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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4 Updated on: 26 Mar 2019, 03:59
Originally posted by TheUltimateWinner on 11 Mar 2019, 08:28.
Last edited by Bunuel on 26 Mar 2019, 03:59, edited 1 time in total.
Edited the OA.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

85% (hard)

Question Stats:

43% (02:06) correct 57% (02:02) wrong based on 119 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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C
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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4 11 Mar 2019, 09:26
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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.
Answer is C


AsadAbu
Is \(\frac{|x|}{|y|} < |x|*|y|\)?

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

(2) \(y^5>y^4\)
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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4 11 Mar 2019, 10:05
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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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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4 11 Mar 2019, 12:32
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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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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4 Updated on: 26 Mar 2019, 03:33
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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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
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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4 26 Mar 2019, 02:57
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Is \(\frac{|x|}{|y|} < |x|×|y|\)?


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

(2) \(y^5>y^4\)
Show more
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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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4 26 Mar 2019, 03:14
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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]
==> \(y^4>1?\)
Show more

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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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4 26 Mar 2019, 03:54
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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?
Show more
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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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4 26 Mar 2019, 03:59
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Bunuel
AsadAbu

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?
Show more

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 26 Mar 2019, 04:08
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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]
==> \(y^4>1?\)
Show more


Bunuel
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).
Show more

But i've deduced the question stem as \(y^4>1\) where there is no existence of x.
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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4 26 Mar 2019, 04:11
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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
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__
Show more

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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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4 26 Mar 2019, 04:16
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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
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.
Show more
Thank you so much. Better if you edit the OA.
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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4 26 Mar 2019, 11:22
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Since division by 0 is not allowed, the prompt should indicate that \(y≠0\):

AsadAbu
If \(y≠0\), is \(\frac{|x|}{|y|} < |x|*|y|\)?


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

(2) \(y^5>y^4\)
Show more

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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Is |x|/|y| < |x||y|? (1) x^3 > y^3 (2) y^5 > y^4 01 Dec 2020, 18:54
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