Maths M.C.Q (1 Marks)

Option (b) is not correct. Also $\{\emptyset\}$ is subset of a set

Option (C) is not correct. The set of an element of given set is subset of given set.

==> ${2^n}({2^{m - n}} - 1) = {2^3} \times 7$, $\therefore $ $n = 3$ and ${2^{m - n}} = 8 = {2^3}$

==> $m - n = 3$ ==> $m - 3 = 3$ ==> $m = 6$; $\therefore \,\,m = 6,\,\,n = 3$.

Hence, $\,A \cap B = \phi $.

==> ${e^{2x}} = 1,\,\,\therefore x = 0,y = 1$

$\therefore A$ and $B$ meet on $(0, 1), $

$\therefore A \cap B = \phi $.

[$\because 5$ and $7$ are relatively prime numbers].

$7N = \{ x \in N:x$ is a multiple of $7\}$

$\therefore 3N \cap 7N$$ = \{ x \in $ is a multiple of $3$ and $7\}$

$ = \{ x \in N:x$ is a multiple of $3$ and $7\}$

$ = \{ x \in N:x$ is a multiple of $21\}=21N.$

$B = \{ 6,\,12,\,18,\,24,\,30,\,....\} $

$\therefore A \cap B = \{ 12,\,24,\,....\}  =    \{x : x$ is a multiple of $12\}.$

$= \{12, 24, 36......\} =$ ${N_{12}}$.

Trick : ${N_3} \cap {N_4} = {N_{12}}$

[ $\because $ $3, 4$ are relatively prime numbers].

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

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

$\therefore $ $bN \cap cN$= the set of positive integral multiples of $bc$
= $b \subset N$, $[ are\, prime]$

$\therefore $ $d = bc$.

= $n(U) - [n(A) + n(B) - n(A \cap B)]$

 $= 700 -[200 + 300 -100] = 300.$

$ = (A \cap A') \cap B'$, (by associative law)

$ = \phi \cap B'$,$(\because A \cap A' = \phi )$

$ = \phi $.

Clearly, $\{ (A - B) \cup (B - C) \cup (C - A)\} ' = A \cap B \cap C$.

$=\mathrm{n}(\mathrm{u})-\mathrm{n}(\mathrm{A} \cup \mathrm{B})$

$=600-(\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B}))$

$=600-100-200+50=350$

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

$n(C) = 10\% \  of \ 10,000 = 1,000$

$n (A \cap B)$ $= 5\% \ of\ 10,000 = 500$

$n (B \cap C)$ $= 3\% \ of\ 10,000 = 300$

$n(C \cap  A)$ $= 4\% \ of \ 10,000 = 400$

$n(A \cap B \cap C)$ $= 2\% \ of \ 10,000 = 200$

We want to find $n(A \cap B^c \cap C^c) = n[A \cap (B \cap C)^c]$

$= n(A) -n[A \cap (B \cup C)] = n(A) -n[(A \cap B) \cup (A \cap C)]$

$= n(A) -[n(A \cap B) + n(A \cap C) -n(A \cap B \cap C)]$

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

$n(M  \cap P) = 12, n(M  \cap C)= 9, n(P  \cap C)= 7 ,$

$n(M  \cap P  \cap C) = 4 $

We have to find $n(M  \cap P'  \cap C'), n(P  \cap M '  \cap C' ),n ( C  \cap M'  \cap P') $

$= n(M)-n(M  \cap (P  \cup C))$

$= n(M) - n[(M \cap P) \cup (M \cap C)]$

$= n(M) -n(M  \cap P)-n(M  \cap C) + n(M  \cap P  \cap C)$

$= 23 -12 -9 + 4 = 27 -21 = 6$

$= n(P)-n[P  \cap (M  \cup C)] = $ $n(P) - n[(P \cap M) \cup (P \cap C)]$

$= n(P) -n(P  \cap M) -n(P  \cap C) + n(P  \cap M  \cap C) $

$= 24 -12 -7 + 4 = 9$

$n(C \cap M' \cap P') = n(C) - n(C \cap P) - n(C \cap M) + n(C \cap P \cap M)$

$= 19 -7 -9 + 4 = 23 -16 = 7.$

So total number of student is $=6+9+7=22$

Then we are given $n\,(B) = 21,\,n\,(H) = 26,n\,(F) = 29$

$n\,(H \cap B) = 14$, $n\,(H \cap F) = 15$, $n\,(F \cap B) = 12$
and $n\,(B \cap H \cap F) = 8$.

