Directions: In the following question, two equations numbered I and II are given. Solve both equations to determine the relationship between x and y.

$\text{I.}{x}^2-9x+20=0$

$\text{II.}{y}^2-15y+56=0$

The correct option is If x < y.

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

Step 1: Solve Equation I

The given equation I is:

$x^2-9x+20=0$

Factorize the quadratic equation by splitting the middle term:

$x^2-5x-4x+20=0$

$x(x-5)-4(x-5)=0$

$(x-4)(x-5)=0$

Thus, the roots for x are:

$x=4$ or $x=5$

Step 2: Solve Equation II

The given equation II is:

$y^2-15y+56=0$

Factorize the quadratic equation by splitting the middle term:

$y^2-8y-7y+56=0$

$y(y-8)-7(y-8)=0$

$(y-7)(y-8)=0$

Thus, the roots for y are:

$y=7$ or $y=8$

Step 3: Compare the values of x and y

Comparing each value of x with the values of y:
- When x = 4: 4 < 7 and 4 < 8.
- When x = 5: 5 < 7 and 5 < 8.

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

Therefore, the relationship between x and y is $x<y$.