AP EAMCET 2022 Mathematics Question Paper with Answer and Solution

(A) Given $r_1 > r_2 > r_3$.
We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Substituting these values,we get $\frac{\Delta}{s-a} > \frac{\Delta}{s-b} > \frac{\Delta}{s-c}$.
Since $\Delta > 0$,we have $s-a < s-b < s-c$.
Subtracting $s$ from all parts,we get $-a < -b < -c$.
Multiplying by $-1$ reverses the inequality signs,resulting in $a > b > c$.

(C) Given,in $\triangle ABC$,$r_1=12, r_2=18, r_3=36$.
We know that $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$.
Substituting the values: $\frac{1}{r} = \frac{1}{12} + \frac{1}{18} + \frac{1}{36} = \frac{3+2+1}{36} = \frac{6}{36} = \frac{1}{6}$.
Thus,$r = 6$.
Also,$\Delta^2 = r_1 r_2 r_3 r = 12 \times 18 \times 36 \times 6 = 46656$.
So,$\Delta = \sqrt{46656} = 216$.
Since $r = \frac{\Delta}{s}$,we have $s = \frac{\Delta}{r} = \frac{216}{6} = 36$.
We know $r_2 = \frac{\Delta}{s-b}$.
Substituting the values: $18 = \frac{216}{36-b}$.
$36-b = \frac{216}{18} = 12$.
$b = 36 - 12 = 24$.

(A) Let $E$ be the set of students who passed in English and $M$ be the set of students who passed in Mathematics.
Given:
Total students $= 205$
$n(E) = 105$
$n(M) = 70$
$n(E \cap M) = 30$
We need to find the number of students who passed in at least one subject,which is $n(E \cup M)$.
Using the formula: $n(E \cup M) = n(E) + n(M) - n(E \cap M)$
$n(E \cup M) = 105 + 70 - 30 = 145$
Now,the number of students who failed in both subjects is the total number of students minus those who passed in at least one subject.
Students failed in both $= 205 - 145 = 60$

(A) Let $y = \tanh^{-1} x$.
Then $x = \tanh y$.
Using the definition of the hyperbolic tangent function,we have $x = \frac{e^y - e^{-y}}{e^y + e^{-y}}$.
Multiplying both sides by $(e^y + e^{-y})$,we get $x(e^y + e^{-y}) = e^y - e^{-y}$.
$xe^y + xe^{-y} = e^y - e^{-y}$.
Rearranging the terms,we get $e^y(1 - x) = e^{-y}(1 + x)$.
Dividing both sides by $(1 - x)$,we get $e^y = e^{-y} \frac{1 + x}{1 - x}$.
Multiplying both sides by $e^y$,we get $e^{2y} = \frac{1 + x}{1 - x}$.
Taking the natural logarithm on both sides,we get $2y = \log \left(\frac{1 + x}{1 - x}\right)$.
Therefore,$y = \frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right)$.

(A) Given equations are:
$x+y+z=12$ ... $(i)$
$x^2+y^2+z^2=50$ ... $(ii)$
$x^3+y^3+z^3=216$ ... $(iii)$
Using the identity $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$:
$12^2 = 50 + 2(xy+yz+zx)$
$144 - 50 = 2(xy+yz+zx)$
$xy+yz+zx = 47$
Using the identity $x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$:
$216 - 3xyz = 12(50 - 47)$
$216 - 3xyz = 12(3) = 36$
$3xyz = 180 \Rightarrow xyz = 60$
Now,$x, y, z$ are roots of the cubic equation $t^3 - (x+y+z)t^2 + (xy+yz+zx)t - xyz = 0$:
$t^3 - 12t^2 + 47t - 60 = 0$
Testing roots,for $t=3$: $27 - 12(9) + 47(3) - 60 = 27 - 108 + 141 - 60 = 0$. So $(t-3)$ is a factor.
$(t-3)(t^2-9t+20) = 0$
$(t-3)(t-4)(t-5) = 0$
The roots are $3, 4, 5$.
Since $x, y, z$ can be any permutation of ${3, 4, 5}$,the number of solutions is $3! = 6$.

(D) Given the relation $f: R_{+} \rightarrow R_{+}$ defined by $f(x) = 3x^2 - 2$.
For $f$ to be a function,every element $x$ in the domain $R_{+}$ must have a unique image $f(x)$ in the codomain $R_{+}$.
Given $x \in (0, \infty)$,we have $x^2 > 0$,so $3x^2 > 0$.
Thus,$f(x) = 3x^2 - 2 > -2$.
This means the range of $f$ is $(-2, \infty)$.
Since the codomain is given as $R_{+} = (0, \infty)$,and the range $(-2, \infty)$ is not a subset of the codomain $(0, \infty)$ (for example,if $x=0.5$,$f(0.5) = 3(0.25) - 2 = 0.75 - 2 = -1.25$,which is not in $R_{+}$),the relation $f$ does not map every element of the domain to an element in the codomain.
Therefore,$f$ is not a function.

