Sets Test

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Sets Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

Which is the simplified representation of $$(A' \cap B'\cap C)\cup(B\cap C) \cup (A\cap C)$$ where A,B,C are subsets of set X

A
$$A$$

B
$$B$$

C
$$C$$

D
$$X\cap(A \cup B\cup C)$$
Question 2
1 / -0

If $$A=\{1,2,3\}, B=\{3,4\}$$ and $$C=\{1,3,5\}$$, then $$A \times (B - C) =$$

A
$$(A \times B ) - (A \times C )$$

B
$$(A \times B ) + (A \times C )$$

C
$$(A \times B ) - (B \times C )$$

D
$$(A \times B ) - (C \times A )$$

Solution

$$(B-C)$$ is set of all elements present in $$B$$ and not in $$C$$

$$\therefore (B-C)=\{4 \}$$

$$A\times(B-C)=\{1,2,3\}\times\{4\}=\{(1,4),(2,4),(3,4)\}$$

Lets find $$(A\times B)$$ and $$(A\times C)$$

$$(A\times B)=\{1,2,3\}\times \{3,4\}$$
$$=\{(1,3),(2,3),(3,3),(1,4),(2,4),(3,4)\}$$

$$(A\times C)=\{1,2,3\}\times \{1,3,5\}$$
$$=\{(1,1),(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,3),(3,5)\}$$

$$(A\times B)-(A\times C)=\{(1,3),(2,3),(3,3),(1,4),(2,4),(3,4)\}-\{(1,1),(1,3),(1,5),(2,1),(2,3),$$
$$(2,5),(3,1),(3,3),(3,5)\}$$

$$(A\times B)-(A\times C)=\{(1,4),(2,4),(3,4)\}$$

$$\therefore A\times(B-C)=(A\times B)-(A\times C)$$
Question 3
1 / -0

In certain town, $$25\%$$ families own a cell phone, $$15\%$$ families own a scooter and $$65\%$$ families own neither a cell phone nor a scooter. If $$1500$$ families own both a cell phone and a scooter, then the total number of families in the town is:

A
$$10000$$

B
$$20000$$

C
$$30000$$

D
$$50000$$

Solution

65% of families own neither a cell phone nor a scooter
35% of families own either a phone or a scooter
(A$$\cup$$B)=A+B-(A$$\cap$$B)
35%x=25%x+15%x-(A$$\cap$$B)
(A$$\cap$$B)=5%x=$$\dfrac{5}{100}x=1500$$
number of families=30000
Question 4
1 / -0

If $$X =\{1,2,3,4,5,6,7,8,9, 10\}$$ is the universal set and$$ A= \{1, 2, 3,4\}, B= \{2,4,6,8\}, C= \{3,4,5,6\}$$ verify the following.
(a) $$A \cup (B\cup C) = (A \cup B) \cup C$$
(b)$$A \cap (B\cup C) = (A \cap B) \cup (A \cap C)$$
(c) $$(A')' =A$$

A
Only a is true

B
Only b and c are true

C
Only a and b are true

D
All three a,b and c are true.

Solution

given
$$X=$${$$1,2,3,4,5,6,7,8,9,10$$},$$A=\{1,2,3,4\},B=\{2,4,6,8\},C=\{3,4,5,6\}$$

now from the given data

$$A\cup B=$$$$\{1,2,3,4,6,8\}$$,$$B\cup C=$$$$\{2,3,4,5,6,8\}$$

Now $$A\cup (B\cup C)=\{1,2,3,4,5,6,8\}\quad and\quad(A\cup B)\cup C=$$$$\{1,2,3,4,5,6,8\}$$

$$\therefore A\cup (B\cup C)=(A\cup B)\cup C=\{1,2,3,4,5,6,8\}$$

$$A\cap B=$$$$\{2,4\}$$,$$A\cap C=$$$$\{3,4\}$$,$$B\cup C=$$$$\{2,3,4,5,6,8\}$$

Now $$A\cap(B\cup C)=\{2,3,4\}\quad and \quad(A\cap B)\cup (A\cap C)=$$$$\{2,3,4\}$$

$$\therefore A\cap(B\cup C)=(A\cap B)\cup (A\cap C)=$$$$\{2,3,4\}$$

$$A'=$$$$\{5,6,7,8,9,10\}$$

therefore $$(A')'=$$$$\{1,2,3,4\}$$$$=A$$

therefore all three a,b,c are true
Question 5
1 / -0

If two sets $$A$$ and $$B$$ are having $$99$$ elements in common, then the number of ordered pairs common to each of the sets $$AxB$$ and $$BxA$$ are

A
$$2^{99}$$

B
$$99^{2}$$

C
$$100$$

D
$$18$$

Solution

Let us the $$n\left( {\left( {A \times B} \right) \cap \left( {B \times A} \right)} \right)$$
$$ = n\left( {\left( {A \cap B} \right) \times \left( {B \cap A} \right)} \right)$$
$$ = n\left( {A \cap B} \right).\,n\left( {B \cap A} \right)$$
$$ = n\left( {A \cap B} \right).\,n\,\left( {A \cap B} \right)$$
$$ = \left( {99} \right)\left( {99} \right)$$
$$ = {99^2}$$
Question 6
1 / -0

If $$A \subset B$$, then

A
$$C - B \subset C - A$$.

