3 Marks Question

Question 13 Marks

Let I be the set of all integers. A relation R on I, such that $x R y$ holds iff $(x-y)$ is divisible by $5, x \in$ $I, y \in I$, i.e., $R =\{(x, y): x \in I, y \in I, x-y$ is divisible by 5$\}$. Prove that it is an equivalence relation.

Answer

Here we observe that,
(i) For each $x \in I, x-x=0$ and 0 is divisible by 5 . Thus, $\forall x \in I$, we have $x R x$. Therefore, $R$ is reflexive.
(ii) Suppose $x R y$; then $x-y$ is divisible by 5 and hence $(y-x)=-(x-y)$ is also divisible by 5 . Thus, $x R y$ $\Rightarrow y R x$. Therefore, $R$ is symmetric.
(iii) Suppose $x R y$ and $y R z$; then $(x-y)$ and $(y-z)$ are both divisible by 5 . Hence 5 is also a divisor of $(x-y)$ $+(y-z)$, i.e., 5 is also a divisor of $(x-z)$. Thus, $x R y$ and $y R z \Rightarrow x R z$.
Therefore, $R$ is transitive.
Since, $R$ is reflexive, symmetric and transitive, therefore $R$ is an equivalence

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Question 23 Marks

Let a be the set of all triangle in a plane and $R$ be a relation in a defined as iff $x$ is congruent to $y, x \in$ A, $y \in B$. Prove that it is an equivalence relation.

Answer

Here we observe that,
(i) $x R x$ for every $x \in A$, since every triangle is congruent to itself. Therefore, R is reflexive.
(ii) $x R y \Rightarrow y R x$, since if triangle $x$ is congruent to triangle $y$, then triangle $y$ is congruent to triangle $x$. Therefore, R is symmetric.
(iii) $x R y$ and $y R z$; since if triangle $x$ is congruent to triangle $y$ and triangle $y$ is congruent to triangle $z$, then triangle $x$ is congruent to triangle $z$. Thus, $x R y$ and $y R z \Rightarrow x R z$.
Therefore, R is transitive.
Since, $R$ is reflexive, symmetric and transitive, therefore $R$ is an equivalence relation.

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Question 33 Marks

In a survey it was found that 21 persons 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:
(a) The number of people who liked at least one product.
(b) The number of people who liked product $C$ only.

Answer

$
\begin{aligned}
n(A)=21, n(B) & =26, n(C)=29 \\
n(A \cap B) & =14, n(C \cap A)=12 \\
n(B \cap C) & =14, n(A \cap B \cap C)=8
\end{aligned}
$
(a)
$
\begin{array}{l}
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) \\
=21+26+29-14-12-14+8 \\
=44
\end{array}
$

(b)
$\begin{array}{l} n(C \text { only })=n(C)-n(A \cap C)-n(B \cap C)+n(A \cap B \cap C) \\ = 29-12-14+8 \\ = 11\end{array}$

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Question 43 Marks

Let $A=\{1,2,3\}, B=\{3,4\}$ and $C=\{4,5,6\}$. Find (i) $A \times(B \cap C)$ and (ii) $(A \times B) \cap(A \times C)$.

Answer

Given, $A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}$,
$\therefore B \cap C=\{4\}$
(i) $A \times(B \cap C)=\{1,2,3\} \times\{4\}=\{(1,4),(2,4),(3,4)\}$
(ii)
$
\begin{array}{l}
(A \times B) \cap(A \times C) \\
=(\{1,2,3\} \times\{3,4\}) \cap(\{1,2,3\} \times\{4,5,6\}) \\
=\{(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)\} \cap\{(1,4),(1,5), \\
(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)\} \\
=\{(1,4),(2,4),(3,4)\}
\end{array}
$

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Question 53 Marks

In a survey of 450 people, it was found that 110 play cricket, 160 play tennis and 70 play both cricket as well tennis. How many play neither cricket nor tennis?