(C) Given $f(x) = ax^2 + bx + c$.
$f(1) = a + b + c$ and $f(2) = 4a + 2b + c$.
From the first condition: $f(1) + 2f(2) = 0$
$(a + b + c) + 2(4a + 2b + c) = 0$
$a + b + c + 8a + 4b + 2c = 0$
$9a + 5b + 3c = 0$ ... $(i)$
From the second condition: $2f(1) + f(2) = 0$
$2(a + b + c) + (4a + 2b + c) = 0$
$2a + 2b + 2c + 4a + 2b + c = 0$
$6a + 4b + 3c = 0$ ... $(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$(9a + 5b + 3c) - (6a + 4b + 3c) = 0 - 0$
$3a + b = 0$.

(D) Given equation: $|x|+|y|=|x-3|+|y-2|$.
Case $I$: $0 \leq x < 3$ and $0 \leq y < 2$.
In this region,$|x|=x$,$|y|=y$,$|x-3|=-(x-3)$,and $|y-2|=-(y-2)$.
Substituting these into the equation: $x+y = -(x-3) - (y-2)$.
$x+y = -x+3-y+2$.
$2x+2y = 5$.
$x+y = 5/2$.
Case $II$: $x \geq 3$ and $y \geq 2$.
In this region,$|x|=x$,$|y|=y$,$|x-3|=x-3$,and $|y-2|=y-2$.
Substituting these into the equation: $x+y = (x-3) + (y-2)$.
$x+y = x+y-5$.
$0 = -5$,which is impossible.
Therefore,the condition $x+y = 5/2$ holds for $0 \leq x < 3$ and $0 \leq y < 2$.

(B) Given equations are:
$1) x+y+z=1$
$2) x^2+y^2+z^2=1$
$3) x^3+y^3+z^3=1$
From $(1)$,we have $z = 1 - x - y$. Substituting this into $(2)$:
$x^2 + y^2 + (1 - x - y)^2 = 1$
$x^2 + y^2 + 1 + x^2 + y^2 - 2x - 2y + 2xy = 1$
$2x^2 + 2y^2 + 2xy - 2x - 2y = 0$
$x^2 + y^2 + xy - x - y = 0$
Multiplying by $2$: $2x^2 + 2y^2 + 2xy - 2x - 2y = 0$,which can be written as $(x+y)^2 + x^2 + y^2 - 2x - 2y = 0$.
Alternatively,note that if two variables are $0$,the third must be $1$. For example,if $x=1, y=0, z=0$,then $1+0+0=1$,$1^2+0^2+0^2=1$,and $1^3+0^3+0^3=1$. These satisfy all equations.
Testing permutations of $(1, 0, 0)$,we get $(1, 0, 0)$,$(0, 1, 0)$,and $(0, 0, 1)$.
Thus,there are $3$ solutions.

(A) The parabolas $y^2 = 4ax$ and $x^2 = 4ay$ are symmetric with respect to the line $y = x$.
Therefore,the line $y = x$ passes through their points of intersection.
Solving $y^2 = 4ax$ and $y = x$:
$x^2 = 4ax \Rightarrow x(x - 4a) = 0 \Rightarrow x = 0$ or $x = 4a$.
When $x = 0$,$y = 0$. When $x = 4a$,$y = 4a$.
Thus,the points of intersection are $(0, 0)$ and $(4a, 4a)$.
Since the line $2bx + 3cy + 4d = 0$ passes through $(0, 0)$:
$2b(0) + 3c(0) + 4d = 0 \Rightarrow 4d = 0 \Rightarrow d = 0$.
Since it also passes through $(4a, 4a)$:
$2b(4a) + 3c(4a) + 4d = 0 \Rightarrow 8ab + 12ac + 4d = 0$.
Since $d = 0$ and $a \neq 0$,we have $8ab + 12ac = 0 \Rightarrow 4a(2b + 3c) = 0 \Rightarrow 2b + 3c = 0$.
Therefore,$d^2 + (2b + 3c)^2 = 0^2 + 0^2 = 0$.

(D) Let the point of intersection be $(x_1, y_1)$.
For the curve $y^2 = 4ax$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4a$,so $\frac{dy}{dx} = \frac{2a}{y}$.
Thus,the slope of the tangent at $(x_1, y_1)$ is $m_1 = \frac{2a}{y_1}$.
For the curve $xy = c^2$,differentiating with respect to $x$ gives $x \frac{dy}{dx} + y = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.
Thus,the slope of the tangent at $(x_1, y_1)$ is $m_2 = -\frac{y_1}{x_1}$.
Since the curves cut orthogonally,$m_1 \times m_2 = -1$.
$\frac{2a}{y_1} \times (-\frac{y_1}{x_1}) = -1 \Rightarrow \frac{2a}{x_1} = 1 \Rightarrow x_1 = 2a$.
Since $(x_1, y_1)$ lies on $y^2 = 4ax$,$y_1^2 = 4a(2a) = 8a^2$.
Since $(x_1, y_1)$ lies on $xy = c^2$,$x_1 y_1 = c^2$,so $(x_1 y_1)^2 = c^4$.
Substituting $x_1 = 2a$ and $y_1^2 = 8a^2$,we get $c^4 = x_1^2 y_1^2 = (2a)^2 (8a^2) = 4a^2 \times 8a^2 = 32a^4$.

