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If x and y are positive integers, is (x/y)^z > 1 ?

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If x and y are positive integers, is (x/y)^z > 1 ?  [#permalink]

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New post 20 Aug 2018, 00:11
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E

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  45% (medium)

Question Stats:

68% (02:59) correct 32% (01:43) wrong based on 62 sessions

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If x and y are positive integers, is (x/y)^z > 1 ?  [#permalink]

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New post 20 Aug 2018, 00:20
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1) as \(x-y=-5\), it means that y is greater than x, so \(x/y\) is less than 1 but as z is unknown u can't determine whether \((x/y)^z\) is less than or greater than 1. e.g if z is positive then \((x/y)^z\) is less than 1, if z is negative then \((x/y)^z\) is greater than 1. INSUFFICIENT

2) still no clue whether z is positive or negative. INSUFFICIENT

by combining both provide no extra information either.

Answer is E
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Re: If x and y are positive integers, is (x/y)^z > 1 ?  [#permalink]

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New post 20 Aug 2018, 00:23
From statement 1 we can conclude that X<Y. Means \(\frac{X}{Y}\) is a fraction.
Since we don't know anything about Z. Statement 1 is insufficient.
From statement 2 we don't have any information about X and Y. So, 2 is insufficient.
Combining both gives two possibilities.
If X=1 and Y=6. and Z = \(\frac{{-1}}{2}\) Then (X/Y)^Z Gives 36. which is greater than 1.
If Z is positive fraction then (X/Y)^Z is less than 1.
Hence combining also is insufficient.
E is the answer.
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Re: If x and y are positive integers, is (x/y)^z > 1 ?  [#permalink]

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New post 21 Aug 2018, 07:57
Bunuel wrote:
If x and y are positive integers, is \((\frac{x}{y})^z > 1\) ?


(1) x - y = -5

(2) z ≠ 0

Here is my approach.

(1) x - y = -5 --> x < y --> \(\frac{x}{y}\) < 1 . No info about z --> not sufficient.
(Number plugging: Let \(\frac{x}{y}\) = \(\frac{1}{6}\). If \(z = 1\) --> Yes, If \(z=-1\) --> No)

(2) z ≠ 0 --> No info about x, y --> not sufficient.

(1) + (2) We still have two different answers --> not sufficient.

Answer E.
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