Try taking some values of A and B
We'll see that  A - B  = A $\cap$ $\overline{B}$ . 

n(A $\cup$ B) = n(A) + n(B) - n(A $\cap B$) = 3 + 6 - 1 $(A\cap B)$

Since maximum number of elements in A $\cap$ B = 3

$\therefore$ Minimum number of elements in A $\cup$ B = 9 - 3 = 6 .

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

We have, 70 = 37 + 52 - $n(A\cap B)$
$n(A\cap B)$ = 19.

Since every rectangle, rhombus and square is parallelogram so

$F_1$=$F_2\cup F_3\cup F_4$

$x^2=16\Rightarrow x=\pm 4$
$2x=6\Rightarrow x=3$

There is no value of $x$ which satisfies both the given equations. The set A is an empty set or a null set.

Thus, A = {}.

n(A) = 40% of 10,000 = 4,000

n(B) = 20% of 10,000 = 2,000

n(C) = 10% of 10,000 = 1,000

n(A ∩ B) = 5% of 10,000 = 500

n(B ∩ C) = 3% of 10,000 = 300

n(C ∩ A) = 4% of 10,000 = 400

n(A ∩ B ∩ C) = 2% of 10,000 = 200

We want to find the number of families which buy only A = n(A) - [n(A ∩ B) + n(A ∩ C) - n(A ∩ B ∩ C)]

=4000 - [500 + 400 - 200] = 4000 - 700 = 3300

Intelligence cannot be defined for students in a class. Hence, the group of intelligent students is not a well defined collection.

Number of subsets of A = ${}^n{C}_0$ + ${}^n{C}_1$  + .............+ ${}^n{C}_n$ = $2^n$

Number of subsets of A = ${}^n{C}_0$ + ${}^n{C}_1$  + .............+ ${}^n{C}_n$ = $2^n$

$x^2=16\Rightarrow x=\pm 4$
$2x=6\Rightarrow x=3$

There is no value of $x$ which satisfies both the given equations. The set A is an empty set or a null set.

Thus, A = {}.