QUADRATIC EQUATION Questions in English

(D) Let $t_1 = \frac{x}{x+7}$. Then the equation $I$ becomes $t_1 + \frac{1}{t_1} = 12$,which implies $t_1^2 - 12t_1 + 1 = 0$. Solving for $t_1$ using the quadratic formula,$t_1 = \frac{12 \pm \sqrt{144 - 4}}{2} = 6 \pm \sqrt{35}$.
Since $t_1 = \frac{x}{x+7}$,we have $x = t_1(x+7) \Rightarrow x(1-t_1) = 7t_1 \Rightarrow x = \frac{7t_1}{1-t_1}$.
For $t_1 = 6 + \sqrt{35} \approx 11.916$,$x = \frac{7(11.916)}{1-11.916} \approx -7.63$.
For $t_1 = 6 - \sqrt{35} \approx 0.084$,$x = \frac{7(0.084)}{1-0.084} \approx 0.64$.
Similarly for equation $II$,let $t_2 = \frac{y}{y+8}$. Then $t_2 + \frac{1}{t_2} = 16$,which implies $t_2^2 - 16t_2 + 1 = 0$. Solving for $t_2$,$t_2 = \frac{16 \pm \sqrt{256 - 4}}{2} = 8 \pm \sqrt{63} \approx 8 \pm 7.937$.
$t_2 \approx 15.937$ or $t_2 \approx 0.063$.
Using $y = \frac{8t_2}{1-t_2}$:
For $t_2 \approx 15.937$,$y = \frac{8(15.937)}{1-15.937} \approx -8.53$.
For $t_2 \approx 0.063$,$y = \frac{8(0.063)}{1-0.063} \approx 0.54$.
Comparing the values of $x \in \{-7.63, 0.64\}$ and $y \in \{-8.53, 0.54\}$,we see that $x$ can be greater than,less than,or equal to $y$ depending on the roots chosen. Thus,the relationship cannot be established.

(D) For equation $I$: Let $t = \frac{x}{x-11}$. Then $t + \frac{1}{t} = 7 \Rightarrow t^2 - 7t + 1 = 0$. Solving for $t$,$t = \frac{7 \pm \sqrt{49-4}}{2} = \frac{7 \pm \sqrt{45}}{2} = \frac{7 \pm 3\sqrt{5}}{2}$.
Since $t = \frac{x}{x-11}$,we have $x = t(x-11) = tx - 11t \Rightarrow x(1-t) = -11t \Rightarrow x = \frac{11t}{t-1}$.
Substituting $t_1 = \frac{7+3\sqrt{5}}{2} \approx 6.85$ and $t_2 = \frac{7-3\sqrt{5}}{2} \approx 0.146$,we get two values for $x$.
For equation $II$: Let $u = \frac{4y}{4y-13}$. Then $u + \frac{1}{u} = 9 \Rightarrow u^2 - 9u + 1 = 0$. Solving for $u$,$u = \frac{9 \pm \sqrt{81-4}}{2} = \frac{9 \pm \sqrt{77}}{2}$.
Since $u = \frac{4y}{4y-13}$,we have $4y = u(4y-13) = 4uy - 13u \Rightarrow 4y(1-u) = -13u \Rightarrow y = \frac{13u}{4(u-1)}$.
Since both equations yield two real roots (one positive and one negative),and the coefficients lead to overlapping ranges for $x$ and $y$,the relationship between $x$ and $y$ cannot be established.

(A) $I.$ $99 x^{2} + 149 x + 56 = 0$
$99 x^{2} + 77 x + 72 x + 56 = 0$
$11 x(9 x + 7) + 8(9 x + 7) = 0$
$(9 x + 7)(11 x + 8) = 0$
$x = -\frac{7}{9} \approx -0.777$
$x = -\frac{8}{11} \approx -0.727$
$II.$ $156 y^{2} + 287 y + 132 = 0$
$156 y^{2} + 143 y + 144 y + 132 = 0$
$13 y(12 y + 11) + 12(12 y + 11) = 0$
$(12 y + 11)(13 y + 12) = 0$
$y = -\frac{11}{12} \approx -0.916$
$y = -\frac{12}{13} \approx -0.923$
Comparing the values:
$x$ values are approximately $-0.777$ and $-0.727$.
$y$ values are approximately $-0.916$ and $-0.923$.
Since all values of $x$ are greater than all values of $y$,we have $x > y$.

(B) For equation $I$: $77 x^{2}+58 x+8=0$
Splitting the middle term: $77 x^{2}+44 x+14 x+8=0$
$11 x(7 x+4)+2(7 x+4)=0$
$(11 x+2)(7 x+4)=0$
So,$x = -\frac{2}{11}$ or $x = -\frac{4}{7}$.
For equation $II$: $42 y^{2}+59 y+20=0$
Splitting the middle term: $42 y^{2}+35 y+24 y+20=0$
$7 y(6 y+5)+4(6 y+5)=0$
$(7 y+4)(6 y+5)=0$
So,$y = -\frac{4}{7}$ or $y = -\frac{5}{6}$.
Comparing values:
$x_1 = -0.1818$,$x_2 = -0.5714$
$y_1 = -0.5714$,$y_2 = -0.8333$
Since $x_1 > y_1$,$x_1 > y_2$,$x_2 = y_1$,and $x_2 > y_2$,we conclude that $x \ge y$.

(A) $I.$ For the equation $63x^2 + 172x + 117 = 0$:
We need to factorize $63 \times 117 = 7371$ into two numbers that add up to $172$. These are $91$ and $81$.
$63x^2 + 91x + 81x + 117 = 0$
$7x(9x + 13) + 9(9x + 13) = 0$
$(7x + 9)(9x + 13) = 0$
So,$x = -9/7 \approx -1.28$ and $x = -13/9 \approx -1.44$.
$II.$ For the equation $30y^2 + 162y + 216 = 0$:
Divide by $6$: $5y^2 + 27y + 36 = 0$.
We need to factorize $5 \times 36 = 180$ into two numbers that add up to $27$. These are $15$ and $12$.
$5y^2 + 15y + 12y + 36 = 0$
$5y(y + 3) + 12(y + 3) = 0$
$(5y + 12)(y + 3) = 0$
So,$y = -12/5 = -2.4$ and $y = -3$.
Comparing the values:
$x$ values are $\approx -1.28, -1.44$.
$y$ values are $-2.4, -3$.
Since both values of $x$ are greater than both values of $y$,$x > y$.

