Directions (42-46) : In each question two equations numbered (I) and (II) are given. You have to solve both the equations and mark appropriate answer.

x²-21x+108=0

y²-14y+48=0

The correct option is If x>y.

Let us solve both quadratic equations step-by-step to determine the relationship between $x$ and $y$.

Step 1: Solve Equation (I) for $x$

The first equation is:

$x^2-21x+108=0$

We need to find two numbers whose product is 108 and whose sum is -21. These numbers are -12 and -9.

Factoring the equation:

$(x-12)(x-9)=0$

Therefore, the values of $x$ are:

$x=12$ or $x=9$

Step 2: Solve Equation (II) for $y$

The second equation is:

$y^2-14y+48=0$

We need to find two numbers whose product is 48 and whose sum is -14. These numbers are -8 and -6.

Factoring the equation:

$(y-8)(y-6)=0$

Therefore, the values of $y$ are:

$y=8$ or $y=6$

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

The roots of $x$ are 12 and 9.

The roots of $y$ are 8 and 6.

Comparing each value of $x$ with each value of $y$:

• For $x=12$: $12>8$ and $12>6$

• For $x=9$: $9>8$ and $9>6$

In all cases, every value of $x$ is strictly greater than every value of $y$.

Thus, the correct relation is $x>y$.