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Set Theory Test 8

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Set Theory Test 8
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  • Question 1
    1 / -0
    If $$A=\left\{2, 4, 6, 8, 10\right\}, B=\left\{1, 3, 5, 7, 9\right\}$$, then $$A-B$$ =____________
    Solution
    $$A=\{ 2,4,6,8,10\} \\ B=\{ 1,3,5,7,9\} \\ A-B=\{ 2,4,6,8,10\} -\{ 1,3,5,7,9\} =\{ 2,4,6,8,10\} $$
  • Question 2
    1 / -0
    If A={1, 2, 4}, B={2, 4, 5}, C={2, 5}, then (A-C)$$\times$$(B-C) is equal to
    Solution
    $$A=\{1, 2, 4\}$$, $$B=\{2, 4, 5\}$$, $$C=\{2, 5\}$$
    $$A-C=\{1, 4\}$$
    $$B-C=\{4\}$$
    $$(A-C)\times (B-C)=\{1, 4\}\times \{4\}$$
    $$=\{1, 4\}, \{4, 4\}$$.

  • Question 3
    1 / -0
    If A and B are two sets such that $$n(A)=17, n(B)=23, n(A \cup B)=38$$, find $$n(A \cap B)$$.
    Solution
    We know,
    $$n(A \cap B)=n(A)+n(B)-n(A \cup B)$$
    $$n(A \cap B)=17+23-38=2$$
  • Question 4
    1 / -0
    Which of the following is set ?
    Solution
    As the collection of months having names starting with J is well defined. So, it's a set. Rest are not well defined , hence are not set.
  • Question 5
    1 / -0
    If X and Y are two sets such that $$n(X)=45, n(X \cup Y)=76, n(X \cap Y)=12,$$ find $$n(Y)$$.
    Solution
    $$n(X \cup Y)=n(X) +n(Y)-n(X \cap Y)$$
    $$76=45+n(Y)-12$$
    $$n(Y)=43$$
  • Question 6
    1 / -0
    If A={$$x\in N$$ : xis a multiple of 3} and
    B={$$x\in N$$ : is a multiple of 6}, then A-B is equal to
    Solution
    $$A=x\epsilon \quad N:x$$ is a multiple of $$3$$.
    $$A=3,6,9,12,15,18,......$$
    $$B=(x\quad \epsilon N:x$$ is a multiple of $$6)$$
    $$B=6,12,18,24,......$$
    $$\therefore A-B=(3,9,15,21,.....)$$
  • Question 7
    1 / -0
    The number of subsets of the set $$A=\{ { a }_{ 1 },{ a }_{ 2 },.........{ a }_{ n }\} $$ which contain even number of elements is
    Solution
    The total no of subsets of $$A$$ is the cardinality of the power set of $$A$$. 

    So, if $$|A|=n$$ then $$|P(A)|=2^n$$. 

    Therefore total no of subsets of $$A$$ is $$2^n$$.

    Similarly,

    The even number of events is given by, $$2^{n-1}$$
  • Question 8
    1 / -0
    Let $$S=\{2,4,6,8,......20\}$$. What is the maximum number of subsets does $$S$$ have ?
    Solution
    Given,

    $$S={2,4,6,8........,20}$$

    There are a total of $$10$$ elements.

    Therefore we have $$2^{10}=1024$$ subsets.
  • Question 9
    1 / -0
    Which of the following collections is a set?
    Solution
    Collection of all months of a year is a set If we denote the given set, then 
    $$A =\{\text{January, February, March, April, ........December}\}$$
  • Question 10
    1 / -0
    For sets $$\phi  , A = \left \{ 1,3 \right \} = B = \left \{ 1, 5, 9 \right \} , C = \left \{ 1, 5, 7, 9 \right \}$$ True option is 
    Solution
    Option (B) is correct , because all the elements of set $$B = \left \{  1, 5, 9 \right \}$$ is present in set $$ C = (1, 5, 7, 9) $$ Hence , $$ B \subset  C $$
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