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If x and y are positive, is x/y greater than 1?

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If x and y are positive, is x/y greater than 1? [#permalink] New post 29 Sep 2010, 23:49
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If x and y are positive, is x/y greater than 1?

(1) xy > 1
(2) x-y > 1

My question if x and y are both positive, shouldnt x/y always be positive?
[Reveal] Spoiler: OA
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Re: DS question [#permalink] New post 29 Sep 2010, 23:57
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vanidhar wrote:
If x and y are positive, is x/y greater than 1?

1)xy > 1
2)x-y > 1

the official answer is B

My question if x and y are both positive, shouldnt x/y always be positive?


To answer your question: Yes. \frac{x}{y} will always be positive. But being positive could also mean numbers between 0 and 1.

To answer the question:

Statement 1 says xy>1. NO information about \frac{x}{y}. Insufficient.

Statement 2:

x - y > 1

Divide by y on both sides

\frac{x}{y} - 1 > \frac{1}{y}

\frac{x}{y} > 1 + \frac{1}{y}

But. y is a positive integer so \frac{1}{y} > 0 which means that \frac{x}{y} > 1.

B.
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Re: DS question [#permalink] New post 30 Sep 2010, 00:15
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vanidhar wrote:
If x and y are positive, is x/y greater than 1?

1)xy > 1
2)x-y > 1

the official answer is B

My question if x and y are both positive, shouldnt x/y always be positive?


Is \frac{x}{y}>1? --> as given that y is positive we can safely multiply boith parts of inequality by it --> so the question becomes "is x>y?" OR: is x-y>0?

(1) xy>1 --> product of two numbers is more than one we can't say which one is greater. Not sufficient.

(2) x-y>1 --> x>y+1 --> as x is more than y plus 1 then it's obviously more than just y alone: x>y. Sufficient.
Or: as x-y>1 then x-y is obviously more than zero --> x-y>1>0. Sufficient.

Answer: B.

As for your question: yes, if x and y are both positive (or both negative) then \frac{x}{y}>0.

Hope it helps.
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Re: DS question [#permalink] New post 30 Sep 2010, 09:14
[quote="vanidhar"]If x and y are positive, is x/y greater than 1?

1)xy > 1
2)x-y > 1

x,y +ve , is x/y > 1 ie is x>y

from 1

insuff

from 2

suff

B
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Re: DS question [#permalink] New post 30 Sep 2010, 14:08
Bunuel wrote:

(2) x-y>1 --> x>y+1 --> as x is more than y plus 1 then it's obviously more than just y alone: x>y. Sufficient.

Hope it helps.


Nice approach, I didn't went with this approach, but good to know.
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Re: DS question   [#permalink] 30 Sep 2010, 14:08
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