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Set Theory Test 16

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Set Theory Test 16
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  • Question 1
    1 / -0
    Three sets $$A, B, C$$ are such that $$\displaystyle A=B\cap C$$ and $$\displaystyle B=C\cap A$$, then
    Solution
    Since, $$\displaystyle A=B\cap C$$ and $$\displaystyle B=C\cap A$$, then $$\displaystyle A\equiv B$$
  • Question 2
    1 / -0
    If n is a member of both set A$$=\left\{\displaystyle\frac{4}{7}, 1, \frac{5}{2}, 4, \frac{1}{2}, 7\right\}$$ and set B$$=\left\{\displaystyle\frac{4}{7}, \frac{7}{4}, 4, 7\right\}$$, which of the following must be true?
    I. n is an integer.
    II. $$4n$$ is an integer.
    III. $$n=4$$
    Solution
    None of the criteria satisfy all the elements in the intersection.

    $$\dfrac47$$ is not an integer nor is $$\dfrac {4\times 4}{7}$$ an integer.

    Of course 4 is not the only element in the intersection. Hence 3 is untrue.
  • Question 3
    1 / -0
    If the universal set $$\{x\in W ,3<x≤12\} ,A=\{5,7,9\}$$, then $$A'=$$
    Solution
    Given universal set is $$\{x\in W ,3<x≤12\}$$
    It can be written as a set $$\{4,5,6,7,8,9,10,11,12\}$$
    $$A$$ is given as $$A=\left\{5,7,9\right\}$$
    $$\therefore A'=W-A=$$ All the elements in the universal set but not in set $$A$$
             $$ =\{4,6,8,10,11,12\}$$
    Hence, option B is correct
  • Question 4
    1 / -0

    Directions For Questions

    $$\mu = \left \{a, b, c, d, e, f, g, h, i, j\right \}$$
    $$P = \left \{a, b, c, e\right \}$$
    $$Q = \left \{b, c, d, f\right \}$$ and
    $$R = \left \{c, f, h, i, j\right \}$$
    Find the number of elements of the set

    ...view full instructions

    $$(P\cap Q)' \cup R$$
    Solution

  • Question 5
    1 / -0
    For any two sets $$A$$ and $$B$$, $$A-\left( A-B \right) $$ equals
    Solution
    Now, $$A-\left( A-B \right) =A-\left( A\cap { B }^{ C } \right) $$
                           $$=A\cap { \left( A\cap { B }^{ C } \right)  }^{ C }$$
                           $$=A\cap \left( { A }^{ C }\cup B \right) $$
                           $$=\left( A\cap { A }^{ C } \right) \cup \left( A\cap B \right) $$
                           $$=A\cap B$$
  • Question 6
    1 / -0
    In a class, $$20$$ opted for Physics, $$17$$ for Maths, $$5$$ for both and $$10$$ for other subjects. The class contains how many students?
    Solution

    $$n(P) = 20$$

    $$n(M) = 17$$

    $$n(M \cap P) = 5$$

    $$n(other \; subjects)=10$$

    $$n(M\cup P) =n(M)+n(P)-n(M\cap P)$$

    $$n(M\cup P) =17+20-5 = 32$$

    Total students $$= n(P\cup M)+n (other \; subjects )$$

    $$32+10=42$$

  • Question 7
    1 / -0
    If $$A = \{ a, b, p, d\}  B = \{ p, d, e\}  C = \{p, e, f, g\}$$ then find 

    $$A \times (B \cap C ) $$ is equal to 
    Solution

  • Question 8
    1 / -0
    In a class of 250 students, 175 take mathematics and 142 take science. How many take both mathematics

    and science? (All take math and/or science.)
    Solution

  • Question 9
    1 / -0
    If X is a finite set. Let $$P(X)$$ denote the set of all subsets of X and let $$n(X)$$ denote the number of elements in X. If for two finite subsets $$A, B, n(P(A)) = n(P(B)) + 15$$ then $$n(B) = $$ ____, $$n(A) =$$ _____
    Solution
    If X is a finite set.
    Let,
    Number of element in a subset $$A$$ be $$n(A)$$ =$$m$$ and 
    Number of elements in subset $$B$$ be $$n(B)$$=$$n$$
    Then total number of subsets of finite set contaning say $$n$$ elements is $$2^n$$.
    Therefore,
    $$n[P(A)]=2^m$$,  $$n[P(B)]=2^n$$
    Substituting in equations , we get:-
    $$2^m=2^n+15$$
    $$2^m-2^n=15$$

    $$m=4,n=0$$

    Option $$\textbf A$$ is correct
  • Question 10
    1 / -0
    P, Q and R are three sets and $$\xi = P\cup Q\cup R$$. Given that $$n(\xi) = 60, n (P\cap Q) = 5, n(Q\cap R) = 10, n(P) = 20$$ and $$n(Q) = 23$$, find $$n(P\cup R)$$

    Solution

    $$n(U) =60 (Universal \; Set) $$

    $$n(P \cap Q ) = 5$$

    $$n(Q \cap R ) = 10$$

    $$n(P ) = 20$$

    $$n(Q)=23$$

    $$n(P \cup Q ) = n(P) + n(Q)- n(P \cap Q ) = 20+23-5 = 38$$

    $$n(R)=n(U)-n(P \cup Q ) +n(Q\cap R)= 60-38+10 = 32$$

    $$n(P \cup R )  = n(P) + n(R)- n(P \cap R ) = 20+32-0 =52$$

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