Set Theory Test 20

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Set Theory Test 20

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Question 1
1 / -0

If $$n(A)=10,n(B)=6$$ and $$(C)=5$$ for three disjoint sets $$A,B,C$$ then $$n(A\cup B\cup C)$$ equals

A
$$21$$

B
$$11$$

C
$$1$$

D
$$9$$

Solution

Since, $$A,B,C$$ are disjoint sets

$$\therefore n(A\cup B\cup C)=n(A)+n(B)+n(C)$$

$$=10+6+5=21$$
Question 2
1 / -0

The total number of subsets of {1, 2, 6, 7} are?

A
$$16$$

B
$$8$$

C
$$64$$

D
$$32$$

Solution

We have to find the total number of subsets of $$\{1,2,6,7\}$$.

We know that, for a set containing $$n$$ elements, the total number of subsets is $$2^n$$.

Consider $$\{1,2,6,7\}$$, wich has $$4$$ elements.

$$\therefore$$ here $$n=4$$

Hence total number of subsets is $$2^4=16$$.

Thus the total number of subsets of $$\{1,2,6,7\}$$ is $$16$$.
Question 3
1 / -0

If U = {1, 3, 5, 7, 9, 11, 13}, then which of the following are subsets of U.
B = {2, 4}
A = {0}
C = {1, 9, 5, 13}
D = {5, 11, 1}
E = {13, 7, 9, 11, 5, 3, 1}
F = {2, 3, 4, 5}

A
$$A$$ and $$B$$

B
$$C$$, $$D$$ and $$E$$

C
$$A$$ and $$D$$

D
only $$A$$

Solution

Given $$U=\{1,3,5,7,9,11,13\}, A=\{0\}, B=\{2,4\}, C=\{1,9,5,13\}, \\ D=\{5,11,1\}, E=\{13,7,9,11,5,3,1\}, F=\{2,3,4,5\}$$

Now we have to find the subsets of $$U$$

Consider $$A=\{0\}$$
Since $$0\notin U, A\nsubseteq U$$

Consider $$B=\{2,4\}$$
Since $$2,4\notin U, B\nsubseteq U$$

Consider $$C=\{1,9,5,13\}$$
Since $$1,9,5,13\in U, C\subset U$$

Consider $$D=\{5,11,1\}$$
Since $$5,11,1\in U, D\subset U$$

Consider $$E=\{13,7,9,11,5,3,1\}$$
Since $$13,7,9,11,5,3,1\in U, E\subseteq U$$

Consider $$F=\{2,3,4,5\}$$
Since $$2,3,4\notin U, F\nsubseteq U$$

Therefore $$C,D,E$$ are subsets of $$U$$.
Question 4
1 / -0

Let $$A$$ and $$B$$ be two events such that $$P\left( \overline { A\cup B } \right) =\dfrac { 1 }{ 6 } ,$$ $$P\left( A\cap B \right) =\dfrac { 1 }{ 4 } $$ and $$P\left( \overline { A } \right) =\dfrac { 1 }{ 4 } $$, where, $$\overline { A } $$ stands for complement of event $$A$$. Then, event $$A$$ and $$B$$ are

A
Mutually exclusive and independent

B
Independent but not equally likely

C
Equally likely but not independent

D
Equally likely and mutually exclusive

Solution

Given that, $$P\left( \overline { A\cup B } \right) =\dfrac { 1 }{ 6 } ,$$ $$P\left( A\cap B \right) =\dfrac { 1 }{ 4 } ,$$ $$P\left( \overline { A } \right) =\dfrac { 1 }{ 4 } $$

$$P\left( \overline { A\cup B } \right) =\dfrac { 1 }{ 6 } $$
$$\Rightarrow 1-P\left( A\cup B \right) =\dfrac { 1 }{ 6 } $$
$$\Rightarrow \left[ 1-P\left( A \right) \right] -P\left( B \right) +P\left( A\cap B \right) =\dfrac { 1 }{ 6 } $$

$$\Rightarrow P\left( \overline { A } \right) -P\left( B \right) +P\left( A\cap B \right) =\dfrac { 1 }{ 6 } $$
$$\Rightarrow \dfrac { 1 }{ 4 } -P\left( B \right) +\dfrac { 1 }{ 4 } =\dfrac { 1 }{ 6 } $$
$$\Rightarrow P\left( B \right) =\dfrac { 2 }{ 4 } -\dfrac { 1 }{ 6 } $$

$$\Rightarrow P\left( B \right) =\dfrac { 1 }{ 3 } $$
and $$P\left( A \right) =1-P\left( \overline { A } \right) =1-\dfrac { 1 }{ 4 } =\dfrac { 3 }{ 4 } $$