Answer

Let $C$ and $T$ denotes the students who play cricket and tennis, respectively.
Given, $n(C)=110, n(T)=160, n(C \cap T)=70, n(U)=$ 450.
Using identity,
$
\begin{aligned}
n(C \cup T) & =n(C)+n(T)-n(C \cap T) \\
& =110+160-70 \\
& =200
\end{aligned}
$
$\therefore$ No. of students play neither cricket nor tennis
$=n(U)-n(C \cap T)$
$\begin{array}{l}=450-200 \\ =250\end{array}$

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Question 63 Marks

Two finite sets have $m$ and $n$ elements. The total number of subsets of first set is 56 more than the total number of subsets of the second set. Find the value of $m$ and $n$.

Answer

Let $A$ and $B$ be two sets having $m$ and $n$ number of elements respectively.
Number of subsets of $A=2^m$
Number of subsets of $B=2^n$
According to questions
$
\begin{aligned}
2^m-2^n & =56=8 \times 7 \\
2^n\left(2^{m-n}-1\right) & =2^3\left(2^3-1\right)
\end{aligned}
$
Therefore, $n=3$ and $m-n=3$
$
\therefore \quad m=6
$

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Question 73 Marks

Let $A=\{1,2\}$ and $B=\{3,4\}$. Write $A \times B$. How many subsets will $A \times B$ have ? List them.

Answer

Given,
$\begin{array}{l}A=\{1,2\} \text { and } B=\{3,4\} \\ \Rightarrow \quad A \times B=\{1,2\} \times\{3,4\}\end{array}$
$\Rightarrow \quad A \times B=\{(1,3),(1,4),(2,3),(2,4)\}$
$\Rightarrow \quad n(A \times B)=4$
Total subsets of $A \times B=2^4=16$
Subsets of $A \times B$ are :
$\phi$,{(1, 3)} {(1, 4)} {(2, 3)} {(2, 4)} {(1, 3), (1, 4)} {(1, 3), (2, 3)} {(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4), {(1, 3), (2, 3), (2, 4)} {(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}

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Question 83 Marks

For any set $A$ and $B$, show that $(A-B)=\left(A \cap B^{\prime}\right)$

Answer

Let $x \in(A-B)$, then
$
\begin{array}{lc}
\Rightarrow & x \in A \text { and } x \notin B \\
\Rightarrow & x \in A \text { and } x \in B^{\prime} \\
\Rightarrow & x \in A \cap B^{\prime} \ldots(i)\\
\therefore & (A-B) \subseteq A \cap B^{\prime}
\end{array}
$
Again, Let $y \in\left(A \cap B^{\prime}\right)$, then
$
\begin{array}{ll}
\Rightarrow& y \in\left(A \cap B^{\prime}\right) \\
\Rightarrow & y \in A \text { and } y \in B^{\prime} \\
\Rightarrow & y \in A \text { and } y \notin B \\
\Rightarrow & y \in(A-B) \\
\therefore & \left(A \cap B^{\prime}\right) \subseteq(A-B)^{\prime}\ldots(ii)
\end{array}
$
From (i) and (ii), we have $(A-B)=\left(A \cap B^{\prime}\right)$

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Question 93 Marks

Justify for the sets $A, B$ and $C,(A \cap B) \cup C =A \cap$ $(B \cup C)$.

Answer

Let $A=\{a, b\}, B=\{b, c\}$ and $C=\{c, d\}$
Now,
$
\begin{aligned}
(A \cap B) \cup C & =(\{a, b\} \cap\{b, c\}) \cup\{c, d\} \\
& =\{b\} \cup\{c, d\}=\{b, c, d\}
\end{aligned}
$
and
$
\begin{aligned}
A \cap(B \cup C) & =\{a, b\} \cap(\{b, c\} \cup\{c, d\}) \\
& =\{a, b\} \cap\{b, c, d\}=\{b\}
\end{aligned}
$
From above, it is observed that
$
(A \cap B) \cup C \neq A \cap(B \cup C)
$
i.e., given statement is not true.

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Applied Maths 3 Marks Question

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