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yes. question and answers correct. it is from veritas . and it says correct answer is E. neither (1) or (2) not suff. But I think something wrong with this.

2p-q>2p+q => 2p-2p>q+q => 2q<0 => q<0 which is stated in Statement 2. Now statment 1 says p<0

Take values of p=-1 and q=-2 keeping this in our original inqulality, we get 2(-1)-(-2)/2(-1)+(-2) > 1 => -2+2/-2-2 > 1 => 0 > 1 which is not possible

You can check by taking values p=-2 and q=-1 u will get 0.6>1 whihc is not possible so, both the statements are not sufficeint to answer the question So answer E...

2p-q>2p+q => 2p-2p>q+q => 2q<0 => q<0 which is stated in Statement 2. Now statment 1 says p<0

Take values of p=-1 and q=-2 keeping this in our original inqulality, we get 2(-1)-(-2)/2(-1)+(-2) > 1 => -2+2/-2-2 > 1 => 0 > 1 which is not possible

You can check by taking values p=-2 and q=-1 u will get 0.6>1 whihc is not possible so, both the statements are not sufficeint to answer the question So answer E...

I dont know whether my approach is right or not..

Ok. yours almost same approach with me. but this is yes or no question. right? If we can aswer to this question as NO with (b), then b is the answer.

Ok. thanks. I think I have missed that point.So only we could solve this equation as giving by numbers.

Picking numbers may not be the only way to solve it. But it is a very simple way to solve it. After picking numbers, we can see that we need to know wbout an additional parameter ie whether |2p| > |q| to decide on whether the given equation is greater than 1.

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Ok. thanks. I think I have missed that point.So only we could solve this equation as giving by numbers.

No, it could be solved easily algebrically as well.

question is: is (2p-q)/(2p+q)>1 ?

or (2p-q)/(2p+q) -1 >0

=> (2p-q-2p-q) / (2p+q) >0

=> -2q/(2p+q) >0 ?

=> is 2q/(2p+q) <0

Statement 1: p <0 Doesnt tell us anything Statement 2: q >0 doesnt tell anything as we dont know what 2p+q would be

Combining, we know that numerator is positive, but still we dont know : denominator could be positive or negative depending on absolute values of p and q.

ıf we arrange question: 2p-q>2p+q then -q>q and so (B) should be ok. Because if q>0, -q will be always <q.

If 2p not equal to -q, is (2p-q)/(2p+q)>1?[/m]?

Is \(\frac{2p-q}{2p+q}>1\)? --> is \(0>1-\frac{2p-q}{2p+q}\)? --> is \(0>\frac{2p+q-2p+q}{2p+q}\)? --> is \(0>\frac{2q}{2p+q}\)?

(1) \(p<0\). Not sufficient.

(2) \(q>0\). Not sufficient.

(1)+(2) \(p<0\) and \(q>0\) --> the numerator (2q) is positive, but we cannot say whether the denominator {negative (2p)+positive (q)} is positive or negative. Not sufficient.

Answer: E.

The problem with your solution is that when you are writing \(2p-q>2p+q\), you are actually multiplying both sides of inequality by \(2p+q\): never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable). Because if \(2p+q>0\) you should write \(2p-q>2p+q\) BUT if \(2p+q<0\), you should write \(2p-q<2p+q\), (flip the sign when multiplying by negative expression).

Re: If 2p not equal to -q, is (2p-q)/(2p+q)>1? [#permalink]

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08 Nov 2012, 22:00

1

This post received KUDOS

Bunuel wrote:

never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable). Because if \(2p+q>0\) you should write \(2p-q>2p+q\) BUT if \(2p+q<0\), you should write \(2p-q<2p+q\), (flip the sign when multiplying by negative expression).

Hi Bonuel, I multiplied both numerator and denominator on (2p+q), I think we can do that. Thus we have (4p^2-q^2)/(2p+q)^2>1 Now we can get rid of denominator as it is always positive. Eventually it comes to q^2+2pq<0.

Considering (1) and (2) together q^2<2pq or q<2p. And of course we don't know that.

You solution is much faster and better! Thanks _________________

Re: If 2p not equal to -q, is (2p-q)/(2p+q)>1? [#permalink]

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30 Aug 2015, 07:09

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