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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 \(\frac{x}{y}>0\).

Re: If x and y are positive, is x/y greater than 1? [#permalink]

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28 Apr 2012, 22:04

C is the answer. 1. xy=1 says nothing about which is greater, x or y. So insufficient. 2. x-y>O i.e. X>Y So if x and y are integars so x/y >1 but if x is negative or y is negative.? It can be less than 1 also. So Insufficient. 1+2 X and y are of same sign by xy>1 and x>y so sufficient.

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.

Re: If x and y are positive, is x/y greater than 1? [#permalink]

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21 Dec 2012, 19:54

Bunuel wrote:

monir6000 wrote:

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

(1) xy > 1 (2) x – y > 0

ORIGINAL QUESTION READS:

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

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.

Slight correction, even though your answer and your reasoning are correct, is that your reasoning for 2 does not address the question mentioned -- again, it's still correct, but want to make sure it talks about the question. It should be \(x-y>0\) not \(x-y>1\)
_________________

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.

Slight correction, even though your answer and your reasoning are correct, is that your reasoning for 2 does not address the question mentioned -- again, it's still correct, but want to make sure it talks about the question. It should be \(x-y>0\) not \(x-y>1\)

The question posted by monir6000 has typos. Again:

ORIGINAL QUESTION READS: If x and y are positive, is x/y greater than 1? (1) xy>1 (2) x-y>1
_________________

Re: If x and y are positive, is x/y greater than 1? [#permalink]

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17 Aug 2014, 21:14

Referring to the same question. 1) The statement is insufficient without a doubt 2) x-y>0, in case x is 7 and y is 3. But if y is -3, then the solution will be 7-(-3)=10 which is greater than 10.

But in case y is negative, then x/y, will not be greater than 1.

Referring to the same question. 1) The statement is insufficient without a doubt 2) x-y>0, in case x is 7 and y is 3. But if y is -3, then the solution will be 7-(-3)=10 which is greater than 10.

But in case y is negative, then x/y, will not be greater than 1.

Need some help here - Am i missing something?

The stem says: if x and y are positive...
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Re: If x and y are positive, is x/y greater than 1? [#permalink]

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26 Dec 2015, 16:24

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Is \(\frac{x}{y}>1\)? --> as given that \(y\) is positive we can safely multiply both parts of the 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 > 0. Directly answers the questions. Sufficient.