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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, 09:11
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Is (y - 3x)/(y - 2x) > 1?

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

* Kudos for all correct solutions.

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

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30 Jan 2017, 09:52
6
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.

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

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31 Jan 2017, 06:25
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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, 02:12
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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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Is (y - 3x)/(y - 2x) > 1?  [#permalink]

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13 Sep 2019, 12:26
1
GMATPrepNow wrote:
Is (y - 3x)/(y - 2x) > 1?

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

* Kudos for all correct solutions.

Looking at the statements, my first thought is that this isn't really an inequalities/algebra problem - it's probably more like a positive/negative problem. However, before I start looking at positives and negatives, I'm going to simplify the math in the question. Always do this if there's anything you can simplify!

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

The top and bottom of the fraction are similar, so I'll rewrite them to be alike:

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

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

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

Is -(x)/(y-2x) > 0?

Is x/(y-2x) < 0?

In other words, "is x/(y-2x) negative?"

Now let's check out the statements:

Statement 1 already tells us the answer! If a fraction is negative, it stays negative when you flip it. The number properties reason for this is that a negative fraction has a top and bottom with different signs - one is positive, and one is negative. If you switch the top and bottom, this will still be the case, and it doesn't matter which is which - the fraction still comes out negative. This statement is sufficient.

Statement 2 says that y-2x is negative. So, the bottom of the fraction is negative. The answer to the question will depend on whether the top is also negative.

x could be negative as well. For instance, if y = -100 and x = -1, then x and y-2x are both negative.

But x could also be positive. If y = 1 and x = 100, then x is positive and y-2x is negative.

So, this statement is insufficient, and the answer is A.
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Is (y - 3x)/(y - 2x) > 1?   [#permalink] 13 Sep 2019, 12:26
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