We have to find $n\,(B \cup H \cup F)$.

To find this, we use the formula

$n\,(B \cup H \cup F) = n\,(B) + n\,(H) + n\,(F)$

$ - n\,(B \cap H) - n\,(H \cap F) - n\,(F \cap B) + n\,(B \cap H \cap F)$

Hence,$n\,(B \cup H \cup F) = (21 + 26 + 29) - (14 + 15 + 12) + 8 = 43$

Thus these are $43$ members in all.

From, $n(N \cup P \cup H) = n(N) + n(P) + n(H) - n(N \cap P)$

$ - n(P \cap H) - n(N \cap H) + n(N \cap P \cap H)$

 $\therefore n(N \cap P) + n(P \cap H) + n(N \cap H) = 16$

Now, number of pupils taking two subjects

$ = n(N \cap P) + n(P \cap H) + n(N \cap H) - 3n(N \cap P \cap H)$

$ = 16 - 0 = 16$.

$n ( P \cap \overline{ M } \cap \overline{ C })=15$

$n ( C \cap \overline{ M } \cap \overline{ P })=3$

$n ( M \cap \overline{ P } \cap \overline{ C })=45$

$n ( P \cap C )=23$

$n ( P \cap M )=20$

$n ( M \cap C )=12$

Let $n(P \cap C \cap M)=x$

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

$+n(C \cap M \cap P)$

$100= 82-2 x \Rightarrow x=9$

Given relation

$\left(1+a^2\right)\left(1+b^2\right)=4 a b$

$\Rightarrow a^2+b^2-2 a b=2 a b-1-a^2 b^2$

$\Rightarrow \quad (a-b)^2=-(1-a b)^2$

$\because a > 0, a \neq 1 \text { and } b \text { is a positive real number}$

$\therefore(a-b)^2 \neq 0 \neq-(1-a b)^2, \text { because }(a-b)^2$

$\text { and }(1-a b)^2 \text { are non-negative real numbers}$

Let the number of boys and girls in classroom is $x$ and $y$, respectively. Given, $\frac{x-x / 5}{y}=\frac{2}{3} \Rightarrow \frac{4 x}{5 y}=\frac{2}{3}$

From Eqs.$(i)$ and $(ii)$, we get

$x=50, y=60$

Let $z$ number of boy leaves so number of boys and number of girls are equal.

$\therefore \quad 50-10-z=60-44  z=40-16=24$

Let the number of boy $=B$ and number of girls $=G$

Sum of marks obtained by boys $=B x$

$\therefore$ Sum of marks obtained by girls $=G y$

Now, given

$\frac{B x+G y}{B+G}=z$

$\Rightarrow B(x-z)=G(z-y)=\frac{B}{G}=\frac{z-y}{x-z}$

Now, $\frac{G}{B+G}=\frac{1}{\frac{B}{G}+1}=\frac{1}{\frac{z-y}{x-z}+1}=\frac{x-z}{x-y}$

$\frac{G}{B+G}=\frac{z-x}{y-x}$

$x$ and $y$ are positive integer

$x^2-y^2=12345678$

$RHS$ $12345678$ is and even number and last digit is $8 .$

$\therefore$ The last digit of $x$ be $3,7$

and the last digit of $y$ be $1,9$.

$\therefore x$ and $y$ must be odd and square of difference is multiple of $8$ but $RHS$ is not multiple of $8$ .

