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# If X, Y are positive real numbers such that X>Y and A is any positive real number, thenA. $\dfrac{X}{Y}\ge \dfrac{X+A}{Y+A}$B. $\dfrac{X}{Y}$$>$$\dfrac{X+A}{Y+A}$C. $\dfrac{X}{Y}\le \dfrac{X+A}{Y+A}$D. $\dfrac{X}{Y}$$<$$\dfrac{X+A}{Y+A}$

Last updated date: 29th Feb 2024
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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}

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