Sets Test

Result

Sets Test - 31

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Weekly Quiz Competition

Question 1
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.

A
$$n(A)$$

B
$$n(B)$$

C
$$n(A+B)$$

D
None of these

Solution
Question 2
1 / -0

Solve the following inequalities: $$\displaystyle\frac{1}{2-|x|}\geq 1$$.

A
$$x\epsilon (-8, -1]\cup [1,2)$$

B
$$x\epsilon (-2, -1]\cup [1,2)$$

C
$$x\epsilon (-2, -1]\cup [1,6)$$

D
$$x\epsilon (-2, 0]\cup [1,2)$$

Solution
Question 3
1 / -0

Which one of the following is correct?

A
$$\left \{ x:x^{2}=-1,x\in Z \right \}=\phi$$

B
$$\phi=0$$

C
$$\phi=\left \{ 0 \right \}$$

D
$$\phi=\left \{ \phi \right \}$$

Solution

$$x^2=-1$$ is satisfied by $$i $$ and $$-i$$ none of which are integers. Hence required set is null.
Question 4
1 / -0

Which one of the following is a finite set?

A
$$\left \{ x:x\in Z,x< 5 \right \}$$

B
$$\left \{ x:x\in W,x\geq 5 \right \}$$

C
$$\left \{ x:x\in N,x> 10 \right \}$$

D
{ $$x:x$$ is an even prime number }

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 5
1 / -0

If $$X=\left \{ a,\left \{ b,c \right \},d \right \}$$, which of the following is a subset of $$X$$?

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

B
$$\left \{ b,c \right \}$$

C
$$\left \{ c,d \right \}$$

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

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 6
1 / -0

If $$A=\left \{ 5,\left \{ 5,6 \right \},7 \right \}$$, which of the following is correct?

A
$$\left \{ 5,6 \right \}\in A$$

B
$$\left \{ 5 \right \}\in A$$

C
$$\left \{ 7 \right \}\in A$$

D
$$\left \{ 6 \right \}\in A$$

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 7
1 / -0

The number of subsets of the set $$\left \{ 10,11,12 \right \}$$ is

A
$$3$$

B
$$8$$

C
$$6$$

D
$$7$$

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 8
1 / -0

If $$n(A) = 20, n(B) = 30$$ and $$n(A \cup B)= 40$$, then $$n(A \cap B)$$ is equal to:

A
$$50$$

B
$$10$$

C
$$40$$

D
$$70$$

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, $$n(A\cap B)=10$$
Question 9
1 / -0

For any three sets , A B and C, $$B\setminus (A \cup C)$$ is:

A
$$(A\setminus B)\cap (A\setminus C)$$

B
$$(B\setminus A)\cap (B \setminus C)$$

C
$$(B \setminus A)\cap (A\setminus C)$$

D
$$(A\setminus B)\cap (B \setminus C)$$

Solution

In the problem statement we are taking union of $$A$$ and $$C$$ and then taking its difference from $$B$$
This means $$B$$ \ $$ (A\cup C)$$ will contain elements that are in $$B$$ but not in both $$B$$ $$\textbf{and}$$ $$C$$.
$$\therefore B$$ \ $$ (A\cup C) $$ is equivalent to taking difference of $$A,B$$ and $$A,C$$ and then taking there intersection i.e. $$(B$$ \ $$ A)\cap (B$$ \ $$ C)$$
Hence option ($$b$$) is correct.
Question 10
1 / -0

If $$A= \{ p, q, r, s \}$$, $$B = \{ r, s, t, u \}$$, then $$A /B$$ is:

A
{ p, q }

B
{ r, s }

C
{ t, u }

D
{p, q, t, u }

Solution

The given sets are $$A=\left\{ p,q,r,s \right\}$$ and $$B=\left\{ r,s,t,u \right\}$$

$$A/setminus B$$ have those elements that belongs to set $$A$$ but does not belongs to set $$B$$.

Let us find $$A/setminus B$$ as follows:

$$Asetminus B=\left\{ p,q,r,s \right\} /setminus \left\{ r,s,t,u \right\} =\left\{ p,q \right\}$$

Hence, $$A/setminus B=\left\{ p,q \right\}$$

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Re-Attempt Weekly Quiz Competition

Sets Test - 31

Result

Sets Test - 31

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SHARING IS CARING

Question 1

$$n(A)$$

$$n(B)$$

$$n(A+B)$$

None of these

Solution

Question 2

$$x\epsilon (-8, -1]\cup [1,2)$$

$$x\epsilon (-2, -1]\cup [1,2)$$

$$x\epsilon (-2, -1]\cup [1,6)$$

$$x\epsilon (-2, 0]\cup [1,2)$$

Solution

Question 3

$$\left \{ x:x^{2}=-1,x\in Z \right \}=\phi$$

$$\phi=0$$

$$\phi=\left \{ 0 \right \}$$

$$\phi=\left \{ \phi \right \}$$

Solution

Question 4

$$\left \{ x:x\in Z,x< 5 \right \}$$

$$\left \{ x:x\in W,x\geq 5 \right \}$$

$$\left \{ x:x\in N,x> 10 \right \}$$

{ $$x:x$$ is an even prime number }

Solution

Question 5

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

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

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

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

Solution

Question 6

$$\left \{ 5,6 \right \}\in A$$

$$\left \{ 5 \right \}\in A$$

$$\left \{ 7 \right \}\in A$$

$$\left \{ 6 \right \}\in A$$

Solution

Question 7

$$3$$

$$8$$

$$6$$

$$7$$

Solution

Question 8

$$50$$

$$10$$

$$40$$

$$70$$

Solution

Question 9

$$(A\setminus B)\cap (A\setminus C)$$

$$(B\setminus A)\cap (B \setminus C)$$

$$(B \setminus A)\cap (A\setminus C)$$

$$(A\setminus B)\cap (B \setminus C)$$

Solution

Question 10

{ p, q }

{ r, s }

{ t, u }

{p, q, t, u }

Solution

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