Set Theory Test 26

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Set Theory Test 26

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Weekly Quiz Competition

Question 1
1 / -0

If m$$=$$ number of distinct rational numbers p/q $$\in (0, 1)$$ such that p, q $$\in \{1, 2, 3, 4, 5\}$$ and n$$=$$ number of onto mappings from $$\{1, 2, 3\}$$ onto $$\{1, 2\}$$, then $$m-n$$ is?

A
$$1$$

B
$$-1$$

C
$$0$$

D
$$3$$

Solution
Question 2
1 / -0

In a universal set x,$$n\left( x \right) = 50$$ ,$$n\left( A \right) = 35$$ , $$n\left( B \right) = 20$$ , $$n\left( {A' \cap B'} \right) = 5$$ ,then $$n\left( {A \cup B} \right),n\left( {A \cap B} \right)$$ are repsectively

A
$$45,10$$

B
$$10,45$$

C
$$25,30$$

D
$$15,25$$

Solution

Given $$n\left(X\right) = 50$$,
$$n\left(A\right) = 35$$,
$$n\left(B\right) = 20$$
$$n\left({A}^{\prime}\cap\,{B}^{\prime}\right) = 5$$

$$(i)\,n\left(A \cup B\right)$$
$$n\left({A}^{\prime}\cap\,{B}^{\prime}\right) = n\left(X\right) - n\left(A\cup\,B\right)$$
$$n\left(A\cup\, B\right) = n\left(X\right) - n\left({A}^{\prime}\cap\,{B}^{\prime}\right)$$
$$n\left(A\cup\, B\right) = 50 - 5= 45$$

$$(ii) n\left(A \cap B\right)$$
$$n\left(A\cup B\right) = n\left(A\right) + n\left(B\right) - n\left(A\cap B\right)$$
$$n\left(A\cap\, B\right) = n\left(A\right) + n\left(B\right) - n\left(A\cup\, B\right)$$
$$n\left(A\cap\, B\right) = 35 + 20 - 45= 10$$
Question 3
1 / -0

If $$A$$ and $$B$$ are two non empty sets then $$( A \cup B ) ^ { C } = ?$$

A
$$A ^ { C } \cup B ^ { C }$$

B
$$A ^ { C } \cap B ^ { C }$$

C
$$A \cup B ^ { C }$$

D
$$A ^ { C } \cap B$$

Solution

If A and B are two no empty sets then $${ \left( A\cup B \right) }^{ C }={ A }^{ C }\cap { B }^{ C }$$
Question 4
1 / -0

Let $$S={1,2,3,.....10}$$ and $$P={1,2,3,4,5}$$ The number of subsets $$'Q'$$ of $$S$$ such that $$p \cup Q=S$$, are.....

A
$$128$$

B
$$256$$

C
$$32$$

D
$$64$$
Question 5
1 / -0

If 'p' is true and 'q' is false, then which of the following statement is not true?

A
$$p\vee q$$

B
$$p\Rightarrow q$$

C
$$p\wedge (\sim q)$$

D
$$q\Rightarrow p$$

Solution
Question 6
1 / -0

The number of subsets $$ R$$ of $$P=(1,2,3,....,9)$$ which satisfies the property "There exit integers a<b<c with a$$\in $$R, b$$\in $$R,c$$\in $$R" is

A
$$512$$

B
$$466$$

C
$$467$$

D
None of these

Solution

Every subset of P containing at least 3 elements from P will always satisfy the required property since each number is distinct in the set P.