Since, $$P\left( A\cap B \right) =P\left( A \right) \cdot P\left( B \right) $$, so the events $$A$$ and $$B$$ are independent events but not equal likely.
Question 5
1 / -0

Let $$S = \left \{(a, b): a, b\epsilon Z, 0\leq a, b\leq 18\right \}$$. The number of elements $$(x, y)$$ in $$S$$ such that $$3x + 4y + 5$$ is divided by $$19$$ is

A
$$38$$

B
$$19$$

C
$$18$$

D
$$1$$

Solution

Maximum value of $$3x+4y+5$$ will occur at $$(18,18)$$
$$\Rightarrow$$ Maximum value$$=3\times18+4\times18+5=131$$

And, the minimum value will occur at $$0,0$$
$$\Rightarrow$$ Minimum value$$=3\times0+4\times0+5=5$$

Now, the multiples of $$19$$ that lie between minimum and maximum value of $$3x+4y+5$$ are $$19,38,57,76,95,114$$

$$\mathbf{Case 1-}$$ When $$3x+4y+5=19$$
By trial and error the solutions for this case is-
$$(2,2)$$
$$\Rightarrow$$ The number of elements $$(x,y)$$ in $$S$$ for this case is $$1$$

$$\mathbf{Case 2-}$$ When $$3x+4y+5=38$$
By trial and error the solutions for this case are-
$$(11,0);(7,3);(3,6)$$
$$\Rightarrow$$ The number of elements $$(x,y)$$ in $$S$$ for this case are $$3$$

$$\mathbf{Case 3-}$$ When $$3x+4y+5=57$$
By trial and error the solutions for this case are-
$$(16,1);(12,4);(8,7);(4,10);(0,13)$$
$$\Rightarrow$$ The number of elements $$(x,y)$$ in $$S$$ for this case are $$5$$

$$\mathbf{Case 4-}$$ When $$3x+4y+5=76$$
By trial and error the solutions for this case are-
$$(17,5);(13,8);(9,11);(5,14);(1,17)$$
$$\Rightarrow$$ The number of elements $$(x,y)$$ in $$S$$ for this case are $$5$$

$$\mathbf{Case 5-}$$ When $$3x+4y+5=95$$
By trial and error the solutions for this case are-
$$(18,9);(14,12);(10,15);(6,18)$$
$$\Rightarrow$$ The number of elements $$(x,y)$$ in $$S$$ for this case are $$4$$

$$\mathbf{Case 6-}$$ When $$3x+4y+5=114$$
By trial and error the solutions for this case are-
$$(15,16)$$
$$\Rightarrow$$ The number of elements $$(x,y)$$ in $$S$$ for this case are $$1$$

Therefore, the total number of elements $$(x,y)$$ in $$S$$ which are divisible by 19$$=1+3+5+5+4+1=19$$

Hence the correct answer is option $$B$$
Question 6
1 / -0

If $$A=\left\{ a,b,c \right\} ,$$ $$B=\left\{ b,c,d \right\} $$ and $$C=\left\{ a,d,c \right\} $$, then $$\left( A-B \right) \times \left( B\cap C \right) $$ is equal to

A
$$\left\{ \left( a,c \right) ,\left( a,d \right) \right\} $$

B
$$\left\{ \left( a,b \right) ,\left( c,d \right) \right\} $$

C
$$\left\{ \left( c,a \right) ,\left( d,a \right) \right\} $$

D
$$\left\{ \left( a,c \right) ,\left( a,d \right) ,\left( b,d \right) \right\} $$

Solution

Given that;
$$A=\left\{ a,b,c \right\} ,$$ $$B=\left\{ b,c,d \right\} $$ and $$C=\left\{ a,d,c \right\} $$

Now, $$A-B=\left\{ a,b,c \right\} -\left\{ b,c,d \right\} =\left\{ a \right\} $$
and $$B\cap C=\left\{ b,c,d \right\} \cap \left\{ a,d,c \right\} $$
$$=\left\{ c,d \right\} $$

