Maths M.C.Q (1 Marks)

$ = \{ x:x \in A $ and $x \in {B^c}\} = A \cap {B^c}$.

$\therefore (A \cup B) \cap C = \{ 3,\,4,\,6\} $.

$A = (A \cap B) \cup (A - B)$ is correct.

$(3)$ is false.

$\therefore (1)$ and $(2)$ are true.

= $(A \cap {A^c}) \cup (A \cap {B^c})$ = $\phi \cup (A \cap {B^c}) = A \cap {B^c}$.

$\therefore A \cap B' = \{ 1,\,2,\,5\} \cap \{ 1,\,2,\,3,\,4,\,5,\,8,\,9,\,10\} = \{ 1,\,2,\,5\} = A$

Now, $n({(A \cup B)^C}) = n(U) - n(A \cup B) = 20 - 17 = 3$.

Thus power set of $A =\{ x , y \}$ is $\{\phi, x , y ,\{ x , y \}\}$

Now $n(C  \cup B) = n(C) + n(B) -n(C  \cap B) $

$= 20 + 50 -10 = 60.$

Hence, required number of persons $= 60\%.$

$n\,(M)$= Number of teachers in Maths

$n\,(P \cup M) = n(P) + n\,(M) - n\,(P \cap M)$

$20 = n\,(P) + 12 - 4$ ==> $n\,(P) = 12$.

Now, $n\,(M \cup P) = n\,(M) + n\,(P) - n\,(M \cap P)$

$100 = 55 + 67 - n\,(M \cap P)$

$\therefore n\,(M \cap P) = 122 - 100 = 22$

Now $n$ ($P$ only) =$n\,(P) - n(M \cap P)$$ = 67 - 22 = 45$.

We can see that, $2{( \pm 2)^2} + 3{( \pm 3)^2} = 35$ and $2{( \pm 4)^2} + 3{( \pm 1)^2} = 35$

$ \therefore (2, 3), (2, -3), (-2, -3), (-2, 3), (4, 1), (4, -1),$

$(-4, -1), (-4, 1)$ are $8$ elements of the set. 

$ \therefore  n = 8$.

$B = [x:x \in R:x - 1 \le - 1$ or $x - 1 \ge 1]$

= $[x:x \in R:x \le 0{\rm{ or }}x \ge 2]$

$\therefore A \cup B = R - D$, where $D = [x:x \in R,\,1 \le x < 2]$.

$ = {3^n}{ + ^n}{C_1}{3^{n - 1}}{ + ^n}{C_2}{3^{n - 2}} + .....{ + ^n}{C_{n - 1}}3{ + ^n}{C_n} - 3n - 1$

(${ = ^n}{C_0}={ ^n}{C_n},{^n}{C_1}$ = ${^n}{C_{n - 1}}$ etc.)

$ = 9{[^n}{C_2}{ + ^n}{C_3}(3) + .....{ + ^n}{C_n}{3^{n - 1}}]$

${4^n} - 3n - 1$ is a multiple of $9$ for $n \ge 2$.

For $n = 1,$ ${4^n} - 3n - 1$ = $4 - 3 - 1 = 0$,

For $n = 2,$ ${4^n} - 3n - 1$= $16 - 6 - 1 = 9$

${4^n} - 3n - 1$ is a multiple of $9$ for all $n \in N$

$X$ contains elements, which are multiples of $9$, and clearly $Y$ contains all multiples of $9$.

$X \subseteq Y$ i.e., $X \cup Y = Y$.

$ = {7^n}{ + ^n}{C_1}{7^{n - 1}}{ + ^n}{C_2}{7^{n - 2}} + .....{ + ^n}{C_{n - 1}}7{ + ^n}{C_n} - 7n - 1$

${ = ^n}{C_2}{7^2}{ + ^n}{C_3}{7^3} + ..{ + ^n}{C_n}{7^n}$,${(^n}{C_0}{ = ^n}{C_n},{\,^n}{C_1}{ = ^n}{C_{n - 1}}\,{\rm{etc}}{\rm{.)}}$

$ = 49{[^n}{C_2}{ + ^n}{C_3}(7) + ......{ + ^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 all $n \in N.$

$\therefore $ $X$ contains elements which are multiples of $49$ and clearly $Y$ contains all multiplies of $49$. $X \subseteq Y$.