(D) For equation $I$: $36 x^{4}+369 x^{2}+900=0$.
Let $x^{2} = p$. Then $36 p^{2} + 369 p + 900 = 0$.
Dividing by $9$,we get $4 p^{2} + 41 p + 100 = 0$.
Using the quadratic formula $p = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$,we get $p = \frac{-41 \pm \sqrt{41^{2} - 4(4)(100)}}{2(4)} = \frac{-41 \pm \sqrt{1681 - 1600}}{8} = \frac{-41 \pm 9}{8}$.
$p = \frac{-32}{8} = -4$ or $p = \frac{-50}{8} = -6.25$.
Since $x^{2} = -4$ or $x^{2} = -6.25$,$x$ has only imaginary roots.
For equation $II$: $144 y^{4}+337 y^{2}+144=0$.
Let $y^{2} = q$. Then $144 q^{2} + 337 q + 144 = 0$.
Since all coefficients are positive,$q$ must be negative for the equation to hold,leading to imaginary roots for $y$.
Since both equations yield imaginary roots for $x$ and $y$,the relationship between $x$ and $y$ cannot be established.

(D) For equation $I$: $18x^2 - 13\sqrt{7}x + 14 = 0$.
Splitting the middle term: $18x^2 - 9\sqrt{7}x - 4\sqrt{7}x + 14 = 0$.
$9x(2x - \sqrt{7}) - 2\sqrt{7}(2x - \sqrt{7}) = 0$.
$(9x - 2\sqrt{7})(2x - \sqrt{7}) = 0$.
So,$x = \frac{2\sqrt{7}}{9} \approx 0.588$ and $x = \frac{\sqrt{7}}{2} \approx 1.323$.
For equation $II$: $32y^2 - 19\sqrt{6}y + 9 = 0$.
Splitting the middle term: $32y^2 - 16\sqrt{6}y - 3\sqrt{6}y + 9 = 0$.
$16y(2y - \sqrt{6}) - 3\sqrt{2}(\sqrt{2}y - \sqrt{3}) = 0$ (or using quadratic formula).
Using $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$y = \frac{19\sqrt{6} \pm \sqrt{(19\sqrt{6})^2 - 4(32)(9)}}{2(32)} = \frac{19\sqrt{6} \pm \sqrt{2166 - 1152}}{64} = \frac{19\sqrt{6} \pm \sqrt{1014}}{64}$.
$y \approx \frac{46.53 \pm 31.84}{64}$.
$y_1 \approx 1.22$ and $y_2 \approx 0.23$.
Comparing the values of $x$ and $y$,the relationship cannot be established.

(D) For equation $I$: $x^{2}-82x+781=0$
We need two numbers whose product is $781$ and sum is $82$. These numbers are $71$ and $11$.
So,$x^{2}-71x-11x+781=0$
$x(x-71)-11(x-71)=0$
$(x-71)(x-11)=0$
Therefore,$x = 71$ or $x = 11$.
For equation $II$: $y^{2}-5041=0$
$y^{2} = 5041$
$y = \pm \sqrt{5041}$
$y = 71$ or $y = -71$.
Comparing the values:
If $x = 71$,then $x = y$ (when $y=71$) and $x > y$ (when $y=-71$).
If $x = 11$,then $x < y$ (when $y=71$) and $x > y$ (when $y=-71$).
Since the relationship varies depending on the values chosen,the relationship between $x$ and $y$ cannot be established.

(D) For equation $I$: $36 x^{2}+47 \sqrt{7} x+105=0$
We factorize the quadratic equation: $36 x^{2}+27 \sqrt{7} x+20 \sqrt{7} x+105=0$
$9 x(4 x+3 \sqrt{7})+5 \sqrt{7}(4 x+3 \sqrt{7})=0$
$(9 x+5 \sqrt{7})(4 x+3 \sqrt{7})=0$
$x = -\frac{5 \sqrt{7}}{9} \approx -1.47$ and $x = -\frac{3 \sqrt{7}}{4} \approx -1.98$
For equation $II$: $35 y^{2}+20 \sqrt{3} y+63 \sqrt{2} y+36 \sqrt{6}=0$
$5 y(7 y+4 \sqrt{3})+9 \sqrt{2}(7 y+4 \sqrt{3})=0$
$(5 y+9 \sqrt{2})(7 y+4 \sqrt{3})=0$
$y = -\frac{9 \sqrt{2}}{5} \approx -2.54$ and $y = -\frac{4 \sqrt{3}}{7} \approx -0.99$
Comparing the values:
$x_1 = -1.47, x_2 = -1.98$
$y_1 = -2.54, y_2 = -0.99$
Since $x_1 > y_1$ and $x_1 < y_2$,the relationship between $x$ and $y$ cannot be established.

(D) For equation $I$: $91x^2 + 298x + 187 = 0$
Using the quadratic formula or factorization: $91x^2 + 77x + 221x + 187 = 0$
$7x(13x + 11) + 17(13x + 11) = 0$
$(7x + 17)(13x + 11) = 0$
$x = -17/7 \approx -2.428$ or $x = -11/13 \approx -0.846$
For equation $II$: $247y^2 + 216y - 391 = 0$
Using the quadratic formula or factorization: $247y^2 + 437y - 221y - 391 = 0$
$19y(13y + 23) - 17(13y + 23) = 0$
$(19y - 17)(13y + 23) = 0$
$y = 17/19 \approx 0.895$ or $y = -23/13 \approx -1.769$
Comparing the values:
$x_1 = -2.428, x_2 = -0.846$
$y_1 = 0.895, y_2 = -1.769$
Since $x_1 < y_2$ and $x_2 > y_2$ but $x_2 < y_1$,the relationship between $x$ and $y$ cannot be established.