(C) Given equation is $x^{11}-x^7+x^4-1=0$.
Factorizing the expression:
$x^7(x^4-1) + 1(x^4-1) = 0$
$(x^4-1)(x^7+1) = 0$.
Case $(i)$: $x^4-1=0 \implies x^4=1$.
The roots are $x = e^{i(2k\pi/4)}$ for $k=0, 1, 2, 3$. These are $1, i, -1, -i$. There are $4$ distinct roots.
Case $(ii)$: $x^7+1=0 \implies x^7=-1$.
The roots are $x = e^{i((2k+1)\pi/7)}$ for $k=0, 1, 2, 3, 4, 5, 6$. There are $7$ distinct roots.
To find the total number of distinct solutions,we check for common roots:
If $x^4=1$ and $x^7=-1$,then $x^4=1 \implies |x|=1$.
Also $x^7=-1 \implies |x|=1$.
If $x$ is a common root,then $x^4=1$ and $x^7=-1$.
Then $x^8 = (x^4)^2 = 1^2 = 1$.
Since $x^8 = x^7 \cdot x = 1$,we have $(-1) \cdot x = 1$,so $x = -1$.
Check if $x = -1$ satisfies both:
$(-1)^4 = 1$ (True) and $(-1)^7 = -1$ (True).
So,$x = -1$ is the only common root.
Total distinct solutions = (Number of roots of $x^4-1=0$) + (Number of roots of $x^7+1=0$) - (Number of common roots)
Total = $4 + 7 - 1 = 10$.

(B) polynomial $f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$ is defined as a monic polynomial if the leading coefficient,which is the coefficient of the highest degree term $x^n$,is equal to $1$.
Therefore,for the given polynomial $f(x)$ to be monic,we must have $a_n = 1$.

(D) Let $f(\alpha) = \left(1 + \frac{1}{\sin^n \alpha}\right) \left(1 + \frac{1}{\cos^n \alpha}\right)$.
By the $AM$-$GM$ inequality,for positive terms,the minimum value occurs when $\sin^n \alpha = \cos^n \alpha$,which implies $\sin \alpha = \cos \alpha$,so $\alpha = \frac{\pi}{4}$.
At $\alpha = \frac{\pi}{4}$,we have $\sin \alpha = \cos \alpha = \frac{1}{\sqrt{2}} = 2^{-1/2}$.
Then $\sin^n \alpha = \cos^n \alpha = (2^{-1/2})^n = 2^{-n/2}$.
Substituting this into the expression:
$f\left(\frac{\pi}{4}\right) = \left(1 + \frac{1}{2^{-n/2}}\right) \left(1 + \frac{1}{2^{-n/2}}\right) = (1 + 2^{n/2})(1 + 2^{n/2}) = (1 + 2^{n/2})^2$.
Thus,the minimum value is $(1 + 2^{n/2})^2$.

(B) Let the vertices be $A=(-1, 3)$,$B=(2, 5)$,and $C=(a, b)$. The orthocenter is $H=(1, 2)$.
Since $AH \perp BC$,the slope of $AH$ $(m_{AH})$ multiplied by the slope of $BC$ $(m_{BC})$ is $-1$.
$m_{AH} = \frac{2-3}{1-(-1)} = \frac{-1}{2}$.
$m_{BC} = \frac{b-5}{a-2}$.
Since $m_{AH} \times m_{BC} = -1$,we have $\left(\frac{-1}{2}\right) \times \left(\frac{b-5}{a-2}\right) = -1 \Rightarrow b-5 = 2(a-2) \Rightarrow b-5 = 2a-4 \Rightarrow 2a-b = -1$ ... $(i)$
Similarly,since $BH \perp AC$,the slope of $BH$ $(m_{BH})$ multiplied by the slope of $AC$ $(m_{AC})$ is $-1$.
$m_{BH} = \frac{2-5}{1-2} = \frac{-3}{-1} = 3$.
$m_{AC} = \frac{b-3}{a-(-1)} = \frac{b-3}{a+1}$.
Since $m_{BH} \times m_{AC} = -1$,we have $3 \times \left(\frac{b-3}{a+1}\right) = -1 \Rightarrow 3b-9 = -a-1 \Rightarrow a+3b = 8$ ... (ii)
Solving equations $(i)$ and (ii):
From $(i)$,$b = 2a+1$. Substituting into (ii): $a + 3(2a+1) = 8 \Rightarrow a + 6a + 3 = 8 \Rightarrow 7a = 5 \Rightarrow a = \frac{5}{7}$.
Then $b = 2(\frac{5}{7}) + 1 = \frac{10}{7} + \frac{7}{7} = \frac{17}{7}$.
Thus,$(a, b) = \left(\frac{5}{7}, \frac{17}{7}\right)$.

