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.
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" \(p^2 > q^2\)