Sets and Relations

In a survey of 1003 families in Calcutta, we have the following information:

63 families have neither a radio nor a TV.

794 families have a radio.

187 families have a TV.

We need to calculate how many families have both a radio and a TV.

Step-by-step Solution:

Calculate the total number of families with either a radio or a TV. Using the formula for the complement set:

$ n(R \cup T) + n(R \cup T)' = \text{Total families} $

$ n(R \cup T) + 63 = 1003 $

Solving for $ n(R \cup T) $:

$ n(R \cup T) = 1003 - 63 = 940 $

Apply the principle of inclusion-exclusion from set theory:

$ n(R \cup T) = n(R) + n(T) - n(R \cap T) $

Substitute the known values:

$ 940 = 794 + 187 - n(R \cap T) $

Simplify to find $ n(R \cap T) $:

$ 940 = 981 - n(R \cap T) $

$ n(R \cap T) = 981 - 940 = 41 $

Therefore, 41 families have both a radio and a TV.

To determine the set $A \cap B$, which represents the common elements between sets $A$ and $B$, we need to find the multiples of both 8 and 12. In other words, we need to find the least common multiples (LCM) of 8 and 12.

First, let's find the prime factorization of 8 and 12:

- 8 can be factored into prime numbers as: $8 = 2^3$

- 12 can be factored into prime numbers as: $12 = 2^2 \cdot 3$

To find the LCM, we take the highest power of each prime that appears in these factorizations. Thus, the LCM of 8 and 12 is:

$\text{LCM}(8, 12) = 2^3 \cdot 3 = 8 \cdot 3 = 24$

Therefore, the multiples of 8 and 12 are common multiples of 24. Consequently, the set $A \cap B$, which consists of numbers that are multiples of both 8 and 12, consists of multiples of 24.

The correct option is:

Let x $ \in $ A $\cap$ B. Then,

x $\in$ A and x $\in$ B

$\Rightarrow$ x is a multiple of 4 and x is a multiple of 6

$\Rightarrow$ x is a multiple of 4 and 6 both

$\therefore$ x is a multiple of 12.

Let R = {(a, b) : a + b is an even integer a, b $\in$ Z}

For a $\in$ Z, a + a = 2a is an even integer.

$\therefore$ (a, a) $\in$ R $\forall$ a $\in$ Z

$\therefore$ R is reflexive.

Let (a, b) $\in$ R $\Rightarrow$ a + b is an even integer.

$\Rightarrow$ (b + a) is an even integer $\Rightarrow$ (b, a) $\in$ R

$\Rightarrow$ R is symmetric.

Let (a, b), (b, c) $\in$ R

$\Rightarrow$ (a + b) and (b + c) are even integers.

$\Rightarrow$ (a + b) + (b + c) = (a + c + 2b) is an even integer.

$\Rightarrow$ (a + c + 2b) $-$ 2b = (a + c) is an even integer.

$\Rightarrow$ (a, c) $\in$ R is transitive.

$\therefore$ R is an equivalence relation.

Let R = {(a, b) : (a $-$ b) is an even integer, a, b $\in$ Z}.

For a $\in$ Z, a $-$ a = 0 is an even integer.

$\therefore$ (a, a) $\in$ R $\forall$ a $\in$ Z

$\therefore$ R is reflexive.

Let (a, b) $\in$ R $\Rightarrow$ (a $-$ b) is an even integer.

$\Rightarrow$ $-$ (a $-$ b) is an even integer.

$\Rightarrow$ (b $-$ a) is an even integer $\Rightarrow$ (b, a) $\in$R

$\Rightarrow$ R is symmetric.

Let (a, b), (b, c) $\in$R $\Rightarrow$ (a $-$ b), (b $-$ c) are even integers.

$\Rightarrow$ (a $-$ b) + (b $-$ c) is an even integer.

$\Rightarrow$ (a $-$ c) is an even integer.

$\Rightarrow$ (a, c) $\in$R

$\Rightarrow$ R is transitive.

$\therefore$ R is an equivalence relation.

Let R = {(a, b) : a < b, a, b$\in$Z}

Let a $\in$Z . a < a is false.

$\therefore$ R is not reflexive.

$\therefore$ R is not an equivalence relation.

Let R = {(a, b) : a = b, a, b $\in$Z}.

It is quite easy to check that R is an equivalence relation.

Given that, $A = \{ x:{x^2} = 1\} = \{ - 1,1\} $ and $B = \{ x:{x^4} = 1\} = \{ - 1,1, - i,i\} $

Now, $A - B = \phi $ and $B - A = \{ - i,i\} $

$\therefore$ $A\Delta B = (A - B) \cup (B - A) = \{ - i,i\} $

We know that,

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

$\therefore$ P = 1500 $-$ n(A $\cap$ B)

$\Rightarrow$ n(A $\cap$ B) = 1500 $-$ P

Clearly, 1 $\le$n(A $\cap$ B) $\le$ 500

$\Rightarrow$ 1 $\le$ 1500 $-$ P $\le$ 500

$\Rightarrow$ $-$1499 $\le$ $-$P $\le$ $-$1000

$\Rightarrow$ 1000 $\le$ P $\le$ 1499