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

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.

Kudos Please... If my post helped.

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.

Hence E it is.

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).

Hope it helps.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Please pay attention to the rules #2 and 3. Thank you.

Thanks so much Bunuel. very helpful. I am always in trouble with absolute value and inequality problems.

Check our question banks here: viewforumtags.php

INEQUALITIES:
DS questions on inequalities: search.php?search_id=tag&tag_id=184
PS questions on inequalities: search.php?search_id=tag&tag_id=189

Hard inequality and absolute value questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939-40.html

The following threads might also be helpful:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

ABSOLUTE VALUE:
Check Absolute Value chapter of Math Book: math-absolute-value-modulus-86462.html

DS questions on absolute value to practice: search.php?search_id=tag&tag_id=37
PS questions on absolute value to practice: search.php?search_id=tag&tag_id=58

Hope it helps.

Thanks again Bunuel for so much help. Those threads marvellous.

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

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