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

I: $x^2-20x+91=0$
II: $y^2-28y+195=0$

To find the relationship between $x$ and $y$, we need to solve both quadratic equations separately and compare their roots.

Step 1: Solve Equation (I) for $x$
The first equation is:
$x^2-20x+91=0$

We can solve this quadratic equation by factoring. We look for two numbers that multiply to $91$ and add up to $-20$.
These two numbers are $-13$ and $-7$, since:
$(-13)\times (-7)=91$
and
$(-13)+(-7)=-20$

Thus, we can rewrite the equation as:
$(x-13)(x-7)=0$

This gives the roots:
$x=13$ or $x=7$

Step 2: Solve Equation (II) for $y$
The second equation is:
$y^2-28y+195=0$

Similarly, we look for two numbers that multiply to $195$ and add up to $-28$.
These two numbers are $-15$ and $-13$, since:
$(-15)\times (-13)=195$
and
$(-15)+(-13)=-28$

Thus, we can rewrite the equation as:
$(y-15)(y-13)=0$

This gives the roots:
$y=15$ or $y=13$

Step 3: Compare the values of $x$ and $y$
Let us compare each value of $x$ with each value of $y$:
- For $x=7$:
Comparing with $y=13$, we have $7<13$ ($x<y$).
Comparing with $y=15$, we have $7<15$ ($x<y$).
- For $x=13$:
Comparing with $y=13$, we have $13=13$ ($x=y$).
Comparing with $y=15$, we have $13<15$ ($x<y$).

Combining all the comparisons, we find that $x$ is either less than or equal to $y$ in all cases.

Therefore, the relationship is $x\le y$.