$\therefore S$ is the empty set.

We have, $A_1, A_2, A_3 \ldots, A_m$ are non-empty subsets of $\{1,2,3, \ldots, 100\}$ $\left|A_1\right|,\left|A_2\right|, \ldots,\left|A_m\right|$ are distincts.

$A_1 \cap A_2 \cap A_3 \ldots \cap A_m=\phi$

$\therefore A_1 \cap A_2 \cup A_3 \ldots \cup A_m=\{1,2,3, \ldots, 100\}$

Let $\left|A_1\right|=1\left|A_2\right|=2 \ldots\left|A_m\right|=M$

$A_1, A_2, A_3 \ldots, A_m$ are disjoint set.

$\therefore\left|A_1\right|+\left|A_2\right| \ldots+\left|A_m\right|=100$

$1+2+3 \ldots+m=100$

$\frac{m(m+1)}{2}=100$

$m^2+m-200=0$

$=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-1 \pm \sqrt{1+4 \cdot 1 \cdot 200}}{2 \cdot 1}$

$=\frac{-1 \pm \sqrt{1+800}}{2}=\frac{-1+\sqrt{801}}{2}$

$=\frac{-1+28.30}{2}=\frac{27.30}{2}=16.65$

$m=\frac{1+28.30}{2}=\frac{29.30}{2}=14.65$

$\therefore m < 14$

$\therefore$ Maximum possible of $m$ is $13$ . ($14$th set will have same size as that of previous size)

We have,

$S_n=\{n+1, n+2, n+3, \ldots, n+18\},$

$n \geq 10$

(a) $n=19$, then $S_n$ has not a multiple of 19 .

Hence, option (a) is false.

(b) $S_n$ has more than one prime for $n \geq 10$.

(c) $n=10$

11,28

Only $15,20,25$ are multiple of $5 .$

(d) Number of odd integer in $S_n=9$

So, every third integer is multiple of $3$.

$\therefore S_n$ has at most six primes.

We have,

$S=\{1,2,3,4, \ldots, 40\}$

$A$ is subset of $S$ whose sum of two element of $A$ is not divisible by $5$ .

Possible set $A=\{1,2,5,6,7,11,12,16$, $17,21,22,26,27,31,32,36,37\}$

$\therefore$ Maximum number of elements in $A$ is $17 .$

We have,

$S=\{1,2,3,4, \ldots, n\}$

$A=\{(a, b): 1 \leq a, b \leq n\}=S \times S$

$B=\{(x, x): x \in S\}$

$\therefore \quad B=\{(1,1),(2,2),(3,3), \ldots,(n, n)\}$

Number of elements in $B=n$

Total number of subset of $B$ is $2^n$.

$\therefore$ Total number of good subset of $A$ is $2^n$.

Given, $74 \%$ students like cricket $76 \%$ students like football

$82 \%$ students like tennis

$\therefore 26 \%$ student not like cricket

$24 \%$ student not like football

$18 \%$ student not like tennis

Student all the three sports like at least

$=100 \%-$ (sport not likes)

$=100 \%-(26+24+18) \%$

$=100 \%-68 \%=32 \%$

$ 102,105,108, \ldots . ., 699 $

$ \mathrm{~T}_{\mathrm{n}}=699=102+(\mathrm{n}-1)(3) $

$ \mathrm{n}=200 $

$ \mathrm{n}(3)=200 $

$ \because \mathrm{n}(4) \Rightarrow$ multiple of $4$

$ 100,104,108, \ldots ., 700 $

$ T_n=700=100+(n-1)(4) $

$ n=151 $

$ n(4)=151 $

$ \mathrm{n}(3 \cap 4) \Rightarrow \text { multiple of } 3 \& 4 \text { both } $

$ 108,120,132, \ldots ., 696 $

$ T_n=696=108+(n-1)(12) $

$ \mathrm{n}=50 $

$ \mathrm{n}(3 \cap 4)=50 $

$ \mathrm{n}(3 \cup 4)=\mathrm{n}(3)+\mathrm{n}(4)-\mathrm{n}(3 \cap 4) $

$ \quad=200+151-50 $

$ =301$

$\mathrm{n}(\overline{3 \cup 4})=$ Total $-\mathrm{n}(3 \cup 4)=$ neither a multiple of $3$ nor a multiple of $4$

