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# M02 Q11 DS

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M02 Q11 DS [#permalink]

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27 Aug 2008, 17:06
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Is $$p^2 > q^2$$ ?

1. $$p > 0$$
2. $$q > 0$$

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

The given statement simplifies to:
$$p^2 - q^2 > 0$$

The real question, then is this: is $$(p + q) (p-q) > 0$$ ? The statements taken together allow for $$p > q$$ and $$p < q$$ , which makes the sign either positive or negative.

I don't understand the answer. Could someone break it down further please?
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Re: M02:Q11 DS [#permalink]

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27 Aug 2008, 22:40
idleking wrote:
Is $$p^2 > q^2$$ ?

1. $$p > 0$$
2. $$q > 0$$

(C) 2008 GMAT Club - m02#11

The given statement simplifies to:
$$p^2 - q^2 > 0$$

The real question, then is this: is $$(p + q) (p-q) > 0$$ ? The statements taken together allow for $$p > q$$ and $$p < q$$ , which makes the sign either positive or negative.
The correct answer is E.

I don't understand the answer. Could someone break it down further please?

$$p^2 - q^2 > 0$$

1) p>0

p^2 - q^2 ---> +ve when p>q (assume q is also positive)
p^2 - q^2 ---> -ve ve when p<q (assume q is also positive)

two solutions insuffcient

2) q>0

p^2 - q^2 ---> +ve when p>q (assume p is also positive)
p^2 - q^2 ---> -ve when p<q (assume p is also positive)

two solutions insuffcient

combine.
p>0 q>0 AND DON'T know the proper relation between p and q
p>q q>p

p^2 - q^2 --> lead +ve or -ve values depends on p>q or q>p
insuffcient

E.
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Re: M02 Q11 DS [#permalink]

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02 Dec 2009, 11:54
p^2 > q^2 ? can be rephrased as |p| > |q| ?

1. p > 0
Doesn't give any information on relationship between absolute values of p and q

2. q > 0
Doesn't give any information on relationship between absolute values of p and q

1 & 2 Combined also doesn't give any information on the relationship.

Hence E.
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Re: M02 Q11 DS [#permalink]

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03 Dec 2009, 11:26
Yeah I think you're just overthinking it.

In order to know whether $$p^2 > q^2$$ we have to know something about the relationship between p and q -- that one is larger than the other, that one is negative and one is positive, that one is less than 1 and the other isn't, etc.

Neither of the two statements separately nor the two statements combined give you any information about this.
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Re: M02 Q11 DS [#permalink]

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03 Dec 2009, 11:38
Making a little change in question, say:

2. q < 0

The answer would still be E
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Re: M02 Q11 DS [#permalink]

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07 Dec 2009, 17:07
Actually I wouldn't take the pains of factorizing and all that .. keeping it simple

Square of any number is positive .. so unless the answer choices help establishing a relationship between p and q it's impossible to say which is greater. Hence the option E.
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Re: M02 Q11 DS [#permalink]

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09 Dec 2010, 06:01
jeckll wrote:
Yeah I think you're just overthinking it.

In order to know whether $$p^2 > q^2$$ we have to know something about the relationship between p and q -- that one is larger than the other, that one is negative and one is positive, that one is less than 1 and the other isn't, etc.

Neither of the two statements separately nor the two statements combined give you any information about this.

Enough said here. P and Q could be equal in each of the scenarios. No way to tell if they are not, so E
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Re: M02 Q11 DS [#permalink]

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09 Dec 2010, 06:25
no need to expand the equation
just put values of p and q as fractions.

Posted from my mobile device
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Re: M02 Q11 DS [#permalink]

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09 Dec 2010, 09:04
idleking wrote:
Is $$p^2 > q^2$$ ?

1. $$p > 0$$
2. $$q > 0$$

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

The given statement simplifies to:
$$p^2 - q^2 > 0$$

The real question, then is this: is $$(p + q) (p-q) > 0$$ ? The statements taken together allow for $$p > q$$ and $$p < q$$ , which makes the sign either positive or negative.

I don't understand the answer. Could someone break it down further please?

