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

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

by Brent@GMATPrepNow » Mon Jan 30, 2017 9:10 am
Here's a practice question I just created. I'd say the difficulty level is in the 600 to 700 range.
Is (y - 3x)/(y - 2x) > 1?

(1) (y - 2x)/x < 0
(2) y - 2x > 0
Answer: A
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by DavidG@VeritasPrep » Mon Jan 30, 2017 10:00 am
Brent@GMATPrepNow wrote:Here's a practice question I just created. I'd say the difficulty level is in the 600 to 700 range.
Is (y - 3x)/(y - 2x) > 1?

(1) (y - 2x)/x < 0
(2) y - 2x > 0
Answer: A
Brent original!

It's worth taking a moment to consider the characteristics of a fraction that is greater than 1. Take two simple examples.

Case A: 4/3; In this case the numerator and denominator are both positive and the numerator is larger than the denominator.

Case B: (-4)/(-3); In this case the numerator and denominator are both negative and the numerator is smaller than the denominator.

So when we're asked if a fraction is greater than 1, there are two relevant questions: are numerator and denominator both + or -? And do we know which is larger?

Statement 2 we can evaluate quickly. It tells us that the denominator of the fraction (y - 2x) is negative, but it tells us nothing about the numerator (y - 3x.) Alone this is not sufficient.

Statement 1, (y - 2x)/x <0, is interesting. If a fraction is negative, either the numerator is positive and the denominator is negative, or vice versa. So let's examine both scenarios.

It could be the case that y - 2x > 0 and x < 0. If we multiply the second inequality by (-1), it will become - x > 0, and we can add them.
y - 2x > 0
-x > 0
y - 3x > 0. Interesting. In this scenario we see that the numerator (y - 3x) and the denominator (y - 2x) are both positive. Moreover, if x is negative, y - 3x will be greater than y - 2x, as we'll be subtracting a negative, the functional equivalent of adding a positive. This gives us the aforementioned Case A, yielding a YES to the question.

Alternatively, it could be the case that y - 2x < 0 and x > 0. If we multiply the second inequality by (-1), it will become - x < 0, and we can add them.
y - 2x < 0
-x < 0
y - 3x < 0. In this scenario we see that the numerator (y - 3x) and the denominator (y - 2x) are both negative. Moreover, if x is positive, y - 3x will be smaller than y - 2x, as we'll be subtracting a positive. This gives us the aforementioned Case B, again yielding a YES to the question.

Because the answer will always be YES, statement 1 alone is sufficient to answer the question. The correct answer is A
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by GMATGuruNY » Mon Jan 30, 2017 2:11 pm
Is (y - 3x)/(y - 2x) > 1?

(1) (y - 2x)/x < 0
(2) y - 2x > 0
(y-3x)/(y-2x) > 1

(y-3x)/(y-2x) - 1 > 0

[(y-3x)/(y-2x)] - [(y-2x)/(y-2x)] > 0

-x/(y-2x) > 0

x/(y-2x) < 0.

The left side will be negative only if x and (y-2x) have DIFFERENT SIGNS.
Questions stem, rephrased:
Do x and (y-2x) have different signs?

Statement 1: (y-2x)/x < 0
Thus, x and (y-2x) have different signs.
SUFFICIENT.

Statement 2: y-2x > 0
No information about x.
INSUFFICIENT.

The correct answer is A.
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by Brent@GMATPrepNow » Mon Jan 30, 2017 2:14 pm
Brent@GMATPrepNow wrote:
Is (y - 3x)/(y - 2x) > 1?

(1) (y - 2x)/x < 0
(2) y - 2x > 0
David's solution is great - it uses some nice number sense.

Another approach is the 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:
Rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100
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by Mo2men » Tue Jan 31, 2017 6:01 am
Brent@GMATPrepNow wrote:
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
Dear Brent,
I have question about statement 2 specifically.

y - 2x > 0...... y>2x

Let assume y= 2.1x

Applying in question.... -0.9x/0.1x > 1..........As long as x can't be zero so X could be eliminated. Answer to question is NO


Let assume y= 4x

Applying in question.... x/2x > 1..........As long as x can't be zero so X could be eliminated. Answer to question is NO

Let assume y= 6x

Applying in question.... 3x/4x > 1..........As long as x can't be zero so X could be eliminated. Answer to question is NO

So it always No. it should be sufficient.

Where did I go wrong??

Thanks for you help.

