Set Theory Test 1

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Set Theory Test 1

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Weekly Quiz Competition

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
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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
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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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Set Theory Test 1

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Question 1

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

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

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

$$B\cap C \neq \phi$$

Solution

Question 2

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 $$\mathrm{x}\leq 0$$

Solution

Question 3

$$\displaystyle \phi $$

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

$$\displaystyle \{3\}$$

$$\displaystyle \{4\}$$

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

$$\phi$$

2

A

0

Solution

Question 7

45

35

40

none of these

Solution

Question 8

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

$$\dfrac13$$

$$\dfrac45$$

$$\dfrac35$$

Solution

Question 9

only $$1$$

only $$2$$

only $$3$$

both $$2$$ and $$3$$

Solution

Question 10

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

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

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

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

Solution

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