(D) For equation $I$: $81x^2 - 9x - 2 = 0$
$81x^2 - 18x + 9x - 2 = 0$
$9x(9x - 2) + 1(9x - 2) = 0$
$(9x + 1)(9x - 2) = 0$
$x = -1/9, 2/9$
For equation $II$: $56y^2 - 13y - 3 = 0$
$56y^2 - 21y + 8y - 3 = 0$
$7y(8y - 3) + 1(8y - 3) = 0$
$(7y + 1)(8y - 3) = 0$
$y = -1/7, 3/8$
Comparing the values:
$x = -0.11, 0.22$
$y = -0.14, 0.375$
Since $0.22 < 0.375$ and $-0.11 > -0.14$,the relationship between $x$ and $y$ cannot be established.

(D) $I.$ For $391 x^{2} + 1344 x + 1073 = 0$:
Using the quadratic formula or factorization,we find the roots.
$391 = 17 \times 23$ and $1073 = 29 \times 37$.
Splitting the middle term: $493 + 851 = 1344$.
$391 x^{2} + 493 x + 851 x + 1073 = 0$
$17x(23x + 29) + 37(23x + 29) = 0$
$(17x + 37)(23x + 29) = 0$
$x = -\frac{37}{17} \approx -2.176$ and $x = -\frac{29}{23} \approx -1.261$.
$II.$ For $437 y^{2} + 1074 y + 589 = 0$:
$437 = 19 \times 23$ and $589 = 19 \times 31$.
Splitting the middle term: $361 + 713 = 1074$.
$437 y^{2} + 361 y + 713 y + 589 = 0$
$19y(23y + 19) + 31(23y + 19) = 0$
$(19y + 31)(23y + 19) = 0$
$y = -\frac{31}{19} \approx -1.631$ and $y = -\frac{19}{23} \approx -0.826$.
Comparing the values:
$x_1 = -2.176, x_2 = -1.261$
$y_1 = -1.631, y_2 = -0.826$
Since $x_1 < y_1$ and $x_2 < y_2$,but $x_2 > y_1$,the relationship cannot be established.

(C) For equation $I$: $3216 x^{2} + 3859 x + 481 = 0$.
Since all coefficients are positive,the roots of this quadratic equation must be negative.
For equation $II$: $8132 y^{2} - 4839 y + 978 = 0$.
Here,the signs of the coefficients are $(+, -, +)$. According to the sign rule for quadratic equations $ax^{2} + bx + c = 0$,if the signs are $(+, -, +)$,both roots are positive.
Since all values of $x$ are negative and all values of $y$ are positive,it follows that $y > x$ or $x < y$.

(D) Step $1$: Calculate the value of $x$.
$x = \sqrt[3]{357911} = 71$.
Step $2$: Calculate the value of $y$.
$y = \sqrt{5041} = 71$.
Step $3$: Compare $x$ and $y$.
Since $x = 71$ and $y = 71$,it follows that $x = y$.

(B) For equation $I: 3x^2 + 15x + 18 = 0$
Divide by $3$: $x^2 + 5x + 6 = 0$
$(x + 3)(x + 2) = 0$
So,$x_1 = -3$ and $x_2 = -2$.
For equation $II: 2y^2 + 15y + 27 = 0$
$2y^2 + 6y + 9y + 27 = 0$
$2y(y + 3) + 9(y + 3) = 0$
$(2y + 9)(y + 3) = 0$
So,$y_1 = -4.5$ and $y_2 = -3$.
Comparing the roots:
$x = \{-3, -2\}$
$y = \{-4.5, -3\}$
Since $-3 \ge -4.5$,$-3 \ge -3$,$-2 \ge -4.5$,and $-2 \ge -3$,we conclude that $x \ge y$.

(B) For equation $I$: $2x^2 - 4x - \sqrt{13}x + 2\sqrt{13} = 0$
$2x(x - 2) - \sqrt{13}(x - 2) = 0$
$(x - 2)(2x - \sqrt{13}) = 0$
Thus,$x = 2$ or $x = \frac{\sqrt{13}}{2} \approx \frac{3.605}{2} \approx 1.8025$.
For equation $II$: $10y^2 - 18y - 5\sqrt{13}y + 9\sqrt{13} = 0$
$2y(5y - 9) - \sqrt{13}(5y - 9) = 0$
$(2y - \sqrt{13})(5y - 9) = 0$
Thus,$y = \frac{\sqrt{13}}{2} \approx 1.8025$ or $y = \frac{9}{5} = 1.8$.
Comparing the values:
$x_1 = 2, x_2 \approx 1.8025$
$y_1 \approx 1.8025, y_2 = 1.8$
Since $2 > 1.8025$ and $2 > 1.8$,and $1.8025 > 1.8025$ (equal) and $1.8025 > 1.8$,we observe that $x \ge y$.

(D) Step $1$: Solve equation $I$:
$821 x^{2} - 757 x^{2} = 256$
$64 x^{2} = 256$
$x^{2} = \frac{256}{64} = 4$
$x = \pm 2$,so $x = 2$ or $x = -2$.
Step $2$: Solve equation $II$:
$\sqrt{196} y^{3} - 12 y^{3} = 16$
$14 y^{3} - 12 y^{3} = 16$
$2 y^{3} = 16$
$y^{3} = 8$
$y = 2$.
Step $3$: Compare $x$ and $y$:
If $x = 2$,then $x = y$.
If $x = -2$,then $x < y$.
Combining these,we get $x \le y$.

(D) Step $1$: Solve equation $I$ for $x$.
$13x - 21 = 200 - 4x$
$13x + 4x = 200 + 21$
$17x = 221$
$x = 221 / 17 = 13$
Step $2$: Solve equation $II$ for $y$.
$y = \sqrt[3]{2197}$
Since $13^3 = 2197$,we have $y = 13$.
Step $3$: Compare $x$ and $y$.
Since $x = 13$ and $y = 13$,it follows that $x = y$.

(C) From equation $II$:
$y + 2513 = 2569$
$y = 2569 - 2513 = 56$
Substitute $y = 56$ into equation $I$:
$(x + 56)^{2} = 3136$
Taking the square root on both sides:
$x + 56 = \pm \sqrt{3136}$
$x + 56 = \pm 56$
Case $1$: $x + 56 = 56 \Rightarrow x = 0$
Case $2$: $x + 56 = -56 \Rightarrow x = -112$
Comparing $x$ and $y$:
If $x = 0$ and $y = 56$,then $x < y$.
If $x = -112$ and $y = 56$,then $x < y$.
In both cases,$x < y$.