Looking at statement 1) we have information for P but not Q so I eliminated A & D off the bat.
Looking at statement 2) we have information for Q but not for P so I eliminate B as well.

We are only left with C or E. Since P and Q are greater than 0 and there is no mention of integers in the problem I am going to test a low number 1 and a fraction .5

Here is what I have for my table, respectively..

P:1,.5,1,.5
Q:.5,1,1,.5

Putting these numbers in the original question I get the following...
Yes, No, No, No

Cross out C since it is insufficient.

E all the way
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Re: M02 Q11 DS [#permalink]

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09 Dec 2010, 21:09
since each statement alone is insufficient to answer, considering two statements together still not sufficient to answer as we do not whether P or Q is greater. For example : As both P and Q are positives , if P=2,Q=1, then P*P> Q*Q.

if P=1,Q= 2, then P*P<Q*Q. No definite answer, so it is insufficient to answer. answer is E
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Re: M02 Q11 DS [#permalink]

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10 Dec 2010, 00:14
Dear GOD! Please dont do this to me. I am less than a month away from my test need a load of tough questions to practice with.
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Respect,
Vaibhav

PS: Correct me if I am wrong.

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Re: M02 Q11 DS [#permalink]

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11 Dec 2010, 21:30
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E.

from question stem it is apparent that sign of variables does not have any impact on the solution
since both p and q are squared its the absolute value that we are after of each variable

Both Statement 1 and 2 provide only the sign of the number, which is postive for both of them.
None talks about absolute value.
Hence neither alone nor together are they sufficient to answer the question
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Re: M02 Q11 DS [#permalink]

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13 Dec 2010, 01:13
Answer choice is E.
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Re: M02 Q11 DS [#permalink]

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13 Dec 2011, 08:41
I think the best way to go about this is absolute value. Anything squared is the same as taking absolute value. A and B do not help. Putting them together doesn't either. E
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Re: M02 Q11 DS [#permalink]

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15 Dec 2011, 07:22
Easy questions. Just try with test numbers, you should get the correct answer.
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Re: M02 Q11 DS [#permalink]

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18 Feb 2012, 14:03
a) even if we have p = pos, q^2 can still be suff or insuff (ie. p = 2, q = 1 or -3)
b) even if we have q = pos, p^2 can still be suff or insuff (ie. q = 2, p = 1 or -3)
d) we've shown both are insuff
c) well we know both are pos, but if you test fractions or integers they can go either way.

E.
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Re: M02 Q11 DS [#permalink]

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13 Dec 2012, 05:08
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idleking wrote:
Is $$p^2 > q^2$$ ?

1. $$p > 0$$
2. $$q > 0$$

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

The given statement simplifies to:
$$p^2 - q^2 > 0$$

The real question, then is this: is $$(p + q) (p-q) > 0$$ ? The statements taken together allow for $$p > q$$ and $$p < q$$ , which makes the sign either positive or negative.

I don't understand the answer. Could someone break it down further please?

Is $$p^2 > q^2$$ ?

Is $$p^2 > q^2$$ ? --> is $$|p|>|q|$$? So, the question basically asks whether $$p$$ is further from zero, on a number line, than $$q$$.

(1) $$p > 0$$. Not sufficient since there is no info abut $$q$$.
(2) $$q > 0$$. Not sufficient since there is no info abut $$p$$.

(1)+(2) We know that both $$p$$ and $$q$$ are positive, though we don't know which one is further from zero (we don't know their relative position on a number line). Not sufficient.

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Re: M02 Q11 DS [#permalink]

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13 Dec 2012, 05:53
We can directly conclude it to be E since we have not been given any relationship between p and q.

Could be tricked by the problem if someone is in a hurry!!!
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Re: M02 Q11 DS [#permalink]

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13 Dec 2012, 05:56
Since p and q can take the form of a + whole no. or a + fraction.
And any of the numbers can be greater than the other.
Considering p>q and that they are whole nos. will result in different conclusion if they are fractions.

Choice E is the correct ans!
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Re: M02 Q11 DS [#permalink]

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13 Dec 2012, 06:36
pretty easy one... Got it correct..

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Re: M02 Q11 DS   [#permalink] 13 Dec 2012, 06:36

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