Set Theory Test...

Let $$A, B$$ and $$C$$ be sets such that $$\phi = A\cap B \subseteq C$$. Then which of the following statements is not true?

Let $$S$$ be a non-empty subset of $$R$$. Consider the following statement:
$$p$$ : There is a rational number $$x$$  such that $$x > 0$$.
which of the following statements is the negation of the statement P ? 

The set $$\displaystyle A=\left\{ x:x\in { x }^{ 2 }=16\quad and\quad 2x=6 \right\} $$ equals:

Let $$n$$ be a fixed positive integer. Let a relation $$R$$ defined on $$I$$ (the set of all integers) as follows: $$aRb$$ iff $$n/(a-b)$$, that is, iff $$a-b$$ is divisible by $$n$$, then, the relation $$R$$ is

If X $$=$$ (multiples of 2), Y $$=$$ (multiples of 5), Z $$=$$ (multiples of 10), then $$X \cap  ( Y \cap  Z )$$ is equal to

A $$\bigcup \phi = $$

Suppose $${ A }_{ 1 },{ A }_{ 2 },...,{ A }_{ 30 }$$ are thirty sets, each with five elements and $${ B }_{ 1 },{ B }_{ 2 },...,{ B }_{ 30 }$$ are $$n$$ sets ecah with three elements. Let $$\displaystyle \bigcup _{ i=1 }^{ 30 }{ { A }_{ i }= } \bigcup _{ j=1 }^{ n }{ { B }_{ j } } =S$$

If each element of $$S$$ belongs to exactly ten of the $${ A }_{ i }'s$$ and exactly none of the $${ B }_{ j }'s$$ then $$n=$$

In a locality two-thirds of the people have cable TV one-fifth have Dish TV and one-tenth have both What is the fraction of people having either cable TV or Dish TV?

Let $$P =$$ Set of all integral multiples of $$3 $$; $$Q =$$ Set of integral multiples of $$4 $$; $$R =$$ Set of all integral multiples of $$6$$. Consider the following relations :
$$1 $$ $$\displaystyle P\cup Q=R$$
$$2.$$ $$\displaystyle P\subset R$$
$$3.$$ $$\displaystyle R\subset \left ( P\cup Q \right )$$
Which of the relations given above is/are correct ?