(C) Given vertices of the parallelogram $PQRS$ are $P(1, 2)$,$Q(4, 6)$,$R(5, 7)$,and $S(a, b)$.
In a parallelogram,the diagonals bisect each other,meaning they share the same midpoint.
Let $A$ be the midpoint of the diagonals $PR$ and $QS$.
The midpoint $A$ of diagonal $PR$ is given by:
$A = \left( \frac{1 + 5}{2}, \frac{2 + 7}{2} \right) = \left( \frac{6}{2}, \frac{9}{2} \right) = \left( 3, 4.5 \right)$.
The midpoint $A$ of diagonal $QS$ is given by:
$A = \left( \frac{4 + a}{2}, \frac{6 + b}{2} \right)$.
Equating the coordinates of the midpoints:
$\frac{4 + a}{2} = 3 \implies 4 + a = 6 \implies a = 2$.
$\frac{6 + b}{2} = 4.5 \implies 6 + b = 9 \implies b = 3$.
Therefore,the values are $a = 2$ and $b = 3$.

(B) Given that the point $(a, 8, -2)$ divides the line segment joining $(1, 4, 6)$ and $(5, 2, 10)$ in the ratio $m: n$.
By the section formula,the coordinates are given by:
$x = \frac{5m + n}{m + n}$,$y = \frac{2m + 4n}{m + n}$,$z = \frac{10m + 6n}{m + n}$.
Equating the $y$-coordinate:
$8 = \frac{2m + 4n}{m + n} \implies 8m + 8n = 2m + 4n \implies 6m = -4n \implies \frac{m}{n} = -\frac{2}{3}$.
Equating the $z$-coordinate:
$-2 = \frac{10m + 6n}{m + n} \implies -2m - 2n = 10m + 6n \implies -12m = 8n \implies \frac{m}{n} = -\frac{8}{12} = -\frac{2}{3}$.
Since both equations yield the same ratio,we find $a$ using the $x$-coordinate:
$a = \frac{5m + n}{m + n} = \frac{5(\frac{m}{n}) + 1}{(\frac{m}{n}) + 1} = \frac{5(-\frac{2}{3}) + 1}{(-\frac{2}{3}) + 1} = \frac{-\frac{10}{3} + 1}{\frac{1}{3}} = \frac{-\frac{7}{3}}{\frac{1}{3}} = -7$.
Finally,calculating the expression:
$\frac{2m}{n} - \frac{a}{3} = 2(-\frac{2}{3}) - (\frac{-7}{3}) = -\frac{4}{3} + \frac{7}{3} = \frac{3}{3} = 1$.

(B) Given points $A(1, 2, -1)$ and $B(-1, 0, 1)$. Point $P$ divides the line segment $AB$ externally in the ratio $m_1: m_2 = 1: 2$.
For external division,the coordinates of $P$ are given by $\left( \frac{m_1 x_2 - m_2 x_1}{m_1 - m_2}, \frac{m_1 y_2 - m_2 y_1}{m_1 - m_2}, \frac{m_1 z_2 - m_2 z_1}{m_1 - m_2} \right)$.
Substituting the values:
$P = \left( \frac{1(-1) - 2(1)}{1 - 2}, \frac{1(0) - 2(2)}{1 - 2}, \frac{1(1) - 2(-1)}{1 - 2} \right)$
$P = \left( \frac{-1 - 2}{-1}, \frac{-4}{-1}, \frac{1 + 2}{-1} \right)$
$P = \left( \frac{-3}{-1}, \frac{-4}{-1}, \frac{3}{-1} \right) = (3, 4, -3)$.
Now,we find the distance $PQ$ where $Q = (1, 3, -1)$:
$PQ = \sqrt{(3 - 1)^2 + (4 - 3)^2 + (-3 - (-1))^2}$
$PQ = \sqrt{(2)^2 + (1)^2 + (-2)^2}$
$PQ = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$ units.

(C) The coordinates of the centroid $G$ of a triangle with vertices $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ are given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Given $A(4, x, 1)$,$B(y, -5, 2)$,$C(7, 8, 3)$,and $G(3, 5, 2)$,we have:
$G = \left(\frac{4+y+7}{3}, \frac{x-5+8}{3}, \frac{1+2+3}{3}\right) = (3, 5, 2)$
Equating the coordinates:
$\frac{11+y}{3} = 3 \Rightarrow 11+y = 9 \Rightarrow y = -2$
$\frac{x+3}{3} = 5 \Rightarrow x+3 = 15 \Rightarrow x = 12$
Since $CG$ is a median,$F$ is the midpoint of $AB$. The coordinates of $F$ are:
$F = \left(\frac{4+y}{2}, \frac{x-5}{2}, \frac{1+2}{2}\right)$
Substituting $x=12$ and $y=-2$:
$F = \left(\frac{4-2}{2}, \frac{12-5}{2}, \frac{3}{2}\right) = \left(\frac{2}{2}, \frac{7}{2}, \frac{3}{2}\right) = \left(1, \frac{7}{2}, \frac{3}{2}\right)$.