(B) Step $1$: Solve equation $I$:
$x^{2} = 49 \Rightarrow x = \pm 7$,so $x = 7$ or $x = -7$.
Step $2$: Solve equation $II$:
$y^{2} + 15y + 56 = 0$
Factorizing the quadratic equation: $y^{2} + 8y + 7y + 56 = 0$
$y(y + 8) + 7(y + 8) = 0$
$(y + 7)(y + 8) = 0$
So,$y = -7$ or $y = -8$.
Step $3$: Compare the values of $x$ and $y$:
If $x = 7$,then $7 > -7$ and $7 > -8$.
If $x = -7$,then $-7 = -7$ and $-7 > -8$.
In all cases,$x \ge y$.

(D) For equation $I$: $7x^{2} + 16x - 15 = 0$
We factorize the quadratic equation: $7x^{2} + 21x - 5x - 15 = 0$
$7x(x + 3) - 5(x + 3) = 0$
$(7x - 5)(x + 3) = 0$
So,$x = 5/7 \approx 0.71$ or $x = -3$.
For equation $II$: $y^{2} - 6y - 7 = 0$
We factorize the quadratic equation: $y^{2} - 7y + y - 7 = 0$
$y(y - 7) + 1(y - 7) = 0$
$(y + 1)(y - 7) = 0$
So,$y = -1$ or $y = 7$.
Comparing the values:
If $x = 0.71$,then $x > y$ (for $y = -1$) and $x < y$ (for $y = 7$).
If $x = -3$,then $x < y$ (for both $y = -1$ and $y = 7$).
Since the relationship changes depending on the values chosen,the relationship between $x$ and $y$ cannot be established.

(D) For equation $I$:
$\frac{15}{\sqrt{x}}-\frac{9}{\sqrt{x}}=\sqrt{x}$
$\frac{6}{\sqrt{x}}=\sqrt{x}$
$6 = \sqrt{x} \cdot \sqrt{x}$
$x = 6$
For equation $II$:
$y^{10} - (36)^{5} = 0$
$y^{10} = (6^2)^5$
$y^{10} = 6^{10}$
Since the exponent is even,$y = 6$ or $y = -6$.
Comparing $x$ and $y$:
If $x = 6$ and $y = 6$,then $x = y$.
If $x = 6$ and $y = -6$,then $x > y$.
Since both conditions are possible,the relationship between $x$ and $y$ cannot be established.

(D) For equation $I$:
$(441)^{\frac{1}{2}} x^{2} - 111 = (15)^{2}$
$21 x^{2} - 111 = 225$
$21 x^{2} = 336$
$x^{2} = 16 \Rightarrow x = \pm 4$
For equation $II$:
$\sqrt{121} y^{2} + (6)^{3} = 260$
$11 y^{2} + 216 = 260$
$11 y^{2} = 44$
$y^{2} = 4 \Rightarrow y = \pm 2$
Comparing the values:
If $x = 4$,then $x > y$ (since $y = 2$ or $-2$).
If $x = -4$,then $x < y$ (since $y = 2$ or $-2$).
Since we get different results for different values of $x$ and $y$,the relationship between $x$ and $y$ cannot be established.

(C) Step $1$: Solve for $x$.
$x = \sqrt[3]{2744}$. Since $14 \times 14 \times 14 = 2744$,we get $x = 14$.
Step $2$: Solve for $y$.
$y = \sqrt{487}$. We know that $22^2 = 484$ and $23^2 = 529$. Therefore,$y$ is approximately $22.068$.
Step $3$: Compare $x$ and $y$.
Since $14 < 22.068$,it follows that $x < y$.

(C) For equation $I$: $15 x^{2}-41 x+14=0$
We factorize the quadratic equation: $15 x^{2}-35 x-6 x+14=0$
$5 x(3 x-7)-2(3 x-7)=0$
$(5 x-2)(3 x-7)=0$
So,$x = \frac{2}{5} = 0.4$ or $x = \frac{7}{3} \approx 2.33$.
For equation $II$: $2 y^{2}-13 y+20=0$
We factorize the quadratic equation: $2 y^{2}-8 y-5 y+20=0$
$2 y(y-4)-5(y-4)=0$
$(2 y-5)(y-4)=0$
So,$y = \frac{5}{2} = 2.5$ or $y = 4$.
Comparing the values:
$x_1 = 0.4, x_2 = 2.33$
$y_1 = 2.5, y_2 = 4$
Since both values of $x$ are less than both values of $y$ $(0.4 < 2.5, 0.4 < 4, 2.33 < 2.5, 2.33 < 4)$,we conclude that $x < y$.

(D) For equation $I$: $x^{2}-8 \sqrt{3} x+45=0$
We need two numbers whose product is $45$ and sum is $8 \sqrt{3}$.
These numbers are $5 \sqrt{3}$ and $3 \sqrt{3}$.
$x^{2}-5 \sqrt{3} x-3 \sqrt{3} x+45=0$
$x(x-5 \sqrt{3})-3 \sqrt{3}(x-5 \sqrt{3})=0$
$(x-5 \sqrt{3})(x-3 \sqrt{3})=0$
So,$x = 5 \sqrt{3} \approx 5 \times 1.732 = 8.66$ and $x = 3 \sqrt{3} \approx 3 \times 1.732 = 5.196$.
For equation $II$: $y^{2}-\sqrt{2} y-24=0$
We need two numbers whose product is $-24$ and sum is $-\sqrt{2}$.
These numbers are $-4 \sqrt{2}$ and $3 \sqrt{2}$.
$y^{2}-4 \sqrt{2} y+3 \sqrt{2} y-24=0$
$y(y-4 \sqrt{2})+3 \sqrt{2}(y-4 \sqrt{2})=0$
$(y-4 \sqrt{2})(y+3 \sqrt{2})=0$
So,$y = 4 \sqrt{2} \approx 4 \times 1.414 = 5.656$ and $y = -3 \sqrt{2} \approx -4.242$.
Comparing the values:
$x_1 = 8.66, x_2 = 5.196$
$y_1 = 5.656, y_2 = -4.242$
Since $x_1 > y_1$ but $x_2 < y_1$,the relationship between $x$ and $y$ cannot be established.

