Sets Test

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Sets Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

In a city, three daily newspapers A, B, C are published, $$42\%$$ read A; $$51\%$$ read B; $$68\%$$ read C; $$30\%$$ read A and B; $$28\%$$ read B and C; $$36\%$$ read A and C; $$8\%$$ do not read any of the three newspapers.
What is the percentage of persons who read only one paper ?

A
$$38\%$$

B
$$48\%$$

C
$$51\%$$

D
None of the above.

Solution

Let total numbers of people be 100, So for newspapers
$$n(A)=42, n(B)=51, n(C)=68, n(A\cap B)=30$$
$$ n(B\cap C)=28$$
$$ n(A\cap C)=36$$

Also given, $$ n(\overline { A\cup B\cup C) } =8$$
$$ n(A\cup B\cup C)=100-n(\overline { A\cup B\cup C) } =100-8=92$$
Also,
$$ n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$$
$$92=42+51+68-30-28-36+n(A\cap B\cap C)$$
$$n(A\cap B\cap C)=25$$
Now,
People who read only one paper $$=n(A)+n(B)+n(C)-2n(A\cap B)-2n(B\cap C)-2n(A\cap C)+3n(A\cap B\cap C)$$
$$=42+51+68-60-56-72+75$$
$$=48$$
Hence, B is correct.
Question 2
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If $$a\mathbb{N}=(an:n\in \mathbb{N})$$ and $$b\mathbb{N}\cap c\mathbb{N}=d\mathbb{N}$$, where $$a,b,c\in \mathbb{N}$$ and $$b,c$$ are coprime, then

A
$$b=cd$$

B
$$c=bd$$

C
$$d=bc$$

D
None of these

Solution

Given : $$a\mathbb{N}=(an:n\in \mathbb{N})$$ and $$b\mathbb{N}\cap c\mathbb{N}=d\mathbb{N}$$, where $$a,b,c\in \mathbb{N}$$ and $$b,c$$ are coprime
We have,
$$b\mathbb{N}=(bn:n\in \mathbb{N})$$, $$c\mathbb{N}=(cn:n\in \mathbb{N})$$
and $$d\mathbb{N}=(dn:n\in \mathbb{N})$$
Also, $$bc\in b\mathbb{N} \cap c\mathbb{N}=d\mathbb{N}$$
$$\therefore$$ $$bc=d$$ ..... $$[\because b,c$$ are coprime$$]$$.
Question 3
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If A and B are finite sets and $$A \subset B$$, then

A
$$n(A \cap B) = \phi$$

B
$$n(A \cup B) = n(B)$$

C
$$n(A \cap B) = n (B)$$

D
$$n(A \cup B) = n (A)$$

Solution

A and B are finite set and $$A\subset B$$
$$A\subset B$$ means A is a subset of B
If A is subset of B, then all elements of A are present in B
$$\therefore n\left( A\bigcup { B } \right) =n\left( B \right) $$
Question 4
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A college awarded $$38$$ medals in football, $$15$$ in basketball and $$20$$ in cricket. If these medals went to a total of $$58$$ men and only three men got medals in all the three sports. Then the number of students who received medals in exactly two of the three sports, is

