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Is x>x/y?

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by Max@Math Revolution » Tue Sep 25, 2018 2:34 am

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[Math Revolution GMAT math practice question]

Is x>x/y?

1) y>1
2) xy>0
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Source: — Data Sufficiency |

by fskilnik@GMATH » Tue Sep 25, 2018 8:08 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is x>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 [spoiler]__(C)______[/spoiler] .

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

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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by Jay@ManhattanReview » Tue Sep 25, 2018 9:25 pm
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is x > x/y?

1) y > 1
2) xy > 0
We need to determine whether x > x/y.

Let's take each statement one by one.

1) y > 1

Certainly insufficient since we have no information about the value of x; x can be anything 0, positive, or negative.

2) xy > 0

=> x and y both are either positive or both are negative.

Case 1: Say x = y = 1

Is x > x/y => Is 1 > 1/1 => No since 1 = 1

Case 2: Say x = 1; y = 2

Is x > x/y => Is 1 > 1/2 => Yes since 1 > 1/2.

No unique answer.

(1) and (2) together

From (1), we have y > 1 (positive), thus, from (2), x is also positive

From x > x/y, we have 1 > 1/y; we can cancell x since x is positive.

1 > 1/y => y > 1, which is what Statement 1 states. Sufficient.

The correct answer: C

Hope this helps!

-Jay
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by Max@Math Revolution » Wed Sep 26, 2018 11:47 pm
=>

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>x/y is equivalent to xy(y-1) > 0 as shown below:
x>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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