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# Given that x/y < 1, and both x and y are positive integers, which one

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Joined: 02 Sep 2009
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Given that x/y < 1, and both x and y are positive integers, which one  [#permalink]

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02 Sep 2016, 04:39
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Given that x/y < 1, and both x and y are positive integers, which one of the following must be greater than 1?

A. x/y^2
B. x^2/y
C. x^2/y^2
D. y/x
E. √(x/y)

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Re: Given that x/y < 1, and both x and y are positive integers, which one  [#permalink]

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02 Sep 2016, 07:02
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Bunuel wrote:
Given that x/y < 1, and both x and y are positive integers, which one of the following must be greater than 1?

A. x/y^2
B. x^2/y
C. x^2/y^2
D. y/x
E. √(x/y)

NOTE: since x and y are both positive, we can safely multiply and divide both sides of the given inequality by x and y.

Given: x/y < 1
Multiply both sides by y to get: x < y
Divide both sides by x to get: 1 < y/x

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Re: Given that x/y < 1, and both x and y are positive integers, which one  [#permalink]

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02 Sep 2016, 05:21
Bunuel wrote:
Given that x/y < 1, and both x and y are positive integers, which one of the following must be greater than 1?

A. x/y^2
B. x^2/y
C. x^2/y^2
D. y/x
E. √(x/y)

Since x/y is a fraction y must always be > 1

Given -

 Which one of the following must be greater than 1

We can get the result one only when the denominator in x/y ( Which is less than 1 ) becomes numerator..

Among the given options only (D) has the required characteristic we are looking for...

PS : In case of doubt , plug in some value for x and Y

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Re: Given that x/y < 1, and both x and y are positive integers, which one  [#permalink]

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02 Sep 2016, 07:14
Bunuel wrote:
Given that x/y < 1, and both x and y are positive integers, which one of the following must be greater than 1?

A. x/y^2
B. x^2/y
C. x^2/y^2
D. y/x
E. √(x/y)

Since x & y are positive the inequality becomes x<y

Hence by plugging numbers, if x=1/3 and y=1/2

the only option is D
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Re: Given that x/y < 1, and both x and y are positive integers, which one  [#permalink]

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24 Jan 2019, 05:27
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Re: Given that x/y < 1, and both x and y are positive integers, which one   [#permalink] 24 Jan 2019, 05:27
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