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Is p^2q > pq^2?

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Is p^2q > pq^2?  [#permalink]

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New post Updated on: 07 Jun 2013, 00:18
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Is p^2q > pq^2?

(1) pq < 0
(2) p < 0

got confused in C & E

and guessed C

Please explain how to eliminate this doubt, as I always get wrong in these two options.

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Originally posted by Manhnip on 06 Jun 2013, 11:47.
Last edited by Bunuel on 07 Jun 2013, 00:18, edited 3 times in total.
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Re: Is p^2q>pq^2? (1)pq<0 (2) p<0  [#permalink]

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New post 06 Jun 2013, 11:59
5
Is \(p^2q>pq^2?\)

Rewrite as \(p^2q-pq^2>0\) or \(pq(p-q)>0\)?

(1)\(pq<0\)
The first term is negative, no info about the sign of \((p-q)\).
Not sufficient

(2)\(p<0\)
Clearly not sufficient

1+2)Now we know that \(pq<0\) and that \(p<0\), so \(q>0\). From those we get \(q>p\) so \(p-q<0\).
The first term \(pq\) is negative, the second is negative => their product is positive.
Sufficient
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Re: Is p^2q > pq^2?  [#permalink]

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New post 10 Sep 2013, 19:02
1
Rephrase the question as:

p * p * q > p * q * q

This simplifies to

is p > q ?

I) only tells us that p & q have opposite signage
II) specifically tells us p < 0 , but tells us nothing about q

using I & II:

we know that p < 0 and p&q have opp signage, so q must be > 0 and so p<q
This is sufficient, so the right answer is C
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Re: Is p^2q > pq^2?  [#permalink]

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New post 10 Sep 2013, 21:36
3
labheshr wrote:
Rephrase the question as:

p * p * q > p * q * q

This simplifies to

is p > q ?

I) only tells us that p & q have opposite signage
II) specifically tells us p < 0 , but tells us nothing about q

using I & II:

we know that p < 0 and p&q have opp signage, so q must be > 0 and so p<q
This is sufficient, so the right answer is C


There is a problem in this approach.

'Is \(p^2q > pq^2\)?' does not get simplified to 'Is \(p > q\)?'

You cannot reduce the expression by p and q without knowing the signs of p and q. If a variable is negative, the inequality sign flips when you divide both sides by it.
Note Zarrolou's method above: Is \(pq( p - q) > 0?\)
This is how you should proceed.

The final answer was not compromised here because it is a DS question. Since you know that pq < 0, actually the question is now 'Is p < q?' so whether you answer with yes or no, it doesn't matter in DS as long as you have a clear cut answer. But in some other question, this distinction could be the difference between the correct and incorrect answer.
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Re: Is p^2q > pq^2?  [#permalink]

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New post 11 Sep 2013, 22:15
VeritasPrepKarishma wrote:
labheshr wrote:
Rephrase the question as:

p * p * q > p * q * q

This simplifies to

is p > q ?

I) only tells us that p & q have opposite signage
II) specifically tells us p < 0 , but tells us nothing about q

using I & II:

we know that p < 0 and p&q have opp signage, so q must be > 0 and so p<q
This is sufficient, so the right answer is C


There is a problem in this approach.

'Is \(p^2q > pq^2\)?' does not get simplified to 'Is \(p > q\)?'

You cannot reduce the expression by p and q without knowing the signs of p and q. If a variable is negative, the inequality sign flips when you divide both sides by it.
Note Zarrolou's method above: Is \(pq( p - q) > 0?\)
This is how you should proceed.

The final answer was not compromised here because it is a DS question. Since you know that pq < 0, actually the question is now 'Is p < q?' so whether you answer with yes or no, it doesn't matter in DS as long as you have a clear cut answer. But in some other question, this distinction could be the difference between the correct and incorrect answer.


From st. 1&2 we know that pq < 0 & p < 0 therefore q>0 and p<q
but what if : p= -1/2 & q is 1/4 then we have p^ 2q =(-1/2) ^1/2 (square root of a negative number) & pq^2 = -1/2 *1/16= -1/32 , how do we decide?
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Re: Is p^2q > pq^2?  [#permalink]

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New post 12 Sep 2013, 02:16
dentobizz wrote:
VeritasPrepKarishma wrote:
labheshr wrote:
Rephrase the question as:

p * p * q > p * q * q

This simplifies to

is p > q ?

