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

Question

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

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

* Kudos for all correct solutions.

Kurtosis · Accepted Answer

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.

Answer: A

BrentGMATPrepNow · Answer

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 Answer: A RELATED VIDEO FROM OUR COURSE https://www.youtube.com/watch?v=MyG1K3ee69w

RMD007 · Answer

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.

ccooley · Answer

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 .

khera · Answer

What level of que is this ?

Posted from my mobile device

Bunuel · Answer

You can check difficulty level of a question in the tags just above the first post:

BrentGMATPrepNow · Answer

You can also see that users correctly answer the question only 34% of the time, which means it's a super difficult question.

bumpbot · Answer

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

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

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