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

Question

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

(1) p < 0
(2) q > 0

Please help me with this. According to me:

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.

Show Answer

Official Answer

E

MacFauz · Accepted Answer

The problem is that the given expression is not the same as 2p-q > 2p+q. If (2p+q) is negative, 2p - q < 2p +q Suppose (2p+q) = -1, (2p-q) = -5 \frac{2p-q}{2p+q} = 5 > 1 But, -5 < -1 ie (2p-q) < (2p+q) An inequality cannot be multiplied by an unknown variable if the polarity of the variable is not known. Kudos Please... If my post helped.

bhavinshah5685 · Answer

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..

eeakkan · Answer

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.

eeakkan · Answer

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

MacFauz · Answer

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.

Vips0000 · Answer

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.

Bunuel · Answer

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

eeakkan · Answer

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

Bunuel · Answer

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.

eeakkan · Answer

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

fukirua · Answer

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

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