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by Brent@GMATPrepNow » Tue Jan 31, 2017 6:22 am
Mo2men wrote:
Dear Brent,
I have question about statement 2 specifically.

y - 2x > 0...... y>2x

Let assume y= 2.1x

Applying in question.... -0.9x/0.1x > 1..........As long as x can't be zero so X could be eliminated. Answer to question is NO


Let assume y= 4x

Applying in question.... x/2x > 1..........As long as x can't be zero so X could be eliminated. Answer to question is NO

Let assume y= 6x

Applying in question.... 3x/4x > 1..........As long as x can't be zero so X could be eliminated. Answer to question is NO

So it always No. it should be sufficient.

Where did I go wrong??

Thanks for you help.
Your conclusion that y > 2x is correct.
However, we can't then conclude that y = 2.1x, because this does not account for negative values of x.
For example, if x = -1, then 2x = -2 and y = -2.1.
In this case, is it not the case that y > 2x

Does that help?

Cheers,
Brent
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by Mo2men » Tue Jan 31, 2017 6:38 am
Brent@GMATPrepNow wrote: Your conclusion that y > 2x is correct.
However, we can't then conclude that y = 2.1x, because this does not account for negative values of x.
For example, if x = -1, then 2x = -2 and y = -2.1.
In this case, is it not the case that y > 2x

Does that help?

Cheers,
Brent
Thanks Brent.

Other general question please. If I have the following:

Can I safely that 3x/2x >1,as long as X is not equal zero so that I can eliminate X???

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by Brent@GMATPrepNow » Tue Jan 31, 2017 6:40 am
Mo2men wrote:
Thanks Brent.

Other general question please. If I have the following:

Can I safely that 3x/2x >1,as long as X is not equal zero so that I can eliminate X???
Yes, if you're told that 3x/2x > 1, then you can be certain that x ≠ 0, which means you can simplify 3x/2x to be 3/2
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by Mo2men » Tue Jan 31, 2017 7:15 am
Brent@GMATPrepNow wrote:
Mo2men wrote:
Thanks Brent.

Other general question please. If I have the following:

Can I safely that 3x/2x >1,as long as X is not equal zero so that I can eliminate X???
Yes, if you're told that 3x/2x > 1, then you can be certain that x ≠ 0, which means you can simplify 3x/2x to be 3/2
Thanks Brent,

I did not phrased my question correctly.

We are not told that 3x/2x >1. Instead, we want to conclude if 3x/2x >1. So my question still the same. Can I eliminate X in that case?

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by Brent@GMATPrepNow » Tue Jan 31, 2017 8:00 am
Mo2men wrote:
Brent@GMATPrepNow wrote:
Mo2men wrote:
Thanks Brent.

Other general question please. If I have the following:

Can I safely that 3x/2x >1,as long as X is not equal zero so that I can eliminate X???
Yes, if you're told that 3x/2x > 1, then you can be certain that x ≠ 0, which means you can simplify 3x/2x to be 3/2
Thanks Brent,

I did not phrased my question correctly.

We are not told that 3x/2x >1. Instead, we want to conclude if 3x/2x >1. So my question still the same. Can I eliminate X in that case?
If you can be guaranteed that x does not equal zero, then 3x/2x = 3/2, so we can be certain that 3x/2x > 1

Cheers,
Brent
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by Matt@VeritasPrep » Fri Mar 03, 2017 4:23 am
Mo2men wrote: We are not told that 3x/2x >1. Instead, we want to conclude if 3x/2x >1. So my question still the same. Can I eliminate X in that case?
If x were 0 it wouldn't be in the denominator of a fraction anyway, since you'd be dividing by 2x, or by 2*0.

This is one reason you want to be very careful dividing any equation by an unknown. If x = 0, your division will merrily play along, but your equation might break! Here's a famous example.

Let's say a = b. From there, we can multiply both sides by a:

a² = ab

then add a² to both sides:

a² + a² = a² + ab

then subtract 2ab from both sides:

2a² - 2ab = a² - ab

then factor out:

2a * (a - b) = a * (a - b)

then divide both sides by a * (a - b):

2 = 1

Wow! Alert the Nobel Committee, I've made a discovery that will destroy math as we know it. :D

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by Brent@GMATPrepNow » Fri Mar 03, 2017 6:28 am
Matt@VeritasPrep wrote: .
.
.

then divide both sides by a * (a - b):

2 = 1
And that's how Matt created the rift in the space-time continuum!

Plus, I have to go back and edit ALL of my posts to reflect this new discovery!
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