(B) Let the points be $P(1,2,3)$ and $R(4,5,6)$. The point $Q(3,4,5)$ divides $PR$ in the ratio $\lambda: 1$.
Using the section formula,the coordinates of $Q$ are given by:
$Q = \left( \frac{\lambda(4) + 1(1)}{\lambda + 1}, \frac{\lambda(5) + 1(2)}{\lambda + 1}, \frac{\lambda(6) + 1(3)}{\lambda + 1} \right) = (3,4,5)$
Comparing the $x$-coordinates:
$\frac{4\lambda + 1}{\lambda + 1} = 3$
$4\lambda + 1 = 3\lambda + 3$
$\lambda = 2$
Now,we need to find the point that divides the line segment joining $Q(3,4,5)$ and $P(1,2,3)$ in the ratio $-1: \lambda$ (which is $-1: 2$).
Let the required point be $S(x, y, z)$. Using the section formula for internal division with ratio $m:n = -1: 2$:
$x = \frac{-1(1) + 2(3)}{-1 + 2} = \frac{-1 + 6}{1} = 5$
$y = \frac{-1(2) + 2(4)}{-1 + 2} = \frac{-2 + 8}{1} = 6$
$z = \frac{-1(3) + 2(5)}{-1 + 2} = \frac{-3 + 10}{1} = 7$
Thus,the required point is $(5,6,7)$.

(D) The centroid of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
For the given vertices $(4, p, -3)$,$(-1, -1, 2)$,and $(3, 5, -8)$,the centroid is $\left(\frac{4-1+3}{3}, \frac{p-1+5}{3}, \frac{-3+2-8}{3}\right) = \left(2, \frac{p+4}{3}, -3\right)$.
The mid-point of $(1, 4, -2)$ and $(q, 2, -4)$ is $\left(\frac{1+q}{2}, \frac{4+2}{2}, \frac{-2-4}{2}\right) = \left(\frac{1+q}{2}, 3, -3\right)$.
Equating the coordinates of the centroid and the mid-point:
$2 = \frac{1+q}{2} \Rightarrow 4 = 1+q \Rightarrow q = 3$.
$\frac{p+4}{3} = 3 \Rightarrow p+4 = 9 \Rightarrow p = 5$.
Thus,$p^2 + q^2 = 5^2 + 3^2 = 25 + 9 = 34$.

(D) Total number of tickets = $35$.
Number of prize-bearing tickets = $10$.
Number of tickets without a prize = $35 - 10 = 25$.
Probability of not getting a prize = $\frac{\text{Number of tickets without a prize}}{\text{Total number of tickets}} = \frac{25}{35} = \frac{5}{7}$.

(D) Total number of balls $= 7 + 5 = 12$.
Probability of drawing the first green ball $= \frac{7}{12}$.
Since the balls are not replaced,the number of green balls remaining is $6$ and the total number of balls remaining is $11$.
Probability of drawing the second green ball $= \frac{6}{11}$.
After drawing two green balls,the number of green balls remaining is $5$ and the total number of balls remaining is $10$.
Probability of drawing the third green ball $= \frac{5}{10} = \frac{1}{2}$.
The probability that all three balls are green is the product of these individual probabilities:
$P = \frac{7}{12} \times \frac{6}{11} \times \frac{5}{10} = \frac{7 \times 6 \times 5}{12 \times 11 \times 10} = \frac{210}{1320} = \frac{7}{44}$.

(D) The set for $x$ is $\{1, 2, 3, 4\}$ and the set for $y$ is $\{5, 6, 7\}$.
Total number of possible outcomes is $4 \times 3 = 12$.
The product $xy$ is even if at least one of $x$ or $y$ is even.
Alternatively,$xy$ is odd only if both $x$ and $y$ are odd.
Odd values in $x$ are $\{1, 3\}$ (count = $2$).
Odd values in $y$ are $\{5, 7\}$ (count = $2$).
Number of outcomes where $xy$ is odd = $2 \times 2 = 4$.
Number of outcomes where $xy$ is even = $\text{Total outcomes} - \text{Odd outcomes} = 12 - 4 = 8$.
Probability that $xy$ is even = $\frac{8}{12} = \frac{2}{3}$.

(C) It can be noted that if the ball is neither red nor green,then it must be blue.
Number of blue balls $= 7$.
Total number of balls $= 8 + 7 + 6 = 21$.
Thus,the probability of picking a blue ball is $= \frac{7}{21} = \frac{1}{3}$.