$=601-301=300$

$ \Rightarrow x=45, y=1$

$ \sum_{(x, y) \in C}(x+y)=46 $

$z=0$        $x+2 y=42 \Rightarrow 22$

$z=1$        $x+2 y=39 \Rightarrow 20$

$z=2$         $x+2 y=36 \Rightarrow 19$

$z=3$         $x+2 y=33 \Rightarrow 17$

$z=4$          $x+2 y=30 \Rightarrow 16$

$z=5$          $x+2 y=27 \Rightarrow 14$

$z=6$          $x+2 y=24 \Rightarrow 13$

$z=7$          $x+2 y=21 \Rightarrow 11$

$z=8$          $x+2 y=18 \Rightarrow 10$

$z=9$          $x+2 y=15 \Rightarrow 8$

$z=10$        $x+2 y=12 \Rightarrow 7$ 

$z=11$         $x+2 y=9 \Rightarrow 5$

$z=12$         $x+2 y=6 \Rightarrow 4$

$z=13$         $x+2 y=3 \Rightarrow 2$

$z=14$         $x+2 y=0 \Rightarrow 1$

Total : $169$

$ |2 \mathrm{a}-1|=[\mathrm{a}]+2 \mathrm{a}$

Case $-1$ : $\mathrm{a}>\frac{1}{2} $

$ 2 \mathrm{a}-1=[\mathrm{a}]+2 \mathrm{a} $

$ {[\mathrm{a}]=-1 \quad \therefore \mathrm{a} \in[-1,0) \text { Reject }} $

Case-$2$: $\mathrm{a}<\frac{1}{2} $

$ -2 \mathrm{a}+1=[\mathrm{a}]+2 \mathrm{a} $

$ \mathrm{a}=\mathrm{I}+\mathrm{f} $

$ -2(\mathrm{I}+\mathrm{f})+1=\mathrm{I}+2 \mathrm{I}+2 \mathrm{f} $

$ \mathrm{I}=0, \mathrm{f}=\frac{1}{4} \quad \therefore \mathrm{a}=\frac{1}{4} $

 Hence $ \mathrm{a}=\frac{1}{4} $

$ 72 \sum_{\mathrm{a} \in \mathrm{S}} \mathrm{a}=72 \times \frac{1}{4}=18$

$\mathrm{x} \leq 11$

$\mathrm{x}=11$ does not satisfy the data.

$ 11+z \leq 15 \Rightarrow z \leq 4$

$ 11+y \leq 15 \Rightarrow y \leq 4$

Now

$ 9-z+0+14-y+z+11+y+5-y-z=40$

$ \Rightarrow y+z=-1$

Not possible

$\Rightarrow \mathrm{x} \leq 10$

For $\mathrm{x}=10$

Hence maximum number of students passed in all the three subjects is $10$.

$ 85 \leq \mathrm{P}+70-\mathrm{x} \leq 95 $

$ 75 \leq \mathrm{C}+80-\mathrm{x} \leq 90 $

$ \mathrm{~m}+\mathrm{P}+\mathrm{C}+120-2 \mathrm{x}=210 $

$ \Rightarrow 15 \leq \mathrm{x} \leq 45 \& 30-\mathrm{x} \geq 0 $

$ \Rightarrow 15 \leq \mathrm{x} \leq 30 $

$ 30+15=45$

$3^{ n }-3 \text { is multiple of } 7$

$3^{ n }=7 \lambda+3$

$n =1,7,13,20, \ldots .97$

Number of possible values of $n=15$

$\Rightarrow n^2-10 n+25 > 0 \text { and } n ^2-10 n +13 < ( n -5)^2 > 0 n \in[5-2 \sqrt{3}, 5+2 \sqrt{3}]$

