Set Theory Test 1

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Set Theory Test 1

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Question 1
1 / -0

Let $A, B$ and $C$ be sets such that $\phi = A\cap B \subseteq C$ . Then which of the following statements is not true?

A
If $(A - C) \subseteq B$ , then $A\subseteq B$

B
$(C \cup A)\cap (C\cup B) = C$

C
If $(A - B)\subseteq C$ , then $A\subseteq C$

D
$B\cap C \neq \phi$

Solution

for $A = C, A - C = \phi$
$\Rightarrow \phi \subseteq B$
But $A\not {\subseteq} B$
$\Rightarrow$ option $1$ is NOT true
Let $x \epsilon (C x \epsilon (C \cup A)\cap (C\cup B)$
$\Rightarrow x\epsilon (C\cup A)$ and $x \epsilon (C\cup B)$
$\Rightarrow (x \epsilon C$ or $x \epsilon A)$ and $(x\epsilon C$ or $x \epsilon B)$
$\Rightarrow x \epsilon C$ or $x \epsilon (A\cap B)$
$\Rightarrow x \epsilon C$ or $x\epsilon C$ (as $A\cup B\subseteq C)$
$\Rightarrow x \epsilon C$
$\Rightarrow (C \cup A)\cap (C\cup B)\subseteq ....(1)$
Now $x \epsilon C\Rightarrow x \epsilon (C\cup A)$ and $x \epsilon (C \cup B)$
$\Rightarrow x\epsilon (C\cup A)\cap (C\cup B)$
$\Rightarrow C\subseteq (C\cup A)\cap (C \cup B) .....(2)$
$\Rightarrow$ from (1) and (2)
$C = (C\cup A)\cap (C\cup B)$
$\Rightarrow$ option 2 is true
Let $x \epsilon A$ and $x \not {\epsilon} B$
$\Rightarrow x \epsilon (A - B)$
$\Rightarrow x \epsilon C$ (as $A - B \subseteq C)$
Let $x \epsilon A$ and $x \epsilon B$
$\Rightarrow x \epsilon (A\cap B)$
$\Rightarrow x \epsilon C$ (as $A\cap B\subseteq C)$
Hence $x \epsilon A \Rightarrow x \epsilon C$
$\Rightarrow A \subseteq C$
$\Rightarrow$ Option 3 is true
as $C\supseteq (A\cap B)$
$\Rightarrow B\cap C\supseteq (A\cap B)$
as $A\cap B\neq \phi$
$\Rightarrow B\cap C \neq \phi$
$\Rightarrow$ Option 4 is true.
Question 2
1 / -0

Let $S$ be a non-empty subset of $R$ . Consider the following statement:
$p$ : There is a rational number $x$ such that $x > 0$ .
which of the following statements is the negation of the statement P ?

A
There is no rational number $\mathrm{x}\in \mathrm{S}$ such that $\mathrm{x}\leq 0$

B
Every rational number $\mathrm{x}\in \mathrm{S}$ satisfies $\mathrm{x}\leq 0$

C
$\mathrm{x}\in \mathrm{S}$ and $\mathrm{x}\leq 0= \mathrm{x}$ is not rational

D
There is a rational number $\mathrm{x}\in \mathrm{S}$ such that $\mathrm{x}\leq 0$

Solution

$\mathrm{P}$ : there is a rational number $\mathrm{x}\in \mathrm{S}$ such that $\mathrm{x}>0$
$\sim \mathrm{P}$ : Every rational number $\mathrm{x}\in \mathrm{S}$ satisfies $\mathrm{x}\leq 0$
Hence, option 'B' is correct.
Question 3
1 / -0

The set $\displaystyle A=\left\{ x:x\in { x }^{ 2 }=16\quad and\quad 2x=6 \right\}$ equals:

A
$\displaystyle \phi$

B
$\displaystyle \{14,3,4\}$

C
$\displaystyle \{3\}$

D
$\displaystyle \{4\}$

Solution

Since, $\displaystyle { x }^{ 2 }=16\Rightarrow x=\pm 4$
and $\displaystyle 2x=6\Rightarrow x=3$
Hence, no value of $x$ is satisfied.
$\displaystyle \therefore A=\phi$
Question 4
1 / -0

Let $n$ be a fixed positive integer. Let a relation $R$ defined on $I$ (the set of all integers) as follows: $aRb$ iff $n/(a-b)$ , that is, iff $a-b$ is divisible by $n$ , then, the relation $R$ is

A
Reflexive only

B
Symmetric only

C
Transitive only

D
An equivalence relation

Solution

$R$ is reflexive since for any integer $a$ we have $a-a=0$ and $0$ is divisible by $n$ .
Hence, $aRa\quad \forall a\in I$ .
$R$ is symmetric, let $aRb$ .
Then by definition of $R$ , $a-b=nk$ where $k\in I$ .
Hence $b-a=(-k)n$ where $-k\in I$ and so $bRa$ .
Thus we have shown that $aRb\Rightarrow bRa$ .
$R$ is transitive, let $aRb$ and $bRc$ .
Then by definition of $R$ , we have $a-b={ k }_{ 1 }n$ and $b-c={ k }_{ 2 }n$ , where ${ k }_{ 1 },{ k }_{ 2 }\in I$ .
It then follows that
$a-c=\left( a-b \right) +\left( b-c \right) ={ k }_{ 1 }n+{ k }_{ 2 }n=\left( { k }_{ 1 }+{ k }_{ 2 } \right) n$
where ${ k }_{ 1 }+{ k }_{ 2 }\in I$
Question 5
1 / -0

