In each of these questions, two equation (I) and (II) are given. You have to solve both the equations and give answer

$\begin{array}{ll}\text{I.}& 3x^2-5x-28=0\\ \text{II.}& y^2-10y+25=0\end{array}$

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

Correct Option: If x ≤ y

Step 1: Solve Equation I for $x$

The first equation is:

$3x^2-5x-28=0$

We split the middle term $-5x$ into two terms such that their product is equal to $3\times (-28)=-84$ and their sum is equal to $-5$. The numbers are $-12$ and $7$.

$3x^2-12x+7x-28=0$

Factoring by grouping:

$3x(x-4)+7(x-4)=0$

$(3x+7)(x-4)=0$

This gives the values for $x$:

$x=4$
or
$x=-\frac{7}{3}\approx -2.33$

Step 2: Solve Equation II for $y$

The second equation is:

$y^2-10y+25=0$

This can be recognized as a perfect square expansion $(y-5{)}^2=0$:

$(y-5)(y-5)=0$

This gives the value for $y$:

$y=5$

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

Comparing each value of $x$ with $y=5$:

1. For $x=4$: $4<5$, so $x<y$
2. For $x=-2.33$: $-2.33<5$, so $x<y$

In both cases, $x<y$, which satisfies the condition If x ≤ y.