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The question is taken from a general GRE math quantitative comparison question.

Problem: $x>y$ and $xy\ne0$
Quantity A:
$ \displaystyle\frac{x^2}{{1+ \frac{1}{y}} }$
Quantity B:
$ \displaystyle\frac{y^2}{{1+ \frac{1}{x}} }$

How can I definitivly tell without plotting graphs whether Quantity A or B is greater or smaller or different in different ranges, for all Real values?

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  • $\begingroup$ multiply by the denominator since it has the same denominator. Rest is trivial. $\endgroup$ – ZirconCode Jan 17 '17 at 9:00
  • $\begingroup$ @ZirconCode not that easy, different answer whether denominator is greater than or less than zero. $\endgroup$ – Mark Jan 17 '17 at 9:01
  • $\begingroup$ @Mark that is true, I missed that $\endgroup$ – ZirconCode Jan 17 '17 at 9:02
  • $\begingroup$ Edit for us old, blind people. $\endgroup$ – David Mitra Jan 17 '17 at 9:08
  • $\begingroup$ I am incredibly sorry for my mistake in the question. The two denominators aren't equal. I have fixed it now. Thanks. $\endgroup$ – Rio1210 Jan 17 '17 at 9:14
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After some implications you should examine the below two parts \begin{align} x(x+1) ? y(y+1) \end{align}

x > y and (x+1)>(y+1) for each x,y > 0. Then Quantity A > Quantity B

However, this cannot be always valid if x,y < 0 or x>0 and y<0.

The answer to your question is that you cannot define a relation between A and B for all Real values nor different ranges.

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  • $\begingroup$ I am Incredibly sorry, There has been a typing mistake. The two denominators are not equal. I am fixing it at once. $\endgroup$ – Rio1210 Jan 17 '17 at 9:12
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For this kind of GRE quant questions, the best thing to do is plug-in numbers.

For $x=2$ and $y=1$,

$A=3$ and $B=\dfrac{3}{2}$; so Quantity A is greater than Quantity B.

However, if $x=1$ and $y=-1$; Quantity A takes $0$ in the denominator, which is undefined.

So the relationship cannot be determined from the information given.

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I would simplify the expression:

$$\frac{x^2}{1+\frac{1}{y}}?\frac{y^2}{1+\frac{1}{x}}\to \frac{x^2y}{y+1}?\frac{y^2x}{x+1}\to \frac{x}{y+1}?\frac{y}{x+1}\to (x^2+x)?(y^2+y)$$

For positive x and y, it is trivially true that $(x^2+x)>(y^2+y)$ since x>y

For negative x and y, we can quickly find an example where the opposite is true. $(-1)^2+-1=0<2=(-2)^2+-2$

This shows that they are different for different ranges.

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