If X $=$ (multiples of 2), Y $=$ (multiples of 5), Z $=$ (multiples of 10), then $X \cap ( Y \cap Z )$ is equal to

A
multiples of 10

B
multiples of 5

C
multiples of 2

D
multiples of 7

Solution

$We\quad can\quad write\quad the\quad given\quad sets\quad as\quad \\ X=\quad \left\{ 2,4,6,8,10,-----18,20,----28,30,---- \right\} \\ Y=\quad \left\{ 5,10,15,20,----30----40---- \right\} \\ Z=\quad \left\{ 10,20,30,40,----- \right\} \\ \therefore \quad Y\cap Z=\left\{ 10,20,30,---- \right\} \\ \therefore \quad X\cap \left( Y\cap Z \right) =\left\{ 10,20,30,----- \right\} =Z\\ \therefore \quad X\cap \left( Y\cap Z \right) =Z\quad \quad (Ans)$
Question 6
1 / -0

A $\bigcup \phi =$

A
$\phi$

B
2

C
A

D
0

Solution

We've,
$A\cup \phi$
$=A$ . [ Using direct formula (consistency property), since $\phi \subset A$ ]
Question 7
1 / -0

Suppose ${ A }_{ 1 },{ A }_{ 2 },...,{ A }_{ 30 }$ are thirty sets, each with five elements and ${ B }_{ 1 },{ B }_{ 2 },...,{ B }_{ 30 }$ are $n$ sets ecah with three elements. Let $\displaystyle \bigcup _{ i=1 }^{ 30 }{ { A }_{ i }= } \bigcup _{ j=1 }^{ n }{ { B }_{ j } } =S$
If each element of $S$ belongs to exactly ten of the ${ A }_{ i }'s$ and exactly none of the ${ B }_{ j }'s$ then $n=$

A
45

B
35

C
40

D
none of these

Solution

Given ${ A }_{ i }$ 's are thirty sets with five elements each, so $\displaystyle \sum _{ i=1 }^{ 30 }{ n\left( { A }_{ i } \right) } =5\times 30=150$ ...(1)
If there are $m$ distinct elements in $S$ and each element of $S$ belongs to exactly $10$ of the ${ A }_{ i }$ 's, we have
$\displaystyle \sum _{ i=1 }^{ 30 }{ n\left( { A }_{ i } \right) } =10m$ ...(2)
$\therefore$ from (1) and (2), we get $10m=150\Rightarrow m=15$ ...(3)
Similarly $\displaystyle \sum _{ j=1 }^{ 30 }{ n\left( { B }_{ j } \right) } =3n$ and $\displaystyle \sum _{ j=1 }^{ 30 }{ n\left( { B }_{ j } \right) } =9m$
$\therefore 3n=9m \displaystyle \Rightarrow \frac { 9m }{ 3 } =3m=3\times 15=45$ (from (3))
Hence, $n=45$
Question 8
1 / -0

In a locality two-thirds of the people have cable TV one-fifth have Dish TV and one-tenth have both What is the fraction of people having either cable TV or Dish TV?

A
$\displaystyle \frac{2}{3}$

B
$\dfrac13$

C
$\dfrac45$

D
$\dfrac35$

Solution

Fraction of people who watch cable only $\displaystyle =\left ( \frac{2}{3}-\frac{1}{10} \right )=\frac{17}{30}$
Fraction of people who watch Dish TV only $\displaystyle =\left ( \frac{1}{5}-\frac{1}{10} \right )=\frac{1}{10}$
$\displaystyle\therefore$ Fraction of people who watch either cable of Dish TV $\displaystyle =\frac{17}{30}+\frac{1}{10}=\frac{20}{30}=\frac{2}{3}$
Question 9
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Let $P =$ Set of all integral multiples of $3$ ; $Q =$ Set of integral multiples of $4$ ; $R =$ Set of all integral multiples of $6$ . Consider the following relations :
$1$ $\displaystyle P\cup Q=R$
$2.$ $\displaystyle P\subset R$
$3.$ $\displaystyle R\subset \left ( P\cup Q \right )$
Which of the relations given above is/are correct ?

