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• $450 Tuition Credit & Official CAT Packs FREE February 15, 2019 February 15, 2019 10:00 PM EST 11:00 PM PST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) Is p^2q > pq^2?  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Hide Tags Intern Joined: 23 May 2013 Posts: 19 Location: United Kingdom WE: Project Management (Real Estate) Is p^2q > pq^2? [#permalink] Show Tags Updated on: 06 Jun 2013, 23:18 2 2 00:00 Difficulty: 65% (hard) Question Stats: 58% (01:51) correct 42% (01:56) wrong based on 401 sessions HideShow timer Statistics 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. _________________ Correct me If I'm wrong !! looking for valuable inputs Originally posted by Manhnip on 06 Jun 2013, 10:47. Last edited by Bunuel on 06 Jun 2013, 23:18, edited 3 times in total. Edited the question. Most Helpful Community Reply VP Status: Far, far away! Joined: 02 Sep 2012 Posts: 1057 Location: Italy Concentration: Finance, Entrepreneurship GPA: 3.8 Re: Is p^2q>pq^2? (1)pq<0 (2) p<0 [#permalink] Show Tags 06 Jun 2013, 10:59 6 1 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 _________________ It is beyond a doubt that all our knowledge that begins with experience. Kant , Critique of Pure Reason Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b] General Discussion Intern Joined: 01 Jan 2012 Posts: 2 Location: United States WE: Information Technology (Investment Banking) Re: Is p^2q > pq^2? [#permalink] Show Tags 10 Sep 2013, 18:02 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 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8880 Location: Pune, India Re: Is p^2q > pq^2? [#permalink] Show Tags 10 Sep 2013, 20:36 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. _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Retired Moderator Joined: 23 Jul 2010 Posts: 465 GPA: 3.4 WE: General Management (Non-Profit and Government) Re: Is p^2q > pq^2? [#permalink] Show Tags 11 Sep 2013, 21:15 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? _________________ How to select your BSchool? General Mistakes to Avoid on the GMAT TOP 10 articles on Time Management on the GMAT Thanks = Kudos. Kudos are appreciated Rules for posting on the verbal forum Math Expert Joined: 02 Sep 2009 Posts: 52905 Re: Is p^2q > pq^2? [#permalink] Show Tags 12 Sep 2013, 01:16 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. _________________ Intern Joined: 21 Mar 2013 Posts: 18 Concentration: Operations, Entrepreneurship GMAT 1: 620 Q47 V28 GMAT 2: 680 Q45 V38 WE: Engineering (Manufacturing) Re: Is p^2q > pq^2? [#permalink] Show Tags 12 Sep 2013, 04:23 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. Intern Joined: 10 Apr 2012 Posts: 22 Concentration: Finance, Economics GMAT 1: 760 Q50 V44 Re: Is p^2q > pq^2? [#permalink] Show Tags 12 Sep 2013, 14:01 1 Alternate Soln Attachments If you lack concept clarity.jpg [ 81.65 KiB | Viewed 6316 times ] If you lack concept clarity.jpg [ 81.65 KiB | Viewed 6315 times ] Manager Joined: 06 Aug 2013 Posts: 71 Re: Is p^2q > pq^2? [#permalink] Show Tags 29 Nov 2014, 00:33 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 ?? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8880 Location: Pune, India Is p^2q > pq^2? [#permalink] Show Tags 30 Nov 2014, 21:57 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 $$\frac{p^2}{q^2} = 1$$. _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Director Joined: 03 Feb 2013 Posts: 846 Location: India Concentration: Operations, Strategy GMAT 1: 760 Q49 V44 GPA: 3.88 WE: Engineering (Computer Software) Re: Is p^2q > pq^2? [#permalink] Show Tags 12 Dec 2014, 11:21 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. _________________ Thanks, Kinjal My Application Experience : http://gmatclub.com/forum/hardwork-never-gets-unrewarded-for-ever-189267-40.html#p1516961 Linkedin : https://www.linkedin.com/in/kinjal-das/ Please click on Kudos, if you think the post is helpful Intern Joined: 17 Aug 2016 Posts: 48 Re: Is p^2q > pq^2? [#permalink] Show Tags 21 Dec 2016, 00:52 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 Math Expert Joined: 02 Sep 2009 Posts: 52905 Re: Is p^2q > pq^2? [#permalink] Show Tags 21 Dec 2016, 00:59 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. _________________ Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6949 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Is p^2q > pq^2? [#permalink] Show Tags 11 Jan 2018, 11:03 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. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Joined: 16 Jul 2018
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Re: Is p^2q > pq^2?  [#permalink]

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22 Sep 2018, 14:11
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 $$\frac{p^2}{q^2} = 1$$.

Couldn't you further reduce this? You stopped before reaching where arnabs got to.

$$p^2 *q > p* q ^2$$

$$=\frac{q}{q^2} * p^2 > p$$

$$=\frac{q}{q^2} > \frac{p}{p^2}$$

$$=\frac{1}{q} > \frac{1}{p}$$

$$=q > p$$

Since P^2 and Q^2 are squares, you know they are positive. You can raise each side to the -1 to get the variables out of the denominator.

Could you let us know if you disagree with this approach?

Thanks
Director
Joined: 19 Oct 2013
Posts: 508
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)
Re: Is p^2q > pq^2?  [#permalink]

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05 Oct 2018, 17:25
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.

rearrange the inequality in the stem to be$$p^2q - pq^2 > 0$$ ?

$$pq (p - q) > 0$$ ?

since pq < 0 it means either p -ve and q +ve or the other way around.

statement 1)
let p = -1 and q = 1
-1 (-1 -1)= -1 * (-2) = 2 > 0 yes
let p = 1 and q = -1
-1 (1+1) = -1 * 2 = -2 > 0 no.

Insufficient.

Now for statement 2) p < 0
repeat the same case above we get a yes and no.
insufficient.

if we combine both we know that p is negative, thus q must be positive.

C
Re: Is p^2q > pq^2?   [#permalink] 05 Oct 2018, 17:25
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