Question

If X, Y are positive real numbers such that X>Y and A is any positive real number then

• A. $\frac{X}{Y}\ge \frac{X+A}{Y+A}$
• B. $\frac{X}{Y}>\frac{X+A}{Y+A}$
• C. $\frac{X}{Y}\le \frac{X+A}{Y+A}$
• D. $\frac{X}{Y}<\frac{X+A}{Y+A}$

Answer 1

The correct option is A $\frac{X}{Y}\ge \frac{X+A}{Y+A}$

If X and Y are positive real numbers such that X>Y, and A is any positive real number,

Then to solve it, we have deduced a relation between X+A and Y+A.

Given : X>Y

so, $X+A>Y+A$

$\frac{X+A}{Y+A}>1$

Let $\frac{X}{Y}=m$ and $\frac{X+A}{Y+A}=n$

Then, $X+mY$, put this value of X in n,

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

Multiply $Y+A$ on both sides,

$m(Y+A)=n(Y+A)$

$mY+A=nY+nA$

By simplifying,

$mY-nY=nA-A$

$Y(m-n)=A(n-1)$

$\frac{Y}{A}=\frac{n-1}{m-n}$ where, $Y>0,A>0,n>1,n-1>0$

So we can write, $m-n>0$, i.e $m?n$

Substituting values we get,

$\frac{X}{Y}>\frac{X+A}{Y+A}$