$n \in R-[5]$

$\therefore n \in[1.3,8.3]$

$\Rightarrow n=2,3,4,6,7,8$

$| B |=25$

$| C |=18$

$| A \cup B \cup C |=60 \quad[\text { Total }]$

$| A \cap B \cap C |=5$

$|A \cup B \cup C|=\sum|A|-\sum|A \cap B|+|A \cap B \cap C|$

$\Rightarrow \sum|A \cap B|=48+25+18+5-60$

$\quad=36$

No. of men who received exactly 2 medals

$=\sum|A \cap B|-3|A \cap B \cap C|$

$=36-15$

$=21$

$B =\{3,6,7,9\}$

Total subset of $A =2^{7}=128$

$C \cap B =\phi$ when set $C$ contains the element $1,2,4,5$

$\therefore S=\{C \subseteq A ; C \cap B \neq \phi\}$

$=\operatorname{Total}-(C \cap B=\phi)$

$=128-2^{4}=112$

$B - A =(-\infty,-3] \cup[3, \infty)= R -(-3,3)$

$=\underbrace{(2+4+6+\ldots+200)}_{\text {Multiple of } 2}-\underbrace{(6+12+\ldots+198)}_{\text {Multiple of } 2 \; and\; 3 \text { i.e. } 6}$

$\quad-\underbrace{(10+20+\ldots+200)}_{\text {Multiple of } 5 \; and \; 2 \text { i.e. } 10}+\underbrace{(30+60+\ldots+180)}_{\text {Multiple of } 2,5 \; and \;3 \text { i.e. } 30}$

$=5264$

And $C^{\prime}$ is a set containing subsets of $A$ whose sum of elements is not prime.

So, we need to calculate number of subsets of $\{3,4,5,6,7\}$ whose sum of elements plus $1$ is composite.

Number of such $54\,elements\,subset\,=1$

Number of such $4$ elements subset $=3$ (except selecting $3$ or $7$ )

Number of such $3$ elements subset $=6$ (except selecting $\{3,4,5\},\{3,6,7\},\{4,5,7\}$ or $\{5,6,7\}$ )

Number of such $2$ elements subset $=7$ (except selecting $\{3,7\},\{4,6\},\{5,7\})$

Number of such $1$ elements subset $=3$ (except selecting $\{4\}$ or $\{6\}$ )

Number of such $0$ elements subset $=1$

$n\left(B^{\prime} \cap C^{\prime}\right)=21 \Rightarrow n(B \cup C)=2^{7}-21=107$

$A \left\{ a _{1}+ a _{2}+\ldots . .+ a _{ k }: K \in N \right\} \,and\, a _{ i } \in S$

Here by the definition of set '$A$'

$A=\{a: a=4 x+6 y+9 z\}$

Except the element $11$,every element of set $T$ is of of the form $4 x+6 y+9 z$ for some $x, y, z \in W$

$\therefore T - A =\{11\}$

Now, $n^{2}-n \leq 100 \times 100$

$\Rightarrow \mathrm{n}(\mathrm{n}-1) \leq 100 \times 100$

$\Rightarrow \mathrm{A}=\{1,2, \ldots, 100\}$

So, $A \cap(B-C)=\{7,13,19, \ldots, 97\}$

Hence, sum $=\frac{16}{2}(7+97)=832$

$100 \geq 89+98-\mathrm{n}(\mathrm{A} \cap \mathrm{B})$

$\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \geq 87$

$87 \leq \mathrm{n}(\mathrm{A} \cap \mathrm{B}) \leq 89$

$(|x|-3)(|x+4|)=6$

$\Rightarrow \quad|x|-3=\frac{6}{|x+4|}$

No. of solutions $=2$