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Set Theory Test 9

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Set Theory Test 9
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  • Question 1
    1 / -0
    All the students of a batch opted Psychology, Business, or both. 73% of the students opted Psychology and 62% opted Business. If there are  220 students, how many of them opted for both Psychology and business?
    Solution
    By set theory

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

    $$=0.73+0.62-1.00=0.35$$

    $$35\%$$ of $$220 = 77$$
  • Question 2
    1 / -0
    In an examination $$80\%$$  passed in English, $$85\%$$ in Maths, $$75\%$$ in both and $$40$$ students failed in both subjects. Then the number of  students appeared are
    Solution
    $$n(E)=80$$
    $$n(M)=85$$
    $$n(E \cap M)=75$$
    $$n(E \cup M)=n(E)+n(M)-n(E \cap M)=80+85-75=90$$
    $$n(E \cup M)'=10$$
    Let n be the total number of students appeared
    $$\dfrac{10}{100} \times n=40$$
    $$\therefore n=400$$
  • Question 3
    1 / -0
    $$p\cap (q\cup r)=?$$
    Solution

  • Question 4
    1 / -0
    There are 19 hockey players in a club. On a  particular day 14 were wearing the prescribed hockey shirts, while 11 were wearing the  prescribed hockey pants. None of then was without hockey pant  or hockey shirt. How many of  them were in complete  hockey uniform?
    Solution
    We can look at it in 2 ways

    First by set theory

    $$n(A\cap B) = n(A) + n(B)-n(A\cup B)$$

    $$=14+11-19=6$$

    Qualitatively, we know that 14 people are wearing prescribed hockey shirts,which leaves us with 5 players who must be wearing hockey pants. So out of 11 players who are wearing hockey pants, 5 are not wearing hockey shirts while the other 6 are in complete uniform.
  • Question 5
    1 / -0
    In an examination, $$34\%$$ of  the candidates fail in Arithmetic and $$42\%$$ in  English. If $$20\%$$ fail in Arithmetic and English, the  percentage of those passing in both subjects is :
    Solution
    Let $$A$$ denote students fail in Arithmetic, $$B$$ denote students fail in English
    $$n(A)=34$$
    $$n(B)=42$$
    $$n(A \cap B)=20$$
    $$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 34+42-20=56$$
    $$n(A \cup B)' = 100 - n(A \cup B) = 100-56=44$$
  • Question 6
    1 / -0
    If $$A$$ and $$B$$ have some elements in common, then $$n(A \cup B)$$ is:
    Solution
    We know that $$n(A\cup B)+n(A\cap B) = n(A)+n(B)$$

    But $$n(A\cap B) >0$$ always
    $$\Rightarrow n(A)+n(B)=n(A\cup B)+n(A\cap B)>n(A\cup B)$$
    Hence, $$n(A\cup B)<n(A)+n(B)$$.
  • Question 7
    1 / -0
    The smallest set $$A$$ such that $$A\cup \left\{ 1,2 \right\} =\left\{ 1,2,3,5,9 \right\} $$ is 
    Solution
    $$A\cup \{1,2\}=\{1,2,3,5,9\}$$
    Thus 
    $$A=\{1,2,3,5,9\}-\{1,2\}$$

    Hence 
    $$A=\{3,5,9\}$$
  • Question 8
    1 / -0
    If $$A \cap B = \phi$$, then $$A \Delta B$$ =
    Solution
    We know that $$A\Delta B=A\cup B-A\cap B $$

    Given $$A\cap B=\phi $$ 

    $$\therefore A\Delta B=A\cup B-\phi=A\cup B $$

    Hence, the answer is $$A\cup B$$.
  • Question 9
    1 / -0
    The equation x + cosx $$=$$ a has exactly one positive root. Complete set of values of 'a' is

    Solution
    Let f(x) = x + cosx a
    $$\Rightarrow f'(x)=1-sinx\geq 0\forall x \epsilon R.$$
    Thus f(x) is increasing in $$\left ( -\infty ,\infty  \right )$$, as zero of f'(x) don't for an interval. f(0) = 1 a

    For a positive root, $$1-a< 0$$

    $$\Rightarrow a> 1$$
  • Question 10
    1 / -0
    In a science talent examination, 50% of the candidates fail in Mathematics and 50% fail in Physics. If 20% fail in both these subjects, then the  percentage who pass in both Mathematics and Physics is:
    Solution
    By set theory

    $$n(M\cup P) = n(M) + n(P)-n(M\cap P)$$         where M and P are sets of students failing in respective subjects.

    $$=0.5 + 0.5 - 0.2 = 0.8$$

    This indicates $$80\%$$ of the class fails in at least one of the given subjects while $$20\%$$ pass in both.
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