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VP  V
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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 37% (01:59) correct 63% (02:04) wrong based on 54 sessions

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Is $$\frac{|x|}{|y|} < |x|*|y|$$?

(1) $$x^3>y^3$$

(2) $$y^5>y^4$$

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Originally posted by Asad on 11 Mar 2019, 08:28.
Last edited by Bunuel on 26 Mar 2019, 03:59, edited 1 time in total.
Edited the OA.
Retired Moderator V
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Re: Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4  [#permalink]

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2
1
Is |x|/ |y| < |x|*|y| ?
Squaring both sides and cross multiplying, we get
or, x^2 < x^2 *y^4
or, x^2 (1-y^4)<0

This would be true if x is not equal to 0, and y >1 or y <-1
So, the question reduces to " Is x is not equal to 0, and y >1 or y <-1" ?

Statement 1: x^3 > y^3
Here x can be zero, and y any negative number. for which the answer can be NO
(x,y) can be (2,3), Answer is YES

NOT SUFFICIENT

Statement 2:
y^5 > y^4
or y^4(y-1) >0

here y >1, but x can be zero or non zero.
if y >1, x =0 , answer is NO
if y >1, x is non zero, ans is YES

NOT SUFFICIENT

Combing Both statements , we get y >1
and x^3 > y^3 , Hence x is NON zero,

So, We have y >1 ans X is NON ZERO.

Hence SUFFICIENT.

Is $$\frac{|x|}{|y|} < |x|*|y|$$?

(1) $$x^3>y^3$$

(2) $$y^5>y^4$$

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Re: Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4  [#permalink]

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If i wanted to solve this question with the help of numbers how can I solve it? Can anyone help me?
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Re: Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4  [#permalink]

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1
I would go for C.

Statement 1: insufficient
Case 1: X=0 & y=-1 (thus, both sides are equal in main question. So, NO to main question)

Case 2: x=2 & y=3 (yes to main question)

Statement 2: insufficient
Case 1: X=0 & y=2 (thus, both sides are equal in main question. So, NO to main question)

Case 2: x=2 & y=3 (yes to main question)

Combining
From statement 2, it is clear that y is positive and greater than 1.
So in statement 1, x must be positive.

In such situation, multiplication of same direction two number is greater than division of same numbers.

I.e. x=3 & y=2

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GMAT 1: 730 Q47 V42 Re: Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4  [#permalink]

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Can people make sure that their answers are correct when posting questions? Makes it a lot harder to study when you are trying to work out whether the answer is actually the answer or not

Originally posted by strawberryjelly on 25 Mar 2019, 14:38.
Last edited by strawberryjelly on 26 Mar 2019, 03:33, edited 1 time in total.
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Re: Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4  [#permalink]

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Is $$\frac{|x|}{|y|} < |x|×|y|$$?

(1) $$x^3>y^3$$

(2) $$y^5>y^4$$

Hi Bunuel and IanStewart,
Can I rearrange the question stem in the following way?
$$\frac{|x|}{|y|} < |x|×|y|$$?
==> $$\frac{x^2}{y^2}<x^2 × y^2?$$
==> $$x^2<x^2 × y^4$$? [dividing by $$x^2$$ in both side as $$x^2$$ is positive]
==> $$y^4>1?$$
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Math Expert V
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Re: Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4  [#permalink]

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Is $$\frac{|x|}{|y|} < |x|×|y|$$?

(1) $$x^3>y^3$$

(2) $$y^5>y^4$$

Hi Bunuel and IanStewart,
Can I rearrange the question stem in the following way?
$$\frac{|x|}{|y|} < |x|×|y|$$?
==> $$\frac{x^2}{y^2}<x^2 × y^2?$$
==> $$x^2<x^2 × y^4$$? [dividing by $$x^2$$ in both side as $$x^2$$ is positive]
==> $$y^4>1?$$

Yes. You could reduce by |x| directly (provided x is not 0). $$\frac{1}{|y|} < |y|$$ --> |y|^2 > 1 --> y < -1 or y > 1.

BTW, the correct answer is C, not B. (2) give y > 1. If x ≠ 0, then $$\frac{|x|}{|y|} > |x|×|y|$$ (answer YES) but if x = 0, then $$\frac{|x|}{|y|} = |x|×|y|$$ (answer NO).

