppose that$\mathrm{A} \subset \mathrm{B}$
Now, we need to prove $A-B \neq \phi$
If possible, let $A-B \neq \phi$
Thus, there exists $X \in A, X \neq B,$but this is impossible as $A \subset B$
$\therefore A-B=\phi$
And $\mathrm{A} \subset \mathrm{B}\Rightarrow\mathrm{A}-\mathrm{B} \neq \phi$
Let suppose that $\mathrm{A}-\mathrm{B} \neq \phi$
Now, we have to prove: $\mathrm{A} \subset \mathrm{B}$
Let $\mathrm{X} \in \mathrm{A}$
It can be concluded that$\begin{equation} X \in B(\text { if } X \notin B, \text { then } A-B \neq \phi) \end{equation}$
Thus, A - B =$\begin{equation} \phi = A \subset B \end{equation}$
$\therefore$(i)$\Leftrightarrow$(ii)
Let us assume that A$\subset$B
To prove: A$\cup$ B = B
$\begin{equation} \Rightarrow B \subset A \cup B \end{equation}$
Let us assume that,$\begin{equation} x \in A \cup B \end{equation}$
$\begin{equation} \Rightarrow x \in A \text { or } x \in B \end{equation}$
Taking case I:$\begin{equation} X \in B \end{equation}$
$\begin{equation} A \cup B=B \end{equation}$
Taking Case II :$\begin{equation} X \in A \end{equation}$
$\begin{equation} \Rightarrow X \in B(A \subset B) \end{equation}$
$\begin{equation} \Rightarrow A \cup B \subset B \end{equation}$
Let A$\cup$ B = B
Let us assume that$\begin{equation} X \in A \end{equation}$
$\begin{equation} \Rightarrow X \in A \cup B(A \subset A \cup B) \end{equation}$
$\begin{equation} \Rightarrow X \in B(A \cup B=B) \end{equation}$
$\begin{equation} \therefore A \subset B \end{equation}$
Thus,(i)$\Leftrightarrow$(iii)
Now, to prove (i) $\Leftrightarrow$(iv)
Suppose that A$\subset$B
It can be observed that$\begin{equation} A \cap B \subset A \end{equation}$
Let$\begin{equation} X \in A \end{equation}$
To show:$\begin{equation} X \in A \cap B \end{equation}$
Since,$\begin{equation} A \subset B \text { and } X \in B \end{equation}$
Therefore,$\begin{equation} \mathrm{X} \in \mathrm{A} \cap \mathrm{B} \end{equation}$
$\begin{equation} \Rightarrow A \subset A \cap B \end{equation}$
$\begin{equation} \Rightarrow A=A \cap B \end{equation}$
Similarly, let us assume that$\begin{equation} \mathrm{A} \cap \mathrm{B}=\mathrm{A} \end{equation}$
Let$\begin{equation} X \in A \end{equation}$
$\begin{equation} \Rightarrow X \in A \cap B \end{equation}$
$\begin{equation} \Rightarrow X \in B \text { and } X \in A \end{equation}$
$\begin{equation} \Rightarrow A \subset B \end{equation}$
$\therefore$(i)$\Leftrightarrow$(ii)
Therefore, proved that (i)$\Leftrightarrow$(ii)$\Leftrightarrow$(iii)$\Leftrightarrow$(iv)

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Question 23 Marks

In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only?

