In the given questions, two quantities are given, one as ‘Quantity I’ and another as ‘Quantity II’. You have to determine relationship between two quantities and choose the appropriate option. Compare only numerical values

Given a > b and a > 4, while b < 5.

$\text{Quantity I }\frac{1}{a}+\frac{1}{b}=0$

$\text{Quantity II }(\frac{1}{b}-\frac{1}{a}){\mathrm{²}}^{}=0$

The correct option is Quantity I < Quantity II.

Let us analyze the given information step-by-step to find the values or relations for Quantity I and Quantity II.

We are given the following conditions for the variables $a$ and $b$:
1. $a>b$
2. $a>4$
3. $b<5$

Let's first analyze the equations representing each Quantity:

Analysis of Quantity I:
The equation for Quantity I is given as:
$\frac{1}{a}+\frac{1}{b}=0$
From this, we can write:
$\frac{1}{a}=-\frac{1}{b}$
Which simplifies to:
$a=-b$ (or equivalently, $b=-a$)

Since we are given $a>4$, $a$ must be a positive number.
If $a>4$ (positive), then $b=-a$ must be negative ($b<-4$).
This satisfies both $a>b$ (since a positive number is greater than a negative number) and $b<5$.
Thus, for Quantity I, the relation is $a=-b$.

Analysis of Quantity II:
The equation for Quantity II is:
${(\frac{1}{b}-\frac{1}{a})}^2=0$
Taking the square root on both sides:
$\frac{1}{b}-\frac{1}{a}=0$
This simplifies to:
$\frac{1}{b}=\frac{1}{a}$
Which means:
$a=b$

However, the original condition states that $a>b$. Since $a$ cannot be equal to $b$, the expression ${(\frac{1}{b}-\frac{1}{a})}^2$ must be strictly greater than $0$ because the square of any non-zero real number is always positive.

Comparing the numerical values:
The value of Quantity I is $0$.
The value of Quantity II, under the condition $a>b$ (so $a\ne b$), is ${(\frac{1}{b}-\frac{1}{a})}^2>0$.

Since Quantity I is equal to $0$ and Quantity II is greater than $0$, we have:
Quantity I < Quantity II