Question

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

Answer 1

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,

& X+A > Y+A \\\ & \dfrac{X+A}{Y+A} > 1 \\\ \end{aligned}$$ _**Complete step-by-step solution:**_ We have to assume the values of $\dfrac{X}{Y}=m$ and $\dfrac{X+A}{Y+A}=n$. From the equation $\dfrac{X}{Y}$= m we get $X=m\times Y$ Use this relation in $\dfrac{X+A}{Y+A}=n$ and simplify to get the required inequality. In the question, it is given that $$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 $\begin{aligned} & \dfrac{mY+A}{Y+A}\times \left( Y+A \right)=n\times \left( Y+A \right) \\\ & mY+A=nY+nA \\\ \end{aligned}$ By simplifying, we get $\begin{aligned} & mY-nY=nA-A \\\ & Y\left( m-n \right)=A\left( n-1 \right) \\\ \end{aligned}$ 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 $\dfrac{X}{Y}\ge \dfrac{X+A}{Y+A}$.