Answer

Here
n (A) = a + b + c + d = 21 ....(i)
n(B) = b + c + f + g = 26 ....(ii)
n(C) = c + d + e + f = 29 ....(iii)
$n(A \cap B) $= b + c = 14 ....(iv)
$n(C \cap A) $= c + d = 12 ....(v)
$n(B \cap C) $= c + f = 14 ....(vi)
$n(A \cap B \cap C) $= c = 8 ....(vii)

Putting value of c in (iv), (v) and (vi)
b + 8 = 14 $\Rightarrow $ b = 6
8 + d = 12 $\Rightarrow $ d = 4
8 + f = 14 $\Rightarrow $ f = 6
Putting value of c, d, f in (iii),
8 + 4 + e + 6 = 29 $\Rightarrow $ e = 29 - 18 = 11
Number of people who like product C only = 11

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Question 33 Marks

Let A and B are sets. If$A \cap X = B \cap X = \phi $and$A \cup X = B \cup X$for some set X. Show that A = B.
[Hints A = A $\cap$( A $\cup$X ) , B = B $\cap$( B $\cup$X ) and use Distributive law ]

Answer

Here $A \cup X = B \cup X$for some X
$\Rightarrow A \cap (A \cup X) = A \cap (B \cup X)$
$\Rightarrow A = (A \cap B) \cup (A \cap X)$$\,[\because A \cap (A \cup X) = A]$
$ \Rightarrow A = (A \cap B) \cup \phi $$[\because A \cap X = \phi ]$
$ \Rightarrow A = A \cap B$. . . (i)
$ \Rightarrow A \subset B$
Also$A \cup X = B \cup X$
$ \Rightarrow B \cap (A \cup X) = B \cap (B \cup X)$
$\Rightarrow (B \cap A) \cup (B \cap X) = B\,$$[\because B \cap (B \cup X) = B]$
$\Rightarrow (B \cap A) \cup \phi = B$$[\because B \cap X = \phi ]$
$\Rightarrow B \cap A = B$
$\Rightarrow B \cap A$. . . (ii)
From (i) and (ii), we have
A = B.

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Question 43 Marks

Decide among the following sets which sets are subsets of each another:
A = {X : X $\in$ R} and x satisfies x2 - 8x + 12 = 0}, B = {2, 4, 6} , C = {2, 4, 6, 8, ...}, D = {6}

Answer

It is given in the question that,
$A=\left\{x: x \in R \text { and } x \text { satisfy } x^{2}-8 x+12=0\right\}$
As, 2 and 6 are the only solutions of$x^{2}-8 x+12=0$
$\therefore A=\{2,6\}$
Also it is given that,
$B=\{2,4,6\}$
$C=\{2,4,6,8, \ldots .\}$
And, $D=\{6\}$
$\therefore \mathrm{D} \subset \mathrm{A} \subset \mathrm{B} \subset \mathrm{C}$
Hence, $A \subset B, A \subset C, B \subset C, D \subset A, D \subset B, D \subset C$

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Question 53 Marks

A college awarded 38 medals in Football, 15 in Basketball and 20 in Cricket. If these medals went to a total of 58 men and only three men got medals in all three sports, then how many received medals in exactly two of the three sports.

Answer

Suppose$n(F), n(B)\ and\ n(C)$ denote the number of men who received medals in Football, Basketball and Cricket, respectively. Then,
n(F) = 38, n(B) = 15, n(C) = 20, n($F \cup B \cup C$) = 58 and n($F \cap B \cap C$) = 3
$\because n ( F \cup B \cup C )$=n(F) + n(B) + n(C) + $n ( F \cap B \cap C )$ $- n ( F \cap B ) - n ( F \cap C ) - n ( B \cap C )$
58 = 38 + 15 + 20 + 3$- n ( F \cap B ) - n ( F \cap C ) - n ( B \cap C )$
$\Rightarrow n ( F \cap B ) + n ( F \cap C ) + n ( B \cap C )$ = 76 - 58 = 18

Here, 'a'= the number of men who got medals in Football and Basketball only.
‘b' = the number of men who got medals in Football and Cricket only.
'c' = the number of men who got medals in Basketball and Cricket only.
'd ’= the number of men who got medals in all the three games.
Thus, d =$n ( F \cap B \cap C )$ = 3
and$n ( F \cap B ) + n ( F \cap C ) + n ( B \cap C )$ = 18
$\Rightarrow$ (a + d) + (b + d) + (c + d) = 18
$\Rightarrow$a + b + c + 3d = 18
$\Rightarrow$a + b + c+ 3(3) = 18 [Put d = 3, given]
$\therefore$a + b + c = 9
Hence, people who got medals in exactly two of the three sports is 9.