B
$$A \cup B = A$$

C
$$A \cap B = B$$

D
None

Solution
Question 7
1 / -0

The set of all $$x$$ for which $$1 + \log x < x$$ is

A
$$(1, \infty)$$

B
$$(0, 1)$$

C
$$(0, \infty)$$

D
None of these

Solution

$$1+\log{x}<x$$
$$\log{x}<x-1$$
$$x<{e}^{x-1}$$
$$\therefore x\in (1,\infty).$$
Question 8
1 / -0

If $$\alpha,\xi,\eta$$ are non-empty sets then:

A
$$(\alpha \times \beta)\cup (\xi \times \eta)=(\alpha \times \beta)\cap (\xi \times \eta)$$

B
$$(\alpha \times \beta)\cap (\xi \times \eta)=(\alpha \times \beta)\cap (\xi \times \eta)$$

C
$$(\alpha \cap \beta)\cup (\xi \cap \eta)=(\alpha \times \beta)\cup (\xi \times \eta)$$

D
$$(\alpha \cap \beta)= (\xi \cap \eta)=(\alpha \times \beta)\cup (\xi \times \eta)$$

Solution
Question 9
1 / -0

$$S_1:(p\Rightarrow q) V ( q \Rightarrow p )$$ is a tautology.
$$S_2: ((p\Rightarrow q) V ( q \Rightarrow p))$$ is a fallacy

A
$$S_1$$ is true, $$S_2$$ is false

B
$$S_1$$ false

C
$$S_1$$ is false, $$S_2$$ is false

D
$$S_1$$ is true

Solution
Question 10
1 / -0

$$If\,A = \left\{ {\left( {x,y} \right)\,\left| {{x^2} + {y^2} \le \left. 4 \right\}\,and} \right.} \right.$$
$$B = \left\{ {\left( {x,y} \right)\,\left| {{{(x - 3)}^2} + {y^2}} \right. \le \left. 4 \right\}} \right.\,and\,the$$
$$po{\mathop{\rm int}} \,P\left( {a,\frac{1}{2}} \right)\,belongs\,to\,the\,set\,B - A$$ then the set of possible real values of $$a$$ is

A
$$\left( {\frac{{1 + \sqrt {3} }}{4},\frac{{7 + \sqrt 7 }}{4}} \right)$$

B
$$\left( {\frac{{7 - \sqrt 7 }}{4},\frac{{1 + \sqrt 7 }}{4}} \right)$$

C
$$\left( {\frac{{1 - \sqrt {31} }}{4},\frac{{7 - \sqrt 7 }}{4}} \right)$$

D
none of these

Solution

$${x^2} + {y^2} > 4$$
$${a^2} + {\left( {a - \frac{1}{2}} \right)^2} > 4$$
$$2{a^2} - a - \frac{5}{4} > 0$$
$$a = \frac{{1 \pm \sqrt {31} }}{4}$$
$$a \in \left( { - \infty ,\frac{{1 - \sqrt {31} }}{4}} \right)\,V\left( {\frac{{1 + \sqrt {31} }}{4},\infty } \right)$$
$$a \in \left( {\frac{{1 + \sqrt 3 }}{4},\frac{{7 + \sqrt 7 }}{4}} \right)$$

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Re-Attempt Weekly Quiz Competition

Sets Test - 37

Result

Sets Test - 37

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SHARING IS CARING

Question 1

$$A$$

$$B$$

$$C$$

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

Question 2

$$(A \times B ) - (A \times C )$$

$$(A \times B ) + (A \times C )$$

$$(A \times B ) - (B \times C )$$

$$(A \times B ) - (C \times A )$$

Solution

Question 3

$$10000$$

$$20000$$

$$30000$$

$$50000$$

Solution

Question 4

Only a is true

Only b and c are true

Only a and b are true

All three a,b and c are true.

Solution

Question 5

$$2^{99}$$

$$99^{2}$$

$$100$$

$$18$$

Solution

Question 6

$$C - B \subset C - A$$.

$$A \cup B = A$$

$$A \cap B = B$$

None

Solution

Question 7

$$(1, \infty)$$

$$(0, 1)$$

$$(0, \infty)$$

None of these

Solution

Question 8

$$(\alpha \times \beta)\cup (\xi \times \eta)=(\alpha \times \beta)\cap (\xi \times \eta)$$

$$(\alpha \times \beta)\cap (\xi \times \eta)=(\alpha \times \beta)\cap (\xi \times \eta)$$

$$(\alpha \cap \beta)\cup (\xi \cap \eta)=(\alpha \times \beta)\cup (\xi \times \eta)$$

$$(\alpha \cap \beta)= (\xi \cap \eta)=(\alpha \times \beta)\cup (\xi \times \eta)$$

Solution

Question 9

$$S_1$$ is true, $$S_2$$ is false

$$S_1$$ false

$$S_1$$ is false, $$S_2$$ is false

$$S_1$$ is true

Solution

Question 10

$$\left( {\frac{{1 + \sqrt {3} }}{4},\frac{{7 + \sqrt 7 }}{4}} \right)$$

$$\left( {\frac{{7 - \sqrt 7 }}{4},\frac{{1 + \sqrt 7 }}{4}} \right)$$

$$\left( {\frac{{1 - \sqrt {31} }}{4},\frac{{7 - \sqrt 7 }}{4}} \right)$$

none of these

Solution

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