(A) The total number of five-digit numbers formed using the digits $1, 2, 3, 4, 5$ without repetition is $5! = 120$.
$A$ number is divisible by $4$ if its last two digits form a number divisible by $4$.
Using the digits ${1, 2, 3, 4, 5}$,the possible two-digit combinations divisible by $4$ are $12, 24, 32, 52$.
For each of these $4$ cases,the remaining $3$ digits can be arranged in the first $3$ positions in $3! = 6$ ways.
Thus,the total number of favorable outcomes is $4 \times 3! = 4 \times 6 = 24$.
The required probability is $\frac{24}{120} = \frac{1}{5}$.

(D) Let the total amount be $S = 20000$. Let $X$ be the amount given to employee $X$ and $Y$ be the amount given to employee $Y$.
Given $X + Y = 20000$ and $X > Y$.
Since $X$ is the bigger part,$X$ must be in the range $(10000, 20000]$.
The total length of the sample space for $X$ is $20000 - 10000 = 10000$.
We want the probability that $X \geq 2Y$.
Substituting $Y = 20000 - X$,we get $X \geq 2(20000 - X)$.
$X \geq 40000 - 2X \Rightarrow 3X \geq 40000 \Rightarrow X \geq 40000/3$.
Since $X$ must be at most $20000$,the favorable range for $X$ is $[40000/3, 20000]$.
The length of this interval is $20000 - 40000/3 = 20000/3$.
The probability is $\frac{20000/3}{10000} = \frac{2}{3}$.

(C) When two dice are rolled together,the total number of outcomes is $6 \times 6 = 36$.
The pairs that result in a sum of $9$ are $(3,6), (4,5), (5,4), (6,3)$.
Thus,the number of favourable outcomes is $4$.
The probability of getting a sum of $9$ is given by the ratio of favourable outcomes to total outcomes:
$P(\text{sum of } 9) = \frac{4}{36} = \frac{1}{9}$.

(D) The total number of ways to select $3$ cards from $52$ is given by ${}^{52}C_3 = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22100$.
Favourable outcomes involve selecting $1$ ace from $4$,$1$ queen from $4$,and $1$ jack from $4$.
This can be done in ${}^{4}C_1 \times {}^{4}C_1 \times {}^{4}C_1 = 4 \times 4 \times 4 = 64$ ways.
The required probability is $\frac{64}{22100} = \frac{16}{5525}$.

(A) Total number of balls = $2 + 3 + 2 = 7$.
Number of ways to draw $2$ balls from $7$ is $^7C_2 = \frac{7 \times 6}{2} = 21$.
We want the probability that none of the balls drawn is blue. This means both balls must be chosen from the red and green balls.
Total non-blue balls = $2 \text{ (red)} + 3 \text{ (green)} = 5$.
Number of ways to choose $2$ balls from $5$ non-blue balls is $^5C_2 = \frac{5 \times 4}{2} = 10$.
Therefore,the probability is $\frac{10}{21}$.

(B) Total number of toys is $21$. The even numbers between $1$ and $21$ are $2, 4, 6, 8, 10, 12, 14, 16, 18, 20$. There are $10$ even numbers.
Probability of drawing the first even number is $\frac{10}{21}$.
After drawing one even number,there are $9$ even numbers left out of $20$ total toys.
Probability of drawing the second even number is $\frac{9}{20}$.
The probability that both toys show even numbers is $\frac{10}{21} \times \frac{9}{20} = \frac{90}{420} = \frac{3}{14}$.

(B) When two dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
The possible total scores range from $2$ to $12$.
The prime numbers in this range are $2, 3, 5, 7, 11$.
The outcomes resulting in these sums are:
Sum $= 2: (1,1)$
Sum $= 3: (1,2), (2,1)$
Sum $= 5: (1,4), (2,3), (3,2), (4,1)$
Sum $= 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$
Sum $= 11: (5,6), (6,5)$
Total number of favorable outcomes $= 1 + 2 + 4 + 6 + 2 = 15$.
The probability is $\frac{15}{36} = \frac{5}{12}$.

(A) The total number of outcomes when a die is rolled $4$ times is $6^4 = 1296$.
For each roll,the face value must be in the set $\{2, 3, 4, 5\}$ to satisfy the condition that the minimum is $\ge 2$ and the maximum is $\le 5$.
There are $4$ such favorable outcomes for each roll.
The number of favorable outcomes for $4$ rolls is $4^4 = 256$.
The required probability is $\frac{4^4}{6^4} = \left(\frac{4}{6}\right)^4 = \left(\frac{2}{3}\right)^4 = \frac{16}{81}$.

(C) Total cards $= 30$.
Number of white cards $= 17$.
Number of green cards $= 30 - 17 = 13$.
Number of $IMPORTANT$ cards $= 4 + 5 = 9$.
Number of cards that are both green and $IMPORTANT = 5$.
Let $G$ be the event of choosing a green card and $I$ be the event of choosing an $IMPORTANT$ card.
We need to find $P(G \cup I) = P(G) + P(I) - P(G \cap I)$.
$P(G) = \frac{13}{30}$,$P(I) = \frac{9}{30}$,$P(G \cap I) = \frac{5}{30}$.
$P(G \cup I) = \frac{13}{30} + \frac{9}{30} - \frac{5}{30} = \frac{17}{30}$.

