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Set Theory Test 17

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Set Theory Test 17
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  • Question 1
    1 / -0
    If A is a finite set, let $$P(A)$$ denote the set all subsets of A and $$n(A)$$ denote the number of elements in A. If for two finite sets X and Y, $$n[P(X)] = n[P(Y)] + 15$$ then find $$n(X)$$ and $$n(Y)$$
    Solution
    If $$X$$ is a finite set.
    Let,
    Number of elements in subset $$A$$ be $$n(A)=m$$
    Number of elements in subset $$B$$ be $$n(B)=n$$
    Then total number of subsets of finitr set containing 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 2
    1 / -0
    If A and B are two sets, where A has more elements than B. Calculate the least possible value of n(A) + n(B), where n(A) is the number of elements in A and n(B) is the number of elements in B.
    Solution

  • Question 3
    1 / -0
    Solve the following inequalities: $$\displaystyle\frac{1}{2-|x|}\geq 1$$.
    Solution

  • Question 4
    1 / -0
    Let $$S = \left \{(a, b, c)\epsilon N\times N\times N : a + b + c = 21. a \leq b\leq c\right \}$$ and $$T = \left \{a, b, c)\epsilon N\times N\times N : a, b, c,\ are\ in\ A.P.\right \}$$, where $$N$$ is the set of all natural numbers. Then the number of elements in the set $$S\cap T$$ is
    Solution
    $$a + b + c = 21$$ and $$b = \dfrac {a + c}{2}$$
    $$\Rightarrow a + c = 14$$ and $$b = 7$$
    So, a can take values from $$1$$ to $$6$$, when $$c$$ ranges from $$13$$ to $$8$$, or $$a = b = c = 7$$
    So, $$7$$ triplets
  • Question 5
    1 / -0
    Which one of the following is correct?
    Solution
    $$x^2=-1$$ is satisfied by $$i $$ and $$-i$$ none of which are integers. Hence required set is null.
  • Question 6
    1 / -0
    Which one of the following is a finite set?
    Solution
    (A)
    Let $$A=\{x:x\in Z, x<5\}$$
    $$A$$ can be represented in roster form as
    $$A=\{........,-3,-2,..,3,4\}$$ which is an infinite set

    (B)
    Let $$B=\{x:x\in W, x\ge 5\}$$
    $$B$$ can be represented in roster form as
    $$B=\{5,6,7,......\}$$ which is an infinite set

    (C)
    Let $$C=\{x:x\in N, x>10\}$$
    $$C$$ can be represented in roster form as
    $$C=\{11,12,13,14,........\}$$ which is an infinite set

    (D)
    Let $$D=\{x:x \text{ is an even prime number}\}$$
    The only prime number which is even is $$2$$
    So, $$D$$ can be represented in roster form as
    $$D=\{2\}$$ which is a finite set

    Hence, the set in option D is a finite set.
  • Question 7
    1 / -0
    If $$X=\left \{ a,\left \{ b,c \right \},d \right \}$$, which of the following is a subset of $$X$$?
    Solution
    Given $$X = \{a,\{b,c\},d\}$$. Hence the elements of $$X$$ are $$a, \{b,c\}$$ and $$d$$.
    Subset of $$X$$ are formed using elements of $$X$$.

    Considering $$\{a,b\}, a \in X$$ but $$b \notin X$$. Hence $$\{a,b\}$$ is not a subset of $$X$$.

    Considering $$\{b,c\}, b \notin X$$ and  $$c \notin X$$. Hence $$\{b,c\}$$ is not a subset of $$X$$.

    Considering $$\{c,d\}, c \notin X$$ and  $$d \in X$$. Hence $$\{c,d\}$$ is not a subset of $$X$$.

    Considering $$\{a,d\}, a \in X$$ and  $$d \in X$$. Hence $$\{a,d\}$$ is a subset of $$X$$.
  • Question 8
    1 / -0
    If $$A=\left \{ 5,\left \{ 5,6 \right \},7 \right \}$$, which of the following is correct? 
    Solution
    If $$A =\{5,\{5,6\},7\}$$ then the elements of $$A$$ are $$5, \{5,6\},7$$

    Hence $$5\in A, \{5,6\}\in A$$ and $$ 7\in A$$

    Note: $$5\ne\{5\}$$
    5 is an element of $$A$$
    i.e, $$5\in A$$

    where as $$\{5\}$$ is a subset of $$A$$
    i.e, $$\{5\}\subset A$$
  • Question 9
    1 / -0
    The number of subsets of the set $$\left \{ 10,11,12 \right \}$$ is
    Solution
    No. of subsets$$ = 2^3=8$$.

    We have 2 choices with each of the elements: either put them in a subset or to not put them in a subset. Hence there are 8 subsets.
  • Question 10
    1 / -0
    If $$n(A) = 20, n(B) = 30$$ and $$n(A \cup B)= 40$$, then $$n(A \cap B)$$ is equal to: 
    Solution
    We know that for any two finite sets $$A$$ and $$B$$, $$n(A\cup B)=n(A)+n(B)-n(A\cap B)$$.

    Here, it is given that $$n(A)=20,n(B)=30$$ and $$n(A\cup B)=40$$, therefore,

    $$n(A\cup B)=n(A)+n(B)-n(A\cap B)\\ \Rightarrow 40=20+30-n(A\cap B)\\ \Rightarrow 40=50-n(A\cap B)\\ \Rightarrow n(A\cap B)=50-40\\ \Rightarrow n(A\cap B)=10$$

    Hence, $$n(A\cap B)=10$$

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