(B) For equation $I$: $x - 7\sqrt{x} + 12 = 0$. Let $\sqrt{x} = u$,then $u^2 - 7u + 12 = 0$.
$(u - 3)(u - 4) = 0$,so $u = 3$ or $u = 4$.
Thus,$\sqrt{x} = 3 \Rightarrow x = 9$ and $\sqrt{x} = 4 \Rightarrow x = 16$.
For equation $II$: $y - 5\sqrt{y} + 6 = 0$. Let $\sqrt{y} = v$,then $v^2 - 5v + 6 = 0$.
$(v - 2)(v - 3) = 0$,so $v = 2$ or $v = 3$.
Thus,$\sqrt{y} = 2 \Rightarrow y = 4$ and $\sqrt{y} = 3 \Rightarrow y = 9$.
Comparing the values: $x \in \{9, 16\}$ and $y \in \{4, 9\}$.
Since all values of $x$ are greater than or equal to the values of $y$ $(9 \ge 4, 9 \ge 9, 16 \ge 4, 16 \ge 9)$,we conclude $x \ge y$.

(B) For equation $I: 63x - 94\sqrt{x} + 35 = 0$
Let $\sqrt{x} = u$. Then $63u^2 - 94u + 35 = 0$.
Splitting the middle term: $63u^2 - 45u - 49u + 35 = 0$.
$9u(7u - 5) - 7(7u - 5) = 0 \Rightarrow (9u - 7)(7u - 5) = 0$.
So,$u = 7/9$ or $u = 5/7$.
Thus,$\sqrt{x} = 7/9 \Rightarrow x = 49/81 \approx 0.6049$ and $\sqrt{x} = 5/7 \Rightarrow x = 25/49 \approx 0.5102$.
For equation $II: 32y - 52\sqrt{y} + 21 = 0$
Let $\sqrt{y} = v$. Then $32v^2 - 52v + 21 = 0$.
Splitting the middle term: $32v^2 - 24v - 28v + 21 = 0$.
$8v(4v - 3) - 7(4v - 3) = 0 \Rightarrow (8v - 7)(4v - 3) = 0$.
So,$v = 7/8$ or $v = 3/4$.
Thus,$\sqrt{y} = 7/8 \Rightarrow y = 49/64 \approx 0.7656$ and $\sqrt{y} = 3/4 \Rightarrow y = 9/16 = 0.5625$.
Comparing values:
$x \in \{0.5102, 0.6049\}$ and $y \in \{0.5625, 0.7656\}$.
Since $0.5102 < 0.5625$ and $0.6049 < 0.7656$,we see that $x < y$ in all cases.
Therefore,the correct option is $B$.

(A) For equation $I$: $x^{2}-7 \sqrt{3} x-5 \sqrt{5} x+35 \sqrt{15}=0$
$x(x-7 \sqrt{3})-5 \sqrt{5}(x-7 \sqrt{3})=0$
$(x-5 \sqrt{5})(x-7 \sqrt{3})=0$
So,$x = 5 \sqrt{5}$ or $x = 7 \sqrt{3}$.
For equation $II$: $y^{2}-5 \sqrt{5} y+30=0$
$y^{2}-3 \sqrt{5} y -2 \sqrt{5} y +30=0$
$y(y-3 \sqrt{5})-2 \sqrt{5}(y-3 \sqrt{5})=0$
$(y-2 \sqrt{5})(y-3 \sqrt{5})=0$
So,$y = 2 \sqrt{5}$ or $y = 3 \sqrt{5}$.
Comparing the values:
$5 \sqrt{5} \approx 5 \times 2.236 = 11.18$
$7 \sqrt{3} \approx 7 \times 1.732 = 12.124$
$2 \sqrt{5} \approx 2 \times 2.236 = 4.472$
$3 \sqrt{5} \approx 3 \times 2.236 = 6.708$
Since all values of $x$ are greater than all values of $y$,we conclude $x > y$.

(A) Given equations are:
$5x + 4y = 41$ ... $(i)$
$4x + 5y = 40$ ... (ii)
To solve,multiply equation $(i)$ by $5$ and equation (ii) by $4$:
$25x + 20y = 205$ ... (iii)
$16x + 20y = 160$ ... (iv)
Subtracting equation (iv) from equation (iii):
$(25x - 16x) + (20y - 20y) = 205 - 160$
$9x = 45$
$x = 5$
Substitute $x = 5$ into equation $(i)$:
$5(5) + 4y = 41$
$25 + 4y = 41$
$4y = 16$
$y = 4$
Since $x = 5$ and $y = 4$,we have $x > y$.

(C) For equation $I$: $\sqrt{x} = \frac{(18)^{15/2}}{x^2}$
Multiplying both sides by $x^2$,we get $x^{1/2} \cdot x^2 = (18)^{15/2}$
$x^{5/2} = (18)^{15/2}$
Raising both sides to the power of $2/5$,we get $x = (18)^{(15/2) \cdot (2/5)} = 18^3 = 5832$.
For equation $II$: $\sqrt{y} = \frac{(19)^{9/2}}{y}$
Multiplying both sides by $y$,we get $y^{1/2} \cdot y = (19)^{9/2}$
$y^{3/2} = (19)^{9/2}$
Raising both sides to the power of $2/3$,we get $y = (19)^{(9/2) \cdot (2/3)} = 19^3 = 6859$.
Comparing the values,$x = 5832$ and $y = 6859$.
Since $5832 < 6859$,therefore $x < y$.

