In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answers.

I. $x^2-32x+112=0$

II. $y^2-7y+12=0$

The correct option is:
$x\ge y$

To determine the relationship between $x$ and $y$, let us solve both quadratic equations step-by-step.

Step 1: Solve Equation (I) for $x$
The first equation is:

$x^2-32x+112=0$

We need to find two numbers that multiply to 112 and add up to -32. These numbers are -28 and -4.
Splitting the middle term:

$x^2-28x-4x+112=0$

$x(x-28)-4(x-28)=0$

$(x-28)(x-4)=0$

Thus, the values of $x$ are:
$x=28$ or $x=4$

Step 2: Solve Equation (II) for $y$
The second equation is:

$y^2-7y+12=0$

We need to find two numbers that multiply to 12 and add up to -7. These numbers are -4 and -3.
Splitting the middle term:

$y^2-4y-3y+12=0$

$y(y-4)-3(y-4)=0$

$(y-4)(y-3)=0$

Thus, the values of $y$ are:
$y=4$ or $y=3$

Step 3: Compare the values of $x$ and $y$
Let us compare each value of $x$ with every value of $y$:
- When $x=28$:
  28 > 4 and 28 > 3 (so, $x>y$)
- When $x=4$:
  4 = 4 and 4 > 3 (so, $x\ge y$)

Combining all the cases, we establish that $x$ is always greater than or equal to $y$.

Hence, the relationship is:
$x\ge y$