So, Total no. of ways = $$^{9}C_3 + ^{9}C_4 + .... + ^{9}C_9 $$
$$= (^{9}C_0 + ^{9}C_1 + ..... + ^{9}C_9) - ^{9}C_0 - ^{9}C_1 - ^{9}C_2$$
$$=2^9 - 46$$
$$=512 -46$$
$$=466$$
Question 7
1 / -0

If the number of $$5$$ elements subsets of the set $$A\left\{\ a_{1},a_{2}.....a_{20}\right\}$$ of $$20$$ distinct elements is $$k$$ times the number of $$5$$ elements subsets containing $$a_{4}$$, then $$k$$ is

A
$$5$$

B
$$\dfrac{20}{7}$$

C
$$4$$

D
$$\dfrac{10}{3}$$

Solution

No of $$5$$ elements subset $$=^{20}C_{5}=\dfrac {20!}{15! \, 15!}=\dfrac {20 \times 19\times 18\times 17\times 16\times 15!}{15! \, 5\times 4\times 3\times 2\times 1}=15504$$

No of $$5$$ elements subset containing $$a_{4}$$ $$=^{19}C_{4}=\dfrac {19!}{4! \, 15!}$$

$$\Rightarrow \dfrac {20!}{15! \, 15!}=k\dfrac {19!}{4!\, 15!}$$

$$\Rightarrow \dfrac {20\times 19!}{5\times 4!}=k\dfrac {19!}{4!}$$

$$\Rightarrow k=4$$
Question 8
1 / -0

If $$p$$ and $$q$$ are two proposition, then $$\sim(p\leftrightarrow q)$$ is

A
$$\sim p \wedge \sim q$$

B
$$\sim p \vee \sim q$$

C
$$(p \wedge \sim q) \vee (\sim p \wedge q)$$

D
$$none of these$$

Solution
Question 9
1 / -0

If two sets $$P$$ and $$Q,n\left(P\right)=5,n\left(Q\right)=4$$ then $$n\left(P\times Q\right)=$$

A
$$20$$

B
$$9$$

C
$$25$$

D
$$5/4$$

Solution

$$\begin{aligned} n(P \times Q) &=n(P) \times n(Q) \\ &=5 \times 4 \\ &=20 \end{aligned}$$
$$\therefore$$ option $$A$$ is correct.
Question 10
1 / -0

The statement that is true among the following is

A
$$p\Longrightarrow q$$ is equivalent to $$p\wedge -q$$

B
$$p\vee q$$ and $$p\wedge q$$ have the same truth value

C
The converse of $$tanx=0\Longrightarrow x=0$$ is $$x\neq 0\Longrightarrow tanx=0$$

D
The contrapositive of $$3x+2=8\Longrightarrow x=2$$ is $$x\neq 2\Longrightarrow 3x+2\neq 8$$

Solution

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Re-Attempt Weekly Quiz Competition

Set Theory Test 26

Result

Set Theory Test 26

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SHARING IS CARING

Question 1

$$1$$

$$-1$$

$$0$$

$$3$$

Solution

Question 2

$$45,10$$

$$10,45$$

$$25,30$$

$$15,25$$

Solution

Question 3

$$A ^ { C } \cup B ^ { C }$$

$$A ^ { C } \cap B ^ { C }$$

$$A \cup B ^ { C }$$

$$A ^ { C } \cap B$$

Solution

Question 4

$$128$$

$$256$$

$$32$$

$$64$$

Question 5

$$p\vee q$$

$$p\Rightarrow q$$

$$p\wedge (\sim q)$$

$$q\Rightarrow p$$

Solution

Question 6

$$512$$

$$466$$

$$467$$

None of these

Solution

Question 7

$$5$$

$$\dfrac{20}{7}$$

$$4$$

$$\dfrac{10}{3}$$

Solution

Question 8

$$\sim p \wedge \sim q$$

$$\sim p \vee \sim q$$

$$(p \wedge \sim q) \vee (\sim p \wedge q)$$

$$none of these$$

Solution

Question 9

$$20$$

$$9$$

$$25$$

$$5/4$$

Solution

Question 10

$$p\Longrightarrow q$$ is equivalent to $$p\wedge -q$$

$$p\vee q$$ and $$p\wedge q$$ have the same truth value

The converse of $$tanx=0\Longrightarrow x=0$$ is $$x\neq 0\Longrightarrow tanx=0$$

The contrapositive of $$3x+2=8\Longrightarrow x=2$$ is $$x\neq 2\Longrightarrow 3x+2\neq 8$$

Solution

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