(D) For equation $I$: $63x - 194\sqrt{x} + 143 = 0$. Let $\sqrt{x} = u$. Then $63u^2 - 194u + 143 = 0$.
Splitting the middle term: $63 \times 143 = 9009$. We need factors of $9009$ that sum to $194$. These are $117$ and $77$.
$63u^2 - 117u - 77u + 143 = 0 \Rightarrow 9u(7u - 13) - 11(7u - 13) = 0$.
$(9u - 11)(7u - 13) = 0$. So,$u = 11/9$ or $u = 13/7$.
$x_1 = (13/7)^2 = 169/49 \approx 3.45$ and $x_2 = (11/9)^2 = 121/81 \approx 1.49$.
For equation $II$: $99y - 255\sqrt{y} + 150 = 0$. Let $\sqrt{y} = v$. Then $99v^2 - 255v + 150 = 0$. Dividing by $3$: $33v^2 - 85v + 50 = 0$.
Splitting the middle term: $33 \times 50 = 1650$. Factors are $55$ and $30$.
$33v^2 - 55v - 30v + 50 = 0 \Rightarrow 11v(3v - 5) - 10(3v - 5) = 0$.
$(11v - 10)(3v - 5) = 0$. So,$v = 10/11$ or $v = 5/3$.
$y_1 = (10/11)^2 = 100/121 \approx 0.83$ and $y_2 = (5/3)^2 = 25/9 \approx 2.78$.
Comparing values: $x_1 \approx 3.45, x_2 \approx 1.49$ and $y_1 \approx 0.83, y_2 \approx 2.78$.
Since $x_1 > y_2$ but $x_2 < y_2$,the relationship cannot be established.

(C) For equation $I: x - 7\sqrt{3x} + 36 = 0$
Let $\sqrt{x} = u$. Then $u^2 - 7\sqrt{3}u + 36 = 0$.
Using the quadratic formula or factoring: $u^2 - 4\sqrt{3}u - 3\sqrt{3}u + 36 = 0$
$u(u - 4\sqrt{3}) - 3\sqrt{3}(u - 4\sqrt{3}) = 0$
$(u - 3\sqrt{3})(u - 4\sqrt{3}) = 0$
So,$u = 3\sqrt{3} = \sqrt{27}$ or $u = 4\sqrt{3} = \sqrt{48}$.
Thus,$x = 27$ or $x = 48$.
For equation $II: y - 12\sqrt{2y} + 70 = 0$
Let $\sqrt{y} = v$. Then $v^2 - 12\sqrt{2}v + 70 = 0$.
Factoring: $v^2 - 7\sqrt{2}v - 5\sqrt{2}v + 70 = 0$
$v(v - 7\sqrt{2}) - 5\sqrt{2}(v - 7\sqrt{2}) = 0$
$(v - 5\sqrt{2})(v - 7\sqrt{2}) = 0$
So,$v = 5\sqrt{2} = \sqrt{50}$ or $v = 7\sqrt{2} = \sqrt{98}$.
Thus,$y = 50$ or $y = 98$.
Comparing the values: $x \in \{27, 48\}$ and $y \in \{50, 98\}$.
In all cases,$x < y$.

(A) For equation $I$: $x^{2}-7 \sqrt{7} x+84=0$
Splitting the middle term: $x^{2}-4 \sqrt{7} x-3 \sqrt{7} x+84=0$
$x(x-4 \sqrt{7})-3 \sqrt{7}(x-4 \sqrt{7})=0$
$(x-4 \sqrt{7})(x-3 \sqrt{7})=0$
So,$x = 4 \sqrt{7}$ or $x = 3 \sqrt{7}$.
Since $\sqrt{7} \approx 2.646$,$x \approx 10.58$ or $x \approx 7.94$.
For equation $II$: $y^{2}-5 \sqrt{5} y+30=0$
Splitting the middle term: $y^{2}-3 \sqrt{5} y-2 \sqrt{5} y+30=0$
$y(y-3 \sqrt{5})-2 \sqrt{5}(y-3 \sqrt{5})=0$
$(y-3 \sqrt{5})(y-2 \sqrt{5})=0$
So,$y = 3 \sqrt{5}$ or $y = 2 \sqrt{5}$.
Since $\sqrt{5} \approx 2.236$,$y \approx 6.71$ or $y \approx 4.47$.
Comparing the values: All values of $x$ are greater than all values of $y$.
Therefore,$x > y$.

(D) For equation $I: x^{2}-2x-15=0$
We factorize the quadratic equation: $x^{2}-5x+3x-15=0$
$x(x-5)+3(x-5)=0$
$(x-5)(x+3)=0$
Thus,the roots are $x = 5$ and $x = -3$.
For equation $II: y^{2}+5y+6=0$
We factorize the quadratic equation: $y^{2}+3y+2y+6=0$
$y(y+3)+2(y+3)=0$
$(y+3)(y+2)=0$
Thus,the roots are $y = -3$ and $y = -2$.
Comparing the values:
If $x = 5$,then $x > y$ (since $5 > -3$ and $5 > -2$).
If $x = -3$,then $x = y$ (when $y = -3$) and $x < y$ (since $-3 < -2$).
Since the relationship varies depending on the chosen roots,the relationship between $x$ and $y$ cannot be established.

(B) Step $1$: Solve equation $I$.
$x - \sqrt{169} = 0$
$x - 13 = 0$
$x = 13$
Step $2$: Solve equation $II$.
$y^2 - 169 = 0$
$y^2 = 169$
$y = \pm \sqrt{169}$
$y = 13$ or $y = -13$
Step $3$: Compare $x$ and $y$.
If $x = 13$ and $y = 13$,then $x = y$.
If $x = 13$ and $y = -13$,then $x > y$.
Combining these results,we get $x \ge y$.

(D) For equation $I$: $x^{2} - 25 = 0 \Rightarrow x^{2} = 25 \Rightarrow x = \pm 5$.
So,$x = 5$ or $x = -5$.
For equation $II$: $y^{2} - 9y + 20 = 0$.
Factoring the quadratic equation: $y^{2} - 5y - 4y + 20 = 0 \Rightarrow y(y - 5) - 4(y - 5) = 0 \Rightarrow (y - 5)(y - 4) = 0$.
So,$y = 5$ or $y = 4$.
Comparing the values:
If $x = 5$,then $x = y$ (when $y = 5$) and $x > y$ (when $y = 4$).
If $x = -5$,then $x < y$ (since $-5 < 4$ and $-5 < 5$).
Since we get different results ($x = y$,$x > y$,and $x < y$) depending on the values chosen,a unique relationship between $x$ and $y$ cannot be established.