A
only $1$

B
only $2$

C
only $3$

D
both $2$ and $3$

Solution

Given that P = $\{3, 6, 9, 12, 15, 18, 21,...\}$
Q = $\{4, 8, 12, 16, 20,...\}$
R = $\{6, 12, 18,...\}$
Considering the choices given :
1. P $\displaystyle \cup$ Q = $\{3, 4, 6, 8, 9, 12, 16, 18,...\}$ $\displaystyle \neq$ R
2. All the element of P are not in R so P $\displaystyle \nsubseteq$ R
3. P $\displaystyle \cup$ Q = $\{3, 4, 6, 8, 9, 12, 15, 16, 18, ...\}$
$\displaystyle \Rightarrow$ All the element of R are in P $\displaystyle \cup$ Q
$\displaystyle \Rightarrow$ R $\displaystyle \subset$ (P $\displaystyle \cup$ Q)
Question 10
1 / -0

Given, universal set = { $x \,\,\epsilon\,\, Z$ : $- 6 < x \leq 6$ }, N = { $n$ : $n$ is a non-negative number} and P = { $x$ : $x$ is a nonpositive number}. Find : $P'$

A
$\{-1,-2,-3,-4,-5,-6\}$

B
$\{1,2,3,4,5,6\}$

C
$\{0,1,2,3,4,5,6\}$

D
$\{0,-1,-2,-3,-4,-5,-6\}$

Solution

Universal set includes $\{-5,-4,-3,-2,-1,0,1,2,3,4,5,6\}$

P= $\{-5.-4.-3,-2,-1,0\}$ .

Hence P' only has the elements in option B. It does not contain 0.
P'= $\{1,2,3,4,5,6\}$ .

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Set Theory Test 1

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Question 1

If (A−C)⊆B(A - C) \subseteq B(A−C)⊆B, then A⊆BA\subseteq BA⊆B

(C∪A)∩(C∪B)=C(C \cup A)\cap (C\cup B) = C(C∪A)∩(C∪B)=C

If (A−B)⊆C(A - B)\subseteq C(A−B)⊆C, then A⊆CA\subseteq CA⊆C

B∩C≠ϕB\cap C \neq \phiB∩C=ϕ

Solution

Question 2

There is no rational number x∈S\mathrm{x}\in \mathrm{S}x∈S such that x≤0\mathrm{x}\leq 0x≤0

Every rational number x∈S\mathrm{x}\in \mathrm{S}x∈S satisfies x≤0\mathrm{x}\leq 0x≤0

x∈S\mathrm{x}\in \mathrm{S}x∈S and x≤0=x\mathrm{x}\leq 0= \mathrm{x}x≤0=x is not rational

There is a rational number x∈S\mathrm{x}\in \mathrm{S}x∈S such that x≤0\mathrm{x}\leq 0x≤0

Solution

Question 3

ϕ\displaystyle \phi ϕ

{14,3,4}\displaystyle \{14,3,4\}{14,3,4}

{3}\displaystyle \{3\}{3}

{4}\displaystyle \{4\}{4}

Solution

Question 4

Reflexive only

Symmetric only

Transitive only

An equivalence relation

Solution

Question 5

multiples of 10

multiples of 5

multiples of 2

multiples of 7

Solution

Question 6

ϕ\phiϕ

2

A

0

Solution

Question 7

45

35

40

none of these

Solution

Question 8

23\displaystyle \frac{2}{3}32​

13\dfrac1331​

45\dfrac4554​

35\dfrac3553​

Solution

Question 9

only 111

only 222

only 333

both 222 and 333

Solution

Question 10

{−1,−2,−3,−4,−5,−6}\{-1,-2,-3,-4,-5,-6\}{−1,−2,−3,−4,−5,−6}

{1,2,3,4,5,6}\{1,2,3,4,5,6\}{1,2,3,4,5,6}

{0,1,2,3,4,5,6}\{0,1,2,3,4,5,6\}{0,1,2,3,4,5,6}

{0,−1,−2,−3,−4,−5,−6}\{0,-1,-2,-3,-4,-5,-6\}{0,−1,−2,−3,−4,−5,−6}

Solution

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If $(A - C) \subseteq B$ , then $A\subseteq B$

$(C \cup A)\cap (C\cup B) = C$

If $(A - B)\subseteq C$ , then $A\subseteq C$

$B\cap C \neq \phi$

There is no rational number $\mathrm{x}\in \mathrm{S}$ such that $\mathrm{x}\leq 0$

Every rational number $\mathrm{x}\in \mathrm{S}$ satisfies $\mathrm{x}\leq 0$

$\mathrm{x}\in \mathrm{S}$ and $\mathrm{x}\leq 0= \mathrm{x}$ is not rational

There is a rational number $\mathrm{x}\in \mathrm{S}$ such that $\mathrm{x}\leq 0$

$\displaystyle \phi$

$\displaystyle \{14,3,4\}$

$\displaystyle \{3\}$

$\displaystyle \{4\}$

$\phi$

$\displaystyle \frac{2}{3}$

$\dfrac13$

$\dfrac45$

$\dfrac35$

only $1$

only $2$

only $3$

both $2$ and $3$

$\{-1,-2,-3,-4,-5,-6\}$

$\{1,2,3,4,5,6\}$

$\{0,1,2,3,4,5,6\}$

$\{0,-1,-2,-3,-4,-5,-6\}$