Since, element in the union $S$ belongs to $10$ of $Ai$' s

Also, $O(S)$ = $O\left( {\bigcup\limits_{j\, = 1}^n {{B_j}} } \right) = \frac{{3n}}{9} = \frac{n}{3}$,

$\therefore$ $\frac{n}{3} = 15 \Rightarrow n = 45$.

$ = 100 - 90 = 10$.

$n\,(H \cap B) = 64,\,\,n(B \cap C) = 80$

$n(H \cap C) = 40$, $n(C \cap H \cap B) = 24$

$n\,({C^c} \cap {H^c} \cap {B^C}) = n\,[{(C \cup H \cup B)^c}]$

$ = n( \cup ) - n(C \cup H \cup B)$

$ = 800 - [n(C) + n(H) + n(B) - n(H \cap C)$

$ - n(H \cap B) - n(C \cap B) + n(C \cap H \cap B)]$

$ = 800 - [224 + 240 + 336 - 64 - 80 - 40 + 24]$

$ = 800 - 640 = 160$.

Let Population of American be $100$.

Then $n\,(A) = 63,n\,(B) = 76$

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

$ = 63 + 76 - n(A \cap B)$

 $\therefore n\,(A \cup B) + n(A \cap B) = 139$

==> $n\,(A \cap B) = 139 - n(A \cup B)$

But $n\,(A \cup B) \le 100$

$\therefore - n\,(A \cup B) \ge - 100$

 $\therefore 139 - n\,(A \cup B) \ge 139 - 100 = 39$

 $\therefore n(A \cap B) \ge 39$ i.e., $39 \le n(A \cap B)$.....(i)

Again, $A \cap B \subseteq A,A \cap B \subseteq B$

$ \therefore n\,(A \cap B) \le n\,(A) = 63$ and $n\,(A \cap B) \le n\,(B) = 76$

$\therefore n(A \cap B) \le 63$…..(ii)

Then, $39 \le n\,(A \cap B) \le 63$ ==> $39 \le x \le 63$.

$n\,({P^c} \cap {C^c}) = 65\% ,\,\,n(P \cap C) = 2000$

Since, $n\,({P^c} \cap {C^c}) = 65\% $

$\therefore$ $n\,{(P \cup C)^c} = 65\% $ and $n(P \cup C) = 35\% $

Now, $n(P \cup C) = n(P) + n(C) - n(P \cap C)$

$35 = 25 + 15 - n(P \cap C)$

$\therefore$ $n(P \cap C) = 40 - 35 = 5$. Thus $n\,(P \cap C) = 5\% $

But $n\,(P \cap C) = 2000$

$\therefore$  Total number of families $ = \frac{{2000 \times 100}}{5} = 40,000$

Since, $n(P \cup C) = 35\% $

and total number of families = $40,000$

and $n(P \cap C) = 5\% $. $\therefore$  $(2)$ and $(3)$ are correct.

$\Rightarrow n(A \cup B)=n(A \cup C)$

$\Rightarrow n(A)+n(B)-n(A \cap B)$

$=n(A)+n(C)-n(A \cap C)$

$B-A \geq 1$

$n(A \Delta B)=n((A-B) \cup(B-A)) \geq 1$

minimum value $=1$

$2 n(A)-n(B)=n(A B)$

$\mathrm{n}(\mathrm{A})+3 \mathrm{n}(\mathrm{B})=5 \mathrm{n}(\mathrm{AB})$

$\Rightarrow \mathrm{n}(\mathrm{A})+3 \mathrm{n}(\mathrm{B})=10 \mathrm{n}(\mathrm{A})-5 \mathrm{n}(\mathrm{B})$

$8 n(\mathrm{B})=9 \mathrm{n}(\mathrm{A})$

$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{B})}=\frac{8}{9}$

$\mathrm{n}(\mathrm{A})=8 \mathrm{k} ; \mathrm{n}(\mathrm{B})=9 \mathrm{k}$

$\Rightarrow \mathrm{n}(\mathrm{AB})=7 \mathrm{k}$

$\mathrm{n}(\mathrm{A} \cup \mathrm{B})=10 \mathrm{k} \leq 10$

$\mathrm{k}=1$