What is the source of the question?
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Re: Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4  [#permalink]

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Bunuel wrote:
Is $$\frac{|x|}{|y|} < |x|×|y|$$?

(1) $$x^3>y^3$$

(2) $$y^5>y^4$$

Hi Bunuel and IanStewart,
Can I rearrange the question stem in the following way?
$$\frac{|x|}{|y|} < |x|×|y|$$?
==> $$\frac{x^2}{y^2}<x^2 × y^2?$$
==> $$x^2<x^2 × y^4$$? [dividing by $$x^2$$ in both side as $$x^2$$ is positive]
==> $$y^4>1?$$

Yes. You could reduce by |x| directly (provided x is not 0). $$\frac{1}{|y|} < |y|$$ --> |y|^2 > 1 --> y < -1 or y > 1.

BTW, the correct answer is C, not B. (2) give y > 1. If x ≠ 0, then $$\frac{|x|}{|y|} > |x|×|y|$$ (answer YES) but if x = 0, then $$\frac{|x|}{|y|} = |x|×|y|$$ (answer NO).

What is the source of the question?

Source: Wizako
In this question, if we can determine that $$y^2>1$$ then the data is sufficient.
Conversely, if we can determine that $$y^2≤1$$ then the data is also sufficient.
Isn't it?
Statement 2 says:
(2) $$y^5>y^4$$
Here, y^4 is positive, so we can divide both part by $$y^4$$. We get from statement 2:
==> $$y>1$$
==> $$y^2>1$$ [multiplying by both side]
So, why not B sufficient?
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Math Expert V
Joined: 02 Sep 2009
Posts: 58312
Re: Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4  [#permalink]

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1
Bunuel wrote:
Hi Bunuel and IanStewart,
Can I rearrange the question stem in the following way?
$$\frac{|x|}{|y|} < |x|×|y|$$?
==> $$\frac{x^2}{y^2}<x^2 × y^2?$$
==> $$x^2<x^2 × y^4$$? [dividing by $$x^2$$ in both side as $$x^2$$ is positive]
==> $$y^4>1?$$

Yes. You could reduce by |x| directly (provided x is not 0). $$\frac{1}{|y|} < |y|$$ --> |y|^2 > 1 --> y < -1 or y > 1.

BTW, the correct answer is C, not B. (2) give y > 1. If x ≠ 0, then $$\frac{|x|}{|y|} > |x|×|y|$$ (answer YES) but if x = 0, then $$\frac{|x|}{|y|} = |x|×|y|$$ (answer NO).

What is the source of the question?

Source: Wizako
In this question, if we can determine that $$y^2>1$$ then the data is sufficient.
Conversely, if we can determine that $$y^2≤1$$ then the data is also sufficient.
Isn't it?
Statement 2 says:
(2) $$y^5>y^4$$
Here, y^4 is positive, so we can divide both part by $$y^4$$. We get from statement 2:
==> $$y>1$$
==> $$y^2>1$$ [multiplying by both side]
So, why not B sufficient?

Because we have one more variable in the question:|x|/|y|[/fraction] < |x|*|y|. From (2) we know that y > 1 but know nothing about x. If x ≠ 0, then $$\frac{|x|}{|y|} > |x|×|y|$$ (answer YES) but if x = 0, then $$\frac{|x|}{|y|} = |x|×|y|$$ (answer NO).
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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4  [#permalink]

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Can I rearrange the question stem in the following way?
$$\frac{|x|}{|y|} < |x|×|y|$$?
==> $$\frac{x^2}{y^2}<x^2 × y^2?$$
==> $$x^2<x^2 × y^4$$? [dividing by $$x^2$$ in both side as $$x^2$$ is positive]
==> $$y^4>1?$$

Bunuel wrote:
Because we have one more variable in the question:|x|/|y|[/fraction] < |x|*|y|. From (2) we know that y > 1 but know nothing about x. If x ≠ 0, then $$\frac{|x|}{|y|} > |x|×|y|$$ (answer YES) but if x = 0, then $$\frac{|x|}{|y|} = |x|×|y|$$ (answer NO).

