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# Is (y - 3x)/(y - 2x) > 1?

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Is (y - 3x)/(y - 2x) > 1?  [#permalink]

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30 Jan 2017, 10:11
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95% (hard)

Question Stats:

32% (01:09) correct 68% (01:50) wrong based on 146 sessions

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Is (y - 3x)/(y - 2x) > 1?

(1) (y - 2x)/x < 0
(2) y - 2x > 0

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Re: Is (y - 3x)/(y - 2x) > 1?  [#permalink]

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30 Jan 2017, 10:52
2
Is (y - 3x)/(y - 2x) > 1?
Is (y - 3x)/(y - 2x) - 1 > 0?
Is (-x)/(y - 2x) > 0?
Is (x)/(y - 2x) < negative?

St1: (y - 2x)/x < 0 --> 1/((y - 2x)/x) is negative.
Sufficient.

St2: y - 2x > 0 --> We do not know whether the numerator is positive, negative or zero.
Not Sufficient.

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Is (y - 3x)/(y - 2x) > 1?  [#permalink]

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31 Jan 2017, 07:25
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GMATPrepNow wrote:
Is (y - 3x)/(y - 2x) > 1?

(1) (y - 2x)/x < 0
(2) y - 2x > 0

Let's spend a little time at the beginning to rephrase the target question.

Target question: Is (y - 3x)/(y - 2x) > 1?
We can use a nice property of fractions to simplify this target question.
The property is (a + b)/c = a/c + b/c.
We can also write: (a - b)/c = a/c - b/c

Given: (y - 3x)/(y - 2x) > 1
Rewrite numerator as: (y - 2x - x)/(y - 2x) > 1
Apply fraction property to get: (y - 2x)/(y - 2x) - x/(y - 2x) > 1
Simplify: 1 - x/(y - 2x) > 1
Subtract 1 from both sides: -x/(y - 2x) > 0
Add -x/(y - 2x) to both side to get: 0 > x/(y - 2x)

Great, we've taken the inequality (y - 3x)/(y - 2x) > 1 and rewritten it as x/(y - 2x) < 0

REPHRASED target question: Is x/(y - 2x) < 0?

With this easier (REPHRASED) target question, it will be very easy to handle the statement...

Statement 1: (y - 2x)/x < 0
PERFECT!
If (y - 2x)/x is negative, then the reciprocal, x/(y - 2x) must also be negative.
In other words, we can be certain that x/(y - 2x) < 0
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: y - 2x > 0
This tells us that the DENOMINATOR in x/(y - 2x) is positive, but we don't know anything about the numerator x.
So, we can't determine whether or not x/(y - 2x) < 0
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

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Re: Is (y - 3x)/(y - 2x) > 1?  [#permalink]

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06 Jul 2017, 03:12
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GMATPrepNow wrote:
Is (y - 3x)/(y - 2x) > 1?

(1) (y - 2x)/x < 0
(2) y - 2x > 0

* Kudos for all correct solutions.

Statement 1 : $$\frac{(y - 2x)}{x} < 0$$

$$\frac{y}{x} - 2 < 0$$

$$\frac{y}{x} < 2$$

Lets rephrase the given equation: $$\frac{(y - 3x)}{(y - 2x)} > 1$$

$$(\frac{y}{x} - 3)/(\frac{y}{x} - 2) > 1$$

In order to get the fraction greater than 1 we need same sign for numerator and denominator and the numerator > denominator

From Statement 1 :$$\frac{y}{x}< 2$$, Both the num and denom are negative and num>denom Sufficient

Statement 2 : $$y - 2x > 0$$

$$\frac{y}{x} - 2 >0$$

$$\frac{y}{x} > 2$$

This statement will not work if $$2 < \frac{y}{x} <3$$ Not Sufficient.
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Re: Is (y - 3x)/(y - 2x) > 1?  [#permalink]

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02 Aug 2018, 23:04
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