(A) The total number of elements in the set is $n(S) = 100$.
The perfect cubes between $1$ and $100$ are $1^3 = 1$,$2^3 = 8$,$3^3 = 27$,and $4^3 = 64$.
Thus,the set of favorable outcomes is $E = \{1, 8, 27, 64\}$,so $n(E) = 4$.
The required probability is $P(E) = \frac{n(E)}{n(S)} = \frac{4}{100} = \frac{1}{25}$.

(D) Total number of cards $= 52$.
Total number of kings in a deck $= 4$.
The number of ways to choose $2$ cards out of $52$ is given by $^{52}C_2 = \frac{52 \times 51}{2 \times 1} = 1326$.
The number of ways to choose $2$ kings out of $4$ is given by $^4C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
Therefore,the probability that both cards are kings is $P = \frac{^4C_2}{^{52}C_2} = \frac{6}{1326} = \frac{1}{221}$.

(C) Total number of balls = $3 + 2 + 5 = 10$.
Number of balls that are not red = $3$ (white) + $2$ (blue) = $5$.
Probability of drawing a ball that is not red = $\frac{\text{Number of non-red balls}}{\text{Total number of balls}} = \frac{5}{10} = \frac{1}{2}$.

(A) Let $P(P)$ be the probability that person $P$ speaks the truth and $P(R)$ be the probability that person $R$ speaks the truth.
Given: $P(P) = \frac{75}{100} = \frac{3}{4}$ and $P(R) = \frac{80}{100} = \frac{4}{5}$.
Consequently,the probabilities of them lying are $P(P') = 1 - \frac{3}{4} = \frac{1}{4}$ and $P(R') = 1 - \frac{4}{5} = \frac{1}{5}$.
They contradict each other if one speaks the truth and the other lies.
Case $I$: $P$ speaks the truth and $R$ lies: $P(P) \times P(R') = \frac{3}{4} \times \frac{1}{5} = \frac{3}{20}$.
Case $II$: $R$ speaks the truth and $P$ lies: $P(R) \times P(P') = \frac{4}{5} \times \frac{1}{4} = \frac{4}{20}$.
Total probability of contradiction $= \frac{3}{20} + \frac{4}{20} = \frac{7}{20}$.

(D) Total number of bottles = $6 + 3 + 4 = 13$.
Number of ways to choose $3$ bottles out of $13$ is $^{13}C_3 = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286$.
Let $E$ be the event that all three bottles are of the same variety.
$E$ occurs if we choose $3$ bottles of $V_1$,or $3$ bottles of $V_2$,or $3$ bottles of $V_3$.
Number of ways to choose $3$ bottles of the same variety = $^6C_3 + ^3C_3 + ^4C_3$.
$^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
$^3C_3 = 1$.
$^4C_3 = 4$.
Total favorable ways for $E = 20 + 1 + 4 = 25$.
$P(E) = \frac{25}{286}$.
The probability that the three bottles are not of the same variety is $1 - P(E) = 1 - \frac{25}{286} = \frac{286 - 25}{286} = \frac{261}{286}$.

(B) The total number of cards in a pack is $52$.
There are $4$ Jacks,$4$ Queens,and $4$ Kings in a pack.
Total number of face cards = $4 + 4 + 4 = 12$.
The probability of drawing a face card is given by the ratio of the number of face cards to the total number of cards.
$P(\text{Face Card}) = \frac{12}{52} = \frac{3}{13}$.

(D) The total number of ways to draw $4$ cards from $52$ is $^{52}C_{4}$.
There are $4$ suits in a pack,each containing $13$ cards.
The number of ways to choose $4$ cards of the same suit is $4 \times ^{13}C_{4}$.
The probability $P$ is given by:
$P = \frac{4 \times ^{13}C_{4}}{^{52}C_{4}}$
$P = \frac{4 \times \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}}{\frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1}}$
$P = \frac{4 \times 715}{270725} = \frac{2860}{270725}$
Simplifying the fraction:
$P = \frac{44}{4165}$

(A) Total number of shirts $= 6 + 4 + 2 + 3 = 15$.
Number of ways to pick $2$ shirts from $15$ is given by $^{15}C_2 = \frac{15 \times 14}{2 \times 1} = 105$.
Number of ways to pick $2$ white shirts from $2$ is $^{2}C_2 = 1$.
Number of ways to pick $2$ blue shirts from $3$ is $^{3}C_2 = 3$.
Since these are mutually exclusive events,the probability is:
$P = \frac{^{2}C_2 + ^{3}C_2}{^{15}C_2} = \frac{1 + 3}{105} = \frac{4}{105}$.

(C) Given that $A$ is the event that $X$ passes the exam and $B$ is the event that $Y$ passes the exam.
$P(A) = \frac{1}{7}$ and $P(B) = \frac{2}{9}$.
Since the events $A$ and $B$ are independent,the probability that both pass the exam is given by $P(A \cap B) = P(A) \times P(B)$.
Substituting the values,we get $P(A \cap B) = \frac{1}{7} \times \frac{2}{9} = \frac{2}{63}$.