I) only tells us that p & q have opposite signage
II) specifically tells us p < 0 , but tells us nothing about q

using I & II:

we know that p < 0 and p&q have opp signage, so q must be > 0 and so p<q
This is sufficient, so the right answer is C


There is a problem in this approach.

'Is \(p^2q > pq^2\)?' does not get simplified to 'Is \(p > q\)?'

You cannot reduce the expression by p and q without knowing the signs of p and q. If a variable is negative, the inequality sign flips when you divide both sides by it.
Note Zarrolou's method above: Is \(pq( p - q) > 0?\)
This is how you should proceed.

The final answer was not compromised here because it is a DS question. Since you know that pq < 0, actually the question is now 'Is p < q?' so whether you answer with yes or no, it doesn't matter in DS as long as you have a clear cut answer. But in some other question, this distinction could be the difference between the correct and incorrect answer.


From st. 1&2 we know that pq < 0 & p < 0 therefore q>0 and p<q
but what if : p= -1/2 & q is 1/4 then we have p^ 2q =(-1/2) ^1/2 (square root of a negative number) & pq^2 = -1/2 *1/16= -1/32 , how do we decide?


If \(p=-\frac{1}{2}\) and \(q=\frac{1}{4}\), then \(p^2q=\frac{1}{16} >-\frac{1}{32}=pq^2\)

Is p^2q > pq^2?

The question asks whether \(pq(p-q) >0\).

(1) pq < 0. The question becomes whether \(p-q<0\) or whether \(p<q\). We don't know that. Not sufficient.

(2) p < 0. Clearly insufficient.

(1)+(2) Since from (2) \(p < 0\), then from (1) we'd have that \(q>0\), thus \(p=negative<q=positive\). Sufficient.

Answer: C.
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Re: Is p^2q > pq^2?  [#permalink]

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New post 12 Sep 2013, 05:23
Is p^2q > pq^2?

(1) pq < 0
(2) p < 0

Here is my method for this one:
1). p x q < 0. This means, we have two cases. Either p<0 & q>0 or q<0 & p>0. Going ahead with the first case. Is (-ve)^2 x (+ve) > (-ve) x (+ve)^2 ? --> +ve x +ve > -Ve. So YES. Going ahead with second case. Is (+ve)^2 x (-ve) > (+ve) x (-ve)^2 ? --> -ve > +Ve. This cannot be true. We get two answers. Insufficient.

2). p < 0. This doesnt tell us anything about q. Insufficient.

Together => We identified two cases from statement 1. Considering statement 2, We can only use the first case from statement 1. So, YES. Sufficient.
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Re: Is p^2q > pq^2?  [#permalink]

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Re: Is p^2q > pq^2?  [#permalink]

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New post 29 Nov 2014, 01:33
VeritasPrepKarishma wrote:
labheshr wrote:
Rephrase the question as:

p * p * q > p * q * q

This simplifies to

is p > q ?

I) only tells us that p & q have opposite signage
II) specifically tells us p < 0 , but tells us nothing about q

using I & II:

we know that p < 0 and p&q have opp signage, so q must be > 0 and so p<q
This is sufficient, so the right answer is C


There is a problem in this approach.

'Is \(p^2q > pq^2\)?' does not get simplified to 'Is \(p > q\)?'

You cannot reduce the expression by p and q without knowing the signs of p and q. If a variable is negative, the inequality sign flips when you divide both sides by it.
Note Zarrolou's method above: Is \(pq( p - q) > 0?\)
This is how you should proceed.

The final answer was not compromised here because it is a DS question. Since you know that pq < 0, actually the question is now 'Is p < q?' so whether you answer with yes or no, it doesn't matter in DS as long as you have a clear cut answer. But in some other question, this distinction could be the difference between the correct and incorrect answer.

hi karishma,

i wanted a clarification here

p^2 .q > p. q ^2
now p^2 > 0 always and so is q^2 > 0
hence, the question boils down to

q > p ??
is this approach correct ??
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Is p^2q > pq^2?  [#permalink]

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New post 30 Nov 2014, 22:57
arnabs wrote:
hi karishma,

i wanted a clarification here

p^2 .q > p. q ^2
now p^2 > 0 always and so is q^2 > 0
hence, the question boils down to

q > p ??
is this approach correct ??