Therefore, $$ \left( A-B \right) \times \left( B\cap C \right) =\left\{ a \right\} \times \left\{ c,d \right\} $$
$$=\left\{ \left( a,c \right) ,\left( a,d \right) \right\} $$
Question 7
1 / -0

$$ A={1 , 11 , 21 , 31 ,....... , 541 , 551}$$. B is a subset of A such that $$ x+y\neq552$$ , for any $$x , y \epsilon B.$$ The maximum number of elements in B is

A
$$26$$

B
$$30$$

C
$$29$$

D
$$28$$

Solution
Question 8
1 / -0

Let $$A,B$$ and $$C$$ be three events such that $$P(A)=0.3,P(B)=0.4,P(C)=0.8,P(A\cup B)=0.08,\quad P(A\cap C)=0.28,P(A\cap B\cap C)=0.09$$. If $$P(A\cup B\cup C)\ge 0.75$$, then $$P(B\cap C)$$ satisfies

A
$$P(B\cap C)\le 0.23$$

B
$$P(B\cap C)\le 0.48\quad $$

C
$$0.23\le P(B\cap C)\le 0.48$$

D
$$0.23\le P(B\cap C)\ge 0.48$$

Solution

Since $$P(A\cup B\cup C)\ge 0.75\quad $$
$$\quad \therefore 0.75\le P(A\cup B\cup C)\le 1$$
$$\Rightarrow 0.75\le P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)\le 1\quad \quad $$
$$\Rightarrow 0.75\le 1.23-P(B\cap C)\le 1\Rightarrow -0.48\le -P(B\cap C)\le -0.23$$
$$\Rightarrow 0.23\le P(B\cap C)\le 0.48\quad$$
Question 9
1 / -0

If $$a, b, c, d$$ are four distinct numbers chosen from the set $$\left \{1, 2, 3, ..., 9\right \}$$, then the minimum value of $$\dfrac {a}{b} + \dfrac {c}{d}$$ is

A
$$\dfrac {3}{8}$$

B
$$\dfrac {1}{3}$$

C
$$\dfrac {13}{36}$$

D
$$\dfrac {25}{72}$$

Solution

To minimize the value of $$\dfrac{a}{b}+\dfrac{c}{d}$$, $$b$$ and $$d$$ must highest numbers chosen from the set and $$a$$ and $$c$$ must be the lowest numbers of given set.
Therefore, $$b=9,\, d=8,\, a=2$$ and $$c=1$$.
Thus, the minimum value for $$\dfrac{a}{b}+\dfrac{c}{d}$$ $$=\dfrac{2}{9}+\dfrac{1}{8}=\dfrac{25}{72}$$
Question 10
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The relation R defined on set A = {1, 2, 3, 4, 5} is defined by R = $${(x, y): |x^2 - y^2| > 4}$$. Which of the following could be the range of relation R?

A
{4, 5}

B
{3, 4, 5 }

C
{1, 2, 3, 4, 5 }

D
{1, 2}

Solution

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Re-Attempt Weekly Quiz Competition

Set Theory Test 20

Result

Set Theory Test 20

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SHARING IS CARING

Question 1

$$21$$

$$11$$

$$1$$

$$9$$

Solution

Question 2

$$16$$

$$8$$

$$64$$

$$32$$

Solution

Question 3

$$A$$ and $$B$$

$$C$$, $$D$$ and $$E$$

$$A$$ and $$D$$

only $$A$$

Solution

Question 4

Mutually exclusive and independent

Independent but not equally likely

Equally likely but not independent

Equally likely and mutually exclusive

Solution

Question 5

$$38$$

$$19$$

$$18$$

$$1$$

Solution

Question 6

$$\left\{ \left( a,c \right) ,\left( a,d \right) \right\} $$

$$\left\{ \left( a,b \right) ,\left( c,d \right) \right\} $$

$$\left\{ \left( c,a \right) ,\left( d,a \right) \right\} $$

$$\left\{ \left( a,c \right) ,\left( a,d \right) ,\left( b,d \right) \right\} $$

Solution

Question 7

$$26$$

$$30$$

$$29$$

$$28$$

Solution

Question 8

$$P(B\cap C)\le 0.23$$

$$P(B\cap C)\le 0.48\quad $$

$$0.23\le P(B\cap C)\le 0.48$$

$$0.23\le P(B\cap C)\ge 0.48$$

Solution

Question 9

$$\dfrac {3}{8}$$

$$\dfrac {1}{3}$$

$$\dfrac {13}{36}$$

$$\dfrac {25}{72}$$

Solution

Question 10

{4, 5}

{3, 4, 5 }

{1, 2, 3, 4, 5 }

{1, 2}

Solution

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