In each question two equations number (I) and (II) are given. You should solve both the equations and mark appropriate answer.

I. $2x^2-9x+10=0$
II. $y^2-12y+35=0$

To determine the relationship between $x$ and $y$, we need to solve both quadratic equations.

Step 1: Solve Equation I for $x$
Equation I is given by:
$2x^2-9x+10=0$

We can solve this quadratic equation using factorization. We need to find two numbers that multiply to $2\times 10=20$ and add up to $-9$. These numbers are $-4$ and $-5$.
Rewriting the middle term:
$2x^2-4x-5x+10=0$

Now, we group the terms and factor them:
$2x(x-2)-5(x-2)=0$
$(2x-5)(x-2)=0$

Thus, the roots for equation I are:
$2x-5=0\Rightarrow x=2.5$
and
$x-2=0\Rightarrow x=2$

So, the values of $x$ are $2$ and $2.5$.

Step 2: Solve Equation II for $y$
Equation II is given by:
$y^2-12y+35=0$

We need to find two numbers that multiply to $35$ and add up to $-12$. These numbers are $-5$ and $-7$.
Rewriting the middle term:
$y^2-5y-7y+35=0$

Now, group the terms and factor:
$y(y-5)-7(y-5)=0$
$(y-7)(y-5)=0$

Thus, the roots for equation II are:
$y-7=0\Rightarrow y=7$
and
$y-5=0\Rightarrow y=5$

So, the values of $y$ are $5$ and $7$.

Step 3: Compare the values of $x$ and $y$
We have:
$x=\{2,2.5\}$
$y=\{5,7\}$

Comparing each value of $x$ with each value of $y$:
- For $x=2$, both $y=5$ and $y=7$ are greater (i.e., $x<y$).
- For $x=2.5$, both $y=5$ and $y=7$ are greater (i.e., $x<y$).

Therefore, we can conclude that $x<y$ is always true.