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Question 63 Marks

Out of 500 car owners investigated, 400 owned Maruti car and 200 owned Hyundai car, 50 owned both cars. Is this data correct?

Answer

Let U be the set of the total number of car owners investigated, M be the set of persons who owned Maruti cars and H be the set of persons who owned Hyundai cars.
According to Question,
n (U) = 500, n(M) = 400, n (H) = 200 and n(M $\cap$ H) = 50.
Now, We know that
$n ( M \cup H ) = n ( M ) + n ( H ) - n ( M \cap H ) $
= 400 + 200 - 50
= 550
But,
$M \cup H \subseteq U$
$\therefore n ( M \cup H ) \leq n ( U ) $
⇒ n(M∪H) ≤ 500
but 550 > 500
Therefore, itis a contradiction. Hence, the given data is incorrect.

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Question 73 Marks

For sets A and B, showthat:$P(A \cap B) = P(A) \cap P(B)$

Answer

Let$x \in P(A \cap B)$
$\Rightarrow x \subset (A \cap B)$
$\Rightarrow x \subset A$and$x \subset B$
$\Rightarrow x \in P(A)$and$x \in P(B)$
$\Rightarrow x \in P(A) \cap P(B)$
$\Rightarrow x \subset P(A) \cap P(B)$
$\therefore P(A \cap B) \subset P(A) \cap P(B)$. . . (i)
Let$x \in P(A) \cap P(B)$
$\Rightarrow x \in P(A)$and$x \in P(B)$
$ \Rightarrow x \subset A$and$\Rightarrow x \subset B$
$\Rightarrow x \subset A \cap B$
$ \Rightarrow x \subset P(A \cap B)$
$\therefore P(A) \cap P(B) \subset P(A \cap B)$. . . . (ii)
From (i) and (ii), we have
$P(A \cup B) = P(A) \cap P(B)$

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Question 83 Marks

Show that A $\cup$B = A $\cap$B implies A = B

Answer

Suppose a $\in$A.
Then a $\in$A $\cup$ B. Since A $\cup$ B = A $\cap$ B , a $\in$A $\cap$B. So a $\in$B.
Thus,A ⊂ B. Similarly, if b ∈ B, then b ∈ A ∪ B.
A $\cup$ B = A $\cap$B, b $\in$A $\cap$B. So, b $\in$A.
Thus, B $\subset$ A. $\Rightarrow$A = B

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Question 93 Marks

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 juice nor orange juice.

Answer

Let A denote the set of students taking apple juice and B denote the set of students taking orange juice
n(U) = 400, n(A) = 100, n (B) = 150 n(A $ \cap $) =75
$n\left( {A' \cap B'} \right) = n\left( {A \cup B} \right)'$
$ = n\left( \cup \right) - n\left( {A \cup B} \right)$
$ = n\left( \cup \right) - [n(A) + n(B) - n(A \cap B)]$
=400-100-150+75=225

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Question 103 Marks

Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. Find A′, B′ , A′ $\cap$ B′, A $\cup$B and hence show that (A $\cup$B)′ = A′ $\cap$B′

Answer

Clearly A′ = {1, 4, 5, 6}, B′ = { 1, 2, 6 }. Therefore, A′ $\cap$B′ = { 1, 6 }
Also A $\cup$B = { 2, 3, 4, 5 }, so that (A $\cup$B )′ = { 1, 6 }
(A $\cup$B)′ = {1, 6} = A′ $\cap$ B′
It can be shown that the above result is true in general. If A and B are any two
subsets of the universal set U, then ,we have
(A $\cup$B)′ = A′ $\cap$B′. Similarly, (A $\cap$B)′ = A′ $\cup$B′.

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