Directions: In the following question, two equations numbered I and II are given. Solve both equations and choose the correct option.

I. $3x^2=432$

II. $y=\sqrt{144}$

The correct option is If x < y.

Step 1: Solve Equation I

The given equation I is:

$3x^2=432$

Divide both sides of the equation by 3:

$x^2=\frac{432}{3}$

$x^2=144$

Taking the square root on both sides gives two values for x:

$x=12$

or

$x=-12$

Step 2: Solve Equation II

The given equation II is:

$y=\sqrt{144}$

Evaluating the principal square root gives:

$y=12$

Step 3: Compare the values of x and y

Comparing the possible values of x with y:

1. When $x=-12$ and $y=12$, we have $-12<12$, so $x<y$.

2. When $x=12$ and $y=12$, we have $x=y$.

Considering the negative root, we obtain the relationship $x<y$.

Hence, the correct option is If x < y.