Answer: Option A. -> 80
:
A
Let the no. of people who do not play NFS be n(N^I) = 50 (Given)
Similarlyno. of people who do not play Dota be n(D^I) = 40 (Given)
And theno. of people who do not play any gamen(N^I ∩ D^I) = 10 (Given)
We have to find the no. of people who do not play both the games = n(N∩ D)^{I
We know from the Demorgan's law}
(A∩ B)^{I =} A^I ∪B^I
So,n(N∩ D)^{I =} n(N^I) ∪n(D^I)
n(N^I) ∪n(D^I)= n(N^I) +n(D^I)-n(N^I ∩ D^I)
n(N^I) ∪n(D^I) = 50 + 40 - 10
= 80

Answer: Option C. -> A ∩ B = ∅
:
C
Since y = $\frac{1}{x}$, y = -x meet when -x = $\frac{1}{x}$ ⇒$x^2$ = -1,
which does not give any real value of x.
Hence, A ∩ B =∅.

Answer: Option C. -> 19
:
C
$n(A\cup B)$ = $n\left(A\right)+n\left(B\right)-n(A\cap B)$
We have, 70 = 37 + 52 - $n(A\cap B)$
$n(A\cap B)$ = 19.

Answer: Option A. -> 60
:
A
$n(S\cup F)$ = $n\left(S\right)+n\left(F\right)-n(S\cap F)$
$\Rightarrow n(S\cup F)$ = 20 + 50 -10 = 60

Answer: Option A. -> A = {}
:
A
$x^2=16\Rightarrow x=\pm 4$
$2x=6\Rightarrow x=3$
There is no value of $x$ which satisfies both the givenequations. The set A is an empty set or a null set.
Thus, A = {}.

Answer: Option B. -> F2∪F3∪F4
:
B
Since every rectangle, rhombus and square is parallelogram so
$F_1$=$F_2\cup F_3\cup F_4$

Answer: Option B. -> 10
:
B
Number of candidates who had none of the three = Total number of candidates - number of candidates who had at least one of three devices.
Total number of candidates = 200 =n(U), where U is the universal set
Number of candidates who had at least one of the three $=n(A\cup B\cup C)$, where A is the set of those who have a two wheeler, B the set of those who have a credit card and C the set of those who have a mobile phone.
We know that $n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(C\right)$
$-\left[n\right(A\cap B)+n(B\cap C)+n(C\cap A\left)\right]+n(A\cap B\cap C)$
Therefore, $n(A\cup B\cup C)=100+70+140-\{40+30+60\}+10$
Or $n(A\cup B\cup C)=190$.
As 190 candidates who attended the interview had at least one of the three gadgets, $n\left(U\right)-n(A\cup B\cup C)$=200 - 190 = 10 candidates had none of three.

Answer: Option D. -> Four points
:
D
A = Set of all values (x,y) : $x^2+y^2=25=5^2$

B = $\frac{x^2}{144}+\frac{y^2}{16}=1$ i.e., $\frac{x^2}{(12{)}^2}+\frac{y^2}{(4{)}^2}=1$
Clearly , A $\cap$ B consists of four points.

Answer: Option B. -> 6
:
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 .

Answer: Option A. -> X ⊆ Y
:
A
Since $8^n-7n-1=(7+1{)}^n-7n-1$
= $7^n+{}^n{C}_1{7}^{n-1}+{}^n{C}_2{7}^{n-2}+.....+{}^n{C}_{n-1}7+{}^n{C}_n-7n-1$
= ${}^n{C}_2{7}^2+{}^n{C}_3{7}^3+...+{}^n{C}_n{7}^n,({}^n{C}_0={}^n{C}_n{}^n{C}_1={}^n{C}_{n-1}etc,)$
= 49[${}^n{C}_2+{}^n{C}_3\left(7\right)+........+{}^n{C}_n{7}^{n-2}]$
$\therefore$ $8^n-7n-1$ is a multiple of 49 for n $\ge$ 2
For n = 1 , $8^n-7n-1=8-7-1=0;$
For n = 2, $8^n-7n-1=64-14-1=49$
$\therefore$ $8^n-7n-1$ is a multiple of 49 for n $\ge$ N
$\therefore$ X contains elements which are multiples of 49 and clearly$\gamma$
contains all multiplies of 49. $\therefore$ X $\subseteq$ Y.