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

I. $2x^2+13x-150=0$
II. $y^2-17y+66=0$

The correct answer is If x ≤ y.

Let's solve both quadratic equations step-by-step to find the values of x and y, and then compare them.

Step 1: Solve Equation I for x

The first equation is:

$2x^2+13x-150=0$

We need to find two numbers whose product is equal to $2\times (-150)=-300$ and whose sum is equal to $13$.

These two numbers are $25$ and $-12$, because:

$25\times (-12)=-300$

$25+(-12)=13$

Splitting the middle term using these factors:

$2x^2-12x+25x-150=0$

$2x(x-6)+25(x-6)=0$

$(2x+25)(\mathrm{x<mo>-</mo><mn>6</mn><mo>)</mo><mo>=</mo><mn>0</mn>}$

Now, set each factor equal to zero:

$2x+25=0\Rightarrow x=-\frac{25}{2}=-12.5$

$x-6=0\Rightarrow x=6$

So, the values of x are $x=-12.5$ and $x=6$.

Step 2: Solve Equation II for y

The second equation is:

$y^2-17y+66=0$

We need to find two numbers whose product is $66$ and whose sum is $-17$.

These two numbers are $-11$ and $-6$, because:

$(-11)\times (-6)=66$

$(-11)+(-6)=-17$

Factoring the equation:

$(y-11)(y-6)=0$

Setting each factor to zero:

$y-11=0\Rightarrow y=11$

$y-6=0\Rightarrow y=6$

So, the values of y are $y=11$ and $y=6$.

Step 3: Compare the values of x and y

Let's compare every possible pair of x and y values:

1. When $x=-12.5$ and $y=11$: $x<y$

2. When $x=-12.5$ and $y=6$: $x<y$

3. When $x=6$ and $y=11$: $x<y$

4. When $x=6$ and $y=6$: $x=y$

Combining all cases, we find that $x\le y$.