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Is x>x/y?

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Math Revolution GMAT Instructor
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Is x>x/y?  [#permalink]

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New post 25 Sep 2018, 03:40
1
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

54% (01:19) correct 46% (01:09) wrong based on 37 sessions

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[Math Revolution GMAT math practice question]

\(Is x>\frac{x}{y}?\)

\(1) y>1\)
\(2) xy>0\)

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Re: Is x>x/y?  [#permalink]

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New post 25 Sep 2018, 04:26
Here is my take on this question

x > x/y

St 1: y >1
No information about x
Case 1: x=2, y=2
2>2/2
2>1 (Yes)

Case 2: x=-2, y=2
-2>-2/2
-2<-1 (No)

AD/BCE

St 2: xy>0

Case 1: x=2, y=2
2>2/2
2>1 (Yes)

Case 2: x=-1,y=-2
-1>-1/-2
-1<1/2 (No)

Hence Insufficient BCE

St 1 + St2:
y>1 and xy>0
so x and y have to be positive.

So x will be always greater than x/y. Hence C is the answer
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Re: Is x>x/y?  [#permalink]

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New post 25 Sep 2018, 06:47
Change the original condition
x>x/y

=> x - x/y > 0
=> x(1 - 1/y) > 0

(1) y > 1 -> we don't know anything about x so INSUFFICIENT

(2) xy > 0, so we knkow x and y have same signs. Looking at the inequality we can for sure say that there must be cases when inequality is less than 0 for both positive or both negative x and y, and there must be cases when inequality is larger than 0 for both positive or both negative x and y.
-> As both less and larger than 0 case happen, this option is insufficient.

(1) (2) together, looking at 1 - 1/y, obviously as y is larger than 1, 1 - 1/y is always positive. Plus we have x is positive too as xy >0
-> Sufficient
C is the answer

Plz let rain the kudosssss
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Re: Is x>x/y?  [#permalink]

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New post 25 Sep 2018, 09:13
MathRevolution wrote:
[Math Revolution GMAT math practice question]

Is \(x>\frac{x}{y}?\)

\(1) y>1\)
\(2) xy>0\)

\(x\,\,\mathop > \limits^? \,\,\frac{x}{y}\)

\(\left( 1 \right)\,\,\,y > 1\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {0,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,\left( {1,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\end{gathered} \right.\)

\(\left( 2 \right)\,\,\,xy > 0\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {1,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\,{\text{Take}}\,\,\left( {1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\, \hfill \\
\end{gathered} \right.\)


\(\left( {1 + 2} \right)\,\,\,\, \Rightarrow \,\,\,\,x,y\,\, > 0\,\,\,\)

\(x\,\,\mathop > \limits^? \,\,\frac{x}{y}\,\,\,\,\,\mathop \Leftrightarrow \limits^{:\,\,x\,\, > \,\,0} \,\,\,\,1\,\,\mathop > \limits^? \,\,\,\frac{1}{y}\,\,\,\,\,\mathop \Leftrightarrow \limits^{ \cdot \,\,y\,\, > \,\,0} \,\,\,\,y\,\,\mathop > \limits^? \,\,\,1\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)


The correct answer is therefore (C).


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Is x>x/y?  [#permalink]

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New post 27 Sep 2018, 00:53
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The question \(x>\frac{x}{y}\) is equivalent to \(xy(y-1) > 0\) as shown below:
\(x>\frac{x}{y}\)
\(=> xy^2 > xy\) since \(y^2 > 0\)
\(=> xy^2 – xy > 0\)
\(=> xy(y – 1) > 0\)

Condition 1) tells us that \(y – 1 > 0\), and condition 2) tells us that \(xy > 0\). Thus, both conditions together are sufficient.

Therefore, C is the answer.
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Re: Is x>x/y? &nbs [#permalink] 27 Sep 2018, 00:53
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