(D) When three unbiased coins are tossed,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
The total number of outcomes is $n(S) = 8$.
Let $E$ be the event of getting at most two heads.
It is easier to calculate the complement event $E'$,which is the event of getting three heads.
$E' = \{HHH\}$
The number of outcomes in $E'$ is $n(E') = 1$.
The probability of getting three heads is $P(E') = \frac{n(E')}{n(S)} = \frac{1}{8}$.
The probability of getting at most two heads is $P(E) = 1 - P(E') = 1 - \frac{1}{8} = \frac{7}{8}$.

(B) When two dice are thrown simultaneously,the total number of outcomes is $n(S) = 6 \times 6 = 36$.
Let $A$ be the event of getting a multiple of $2$ on the first die,i.e.,$A \in \{2, 4, 6\}$.
Let $B$ be the event of getting a multiple of $3$ on the second die,i.e.,$B \in \{3, 6\}$.
We want to find the probability of getting a multiple of $2$ on one die and a multiple of $3$ on the other.
Let $E$ be the set of favorable outcomes:
$E = \{(2,3), (4,3), (6,3), (2,6), (4,6), (6,6), (3,2), (3,4), (3,6), (6,2), (6,4)\}$.
Note that $(6,6)$ is included because it satisfies the condition of having a multiple of $2$ on one die and a multiple of $3$ on the other.
Counting the elements in $E$,we get $n(E) = 11$.
Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{11}{36}$.

(A) Let $P(X), P(Y), P(Z)$ be the probabilities of students $X, Y, Z$ passing the exam respectively.
Given:
$P(X) = \frac{1}{5} \implies P(X') = 1 - \frac{1}{5} = \frac{4}{5}$
$P(Y) = \frac{1}{4} \implies P(Y') = 1 - \frac{1}{4} = \frac{3}{4}$
$P(Z') = \frac{2}{3} \implies P(Z) = 1 - \frac{2}{3} = \frac{1}{3}$
We need the probability that at least two of them pass,which is the sum of probabilities of exactly two passing and all three passing.
$P(\text{at least 2 pass}) = P(X)P(Y)P(Z') + P(X)P(Y')P(Z) + P(X')P(Y)P(Z) + P(X)P(Y)P(Z)$
$= (\frac{1}{5} \times \frac{1}{4} \times \frac{2}{3}) + (\frac{1}{5} \times \frac{3}{4} \times \frac{1}{3}) + (\frac{4}{5} \times \frac{1}{4} \times \frac{1}{3}) + (\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3})$
$= \frac{2}{60} + \frac{3}{60} + \frac{4}{60} + \frac{1}{60} = \frac{10}{60} = \frac{1}{6}$

(B) The total number of outcomes when two dice are thrown is $6 \times 6 = 36$.
The product of two numbers is odd only if both numbers are odd.
The odd numbers on a die are $\{1, 3, 5\}$.
The number of outcomes where both dice show an odd number is $3 \times 3 = 9$.
The product is even if it is not odd.
Therefore,the number of favorable outcomes where the product is even is $36 - 9 = 27$.
The probability of getting an even product is $\frac{27}{36} = \frac{3}{4}$.

(C) Total number of ways in which $3$ men can check into $4$ hotels is $4 \times 4 \times 4 = 64$.
If each man has to check into a different hotel,the number of ways is $4 \times 3 \times 2 = 24$.
(The first man has $4$ choices,the second man has $3$ choices,and the third man has $2$ choices).
Therefore,the required probability is $\frac{24}{64} = \frac{3}{8}$.

(C) Let $A$ be the event that person $A$ solves the problem and $B$ be the event that person $B$ solves the problem.
Given,$P(A) = 0.90$ and $P(B) = 0.70$.
The probability that person $A$ fails to solve the problem is $P(\bar{A}) = 1 - 0.90 = 0.10$.
The probability that person $B$ fails to solve the problem is $P(\bar{B}) = 1 - 0.70 = 0.30$.
The probability that at least one of them solves the problem is given by $1 - P(\text{none solves the problem})$.
Since the events are independent,$P(\text{none solves}) = P(\bar{A}) \times P(\bar{B}) = 0.10 \times 0.30 = 0.03$.
Therefore,the probability that at least one of them solves the problem is $1 - 0.03 = 0.97$.

(D) Total number of slips for $III$ year students $= 3 \times 100 = 300$.
Total number of slips for $II$ year students $= 2 \times 150 = 300$.
Total number of slips for $I$ year students $= 1 \times 200 = 200$.
Total number of slips in the lottery $= 300 + 300 + 200 = 800$.
The probability that a $III$ year student gets the room is the ratio of the number of $III$ year student slips to the total number of slips.
Required probability $= \frac{300}{800} = \frac{3}{8}$.