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

I. $2x^2-11x+15=0$
II. $y^2-11y+30=0$

The correct option is If x < y.

We are given two quadratic equations. Let us solve them step-by-step to determine the relationship between $x$ and $y$.

Step 1: Solve Equation I for x

The first equation is:

$2x^2-11x+15=0$

We need to find two numbers whose product is $2\times 15=30$ and whose sum is $-11$. These two numbers are $-5$ and $-6$.

Splitting the middle term:

$2x^2-6x-5x+15=0$

$2x(x-3)-5(x-3)=0$

$(2x-5)(x-3)=0$

Setting each factor equal to zero:

$2x-5=0\Rightarrow x=\frac{5}{2}=2.5$

$x-3=0\Rightarrow x=3$

So, the values of $x$ are $2.5$ and $3$.

Step 2: Solve Equation II for y

The second equation is:

$y^2-11y+30=0$

We need to find two numbers whose product is $30$ and whose sum is $-11$. These two numbers are $-5$ and $-6$.

Splitting the middle term:

$y^2-5y-6y+30=0$

$y(y-5)-6(y-5)=0$

$(y-5)(y-6)=0$

Setting each factor equal to zero:

$y-5=0\Rightarrow y=5$

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

So, the values of $y$ are $5$ and $6$.

Step 3: Compare the values of x and y

Let us compare each value of $x$ with each value of $y$:

• When $x=2.5$, both $2.5<5$ and $2.5<6$ ($x<y$).

• When $x=3$, both $3<5$ and $3<6$ ($x<y$).

Since $x$ is strictly less than $y$ in all cases, the correct relationship is $x<y$.