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705-805 (Hard)|   Algebra|      
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vikrantgulia
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If p = 3q, is p^2 > q^2 Updated on: 03 Jan 2014, 08:01
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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75% (hard)

Question Stats:

51% (02:11) correct 49% (02:20) wrong based on 274 sessions

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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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A
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vikrantgulia
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If p = 3q, is p^2 > q^2 03 Jan 2014, 08:02
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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 03 Jan 2014, 08:10
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If p = 3q, is p^2 > q^2

Substitute \(p = 3q\) into the question: is \(9q^2>q^2\)? --> is \(8q^2>0\)? --> is \(q\neq{0}\)?

(1) q + p < q - p --> \(p<0\). Since \(p = 3q\), then q is less than 0, so not equal to 0. Therefore we have an YES 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.
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If p = 3q, is p^2 > q^2 03 Jan 2014, 11:28
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vikrantgulia
If p = 3q, is p^2 > q^2

(1) q + p < q - p

(2) p^2 = 9q^2
Show more

p2>q2?

statement A: q+p<q-p ======> 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 18 Jun 2015, 12:43
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Bunuel
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.
Show more

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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If p = 3q, is p^2 > q^2 18 Jun 2015, 22:14
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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 18 Jun 2015, 23:32
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cledgard
Bunuel
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.
Show more

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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If p = 3q, is p^2 > q^2 18 Jun 2015, 23:41
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NehaBhargava
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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Hello NehaBhargava
You 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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If p = 3q, is p^2 > q^2 19 Jun 2015, 12:44
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Harley1980
cledgard
Bunuel
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" \(p^2 > q^2\)
Show more


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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If p = 3q, is p^2 > q^2 01 Sep 2018, 20:02
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Mo2men
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
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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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If p = 3q, is p^2 > q^2 01 Sep 2018, 22:34
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vikrantgulia
If p = 3q, is p^2 > q^2

(1) q + p < q - p

(2) p^2 = 9q^2
Show more

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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If p = 3q, is p^2 > q^2 16 Nov 2018, 04:14
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Given : p = 3q

From 1: q+p<q-p
Solving Statement 1 , 2p<0 therefore, p<0 which means that q<0.
Hence, 1 is sufficient.

From 2: p^2= 9q^2
Solving Statement 2, p^2-9q^2=0 or (p-3q)(p+3q)=0 or p=3q or p=-3q.
We are not sure of the value p holds and hence cannot conclude about p^2>q^2.
Therefore, 2 is not sufficient.

A is the answer.
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