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If p = 3q, is p^2 > q^2
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Updated on: 03 Jan 2014, 08:01
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If p = 3q, is p^2 > q^2 (1) q + p < q  p (2) p^2 = 9q^2
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Originally posted by vikrantgulia on 03 Jan 2014, 07:57.
Last edited by Bunuel on 03 Jan 2014, 08:01, edited 1 time in total.
Renamed the topic and edited the question.



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Re: if p=3q, is p^2 > q^2
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03 Jan 2014, 08:02
I wouldn't find a suitable way to solve this problem.Moreover, can we simplify the algebra problem as we do in Word problems or arithmetic problems. If i simplify it by putting the p=3q then question become 9q^2>q^2 and further simplifying it as q^2 is always positive so we divide it which destroy the complete question.
Please explain how to approach



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If p = 3q, is p^2 > q^2
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03 Jan 2014, 08:10



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Re: If p = 3q, is p^2 > q^2
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03 Jan 2014, 11:28
vikrantgulia wrote: If p = 3q, is p^2 > q^2
(1) q + p < q  p
(2) p^2 = 9q^2 p2>q2? statement A: q+p<qp ======> that means p<0. If p is less than 0 means p is negative.. p=3q...that means q will be negative too. if q will be negative that means p will less than q? square of large negative number will be greater in value than less negative number. so p2 will be greater than q2. Statement B is insufficient.(same as bunuel's explantion) so answer is A. I took more than 3 mints. After luking at bunuel's solution i think thats much easy to do.
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If p = 3q, is p^2 > q^2
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18 Jun 2015, 12:43
Bunuel wrote: If p = 3q, is p^2 > q^2
Substitute \(p = 3q\) into the question: is \(9q^2>q^2\)? > is \(8q^2>0\)? > is \(q>0\)?
(1) q + p < q  p > \(p<0\). As \(p = 3q\), then q is also less than 0, therefore we have a NO answer to the question. Sufficient.
(2) p^2 = 9q^2. We already know this from the stem: \(p = 3q\) > \(p^2 = 9q^2\). Therefore this statement is useless. Not sufficient.
Answer: A.
Hope it's clear. I agree that A is the answer; however, the answer to the question is "yes". You divided both sides of the inequality by q, which is negative, without flipping the inequality sign. p < q (both negative) but P^2 > q^2 As a matter of fact, the question, as you said, is \(9q^2>q^2\)? if you divide both sides by q^2 (positive, no flipping sign) you get 3>1? and the answer is yes. We only need statement 1 in order to eliminate zero as an answer.
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Re: If p = 3q, is p^2 > q^2
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18 Jun 2015, 22:14
I have a confusion. Even if q is less than 0, 8q^2 will be greater than 0. So the condition 8q^2 >0 hold true always.. Isn't it?



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If p = 3q, is p^2 > q^2
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18 Jun 2015, 23:32
cledgard wrote: Bunuel wrote: If p = 3q, is p^2 > q^2
Substitute \(p = 3q\) into the question: is \(9q^2>q^2\)? > is \(8q^2>0\)? > is \(q>0\)?
(1) q + p < q  p > \(p<0\). As \(p = 3q\), then q is also less than 0, therefore we have a NO answer to the question. Sufficient.
(2) p^2 = 9q^2. We already know this from the stem: \(p = 3q\) > \(p^2 = 9q^2\). Therefore this statement is useless. Not sufficient.
Answer: A.
Hope it's clear. I agree that A is the answer; however, the answer to the question is "yes". You divided both sides of the inequality by q, which is negative, without flipping the inequality sign. p < q (both negative) but P^2 > q^2 As a matter of fact, the question, as you said, is \(9q^2>q^2\)? if you divide both sides by q^2 (positive, no flipping sign) you get 3>1? and the answer is yes. We only need statement 1 in order to eliminate zero as an answer. Hello cledgard When \(9q^2>q^2\) transform to \(8q^2>0\) it's not dividing but subtracting \(q^2\) from both sides. During subtracting we shouldn't change sign of inequality. But I think you are right about answer because I too see that answer is "yes" \(p^2 > q^2\)
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Re: If p = 3q, is p^2 > q^2
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18 Jun 2015, 23:41
NehaBhargava wrote: I have a confusion. Even if q is less than 0, 8q^2 will be greater than 0. So the condition 8q^2 >0 hold true always.. Isn't it? Hello NehaBhargavaYou are right but not in the case when q = 0 So this part \(8q2>0\)? > is \(q>0\) should looks like \(8q2>0\)? > is \(q<>0\)
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Re: If p = 3q, is p^2 > q^2
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19 Jun 2015, 12:44
Harley1980 wrote: cledgard wrote: Bunuel wrote: If p = 3q, is p^2 > q^2
Substitute \(p = 3q\) into the question: is \(9q^2>q^2\)? > is \(8q^2>0\)? > is \(q>0\)?
(1) q + p < q  p > \(p<0\). As \(p = 3q\), then q is also less than 0, therefore we have a NO answer to the question. Sufficient.
(2) p^2 = 9q^2. We already know this from the stem: \(p = 3q\) > \(p^2 = 9q^2\). Therefore this statement is useless. Not sufficient.
Answer: A.
Hope it's clear. I agree that A is the answer; however, the answer to the question is "yes". You divided both sides of the inequality by q, which is negative, without flipping the inequality sign. p < q (both negative) but P^2 > q^2 As a matter of fact, the question, as you said, is \(9q^2>q^2\)? if you divide both sides by q^2 (positive, no flipping sign) you get 3>1? and the answer is yes. We only need statement 1 in order to eliminate zero as an answer. Hello cledgard When \(9q^2>q^2\) transform to \(8q^2>0\) it's not dividing but subtracting \(q^2\) from both sides. During subtracting we shouldn't change sign of inequality. But I think you are right about answer because I too see that answer is "yes" \(p^2 > q^2\) The question was simplified: is \(9q^2>q^2\)? > is \(8q^2>0\)? > is \(q>0\)? I was referring to the last step \(q>0\)? To get from \(8q^2>0\) to \(q>0\)? you must either divide by 8q or take the square root from both sides, and then the question would be is q>0? This answer must be yes, unless q is 0. It is because the possibility of q being 0 that we need statement 1.
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Re: If p = 3q, is p^2 > q^2?
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01 Sep 2018, 20:02
Mo2men wrote: If p = 3q, is p^2 > q^2?
(1) q + p < q  p
(2) p^2 = 9q^2
kudos for the explanation of the answer Given, p=3q. Question stem: Is \(p^2 > q^2?\) Or, Is \((3q)^2> q^2\) Or, Is \(9q^2>q^2\) Or, Is \(q^2>0\) Does 'q' lie in the interval (inf,0) or (0,+inf)? St1: q + p < q  p Or, 4q<(3q) Or, q<0 Answer to question stem is Yes. Sufficient. St2: \(p^2 = 9q^2\) Or, \(9q^2=9q^2\) This is true for all values of q. Hence insufficient. Ans. (A)
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Re: If p = 3q, is p^2 > q^2
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01 Sep 2018, 22:34
vikrantgulia wrote: If p = 3q, is p^2 > q^2
(1) q + p < q  p
(2) p^2 = 9q^2 From Statement I: p < 0, q < 0. As \(p = 3q\) and both are of same sign, \(p^2 > q^2\). From Statement II: When, \(p =0, q = 0\).. Its Insufficient. Hence, \(A\).
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