But i've deduced the question stem as $$y^4>1$$ where there is no existence of x.
Thanks__
_________________
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Math Expert V
Joined: 02 Sep 2009
Posts: 58312
Re: Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4  [#permalink]

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1
Can I rearrange the question stem in the following way?
$$\frac{|x|}{|y|} < |x|×|y|$$?
==> $$\frac{x^2}{y^2}<x^2 × y^2?$$
==> $$x^2<x^2 × y^4$$? [dividing by $$x^2$$ in both side as $$x^2$$ is positive]
==> $$y^4>1?$$

Bunuel wrote:
Because we have one more variable in the question:|x|/|y|[/fraction] < |x|*|y|. From (2) we know that y > 1 but know nothing about x. If x ≠ 0, then $$\frac{|x|}{|y|} > |x|×|y|$$ (answer YES) but if x = 0, then $$\frac{|x|}{|y|} = |x|×|y|$$ (answer NO).

But i've deduced the question stem as $$y^4>1$$ where there is no existence of x.
Thanks__

Check my post above: You could reduce by |x| directly (provided x is not 0). If x = 0, then we cannot reduce and thus only y^5 > y^4 will not be sufficient.
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Re: Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4  [#permalink]

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Bunuel wrote:
Can I rearrange the question stem in the following way?
$$\frac{|x|}{|y|} < |x|×|y|$$?
==> $$\frac{x^2}{y^2}<x^2 × y^2?$$
==> $$x^2<x^2 × y^4$$? [dividing by $$x^2$$ in both side as $$x^2$$ is positive]
==> $$y^4>1?$$

Bunuel wrote:
Because we have one more variable in the question:|x|/|y|[/fraction] < |x|*|y|. From (2) we know that y > 1 but know nothing about x. If x ≠ 0, then $$\frac{|x|}{|y|} > |x|×|y|$$ (answer YES) but if x = 0, then $$\frac{|x|}{|y|} = |x|×|y|$$ (answer NO).

But i've deduced the question stem as $$y^4>1$$ where there is no existence of x.
Thanks__

Check my post above: You could reduce by |x| directly (provided x is not 0). If x = 0, then we cannot reduce and thus only y^5 > y^4 will not be sufficient.

Thank you so much. Better if you edit the OA.
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Joined: 04 Aug 2010
Posts: 474
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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4  [#permalink]

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Since division by 0 is not allowed, the prompt should indicate that $$y≠0$$:

If $$y≠0$$, is $$\frac{|x|}{|y|} < |x|*|y|$$?

(1) $$x^3>y^3$$

(2) $$y^5>y^4$$

Since an absolute value cannot be negative. we can simplify the question stem by multiplying both sides by |y|:
$$\frac{|x|}{|y|}|y| < |x||y||y|$$
$$|x| < |x|y^2$$?

If $$x=0$$, the answer to the question stem is NO.
If $$x≠0$$, we can divide both sides by $$|x|$$:
$$\frac{|x|}{|x|} < \frac{|x|y^2}{|x|}$$
$$1 < y^2$$?
Here, the answer will be YES if $$y>1$$ or $$y<-1$$.

Question stem, rephrased:
Is it true that $$x≠0$$ and that $$y>1$$ or $$y<-1$$?

Statement 1:
Case 1: x=3 and y=2
Since $$x≠0$$ and $$y>1$$, the answer to the question stem is YES.
Case 2: x=2 and y=1
Since $$y=1$$, the answer to the question stem is NO.
INSUFFICIENT.

Statement 2:
Since the inequality implies that $$y≠0$$, we can safely divide both sides by $$y^4$$, which must be positive:
$$\frac{y^5}{y^4}>\frac{y^4}{y^4}$$
$$y > 1$$
No information about $$x$$.
INSUFFICIENT.

Statements combined:
$$y>1$$ implies that $$y^3>1$$.
Thus:
$$x^3 > y^3 > 1$$
$$x^3 > 1$$
$$x > 1$$
Since $$x≠0$$ and $$y>1$$, the answer to the question stem is YES.
SUFFICIENT.

.
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Available for tutoring in NYC and long-distance. Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4   [#permalink] 26 Mar 2019, 11:22
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