A
$$18$$

B
$$15$$

C
$$9$$

D
$$6$$

Solution

$$\textbf{Step – 1: Formulating the information to use Inclusion-Exclusion Rule.}$$
  $$\text{Let number of men who have got medals in football, basketball, and cricket be represented as}$$
$$n(F),n(B)$$ and $$n(C)$$.
   $$\text{Therefore the number of men who won in all the three sports will be}$$ $$n(F \cap B \cap C)$$
    $$\text{and the number of people who got medals in two of the three sports is}$$
$$n(F \cap B) + n(B \cap C) + n(C \cap F)$$
  $$\text{and the total number of men is represented as}$$ $$n(F \cup B \cup C)$$.
$$\text{From the given data we have:}$$
$$n(F) = 38$$
$$n(B) = 15$$
$$n(C) = 20$$
$$n(F \cup B \cup C) = 58$$
$$n(F \cap B \cap C) = 3$$
$$\textbf{Step – 2: Apply the Inclusion-Exclusion Rule}$$.
  $$\text{Using the Inclusion-Exclusion Rule we get:}$$
$$n(F \cup B \cup C) = n(F) + n(B) + n(C) - n(F \cap B) - n(B \cap C) - n(C \cap F) + n(F \cap B \cap C)$$
$$\Rightarrow 58 = 38 + 15 + 20 - n(F \cap B) - n(B \cap C) - n(C \cap F) + 3 $$
$$\Rightarrow n(F \cap B) + n(B \cap C) + n(C \cap F) = 38 + 15 + 20 + 3 - 58 $$
$$\Rightarrow n(F \cap B) + n(B \cap C) + n(C \cap F) = 18 $$
  $$\therefore\text{ the number of students who have received medals in exactly two of the three sports is}$$
$${= n(F \cap B) + n(B \cap C) + n(C \cap F) - 3 \times n(F \cap B \cap C) }$$
$${= 18 - 3 \times 3 }$$
$${= 18 - 9}$$
$${= 9}$$
$$\textbf{Hence, the number of students who have received medals in exactly two of the three sports is 9}$$
Question 5
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The set $$(A\cup B\cup C) \cap (A\cap B'\cap C')\cap C'$$ is equal to?

A
$$B\cap C'$$

B
$$A\cap C$$

C
$$B' \cap C'$$

D
None of these

Solution

$$(A\cup B\cup C) \cap (A\cap B'\cap C')\cap C'$$
$$=(A\cup B\cup C) \cap (A'\cap B\cap C)\cap C'$$
$$= [(A\cap A')\cup (B\cup C)]\cap C'$$
$$= (\phi \cup B\cup C)\cap C' = (B\cup C)\cap C'$$
$$= (B\cap C') \cup (C\cap C')$$
$$= (B\cap C') \cup \phi = B\cap C'$$.
Question 6
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If $$A/B=\left\{ a,b \right\} ,B\setminus A=\left\{ c,d \right\} $$ and $$A\cap B=\left\{ e,f \right\} $$ then the set $$B$$ is equal to

A
$$\left\{ a,b,c,d \right\} $$

B
$$\left\{ e,f,c,d \right\} $$

C
$$\left\{ a,b,e,f \right\} $$

D
$$\left\{ c,d,a,e \right\} $$

E
$$\left\{ d,e,a,b \right\} $$

Solution

Given, $$A/B=\left\{ a,b \right\} ,B\setminus A=\left\{ c,d \right\} $$

and $$A\cap B=\left\{ e,f \right\} $$

$$\therefore A-B=\left\{ a,b \right\} ...(i)\quad $$
$$B-A=\left\{ c,d \right\} ...(ii)$$

$$A\cap B=\left\{ e,f \right\} ...(iii)$$
From Eqs. (i),(ii),(iii) we get

$$B=\left\{ c,d,e,f \right\} $$
Question 7
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Let $$X$$ and $$Y$$ be two non-empty sets such that $$X\cap A=Y\cap A=\phi$$ and $$X\cup A=Y\cup A$$ for some non-empty set $$A$$. Then which of the following is true?

A
$$X$$ is a proper subset of $$Y$$

B
$$Y$$ is a proper subset of $$X$$

C
$$X=Y$$

D
$$X$$ and $$Y$$ are disjoint sets

E
$${ X }/{ A }=\phi $$

Solution

We have, $$X\cup A=Y\cup A$$
$$\Rightarrow X\cap \left( X\cup A \right) =X\cap \left( Y\cup A \right)$$
$$\Rightarrow X=\left( X\cap Y \right) \cup \left( X\cap A \right) $$ $$\left[ \because X\cap \left( X\cup A \right) =X \right] $$
$$\Rightarrow X=\left( X\cap Y \right) \cup \phi $$ $$\left[\because X\cap A=\phi \right]$$
$$\Rightarrow X=X\cap Y$$ .....(i)
Again, $$X\cup A=Y\cup A$$
$$\Rightarrow Y\cap \left( X\cup A \right) =Y\cap \left( Y\cup A \right)$$
$$\Rightarrow \left( Y\cap X \right) \cup \left( Y\cap A \right) =Y$$
$$\Rightarrow \left( Y\cap X \right) \cup \phi =Y$$
$$\Rightarrow Y\cap X=Y$$
$$\Rightarrow X\cap Y=Y$$ ....(ii)
From equations (i) and (ii), we get
$$X=Y$$
Question 8
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If $$A = \left \{(x, y) : x^{2} + y^{2} \leq 1; x, y \epsilon R\right \}$$ and $$B = \left \{(x, y) : x^{2} + y^{2} \geq 4; x, y \epsilon R\right \}$$, then

