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Hint: To solve this question, we have to deduce the relation between $X+A$ and $Y+A$. From the question, we can write \[X > Y\], and adding A on both sides doesn’t change the inequality. Adding A on both sides, we get,

\[\begin{align}

& X+A > Y+A \\

& \dfrac{X+A}{Y+A} > 1 \\

\end{align}\]

We have to assume the values of $\dfrac{X}{Y}=m$ and $\dfrac{X+A}{Y+A}=n$. From the equation $\dfrac{X}{Y}$= m we get

$X=m\times Y$

Use this relation in $\dfrac{X+A}{Y+A}=n$ and simplify to get the required inequality.

In the question, it is given that \[X > Y\], A is a positive real number we can add A on both sides and the inequality remains the same. By adding A, we get

\[X+A > Y+A\]

As Y + A is a positive number, we can divide the above equation with Y + A and the inequality remains the same.

\[\dfrac{X+A}{Y+A} > 1\to \left( 1 \right)\]

Let us assume that $\dfrac{X}{Y}=m$ and $\dfrac{X+A}{Y+A}=n > 1\to \left( 2 \right)$.

From $\dfrac{X}{Y}=m$, we get

$X=mY$

Using this in the equation $\dfrac{X+A}{Y+A}=n$, we get

$\dfrac{mY+A}{Y+A}=n$

Multiplying with $Y+A$ on both sides, we get

$\begin{align}

& \dfrac{mY+A}{Y+A}\times \left( Y+A \right)=n\times \left( Y+A \right) \\

& mY+A=nY+nA \\

\end{align}$

By simplifying, we get

$\begin{align}

& mY-nY=nA-A \\

& Y\left( m-n \right)=A\left( n-1 \right) \\

\end{align}$

Cross-multiplying gives

$\dfrac{Y}{A}=\dfrac{n-1}{m-n}$

If we observe the equation that is written above, the terms

$Y > 0$ as given in the question,

$A > 0$ as given in the question,

$n > 1\; \text{and}\; n – 1 > 0$ from equation-2.

So, we can write that the term $m-n > 0$ as the remaining terms in the equation are positive.

As $m-n > 0$, we can write that

$m > n$

Substituting the values of m, n in the above inequality, we get

$\dfrac{X}{Y} > \dfrac{X+A}{Y+A}$.

\[\begin{align}

& X+A > Y+A \\

& \dfrac{X+A}{Y+A} > 1 \\

\end{align}\]

**Complete step-by-step solution:**We have to assume the values of $\dfrac{X}{Y}=m$ and $\dfrac{X+A}{Y+A}=n$. From the equation $\dfrac{X}{Y}$= m we get

$X=m\times Y$

Use this relation in $\dfrac{X+A}{Y+A}=n$ and simplify to get the required inequality.

In the question, it is given that \[X > Y\], A is a positive real number we can add A on both sides and the inequality remains the same. By adding A, we get

\[X+A > Y+A\]

As Y + A is a positive number, we can divide the above equation with Y + A and the inequality remains the same.

\[\dfrac{X+A}{Y+A} > 1\to \left( 1 \right)\]

Let us assume that $\dfrac{X}{Y}=m$ and $\dfrac{X+A}{Y+A}=n > 1\to \left( 2 \right)$.

From $\dfrac{X}{Y}=m$, we get

$X=mY$

Using this in the equation $\dfrac{X+A}{Y+A}=n$, we get

$\dfrac{mY+A}{Y+A}=n$

Multiplying with $Y+A$ on both sides, we get

$\begin{align}

& \dfrac{mY+A}{Y+A}\times \left( Y+A \right)=n\times \left( Y+A \right) \\

& mY+A=nY+nA \\

\end{align}$

By simplifying, we get

$\begin{align}

& mY-nY=nA-A \\

& Y\left( m-n \right)=A\left( n-1 \right) \\

\end{align}$

Cross-multiplying gives

$\dfrac{Y}{A}=\dfrac{n-1}{m-n}$

If we observe the equation that is written above, the terms

$Y > 0$ as given in the question,

$A > 0$ as given in the question,

$n > 1\; \text{and}\; n – 1 > 0$ from equation-2.

So, we can write that the term $m-n > 0$ as the remaining terms in the equation are positive.

As $m-n > 0$, we can write that

$m > n$

Substituting the values of m, n in the above inequality, we get

$\dfrac{X}{Y} > \dfrac{X+A}{Y+A}$.

**$\therefore $The answer is $\dfrac{X}{Y} > \dfrac{X+A}{Y+A}$. The answer is option-B.****Note:**There is a possibility of a mistake while selecting the option. The options A and B are quite close to each other. The option A will be true if the condition given in the question is $X\ge Y$, but the condition is $X > Y$, so we can only have the answer as $\dfrac{X}{Y}$$>$$\dfrac{X+A}{Y+A}$ but not $\dfrac{X}{Y}\ge \dfrac{X+A}{Y+A}$.Recently Updated Pages

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