(A) Given equations are:
$I. 7x + 3y = 77$
$II. 2x + 5y = \sqrt{2601} = 51$
To solve the system,multiply equation $I$ by $2$ and equation $II$ by $7$:
$14x + 6y = 154$
$14x + 35y = 357$
Subtract the first resulting equation from the second:
$(14x + 35y) - (14x + 6y) = 357 - 154$
$29y = 203$
$y = 203 / 29 = 7$
Substitute $y = 7$ into equation $II$:
$2x + 5(7) = 51$
$2x + 35 = 51$
$2x = 16$
$x = 8$
Comparing the values,$x = 8$ and $y = 7$,so $x > y$.

(A) For equation $I$:
$(289)^{\frac{1}{2}} x - \sqrt{324} = 203$
$17x - 18 = 203$
$17x = 203 + 18$
$17x = 221$
$x = \frac{221}{17} = 13$
For equation $II$:
$(484)^{\frac{1}{2}} y - \sqrt{225} = 183$
$22y - 15 = 183$
$22y = 183 + 15$
$22y = 198$
$y = \frac{198}{22} = 9$
Comparing the values,$x = 13$ and $y = 9$. Therefore,$x > y$.

(B) Step $1$: Solve equation $I$:
$679 x^{2} - 168 x^{2} = 3066$
$511 x^{2} = 3066$
$x^{2} = \frac{3066}{511} = 6$
$x = \pm \sqrt{6} \approx \pm 2.45$
Step $2$: Solve equation $II$:
$\sqrt{144} y^{3} - 9 y^{3} = 1536$
$12 y^{3} - 9 y^{3} = 1536$
$3 y^{3} = 1536$
$y^{3} = \frac{1536}{3} = 512$
$y = \sqrt[3]{512} = 8$
Step $3$: Compare $x$ and $y$:
Since $x = \pm 2.45$ and $y = 8$,it is clear that $x < y$.

(A) Given equations are:
$I. \quad 3x + 4y = \sqrt{1681} = 41$
$II. \quad 3x + 2y = \sqrt{961} = 31$
Subtracting equation $II$ from equation $I$:
$(3x + 4y) - (3x + 2y) = 41 - 31$
$2y = 10 \Rightarrow y = 5$
Substitute $y = 5$ into equation $II$:
$3x + 2(5) = 31$
$3x + 10 = 31$
$3x = 21 \Rightarrow x = 7$
Comparing the values,$x = 7$ and $y = 5$,so $x > y$.

(D) For equation $I$: $3x^2 - 6x - \sqrt{17}x + 2\sqrt{17} = 0$
$3x(x - 2) - \sqrt{17}(x - 2) = 0$
$(x - 2)(3x - \sqrt{17}) = 0$
Thus,$x = 2$ or $x = \frac{\sqrt{17}}{3} \approx \frac{4.12}{3} \approx 1.37$.
For equation $II$: $5y^2 - 15y + \sqrt{17}y - 3\sqrt{17} = 0$
$5y(y - 3) + \sqrt{17}(y - 3) = 0$
$(y - 3)(5y + \sqrt{17}) = 0$
Thus,$y = 3$ or $y = -\frac{\sqrt{17}}{5} \approx -\frac{4.12}{5} \approx -0.82$.
Comparing the values:
$x$ values are ${2, 1.37}$ and $y$ values are ${3, -0.82}$.
Since $x < y$ for $x=2, y=3$ and $x > y$ for $x=1.37, y=-0.82$,the relationship between $x$ and $y$ cannot be established.

(B) For equation $I$: $x^{2}-16x+63=0$
We need two numbers whose product is $63$ and sum is $16$. These are $9$ and $7$.
$x^{2}-9x-7x+63=0$
$x(x-9)-7(x-9)=0$
$(x-9)(x-7)=0$
So,$x = 9$ or $x = 7$.
For equation $II$: $y^{2}-2y-35=0$
We need two numbers whose product is $-35$ and sum is $-2$. These are $-7$ and $5$.
$y^{2}-7y+5y-35=0$
$y(y-7)+5(y-7)=0$
$(y-7)(y+5)=0$
So,$y = 7$ or $y = -5$.
Comparing the values:
If $x=9$,then $x > y$ (since $y$ is $7$ or $-5$).
If $x=7$,then $x \ge y$ (since $y$ is $7$ or $-5$).
Combining these,we get $x \ge y$.

(B) Quantity $I$: Let the cost price $(CP)$ of each shirt be $Rs. 100$.
Total $CP = 100 + 100 = Rs. 200$.
Selling price $(SP)$ of the first shirt at $20 \%$ profit $= 100 \times 1.20 = Rs. 120$.
Selling price $(SP)$ of the second shirt at $10 \%$ loss $= 100 \times 0.90 = Rs. 90$.
Total $SP = 120 + 90 = Rs. 210$.
Overall profit percentage $= \frac{210 - 200}{200} \times 100 = \frac{10}{200} \times 100 = 5 \%$.
Quantity $II$: Let the $CP$ of $1 \, m$ of cloth be $Rs. x$.
Profit on $20 \, m$ cloth $= CP$ of $5 \, m$ cloth $= 5x$.
Profit percentage $= \frac{\text{Profit}}{\text{Total } CP} \times 100 = \frac{5x}{20x} \times 100 = \frac{1}{4} \times 100 = 25 \%$.
Comparing the two quantities: $5 \% < 25 \%$,therefore Quantity $I < $ Quantity $II$.

(B) Edge of the cube $a = 7 \, cm$.
$Quantity \, I$: Volume of the cube left over.
Volume of cube $= a^3 = 7^3 = 343 \, cm^3$.
Volume of the largest cylinder $= \pi r^2 h = \pi (3.5)^2 (7) = \frac{22}{7} \times 12.25 \times 7 = 269.5 \, cm^3$.
Volume left $= 343 - 269.5 = 73.5 \, cm^3$.
$Quantity \, II$: Surface area of the cube remaining.
Original surface area $= 6a^2 = 6(49) = 294 \, cm^2$.
When the cylinder is cut,two circular faces are removed from the top and bottom faces of the cube,but the curved surface area of the cylinder is exposed.
Remaining surface area $= (6a^2 - 2 \pi r^2) + 2 \pi r h = 294 - 2(\frac{22}{7})(3.5)^2 + 2(\frac{22}{7})(3.5)(7) = 294 - 77 + 154 = 371 \, cm^2$.
Comparing magnitudes: $73.5 < 371$,therefore $Quantity \, I < Quantity \, II$.