Is \(p^2 .q > p. q ^2\)?
does not reduce to 'Is \(q >p\)?'

Note that we could have ignored \(p^2\) and \(q^2\) had we known that they are both equal in magnitude and non zero.

Say, if you divide the inequality by \(q^2\) (assuming q is not 0), you get

Is \(\frac{p^2}{q^2} * q > p\)?

Is this the same as 'Is \(q>p\)?'

No, to make this same, we are assuming that \(\frac{p^2}{q^2} = 1\).
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Re: Is p^2q > pq^2?  [#permalink]

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New post 12 Dec 2014, 12:21
pq(p-q) > 0 actual question.

Statement 1:
pq < 0

p < 0 and q > 0 -> (p-q) < 0
or
p>0 and q < 0 -> (p-q) < 0 or (p-q) > 0

Not sufficient.

Statement 2 :
p < 0
not sure about q. So not sufficient.

Combining the two statements:
p < 0 so q has to be > 0
as pq < 0
so (p-q) < 0 and hence pq(p-q) > 0 Hence Sufficient.
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Re: Is p^2q > pq^2?  [#permalink]

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New post 21 Dec 2016, 01:52
Bunuel wrote:


The question asks whether \(pq(p-q) >0\).

(1) pq < 0. The question becomes whether \(p-q<0\) or whether \(p<q\). We don't know that. Not sufficient.

(2) p < 0. Clearly insufficient.

(1)+(2) Since from (2) \(p < 0\), then from (1) we'd have that \(q>0\), thus \(p=negative<q=positive\). Sufficient.

Answer: C.


Hi Bunuel - I have a fundamental problem when reading these kind of questions, I never get what the questions is i.e. p^2q means (p^2)q or p^(2q)? Since parenthesis are rarely used, is there a convention in the Gmat exam? Thanks
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Re: Is p^2q > pq^2?  [#permalink]

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New post 21 Dec 2016, 01:59
1
bazu wrote:
Bunuel wrote:


The question asks whether \(pq(p-q) >0\).

(1) pq < 0. The question becomes whether \(p-q<0\) or whether \(p<q\). We don't know that. Not sufficient.

(2) p < 0. Clearly insufficient.

(1)+(2) Since from (2) \(p < 0\), then from (1) we'd have that \(q>0\), thus \(p=negative<q=positive\). Sufficient.

Answer: C.


Hi Bunuel - I have a fundamental problem when reading these kind of questions, I never get what the questions is i.e. p^2q means (p^2)q or p^(2q)? Since parenthesis are rarely used, is there a convention in the Gmat exam? Thanks


Good news is that on the exam you'll see formulas so you won't be confused.

To answer your question p^2q ALWAYS means p^2*q if it were p^(2q) it would be written that way.
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Re: Is p^2q > pq^2?  [#permalink]

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New post 11 Jan 2018, 12:03
Manhnip wrote:
Is p^2q > pq^2?

(1) pq < 0
(2) p < 0

got confused in C & E

and guessed C

Please explain how to eliminate this doubt, as I always get wrong in these two options.


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

\(p^2q > pq^2\)
\(⇔ p^2q - pq^2 > 0\)
\(⇔ pq(p-q) > 0\)

Since we have 2 variables (p and a) and 0 equations,C is most likely to be the answer. So, we should consider 1) & 2) first.

Condition 1) & 2):
Since \(pq < 0\) and \(p < 0\), we have \(q > 0\).
\(p - q < 0\) since \(p < 0 < q\) or \(p < q\).
Thus pq(p-q) > 0 since p < 0, q > 0 and p - q < 0.
Both conditions together 1) & 2) are sufficient.

Since this is an inequality question (one of the key question areas), we should also consider choices A and B by CMT(Common Mistake Type) 4(A).

Condition 1)
We don't know that \(p - q\) is positive or negative from the condition 1) only.
The condition 1) is not sufficient.

Condition 2)
We don't have any information about \(q\) from the condition 2)
The condition 2) is not sufficient.

Therefore, the answer is C.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: Is p^2q > pq^2? &nbs [#permalink] 11 Jan 2018, 12:03
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