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

I. $x^2-10x+24=0$
II. $5y^2-16y+12=0$

The correct option is If x>y.

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

Step 1: Solve Equation I for $x$

The first equation is given as:
$x^2-10x+24=0$

To solve this by factorization, we need to find two numbers that multiply to 24 and add up to -10. These numbers are -6 and -4.
By splitting the middle term, we get:
$x^2-6x-4x+24=0$

Factor out the common terms:
$x(x-6)-4(x-6)=0$
$(x-4)(x-6)=0$

This gives the roots:
$x=4$ or $x=6$

Step 2: Solve Equation II for $y$

The second equation is given as:
$5y^2-16y+12=0$

To factor this quadratic equation, we look for two numbers that multiply to $5\times 12=60$ and add up to -16. These numbers are -10 and -6.
By splitting the middle term, we get:
$5y^2-10y-6y+12=0$

Factor out the common terms:
$5y(y-2)-6(y-2)=0$
$(5y-6)(y-2)=0$

This gives the roots:
$y=\frac{6}{5}=1.2$ or $y=2$

Step 3: Compare the values of $x$ and $y$

Let us compare each value of $x$ with each value of $y$:
- Comparing $x=4$ with $y=1.2$ and $y=2$:
$4>1.2$ and $4>2$ (so, $x>y$)
- Comparing $x=6$ with $y=1.2$ and $y=2$:
$6>1.2$ and $6>2$ (so, $x>y$)

Since all values of $x$ are strictly greater than all values of $y$, we conclude that $x>y$.