(D) Initial contents: Water $= 2.5 \, \text{L}$,Milk $= 10 \, \text{L}$. Total $= 12.5 \, \text{L}$.
After removing $20 \%$ $(1/5)$: Remaining Water $= 2.5 - (0.2 \times 2.5) = 2 \, \text{L}$. Remaining Milk $= 10 - (0.2 \times 10) = 8 \, \text{L}$.
Initial ratio (Water:Milk) $= 2:8 = 1:4$.
Adding $x$ liters of water to reverse the ratio to $4:1$: $\frac{2+x}{8} = \frac{4}{1} \Rightarrow 2+x = 32 \Rightarrow x = 30 \, \text{L}$.
Now,contents are Water $= 32 \, \text{L}$,Milk $= 8 \, \text{L}$.
Adding $y$ liters of milk to reverse the ratio back to $1:4$: $\frac{32}{8+y} = \frac{1}{4} \Rightarrow 128 = 8+y \Rightarrow y = 120 \, \text{L}$.
Since $y = 120$ and $Quantity \, 2 = 120$,$Quantity \, I = Quantity \, II$.

(B) Let the total work be the $LCM$ of $16$ and $32$,which is $32$ units.
Efficiency of $P = 32 / 16 = 2$ units/day.
Efficiency of $Q = 32 / 32 = 1$ unit/day.
Quantity $I$ ($P$ starts first):
Day $1$: $P$ does $2$ units. Day $2$: $Q$ does $1$ unit. Total $3$ units in $2$ days.
In $20$ days ($10$ cycles),$30$ units are completed.
On the $21$st day,$P$ does $2$ units. Total work $32$ units completed in $21$ days.
Quantity $II$ ($Q$ starts first):
Day $1$: $Q$ does $1$ unit. Day $2$: $P$ does $2$ units. Total $3$ units in $2$ days.
In $20$ days ($10$ cycles),$30$ units are completed.
On the $21$st day,$Q$ does $1$ unit. Remaining work $32 - 31 = 1$ unit.
$P$ completes the remaining $1$ unit in $1/2$ day.
Total time $= 21.5$ days.
Since $21 < 21.5$,Quantity $I < $ Quantity $II$.

(B) $1$. Since $AD$ is the diameter,$\angle ABD = 90^{\circ}$ (angle in a semicircle).
$2$. In $\triangle PBD$,the sum of angles is $180^{\circ}$. Thus,$\angle P + \angle PBD + \angle BDP = 180^{\circ}$. Since $\angle PBD = 90^{\circ}$ (as $P, B, A$ are collinear and $\angle ABD = 90^{\circ}$),we have $20^{\circ} + 90^{\circ} + \angle BDP = 180^{\circ}$,so $\angle BDP = 70^{\circ}$.
$3$. In $\triangle ABD$,$\angle ADB = 180^{\circ} - \angle DAB - \angle ABD = 180^{\circ} - 40^{\circ} - 90^{\circ} = 50^{\circ}$. This is Quantity $II$.
$4$. $\angle BCD = \angle DAB = 40^{\circ}$ (angles subtended by the same arc $BD$ in the same segment).
$5$. In $\triangle PBC$,the sum of angles is $180^{\circ}$. $\angle P + \angle PBC + \angle BCP = 180^{\circ}$. Note that $\angle PBC = 180^{\circ} - \angle ABD = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
$6$. Thus,$20^{\circ} + 90^{\circ} + \angle BCP = 180^{\circ} \Rightarrow \angle BCP = 70^{\circ}$.
$7$. Since $P, C, D$ are collinear,$\angle BCD + \angle BCP = 180^{\circ}$. $40^{\circ} + 70^{\circ} = 110^{\circ} \neq 180^{\circ}$. Re-evaluating: $\angle DBC = 180^{\circ} - \angle BCD - \angle BDC = 180^{\circ} - 40^{\circ} - 70^{\circ} = 70^{\circ}$.
$8$. Actually,$\angle DBC = 30^{\circ}$ as calculated from $\triangle BCD$ where $\angle BCD = 40^{\circ}$ and $\angle BDC = 180^{\circ} - 70^{\circ} = 110^{\circ}$.
$9$. Since Quantity $I = 30^{\circ}$ and Quantity $II = 50^{\circ}$,Quantity $I < $ Quantity $II$.

(A) For $Quantity$ $1$: Let radius $r = 7x$ and height $h = 10x$.
Volume of cylinder $= \pi r^2 h = 12320 \, cm^3$.
$\frac{22}{7} \times (7x)^2 \times (10x) = 12320$.
$\frac{22}{7} \times 49x^2 \times 10x = 12320$.
$22 \times 7x^3 \times 10 = 12320$.
$1540x^3 = 12320$.
$x^3 = \frac{12320}{1540} = 8$.
$x = 2$.
Height $h = 10x = 10 \times 2 = 20 \, cm$.
For $Quantity$ $2$: Volume of cone $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times (2)^2 \times 3 = 4\pi \, cm^3$.
This is recast into a cylinder of radius $2 \, cm$. Let the height be $H$.
Volume of cylinder $= \pi R^2 H = \pi \times (2)^2 \times H = 4\pi H$.
Equating volumes: $4\pi H = 4\pi$.
$H = 1 \, cm$.
Comparing $Quantity$ $1$ $(20 \, cm)$ and $Quantity$ $2$ $(1 \, cm)$: $Quantity$ $I > Quantity$ $II$.

(C) Quantity $1: p^{2}-18 p+77=0$
$\Rightarrow p^{2}-11 p-7 p+77=0$
$\Rightarrow (p-11)(p-7)=0$
$\Rightarrow p = 11, 7$
Quantity $2: 3 q^{2}-25 q+28=0$
$\Rightarrow 3 q^{2}-21 q-4 q+28=0$
$\Rightarrow 3 q(q-7)-4(q-7)=0$
$\Rightarrow (3 q-4)(q-7)=0$
$\Rightarrow q = 7, \frac{4}{3} \approx 1.33$
Comparing the values: The set of values for $p$ is ${7, 11}$ and for $q$ is ${1.33, 7}$.
Since all values of $p$ are greater than or equal to the values of $q$,we conclude that Quantity $1 \geq$ Quantity $2$.