A
$$A - B = \phi$$

B
$$B - A = \phi$$

C
$$A\cap B \neq \phi$$

D
$$A\cap B = \phi$$

Solution

$$A$$ is the set of all points on the inner circle $$x^{2} + y^{2} = 1. B$$ is the set of all points on the outer circle $$x^{2} + y^{2} = 4$$.
$$\therefore A - B = A, B - A = B, A\cap B = \phi$$.
Question 9
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If $$X = \left \{1, 2, 3, ..., 10\right \}$$ and $$A = \left \{1, 2, 3, 4, 5\right \}$$. Then, the number of subsets $$B$$ of $$X$$ such that $$A - B = \left \{4\right \}$$ is

A
$$2^{5}$$

B
$$2^{4}$$

C
$$2^{5} - 1$$

D
$$1$$

E
$$2^{4} - 1$$

Solution

Given, $$X = \left \{1, 2, 3, ...., 10\right \}$$
and $$A = \left \{1, 2, 3, 4, 5\right \}$$
Now, $$A - B = \left \{4\right \}$$
$$\Rightarrow B = A - \left \{4\right \} = \left \{1, 2, 3, 4, 5\right \} - \left \{4\right \}$$
$$= \left \{1, 2, 3, 5\right \}$$
$$\therefore X - \left \{4\right \} - B$$
$$= \left \{1, 2, 3, ..., 10\right \} - \left \{4\right \} - \left \{1, 2, 3, 5\right \}$$
$$= \left \{6, 7, 8, 9, 10\right \}$$
Therefore,number of subsets $$B$$ of $$X$$ i.e., $$\left \{6, 7, 8, 9, 10\right \}$$ $$= 2^{5}$$.
Question 10
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If $$n(A)=10,n(B)=6$$ and $$(C)=5$$ for three disjoint sets $$A,B,C$$ then $$n(A\cup B\cup C)$$ equals

A
$$21$$

B
$$11$$

C
$$1$$

D
$$9$$

Solution

Since, $$A,B,C$$ are disjoint sets

$$\therefore n(A\cup B\cup C)=n(A)+n(B)+n(C)$$

$$=10+6+5=21$$

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Re-Attempt Weekly Quiz Competition

Sets Test - 33

Result

Sets Test - 33

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SHARING IS CARING

Question 1

$$38\%$$

$$48\%$$

$$51\%$$

None of the above.

Solution

Question 2

$$b=cd$$

$$c=bd$$

$$d=bc$$

None of these

Solution

Question 3

$$n(A \cap B) = \phi$$

$$n(A \cup B) = n(B)$$

$$n(A \cap B) = n (B)$$

$$n(A \cup B) = n (A)$$

Solution

Question 4

$$18$$

$$15$$

$$9$$

$$6$$

Solution

Question 5

$$B\cap C'$$

$$A\cap C$$

$$B' \cap C'$$

None of these

Solution

Question 6

$$\left\{ a,b,c,d \right\} $$

$$\left\{ e,f,c,d \right\} $$

$$\left\{ a,b,e,f \right\} $$

$$\left\{ c,d,a,e \right\} $$

$$\left\{ d,e,a,b \right\} $$

Solution

Question 7

$$X$$ is a proper subset of $$Y$$

$$Y$$ is a proper subset of $$X$$

$$X=Y$$

$$X$$ and $$Y$$ are disjoint sets

$${ X }/{ A }=\phi $$

Solution

Question 8

$$A - B = \phi$$

$$B - A = \phi$$

$$A\cap B \neq \phi$$

$$A\cap B = \phi$$

Solution

Question 9

$$2^{5}$$

$$2^{4}$$

$$2^{5} - 1$$

$$1$$

$$2^{4} - 1$$

Solution

Question 10

$$21$$

$$11$$

$$1$$

$$9$$

Solution

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