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

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If 2p not equal to -q, is (2p-q)/(2p+q)>1? [#permalink]  05 Nov 2012, 12:24
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If 2p not equal to -q, is (2p-q)/(2p+q)>1?

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

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.
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Nov 2012, 03:01, edited 1 time in total.
Renamed the topic, edited the question and moved to DS forum.
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Re: number properties [#permalink]  05 Nov 2012, 22:20
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eeakkan wrote:
ıf 2p not equal to -q, is (2p-q)/(2p+q)>1?

1)p<0
2)q>0

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.

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.

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Re: If 2p not equal to -q, is (2p-q)/(2p+q)>1? [#permalink]  06 Nov 2012, 03:41
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Expert's post
eeakkan wrote:
If 2p not equal to -q, is (2p-q)/(2p+q)>1?

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

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.

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.

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Re: If 2p not equal to -q, is (2p-q)/(2p+q)>1? [#permalink]  06 Nov 2012, 05:37
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eeakkan wrote:
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

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

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Re: If 2p not equal to -q, is (2p-q)/(2p+q)>1? [#permalink]  08 Nov 2012, 21:00
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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
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Re: number properties [#permalink]  05 Nov 2012, 13:44
can you double-check the source of the question/answer ?
i would think statement (2) is enough info to answer the question as well...
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Re: number properties [#permalink]  05 Nov 2012, 13:49
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.
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Re: number properties [#permalink]  05 Nov 2012, 21:59
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..
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Re: number properties [#permalink]  05 Nov 2012, 22:09
bhavinshah5685 wrote:
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.
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Re: number properties [#permalink]  05 Nov 2012, 22:27
Ok. thanks. I think I have missed that point.So only we could solve this equation as giving by numbers.
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Re: number properties [#permalink]  05 Nov 2012, 23:53
eeakkan wrote:
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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Re: number properties [#permalink]  06 Nov 2012, 00:06
eeakkan wrote:
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.

Hence E it is.
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Re: If 2p not equal to -q, is (2p-q)/(2p+q)>1? [#permalink]  06 Nov 2012, 05:28
Thanks so much Bunuel. very helpful. I am always in trouble with absolute value and inequality problems.
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Re: If 2p not equal to -q, is (2p-q)/(2p+q)>1? [#permalink]  06 Nov 2012, 06:41
Thanks again Bunuel for so much help. Those threads marvellous.
Re: If 2p not equal to -q, is (2p-q)/(2p+q)>1?   [#permalink] 06 Nov 2012, 06:41
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