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Set Theory Test 27

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Set Theory Test 27
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  • Question 1
    1 / -0
    Let  $$Q$$  be a non empty subset of  $$N$$  and  $$q$$  is a statement as given below :
    $$q:$$  There exists an even number  $$a \in Q$$  Negation of the statement  $$q$$  will be :
    Solution
    Let $$Q$$ be a non empty subset of $$N$$ and $$q.$$ There exists even number $$a\in Q$$ negation of statement $$q.$$ There does not exist a even number in set Q.
    Hence, the answer is there is no even number in the set $$Q.$$

  • Question 2
    1 / -0
    Let $$A$$ and $$B$$ be two sets. $$A \cap B = \{1, 2\}, A \cup B = \{1, 2, 3, 4, 5\}$$ and if $$n_1$$ is the maximum number of function from $$A$$ to $$B$$ and $$n_2$$ is the minimum number of function form $$A$$ to $$B$$ then $$n_1 - n_2$$ equals
    Solution

  • Question 3
    1 / -0
    In a battle $$70$$% of the combatants lost one eye, $$80$$% an ear, $$75$$% an arm, $$85$$% a leg, $$x$$% lost all the four limbs the minimum value of $$x$$ is
    Solution
    $$70\%$$ of the combatants lost one eye, $$80\%$$ an ear, $$75\%$$ an arm and $$85\%$$ a leg.
    Now, 
    $$\Rightarrow$$  The combatants who lost one eye and one ear $$=(70+80-100)\%$$
                                                                                          $$=50\%$$

    $$\Rightarrow$$  The combatants who lost one eye, one ear and one erm $$=(50+75-100)\%$$
                                                                                                          $$=25\%$$

    $$\Rightarrow$$  The combatants who lost one eye, one ear one arm and one leg $$=(25-85-100)\%$$ 
                                                                                                                        $$=10\%$$
    $$\therefore$$  Combatant who lost all the four limbs $$=10\%$$
    $$\therefore$$  $$x=10\%$$
  • Question 4
    1 / -0
    The value of set $$(A\cup B\cup C)\cap(A\cap B^1\cap C^1)^1\cap C^1$$ is equal to  
    Solution

  • Question 5
    1 / -0
    Two sets A and B are defined as follows
    $$A=\left\{ \left( x,y \right) :y={ e }^{ 2x },x\in R \right\} $$ and 
    $$B=\left\{ \left( x,y \right) :y={ x }^{ 2 },x\in R \right\} $$, then
    Solution
    $$\begin{array}{l} A=\left\{ { \left( { x,\, y } \right) \, :\, y={ e^{ 2x } },\, x\in R } \right\} \, and \\ B=\left\{ { \left( { x,\, y } \right) \, :\, y={ x^{ 2 } },\, x\in R } \right\}  \\ then, \\ A=\left( { { e^{ \circ  } },\, { e^{ 1 } },\, { e^{ 2 } },....,\, { e^{ \infty  } } } \right)  \\ =\left( { 1,\, { e^{ 1 } },\, { e^{ 2 } },....... } \right)  \\ and, \\ B=\left( { 0,\, 1,\, 4,\, ..... } \right)  \\ \therefore A\subset B \\ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}$$
  • Question 6
    1 / -0
    If $$A=\left\{1,2,3,4\right\}; B=\left\{2,4,6,8\right\}; C=\left\{3,4,5,8\right\}$$ then $$A\cap B\cap C=$$
    Solution
    $$\begin{matrix} A=\left\{ { 1,2,3,4 } \right\}  \\ B=\left\{ { 2,4,6,8 } \right\}  \\ C=\left\{ { 3,4,5,8 } \right\}  \\ A\cap B\cap C=\left\{ { 4 } \right\} . \\  \end{matrix}$$
  • Question 7
    1 / -0
    If n(A)=3, n(B)=4, then $$n(A\times B\times C)=36 find \,n(C)$$ is equal to :
    Solution
    $$n(A)=3$$
    $$n(B)=4$$
    $$n(A\times B\times C)=36$$
    $$n(3\times 4\times C)=36$$
    $$n(12\times C)=36$$
    $$\therefore n(C)=3$$
    So, $$n(A\times B\times C)=36$$
    $$n(3\times 4\times 3)=36$$. H.P.

  • Question 8
    1 / -0
    In a group of 800 people, 550 can speak Hindi and 450 can speak English. How many can speak both Hindi and English ?
    Solution
    Let H denote the set of people speaking Hindi and E denote the set of people speaking English. We are given :
    $$n(H)=550, n(E)=450, n(H\cup E)=800$$
    Now,
    $$n(H\cup E)=n(H)+n(E)-n(H\cap E)$$
    $$n(H \cap E)=550+450-800=200$$
    Hence, 200 persons can speak both Hindi and English.
  • Question 9
    1 / -0
    If A= {1, 2, 5} and B= {3, 4, 5, 9}, then $$A \bigcup B$$ is equal to :
    Solution
    $$A=\left\{ 1,2,5 \right\} \quad B=\left\{ 3,4,5,9 \right\} $$
    $$A\cup B=\left\{ 1,2,3,4,5,9 \right\} $$
    Option C.
  • Question 10
    1 / -0
    If P(S) denotes the set of all subsets of a given set S, then the number of one to one function from the set s={1,2,3} to the set of P(S) is 
    Solution
    $$\begin{array}{l} Given \\ P\left( s \right) =\left\{ { 1,\, 2,\, 3 } \right\}  \\ P\left( s \right) =all\, subset\, of\, 5 \\ =\left\{ { \phi ,\left\{ 1 \right\} ,\left\{ 3 \right\} ,\left\{ { 1,2 } \right\} ,\left\{ { 2,3 } \right\} ,\left\{ { 3,1 } \right\} ,\left\{ { 1,2,3 } \right\}  } \right\}  \\ No.\, of\, element\, of\, 5=0\left( 5 \right) =3 \\ similarly \\ O\left( { p\left( s \right)  } \right) =8 \\ Total\, no.\, of\, one-one\, function \\ =\frac { { n! } }{ { \left( { n-m } \right) ! } } \, ,\, where\, n=o\left( { p\left( s \right)  } \right) \, and\, m=o\left( 5 \right)  \\ =\frac { { 8! } }{ { \left( { 8-3 } \right) ! } } =\frac { { 8! } }{ { 5! } }  \\ =336\, \,  \\ No.\, of\, one-one\, function\, is\, 336 \\ Hence,\, \, option\, \, A\, \, is\, correct, \end{array}$$

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